Lo**ryotl
'20. "'
f du f ::r J u\/a+n f
t l
 {Ll *,  ir a > o [1r" "' 16lT \/i + lal +Tn l\/a  
lZr^n_r l!a
L"
i.J ''
b(2n  3) f
\/;T6
itu
2a(n  l)
n(n  7\ynr
J ,nytfifi
q6 ifa b) than a; with symbols we write
if andonlyif b isless
albifandonlyifb b if. and only if either a > b ot a: bT h e s ta te m entsa 1b, a) b, a < b, anda = b are cal l ed i nequal i ti es. 4, 3> 6, Some examplesare 2 < >
2 a n d 3
L.1.30Theorem
>5;
so7*3>2+(5).
If a > b and if c is any positive number, then ac ) bc. o rLLUsrRArroN 12: 3 > 7; so (3)
> eDA.
o
1.1 SETS,REAL NUMBERS,AND INEQUALITIES
1.1.31Theorem lf. a > b and if c is any negative number, then ac I bc. o rLLUsrRArroN 13: 3 > 7; so (3)(4)
< (7)(4).
o
1.1.32Theorem If a > b > }and c > d > 0, then ac > bd. o r L L U S r R A r r o1N4 : 4 > 3 > 0 a n d 7
n
79
tz24
lllttrttllrll 4t 2
0 
zt
4
F i g u r e1 . 1 . 1
>5 ) 0; so 4(7)>3(6).
o
So far we have required the set Rl of real numbers to satisfy the field axioms and the order axiom, and we have stated that because of this requirement Rr is an ordered field. There is one more condition that is imposed upon the set R1. This condition is called the axiom of completeness (Axiom 16.2.5). We defer the statement of this axiom until Section 16.2 because it requires some terminology that is best introduced and discussed later. However, we now give a geometric interpretation to the set of real numbers by associating them with the points on a horizontal line, called an axis. The axiom of completeness guarantees that there is a onetoone correspondence between the set Rl and the set of points on an axis. Refer to Figure 1.1.1. A point on the axis is chosen to represent the number 0. This point is called the origin A unit of distance is selected. Then each positive number x is represented by the point at a distance of r units to the right of the origin, and each negative number x is represented by the point at a distance of x units to the left of the origin (it should be noted that if r is negative, then r is positive). To each real number there corresponds a unique point on the axis, and with each point on the axis there is associated only one real number; hence, we have a onetoone correspondence between Rl and the points on the axis. So the points on the axis are identified with the numbers they represent, and we shall use the same symbol for both the number and the point representing that number on the axis. We identify Rl with the axis, and we call Rl the real number line. We see that a < b if and only if the point representing the number a is to the left of the point representing the number b. Similarly, a > b if and only if the point representing a is to the right of the point representing b. For instance, the number 2 is less than the number 5 and the point 2 is to the left of the point 5. We could also write 5 ) 2 and say that the point 5 is to the right of the point 2. A numberx is between a andbif a ( rand x "1..2.3 Corollary Irl = a if. and only if a < )c < a, where a > \.2.4 Theorem ltl > a if andonly if x > a or x I a,where 7.2.5 Corollary ltl = a if andonly if x > a or x s a,where
0. 0. a > 0. a ) 0.
The proof of a theorem that has ant"if and only if " qualification requires two parts, as illustrated in the following proof of Theorem 1.2.2. plnr l.: Prove that lrl < a if.a < x I a,where a ) O.Here,we haveto consider two cases:.x > 0 and r < 0. Casel: x=0. Then ltl : x. Becauser < a,we concludethat lrl < a. C a s e 2 :x < 0 . and obtain Then lrl : r. Becausea < x, we aPPlyTheorem L.1,.25 a)x or, equivalently,x < a.But becausex: lrl, we have l*l < o. In both cases,then, wherea>0 0. Here we panr 2: Prove that lrl < a only if must show that whenever the inequality ltl < a holds, the inequality a < x 1a also holds. Assume that lrl < a and consider the two cases x>0andr(0. CaseT: x>0. Then ltl : r. Because lxl < a,we conclude that r < a. Also, because a 10 a ) 0, it follows from Theorem 1.1.31 thata < 0. Thus/ we have <x5
or 3x*25 or, €guivalently, x)l Therefore, every number in the interval (1, +oo; is a solution. From the second inequality, we have 3xt2 7
!. lr+41= lzx61 ) a' l.1 3o + x s l r2=l 1 27. Prcve Theorem 1.2.4.
16. l2xsl < 3 1 9 .l 2 x  s l > 3 22. l3rl > 16 }xl v*?l .n 2s.llffil
1 7 .l 3 x  4 1 z 2 0 .1 3 + 2 x l < l a  r l 23.le zxl = l4xl rc I l1_l
=1,rr 26. !ffiI
28. Prove Theorem 1'.2.7.
In Exercises29 through 32, solve for r and use absolute value bars to write the answer.
29. 31..
12'8') thenlcbl = l4l +lbl.(HINI:Writeabasa'+(b) anduseTheorem 33. prcvethatif , andb areany nurnberg, 34. prcve that if a and b are any numben, then lal lbl < la bl. (Hrr.rr:Let lal: l(a b) + bl, and useTheorem1.2.8.) 35. What singleinequalityis equivatentto the followingtwo inequalities:s>b+ ca d a> b c?
1.3 THE NUMBER PLANE AND GRAPHS OF EQUATIONS writing them in parentheses with a comma separating them as (I/ y). Note that the ordered pair (3, 7) is different from the ordered pair (7,3).
Definition 1..3.1.
The set of all ordered pairs of real numbers is called the number Plane' and each ordered Pafi (x, D is called a point in the number plane. The number plane is denoted by R2. fust as we can identify Rl with points on an axis (a onedimensional space), we can identify R2 with points in a geometric plane (a twodimendional space). The method we use with R2 is the one attributed to the French mathematician Rene Descartes (15951650), who is credited with the invention of analytic geometry in L637. A horizontal line is chosen in the geometric plane and is called the r axis. A vertical line is chosen and is called lhe y axis. The point of intersection of the r axis and tlne y axis is called the origin and is denoted by the letter O. A unit of length is chosen (usually the unit length on each axis is the same). We establish the positive direction on the r axis to the right of the origin, and the positive direction on the y axis above the origin.
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
F i g u r e1 . 3 . 1
We now associate an ordered pair of real numbers (x, y) with a point P in the geometric plane. The distance of P from the y axis (considered as positive if P is to the right of the y axis and negative if P is to the left of the y axis) is called the abscissa(or x coordinate) of P and is denoted by *.The distance of P from the r axis (considered as positive if P is above r the x axis and negative if P is below the x axis) is called the ordinate (or y coordinate)of P and is denoted by y. The abscissa and the ordinate of a point are called the rectqngular cartesian coordinatesof the point. There is a onetoone correspondence between the points in a geometric plane 7) and R2; that is, with each point there corresponds a unique ordered pair (x,y), and with each ordered pafu (x, y) there is associated only one point. This onetoone correspondence is called a rectangular cartesian coordinate system, Figure 1.3.1 illustrates a rectangular cartesian coordinate system with some points plotted. The x and y axes are called the coordinateaxes,They divide the plane into four parts, called quadrants, The first quadrant is the one in which the abscissa and the ordinate are both positive, that is, the upper right quadrant. The other quadrants are numbered in the counterclockwise direction, with the fourth, for example, being the lower right quadrant. Because of the onetoone correspondence, we identify R2 with the geometric plane. For this reason we call an ordered pair (x, y) a point. Similarly, we refer to a "line" in R2 as the set of all points corresponding to a line in the geometric plane, and we use other geometric terms for sets of points in R2. Consider the equation A:x22
(1)
where (x,y) is a point inR2. we call this an equation inR2. By u solution of this equation, we mean an ordered pair of numbers, one for r and one for y, which satisfies the equation. Foi example, if r is re p l a c e d by 3 i n E q. (1), w e see that A :7; thus, x:3 and,y:7 const itutes a solution of this equation. If any number is substituted for x in the right side of Eq. (1), we obtain a corresponding value for y.It is seen, then, that Eq. (1) has an unlimited number of solutions. Table 1.3.1 gives a few such solutions. T a b l e1 .3 .7
x
l0
a:x2zlz
F i g u r e1 . 3 . 2
L2
1
3
34
41.Z
2 7 74 1
2
7
14
If we plot the points having as coordinates the number pairs (x,y) satisfying Eq. (1), we have a sketch of the graph of the equation. In Fig. 1'.3.2we have plotted points whose coordinates are the number pairs ob
1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS
tained frorn Table 1.3.1. These points are connected by u smooth curve. Ary point (x, y) on this curye has coordinates satisfying Eq. (L). Also, the coordinates of any point not on this curve do not satisfy the equation. We have the following general definition. 1.g.2 Definition
The graph of an equation in I€ is the set of all points (x, y) in R2 whose coordinates are numbers satisfying the equation. We sometimes call the graph of an equation the locus of the equation. The graph of an equation in R2 is also called a curae. Unless otherrryise stated, an equation with two unknowns, x and y, is considered an equation in R2.
sxelrprn L: Draw a sketch of the graph of the equation' Y'x2:0
Q)
soLUrIoN:
Solving Eq. (2) for !, we have
(3)
y:!\tX+2 Equations (3) are equivalent to the two equations
(4 )
Y:lx+2
(s)
Y_\'x2
The coordinates of all points that satisfy Eq. (3) will satisfy either Eq. (4) or (5), and the coordinates of any point that satisfieseither Eq. (4) or (5) will satisfy Eq. (3). Table 1.3.2 gives some of these values of x and y. Table1.3.2
x
00
v \/2 \/2
1
\/i
"1. 2
\/5
2
2 2
3
3L12
\/5 \tr
1, L
0
Note that for any value of r 2 there are two values for y. A sketch of the graph of Eq. (2) is shown in Fig. 1.3.3.The graph is a parabola.
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
2: Draw sketchesof ExAMPLE the graphs of the equations y:\/x+2 and
soLUrIoN: Equation (5) is the same as Eq. (a). The value of y is nonnegative; hence, the graph of Eq. (6) is the upper half of the graph of Eq. (3). A sketch of this graph is shown in Fig. 1..3.4. (5) Similarly, the graph of the equation
Y:\E+2
Y:1/Yt'2
a sketch of which is shown in Fig. 1.3.5, is the lower half of the parabola o f F i g . 1 .3.3.
F i g u r e1 . 3 . 4
F i g u r e1 . 3 . 5
rxltrpr.n 3: Draw a sketch of the graph of the equation
soLUTroN: From the definition have
y:lx*31
0)
y:x*3
if
of the absolute value of a number, we
r*3>0
and
y(x+3)
if
r*33
and y(x+3)
if
x13
Table 1.3.3 gives some values of x and y satisfying Eq. (Z). T a b l e1 .3.3 Figure1.3.6
x
0
L
2
3
v
3
4
5
6
L 2
2 1
3 0
4 1,
5 2
6
7
3
8
4
5
A sketch of the graph of Eq. (7) is shown in Fig. 1,.9.6. rxetvrplr 4: Draw a sketch of the graph of the equation (x2y+3)(Vx'):0
(8)
soLUTroN: By the property of real numbers that ab:O a : 0 o r b :0, w e have from E q. (8) x2y*3:0
if and
9 6
1.3 THE NUMBERPLANE AND GRAPHSOF EQUATIONS
yxz:o
(10)
The coordinates of all points that satisfy Eq. (8) will satisfy either Eq. (9) or Eq. (L0), and the coordinates of any point that satisfies either Eq. (9) or (10) will satisfy Eq. (8). Therefore, the graph of Eq. (8) will consist of the graphs of Eqs. (9) and (10).Table 1..3.4gives some values of r and y satisfying Eq. (9), and Table 1.3.5 gives some values of x and y satisfying Eq. (10).A sketchof the graph of Eq. (8) is shown in Fig. 7.3.7. L.3.4 Table
F i g u r e1 . 3 . 7
x
0L2312345
v
821r31,+o+1
T a b l e1 .3 .5
1.3.3 Definition
x
01,23723
v
0L49149
An equation of a graph is an equation which is satisfied by the coordinates of those, and only those, points on the graph. is an equation whose graph consists of o ILLUSTRATIoN1: In R2, y:8 those points having an ordinate of 8. This is a line which is parallel to the o x axis, and 8 units above the r axis. In drawing a sketch of the graph of an equation, it is often helpful to consider properties of symmetry of a graph.
1.3.4 Definition
(3'2\o
F i g u r e1 . 3 . 8
Two points P and Q are said to be symmetric with respect to a line if and only if the line is the perpendicular bisector of the line segment PQ. Two points P and Q are said to be symmetric with respectto a thirdpoint if and only if the third point is the midpoint of the line segment PQ. o rLLUsrRArroN 2: The points (3,2) and (3,2) are symmetric with respect to the x axis, the points (3, 2) and (3, 2) are symmetric with respect to the y axis, and the points (3,2) and (3, 2) are symmetricwith o respect to the origin (see Fig. 1.3.8). In general, the points (x, y) and (x, y) are symmetric with respect to the x axis, the points (r, y) and (x, y) are symmetric with respect to the y axLS,and the points (x,y) and (r, y) arc symmetric with respect to the origin.
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
1.3.5 Definition
are symmetric with respect to R. From Definition 1.3.5 it follows that if a point (x, y) is on a graph which is symmetric with respect to the x axis, then the point (x,y) also must be on the graph. And, if both the points (x,y) and (x,y) are on the graph, then the graph is symmetric with respect to the x axis. Therefore, the coordinates of the point (x,y) as well as (x, y) must satisfy an equation of the graph. Hence, we may conclude that the graph of an equation in r and y is symmetric with respect to the r axis if and only if an equivalent equation is obtained when y is replaced by y in the equation. We have thus proved part (i) in the following theorem. The proofs of parts (ii) and (iii) are similar. L.3.6 Theorem (Testst'or Symmetry)
The graph of an equation in r and y is (i) symmetric with respect to the x axis if and only if an equivalent equation is obtained when y is replaced by y in the equation; (ii) symmetric with respect to the y axis if and only if an equivalent equation is obtained when x is replaced by x in the equation; (iii) symmetric with respect to the origin if and only if an equivalent equation is obtained when r is replaced by x and y is replaced by y in the equation. The graph in Fig. 1.3.2 is symmetric with respect to the y axis, and for Eq. (1) an equivalent equation is obtained when x is replaced by r. In Example 1 we have Eq. (2) for which an equivalent equation is obtained when y is replaced byy, and its graph sketched inFig. L.3.3 is symmetric with respect to the r axis. The following example gives a graph which is symmetric with respect to the origin.
rxlrvrpr.r 5: Draw a sketch of the graph of the equation xy:1
solurroN: We see that if in Eq. (1,1)x is replaced by x and y is replaced by y,an equivalent equation is obtained; hence,by Theorem 1.3.6(iii) the graph is symmetric with respect to the origin. Table L.3.6 gives some (11) values of x and y satisfying Eq. (11). T a b l eL .3.6
1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS
'l.lx. We seethat as r increasesthrough From Eq. (11) we obtain y : positive values and gets closer and positive values, y decreasesthrough positive values,y increasesthrough closerto zero.As x decreasesthrough positive values and gets larger and larger. As I increasesthrough negative values (i.e.,r takeson the values4, 3, 2, t, +, etc.),y takes on negative values having larger and larger absolute values. A sketch of the graph is shown in Fig. 1..3.9.
F i g u r e1 . 3 . 9
1.3 Exercises In Exercises 1 through 6, plot the dven Point P and such of the following PoinB as may aPPly: (a) The point O suitr ttrit ttre tnJ through Q and P is perpendicular to the r axis and is bisected by it. Give the coordinates of Q. (b) The poiniR such that the line through P and R is perpmdicular to and i6 bisected by the y axis. Give the coordinates of R. (c) The point s such that the line through P and 5 iE bisected by the origrn. Give the_coordin"F"9f s.by the als' line though the origin 1ai fn" poi"t T such that the line *Eough P and 7 is perpendicular to and is bisected of T. coordinates bisecting the first and thtd quadrants. Give the 1 . P ( 1,  2 ) 4. P(2, 2)
2. P(2,2) 5 . P (  1,  3 )
3. P(2,2) 5 . P ( 0 , 3 )
In Exercises7 through2S, draw a sketch of the graph of the equation.
7'Y:2x*5 l0.y:nffi 1 3 .x :  J 15.y:lx+21 1 9 .4 f * 9 Y 2 : 3 6 22. Y' : 41cs 25. xz*y',:O 2 8 . ( y 2 x * z ) ( V + t E 
8'Y:4x3 11.y':x3 'I.. y2 * 1 7 .y : l r l  5 2 0 . 4 x 2 9 Y ' : 3 6 L4. x:
23. 4x2 Az:0 2 6 .( 2 x + V  l ) ( 4 V + r 2 ) : 0
Y: \81 y:5 y:lx51 ylxl+z A:4x3 3x2 L3xy L0y2: g xa 5x2yI 4y2: g
+):O 29. Draw a sketch of the graph of each of the following equations:
(a) V: lzx
(b\ v : {2x
(c) y' :2x
30. Draw a sketch of the graph of each of the following equations:
(a) Y: !2x
(b\ y : !2x
(c\ y' : 2N
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
3L. Draw a sketch of the graph of each of the following equations:
( a )r * 3 y : 0
(c) r'  9y' :0
(b)r3y:0
32. (a) Write an equation whose gtaPh is the r axis. (b) Write an equation whose graph is the y axis. (c) Write an equafion whose graph is the set of all points on either the .x axis or the y axis. 33. (a) Write a,n__equ{iol whose graph consists of all points having an abscissa of 4. O) Write an equation whose graph consistE of all points having an ordinate of 3. 34. Prove that a graPh that i5 symmetric with respect to both coordinate axes is also symmetric with rcspect to the origin. 35' Proeethat a graPh that is symmeFic with rcspect to any two perpendicular lines is also sym.rretric with respect to their point of intersection
1.4 DISTANCE FORMULA AND MIDPOINT FORMULA
If A is the point (x,.,y) and B is the point (xr,y) (i.e., A and B havethe same ordinate but different abscissas), then the directed distance from A to B, denoted by AB, is defined ds x2  x1. o ILLUSTRATToN 1: Refer to Fig. 1.4.1(a),(b), and (c).
u B(r,z)
A(4,2)
AB:'/.,4 (b)
Figur1 e.4.1 If .4 is the point (3, 4) and B is the point (9, 4), then AB :9  3: 6. rf A is the point (8, 0) and B is the point (6,0), then B :6  (s) : 14. If A is the point (4, 2) and B is the point (L,2), then AB : 1. 4: 3. we see that AB is positive if B is to the right of A, and AB is negative if B is to the left of A. o
v
D(2,4)
C(2, g) CD: 6 (a)
Figure 1.4.2
CD:7 (b)
If C is the point (x,yr) and D is the point (x, yr), then the directed distance from C to D, denoted by CD, is defined as!z!r. o rLLUsrRArroN 2: Refer to Fig. La.2@) and (b). If C is the point (L, 2) and D is the point (1, 8), then CD : g J2)6. If C is the point (2, 3) and D is the point (2,4), then (3) :7. Tl r.enumber C p i s posi ti vei f D i s above C , and CD C D :4 is negative if D is below C. o We consider a directed distance AB as the signed distance traveled by u particle that starts at A(rr, y) and travels to B(x2, !). ln such a case, the abscissa of the particle changes from 11 to x2, dfld we use the notation Ar ("delta x") to denote this change; that is, A,x: xz xt Therefore, AB:
Lx.
1.4 DISTANCEFORMULAAND MIDPOINTFORMULA
Pr(x", vz)
It is important to note that the symbol Ar denotes the difference between the abscissa of B and the abscissa of A, and it doesnot mean "delta multiplied by x." Similarly, if we consider a particle moving along a line parallel to the y axis from a point C(x, yr) to a point D(x, yr), then the ordinate of the particle changes from Ar to Az.We denote this change by Ly ot
Ly: az ar Thus, CD: Ly. Now let Pr(r1, yr) and Pr(xr, a) be any two points in the plane. we wish to obtain a formula for finding the nonrtegative distance between these two points. We shatl denote this distanceby lPtPrl. W" use absolutevalue bars becausewe are concernedonly with the length, which is a nonnegative number, of the line segmentbetween the two points P1and Pr. To derive the formula, we note that lffil is the length of the hypotThis is illustrated in Fig. 1.4.3for P1and of a right triangle PLMP2. "rrrrr" quadrant. first are in the Pr, both of which we have theorem, Using the Pythagorean
F i g u r e1 . 4 . 3
+llvl' EF"f: lArl2 So
lFnl:\MV+WP That is,
(1)
lF:P,l:
Formula (1) holds for all possible positions of Pt and P, in all four qqadrants. The length of the hypotenuse will always be lFfrl, and the lengths of the two legs will always be [Ar[ and lAyl (seeExercisesL and 2). We state this result as a theorem. '1,.4.1 Theorem The undirected distance between the two points Pr(xr, Ar) and Pr(xr, yr) is given by
EXAMPLEl.: If a point P(x, U) is such that its distance from A(3,2) is always twice its distance from B(4, L), find an equation which the coordinates of P must satisfy.
solurroN:
From the statement of the problem
lPTl:2lPBl Using formula (1), we have
@:z@ Squaring on both sides, we have x 2 6 x + 9 + y '  4 y I 4 :
4(xz*8r
or, equivalently, 3 x 2* 3 y ' * 3 8 r  4 y + 5 5 : 0
* t6 + y2 2y * 1)
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
ExAMPLE2: Show that the triangle with verticesat A(2, 4), B(5, L), and C(6,5) is isosceles.
SOLUTION:
The triangle is shown in Fig. L.4.4.
lBel: @ : \ 4 + 1 6 : f r : @:\tr6+1:{lT lae; IEAI:W : \ / N : j f i Therefore,
A(2,4)
Hence, triangle ABC is isosceles.
B(5, i.)
Figure1.4.4
nxervrrrn3: Prove analytically that the lengths of the diagonals of a rectangleare equal.
C(0, b)
(0,0) F i g u r e1 . 4 . 5
B(a,b)
SOLUTION:Draw a generalrectangle.Becausewe can choosethe coordinate axes anywhere in the plane, and becausethe choice of the position of the axes does not affect the truth of the theorem, we take the origin at one vertex, the x axis along one side, and the y axis along another side. This procedure simplifies the coordinates of the vertices on the two axes. Referto Fig. 1..4.5. Now the hypothesis and the conclusion of the theorem can be stated. Hypothesis:OABC is a rectanglewith diagonalsOB and AC. Conclusion: IOB : lrel .
l o B:lf f i : r t + 6 2 l A C: lW : \ f r ] A t Therefore,
lml : ldcl Let P, (xr, yr) and P, (xr, yr) be the endpoints of a line segment. We shall denote this line segment by PtPr. This is not to be confused with the notation PrPr, which denotes the directed distance from P, to Pr. That is, PrP, denotes a number, whereas PrP2is a line segment. Let P(x, y) be the midpoint of the line segment P,Pr. Refer to Fig. L.4.6. In Fig. I.4.6 we see that triangles P'RP andPTPrare congruent. Therefore, lPtRl: lPTl, and so r  xr: xz x, giving us
1.4 DISTANCEFORMULAAND MIDPOINTFORMULA
Similarly, lRPl :
lTPzl.Then A Ar: Az A, andtherefore (3)
Hence, the coordinates of the midpoint of a line segment are, respectively, the average of the abscissasand the average of the ordinates of the endpoints of the line segment. P z( x z ,v z )
P(x,y)
u__ R(t, yt)
T(xr, Y)
__JS(rr,yr)
F i g u re1 .4 .6 In the derivation of formulas (2) and (3) it was assumed that x, ) xt and y, ) yr.The same formulas are obtained by using any orderings of these numbers (see Exercises 3 and 4).
nxaurr,E 4: Prove analytically that the line segments ioining the midpoints of the opposite sides of any quadrilateral bisect each other.
B(b,c)
F i g u r e1 . 4 . 7
Draw a general quadrilateral. Take the origin at one vertex solurroN: and the x axis along one side. This method simplifies the coordinates of the two vertices on the r axis. See Fig. L.4.7. Hypothesis: OABC is a quadrilateral. M is the midpoint of OA, N is the midpoint of CB, R is the midpoint of OC, and S is the midpoint of AB. Conclusion:MN and RS bisect each other. pRooF:
To prove that two line segments bisect each other, we show that they have the same midpoint. Using formulas (2) and (3), we obtain the coordinates of M, N, R, and S. M is the point Ga, 0), N is the point ( + ( b + d ) , t r ( c * e ) ) ,R i s t h e p o i n t 1 ! d , t e ) , a n d S i s t h e p o i n t G @ + b ) , i c ) . + d). T h e a b s c i s s ao f t h e m i d p o i n t o f M N i s i l L a + + ( b + d ) l : * ( a + b : e). The ordinate of the midpoint of MN is +[0 + Lk + e)f ik + Therefore, the midpoint of MN is the point (ifu + b + d), ik + e)). + d). T h e a b s c i s s ao f t h e m i d p o i n t o f R S i s i l i a + * @ + b ) l : * ( a + b The ordinate of the midpoint of RS is tlie + *cf : Ik + e). Therefore, the midpoint of RS is the point (ifu + b + d), i(c + e)). Thus, the midpoint of MN is the same point as the midpoint of RS. I Therefore, MN and RS bisect each other.
92
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
Exercises 7.4 1. Derive distance formula (1) if Pr is in the third quadrant and P, is in the second quadrant. Draw a figure. 2. Derive distance formula (1) if Pr is in the second quadrant and P, is in the fourth quadrant. Draw a figure. 3. Derive midpoint formulas (2) and (3) if Pr is in the first quadrant and p, is in the third quadrant. 4. Derive midpoint formulas (2) and (3) iI4 (rr, yr) and &(rr, yr) are both in the second quadrant and x, > xrand,y, > yr. 5. Find the length of the medians of the rriangle having vertices A(2,3),8(3,g), and C(1,1). 5. Find the midpoints of the diagonals of the quadrilateral whose vertices are (0,0), (0,4), (3,5), and (3, 1). (.\Prove that the trianSle with vertices A(3,6), B(8,2), and C(1, 1) is a right triangle. Find the areaof the triangle. \'l(HrNr: Use the converseof the Pythagoreantheorem.) 8. Prove that the Points A(6, lS) , B(2, 2), C(13, 10), and D(21, 5) are the vertices of a square. Find the tength of a diagonal. 9. By using distance formula (1), prove that the points (3, 2) , (1, 2), and (9, 10) lie on a line. 10. If one end of a line segment is the point (4, 2) and the midpoint is (3, L), find the coordinates of the other end of the line segment.
11. The abscissa of a Point iB 6, and its distance from the point (1, 3) is V74. Find the ordinate of the point. 12. Determine whether or not the poifis (14,7) , (2, 2) , and,(4, 1) lie on a line by using distance formula (1). /;:\ '\3.llf two vertices of an equilateral hiangle are (4,3) and (0,0), find the third vertex. d Findan equation that must be satisfied by the coordinatesof any point that is equidistant from the two points (3, 2) and (4,6). 15. Find an equation that must be satisfied by the coordinates of any point whose distance from the point (5, 3) is atways two units Feater than it8 distance frcm the point (4, 2). 16. Giver the two Points d(3,4) andB(2,5), find the coordinates of a point P on the line through A and B such thatp is (a) twice as far from 4 as frcm B, and (b) twice as far frcm B as from _A. 17. Find the coordinates of the three points that divide the tine segment ftom l(5,
3) to 8(6, g) into four equal parts.
18. If rr. and r, are positiv€_intgg:ers,provethat the coordinates of the point p(r, y), which divides the line segment plp, : r,/rrare given by in the ratio rr/r2that i3, IP,,PUlPrEI ,:
(t"  rt)x, + rrxu fz"t2
ur4 , :
(r,  ,r)y, + ,rv"
In Exercises 19 through 23, use the formulas of Exercise 18 to find the coordinates oI point p. 19' The Point P is on the line segment between points Pr (1 ,3) ?flrd,P26,2) and is three times as far from P, as it b ftom pr. 20. The Point P is on the line segmentbetween points Pl(1 ,3) and Pr(6,2) and is three times as far from Pr as it is frcmpr. 21. The Point P i8 on the line through P' and P, and is three times as far ftom Pr(6, 2) as it is from P, (1,3) but is not between Pr and Pr. 22. The Point P is on the Une though P, and P, and is three times as far from P1(1, 3) as it is ftom Pr(6,2) but is not between P1and P2.
23. The point P is on the line through Pr(3,5) and Pr(L,2) so that ppr : + . p.,pr. 24. Find an equation whose graph is the circle that is the set of all points that are at a distance of 4 units from the point (1, 3).
1.5 EQUATIONSOF A LINE
25. (a) Find an equation whose graph consist8 of all points equidistant frcm the Points (1,2) and (3,4). (b) Draw a skekh of the gnph of the equation found in (a)' 26. prove analytically that the sum of the squares of the distanc$ of any Point from two opposite vertic€s of any rcctangle is equal to the sum of the squares of its distances from the other two verticeE' 22. prcve analytically that the line segment ioining the midpoints of two opposite sides of any quadrilateral and the line segment joining the midpoints of the diagonala of the quadrilateral bieect each other' 28. prove analytically that the midpoint of the hypotenuse of any right triangle is equidistant frcm each of the three vettices. 29. Prove analytically that if the lengths of two of the medians of a triangle arc equal, the triangle is isoeceles.
1..sEQUATIONS OF A LINE
Letl be a nonverticalline and Pr(x' U) andP"(xr, A)be any two distinct points on l. Figure L.5.1shows such a line. In the figure, R is the point Qc,Ur),and the points Pr, Pu and R are vertices of a right triangle; furthermore, P,.R: xzxr and RPr: AzAr The number AzAr gives the measure of the change in the ordinate from Pr to Pr, and it may be positive, negative, or zero.The number x2 xl gives the measureof the change in the abscissafrom P, to Pr, and it may be positive or negative.The number x,  rr may not be zero becaus?x2 * r, since the line I is not vertical. For all choices of the points P, and P, on l, the quotient
R(rz,yJ
Az Ur Xz xr
,
is constanq this quotient is called the "slope" of the line. Following is the formal definition.
F i g u r e1 . 5 . 1
1.S.1 Definition
If Pr(xr, Ar) and Pr(xr, yr) arc any two distinct points on line l, which is not parallel to the y axis, then the slopeof l, denoted by m, is given by
; : y EXz *
at P, (xr, y, R(xr,Yr)
Pr(7r,V,)
(1) ?Cr
In Eq. (L), x, * x1 since / is not parallel to the y axis. The value of m computed from Eq.(t) is independent of the choice of the two pointr_P, and P2 on L To show this, suppose we choose two different points, tr(n, yr) and Pr(ir,yr), and compute a number nr from Ee. (1). ar *m :_ A  zxz xt
7z,Ai x
We shall show tlnilt n  m. Refer to Fig. 1.5.2. Triangles PrRPz and P1RP2 are similar, so the lengths of corresponding sides are proportional. Therefore,
F i g u r e1 . 5 . 2
V:
Y_t Az
Xz
Xr
or m:m

Xz
Ar Xt
34
REAL NUMBERS,INTRODUCTION T O ANALYTICGEOMETRY, AND FUNCTIONS
Hence, we conclude that the value of m computedfrom Eq. (1) is the same number no matter what two points on I are selected. In Fig. 1.5.2, x2 ) x1, Uz) Ar, x, > n, and y2 > fr. The discussion above is valid for any ordering of these pairs of numbers since Definition 1.5.1holds for any ordering. In sec. 1.4 we defined Ly:uzur and Ar:xz rr. substituting thesevalues into Eq. (L), we have F i g u r e1 . 5 . 3
Multiplying on both sidesof this equationby Lx,we obtain Ly: mLx
e)
It is seen from Eq. (2) that if we consider a particle moving along line l, the change in the ordinate of the particle is proportional to the change in the abscissa, and the constant of proportionality is the slope of the line. If the slope of a line is positive, then as the abscissa of a point on the line increases, the ordinate increases. Such a line is shown in Fig. 1.5.3. In Fig. "1..5.4, we have a line whose slope is negative. For this line, as the abscissa of a point on the line increases, the ordinate decreases. Note that if the line is parallel to the x axis, then Uz: Ur and so m:0. If the line is parallel to the y axis, xz: xr, thus, Eq. (1) is meaningless because we cannot divide by zerc. This is the reason that lines paiallel to the y axis, or vertical lines, are excluded in Definition 1.5.L. We Jay that a vertical line does not have a slope.
F i g u r e1 . 5 . 4
. rLLUSrRArroNl.: Let I be the line through the points P,,(2,3) and p2(4,7). The slope of I, by Definition "!..5.'!., is given by
*:7n!:z L
Pr $ , 9 )
Referto Fig. 1.5.5.rf P(x,y) and Q@+ L*,y + Lil are any two points on I, then L A_ . , Ax' Ay :2
P:(2,3)
P+(1
F i g u r e1 . 5 . 5
A,x
Thus, if a particle is moving along the line l, the change in the ordinate is twice the change in the abscissa. That is, if the particle is at pr(4,7) and the abscissa is increased by one unit, then the ordinate is increased by two units, and the particle is at the point Pr(5,9). Similarly, if the particle is at Pt(2,3) and the abscissa is decreased by three units, then the ordinate is decreased by six units, and the particle is at Pr(L, 3). o Since two points Pr(xr, yr) and Pr(xr, y)
determine a unique line, we
1.5 EQUATIONS OF A LINE
should be able to obtain an equation of the line through these two points. Consider P(x,y) any point on the line. We want an equation that is satisfied by r and y if and only rf P(x, D is on the line through P1(rr, yr) and Pr(xr, Ar).We distinguish two cases. C a s e7 : x z : x r . In this case the line through P, and P, is parallel to the y axis, and all points on this line have the same abscissa. So P(x,,U) ts any point on the line if and only if (3)
X: Xr
Equation (3) is an equation of a line parallel to the y axis. Note that this equation is independent of y; that is, the ordinate may have any value whatsoever, and the point P(x,y) is on the line whenever the abscissais rr. Case2: x2 * x1. The slope of the line through Pt and P, is given by Ar ry:Uzxz xr lf P(x, g) is any point on the line except (\, given by
(4) !r), the slope is also
At 1A x xr
(s)
The point P will be on the line through P, and Pz if and only if the value of m fuomEq. (4) is the same as the value of m ftom Eq. (5), that is, if and only if U A t A z  A r X
Xr
Xz Xr
Multiplying on both sides of this equation by (x  xr),we obtain (6) Equation (5) is satisfied by the coordinates of P, as well as by the coordinates of any other point on the line through P1 and Pr. Equation (6) is called tkte twopoint fonn of an equation of the line. It gives an equation of the line if two points on the line are known. 3) o rLLUsrRArroN 2: An equation of the line through the two points (5, and (2,3) is
y(3):*(r5)
y+32@6) 3x*4Y:6
REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
If in Eq. (6) we replace(y,  !r)l(x,  x) by *, we get I i.:r )i' I.'
+ l o :
t I
';
...
} i Yfl:gqt{,&Es'*"atu
e)
Equation (7) is called the pointslope form of an equation of the line. It gives an equation of the line if a point Pr(xr, a) on the line and the slope m of the line are known. . rLLUSrRArroN 3: An equation of the line through the point (4,5) and having a slope of 2 is
y  (s) :Zlx  (4)l 2xy*3:0
o
If we choosethe particular point (0, b) (i.e., the point where the line intersectsthe y axis) for the point (h, A) in Eq. (7), we have yb:m(x0) or, equivalently,
+b
(8)
The number b, which is the ordinate of the point where the line intersects the y axis, is called the y intercept of the line. Consequently, Eq. (8) is called the slopeintercept form of an equation of the line. This is especially important because it enables us to find the slope of a _form line from its equation. It is also important becauseit expressesttre y coordinate explicitly in terms of the r coordinate. rxevrpln L: Given the line having the equation 3r I 4y :7, find the slope of the line.
sol.urroN:
Solving the equation f.or !, we have
,:fix*I Comparing this equation with Therefore, the slope is *.
Eq. (8), we see that
and b:*.
Another form of an equation of a line is the one involving the intercepts of a line. We define the r intercept of a line as the abscissa of the point at which the line intersects the r axis. The r intercept is denoted by o. If the r intercept a and the y intercept b arc given, we have two points (a, 0) and (0, b) on the line. Applying Eq. (6), the twopoint form, we have h 0 .
Yo:6ik
aY:bxab b x* a A : a b
a)
OF A LINE 1.5 EOUATIONS
by ab, if. a * 0 and b + 0, we obtain
Dividing
(e) Equation (9) is called the intercept fonn of an equation of the line. Obviously it does not apply to a line through the origin, becausefor such a line both a and b arc zero. nxlupln 2: The point (2,3) bisects that portion of a line which is cut off by the coordinate axes.Find an equation of the line.
Refer to Fig. 1.5.6. lf. a is the r intercept of the line and b is the y intercept of the line, then the point (2, 3) is the midPoint of the line segment joining (a, 0) and (0, b). By the midpoint formulas/ we have soLUrIoN:
A
andt:ry
atl
A:4
and b:6
The intercept form, Eq. (9), gives us x +!t 4'6 or, equivalently, F i g u r e1 . 5 . 6
3x*2y12 t.5.2 Theorem The graph of the equation (10)
Ax*By*C:0
where A, B, and C are constants and where not both A and B are zero, is a straight line. PRooF: Consider the two cases B + 0 and B : 0. CaseL: B+0. Because B + 0, we divide on both sides of Eq. (10) by B and obtain AC
(11)
yi*,
Equation (11) is an equation of a straight line becauseit is in the slope CIB. intercept form, where m: AlB and b C a s e2 : B : 0 . BecauseB:0, Ax*C:0
we may concludethat A * 0 and thus have
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
or/ equivalently,
*: aC
(12)
Equation (12) is in the form of Eq. (3), and so the graph is a straight line parallel to the y axis. This completes the proof. r Because the graph of Eq. (10) is a straight line, it is called a linear equation.Equation (10) is the general equation of the first degree in x andy. Because two points determine a line, to draw a sketch of the graph of a straight line from its equation we need only determine the cooidit utet of two points on the line, plot the two points, and then draw the line. Aty two points will suffice, but it is usually convenient to plot the two points where the line intersects the two axes (which are given by the intercepts). rxeupm 3: Given line lr, having the equation 2x  3y : 12, and line Ir, having the equation 4x * 3y:6, draw a sketch of each of the lines. Then find the coordinates of the point of intersection of l, and Ir.
solurroN: To draw a sketch of the graph of lr, we find the intercepts a and b. In the equation of Ir, we substitute 0 for r and get b  4.In the equation of Ir, we substitute 0 for y and get a: 6. Similarly, we obtain the intercepts a and b for lr, and for l, we have a:8 and,b: 2. The two lines are plotted in Fig. 1..5.7. To find the coordinates of the point of intersection of /, and Ir, we solve the two equations simultaneously. Because the point must lie on both lines, its coordinates must satisfy both equations. If both equations are put in the slopeintercept form, we have
a:?x4
and
tx*2
Eliminating y gives 3x4:gx*2 2xL2:4x#6 x:3 So
y:BG)_4
F i g u r e1 . 5 . 7
Therefore, the point of intersection is (3, 2). 1'5'3 Theorem
If /r and l, are two distinct nonvertical lines having slopes m, and mr, respectively, then /, and l, are parallel if and only if mr: //t2. PRooF: Let an equation of /, be : tftrx * br, and let an equation of l, A be y : rnzx * br. Because there is an "if and only if " qualiiication, the proof consists of two parts. panr 1: Prove that l, and I, are parallel if ftir:
tnz.
1.5E QU A TIONOF S A LIN E Assume that I, and I, are not parallel. Let us show that this assumPtion leads to a contradiction. If I, and I, are not parallel, then they intersect. Catl this point of intersection P(*o, /o).The coordinates of P must satisfy the equations of 11and Ir, and so we have ls: But mr: ls:
!o:
fti2xo* b2
!o:
nttfio* b2
rnlxs* b1 and rrt2rwhich gives mfis * b1 and
b2,both from which it follows thatbl: bz.Thus, becauserrtr: mrandbl: lines 11and /, have the same equation , A : mrx * br, and so the lines are the same. But this contradicts the hypothesis that l, and I, are distinct lines. Therefore, our assumption is false. So we conclude that /r and 12 are parallel. pARr 2: Prove that l, and I, are parallel only if mt: lftz. Here we must show that if Ir and l, are parallel, then rflt: rtrz. Assume that m, * mr. Solving the equations for \ and I, simultaneously, we get, upon eliminating V, m t x * b r : m 2 x* b 2 from which it follows that ( m ,  m r ) x : b z b t Because we have assumed that m, * *r, this gives   _ _b r  b ,
L _
t7\
ffiz
Hence, 11and l, have a point of intersection, which contradicts the hypothesis that l, and 12are parallel. So our assumption is false, and thereI fore mr: nflz. In Fig. 1.5.8, the two lines Ir and l, are perpendicular. We state and prove the following theorem on the slopes of two perpendicular lines. 1.5.4 Theorem
If neither line /, nor line l, is vertical, then Ir and 12are perpendicular if and only if the product of their slopes is 1.. That is, if m, is the slope of 11and m, is the slope of. Ir, then Ir and I, are perpendicular if and only if n tfl tz  1 . PROOF:
 1' Panr 1: Prove that l, and 12are perPendicular only if mtm2 Let L, be the line through the origin parallel to l, and let L, be the line through the origin parallel to Ir. See Fig. 1.5.8. Therefore,by Theorem 1..5.3,the slope of line L, ts m, and the slope of line L, is m2.Because neither
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY,AND FUNCTIONS
L, nor L, rs vertical, these two lines intersect the line x: ! at points p, and Pr, respectively.The abscissaof both P, and P, is L. Let be the ordi/ nate of Pt. Since Lr contains the points (0,0) and (1.,y) and its slopeis mr, we have from Definition 1.5.1 ttL:
a0 i=
and so y = mr.Similarly, the ordinate of P, is shown to be mr. Applying the Pythagorean theorem to right triangle prop2, we get
l o t r ,+; zl @ f : l P , k l , F i g u r e1 . 5 . 8
(1 3 )
By applyingthe distanceformula,we obtain l o  4 l r : ( 1 0 ) ,f  ( m r  0 ) ,: 7 * m ] l O t r l r : ( 1 o ) , * ( m r  o ) , : r * m z 2 l F J r P : ( 1  1 ) , * ( m r  m r ) , n t r z 2 m r m r t m r , Substituting into Eq. (13)gives (l + mrz)+ (1 + m22) ftizz 2mrm,I.mrt 2:
2mrffiz
tltinz  1 penr 2: Prove that I, and l, are perpendicular if.mrm"  1. Starting with ltfitz  '1, we can reverse the steps of the proof of Part 1 in order to obtain
lO&l'+ l1trl': lPEl'z from which it follows, from the converse of the Pythagorean theorem, that Lt andL, are pe{pendicular; hence, /, and l, are perpendicular. I Theorem I.5.4 states that if two lines are perpendicular and neither one is vertical, the slope of one of the lines is the negative reciprocal of the slope of the other line. o rLLUsrRArroN 4: If line /t is perpendicular to line I, and the slope of /, is 3, then the slope of /, must be 9. o
sxarvrpr,n 4: Prove by meahs of slopes that the four points A(5,2), B(9,6), C(4,g), and DQ,4) are the verticesof a rectangle.
solurroN:
See Fig. 1.5.9. Let
62 ntr: the slope of AB  8  6 86 m, J the slope of BC 
4 I
1.5 EQUATIONSOF A LINE
/fi.:the sloPe of DC:ft:
C(4,8)
B(8,6)
41
''
rn.:the slopeof.AD:=:+. Because
D
Tnr: mr, AB ll DC. /n2: ma,BC ll AD. ntrttrz'1, AB L BC.
F i g u r e1 . 5 . 9
Therefore, quadrilateral ABCD has opposite sides parallel and two adjacent sides perpendicular, and we conclude that ABCD is a rectangle.
EXAMPLE5: Given the line / having the equation
Because the required line is perpendicular to line l, its slope solurroN: must be the negative reciprocal of the slope of l. We find the slope of lby putting its equation into the slopeintercept form. Solving the given' equation for y, we obtain
2x*3Y5:0
find an equation of the line perpendicular to line I and passing through the point
Aet,3).
ytx** Therefore, the slope of I is 3, and the slope of the required line is 9. gecausewe also know that the required line contains the point (L, 3), we use the pointslope form, which gives y3:$(x*t) 2y6:3x*3 3x2Y+9:0
1.5 Exercises In Exercises L through 4, find the slope of the line through the given points.
1 . ( 2 , 3 ) , (  4 , 3 )
2 . ( 5 , 2 ), (  2 ,  3 )
3.(+,+),eE,&)
4 . (  2 . T , 0 . 3 ), ( 2 . 3 , 1 . 4 )
In Exercises5 through14, find an equation of the line satisfying the given conditions. 5. The slope is 4 and through the point (2,3). 6. Through the two points (3, L) and (5,4). 7. Through the point (3, 4) and parallel to the y axis. 8. Through the point (I,7) and parallelto the x axis. 9. The x intercept is 3, and the y intercept is 4.
TO ANALYTICGEOMETRY. REAL NUMBERS,INTRODUCTION AND FUNCTIONS
10. Through (1, 4) and parallel to the line whose equation is 2r  5y + 7:0. 11. Through (2, 5) and having a slope of l/5. 12. Through the origin and bisecting the angle between the axes in the first and third quadranb, 13. Through the origin and bisecting the angle between the axes in the second and fou h quadrants. 14. The slope is 2, and the r intercept is 4. the points (1, 3) and (2, 2), and put the equation in the intercept form. equation of the line through the points (3, 5) and (1, 2), and put the equation in the slopeinterceptform. dg. Fitrd "tr 17. Show by means of slopes that the points (4, 1) , (3, +) , (8, 4) , and (2, 9) are the vertices of a trapezoid. 15. Find an equation of the line thouth
18. Thrce consecutivevertices of a parallelogramare (4, 1), (2,3),and (8,9). Find the coordinatesof the fou h vertex. 19. For each of the following sets of three point6, detemine by means of slopes if the points are on a line: (a) (2, 3), (4,7), (s,8);(b) (3,6), (3,2), (e,2); (cl (2,7), (7,7), (3,t); and (d) (1,6), (1,2), (s,4). 20. Prove by means of slopes that the three points A(3, 1), B(6,0), and C(4, ) are the vertices of a right triangle, and find the area of the triangle. /\2T. Given the line I having the equation 2y  3r : 4 and the point P(1, 3), find (a) an equation of the line through P " and perpendicul to l/(b) the shortest distance from P to line l. \ 22. lf A,B, C, and D arc cdhstarts, show that (a) the lines Ax + By+ C:0 a dAx+By+D:0 are parallel and (b) the ljmesAr+ By + C:0 and Bx Ay * D:0 areperpendicular. 23. Given the line l, having the equation Ar + By + C: O, B + 0, find (a) the slope, (b) the y intercept, (c) the x intercept, (d) an equation of the line through the origin perpendicular to L 24. Find an equation of the line which has equal intercepts and which passesthrouth the point (8, 6). 25. Find equations o{ the three medians of the triangle having vertices A(3, 2) , BG, a), and C(l,1), they m€et in a point.
and prove that
26. Find equations of the perpendicular bisectors of the sides of the triande having vertices A(l , 3) , B (5, 3), and C(5,5), and prove that they meet in a point. 27. Find an equation of each of the lines through the point (3,2), which forms with the coordinateaxesa triangle of area12. 28. Let il be the line having the equation.r4'r* Bly+ C1:0,andlet I'be the line having the equationArx+ 82! + Ca:0. If ll is not parallel to L and iJ & is any constant,the equation A'x 'l Bry I C' * k(A"x * Bry * C") :0 represents an unlimited number of lines. Prove that eadr of these lines contains the point of intersection df lr and t. 29. Given an equation ot It is 2x 'f 3y  5: 0 and an equation of l, is 3x + 5y  8 : 0, by using Exercise 28 and without finding the coordinates of the point of intersection of ,r and lr, find an equation of the line through this point of intersection and (a) passing through the point (1, 3); (b) parallel to the r axis; (c) parallel to the y axis; (d) having slope2; (e) perpendicular to the line having the eqtatio\ 2x + y :7; (f) forming an isoscelestriangle with the coordinate axes. 30. Find an equation of each straight line that is perpendicular to the line having the equation 5r  y: I and that forms with the coordinate axes a triantle having an area of measure 5. 31. Prove analytically that the diatonals of a rhombus are perpendicular. 32. Prove analytically that the line segments joining conEecutivemidpoints of the 6ide8 of any quadrilateral form a parallelogram.
1.6THECIBCLE 43 33. Prove analytically that the diagonals of a parallelogram bisect each other. 34. Prove analytically that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
1.5 THE CIRCLE The simplest curve that is the graph of a quadratic equation in two variables is the "citcle," which we now define.
'1,.6.1 Definition
1.6.2 Theorem
A circle is the set of all points in a plane equidistant from a fixed point. The fixed point is called the center of the circle, and the measure of the constant equal distance is called the radius of the circle. The circle with center at the point C(h, k) and radius r has as an equation ';
l ,: i
:
;,(x *'h}P, * (g1= k)2,* 7zi'1
(1)
pRooF: The point P(x, y) lies on the circle if and only if
l n c l :, that is, if and only if
@:r This is true if and only if (xh)'+
(yk)ztz
which is Eq. (1). Equation (1) is satisfied by the coordinates of those and only those points which lie on the given circle. Hence, (1) is an equation I of the circle. From Definition L.3.2, it follows that the graph of Eq. (1) is the circle with center at (h, k) and radius r. If the center of the circle is at the origin, then h: k: 0; therefore, its equation is x' + y' : 12.
' :LJ'
C(2,  3)
o rLLUsrRArroN 1: Figure 1.6.1 shows the circle with center at (2, 3) and radius equal to 4. For this circle h:2,k:3, and r:4. We obtain an equation of the circle by substituting these values into Eq. (1) and we obtain (x2)'+ly
 (3)12:42
(x2)'+ (y+3)':t6 Squaring and then combining terms, we have x24x+4+y'*5y*9:t6
F i g u r e1 . 6 . 1
)c'+y'4x+6y3:0
REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
EXAMPLE1:
Given the equation
x'+ y'* 6x2y  15: o prove that the graph of this equation is a circle, and find its center and radius.
solurroN:
The given equation may be written as
(xr+6x)t(yr2y):ts Completing the squares of the terms in parentheses by adding 9 and 1 on both sides of the equation, we have
(x' * 6x*9) + (y'  2y I1) : 15+ 9 + 1, (r*3)2+(V1)':25 Comparing this equation with Eq. (1), we see that this is an equation of a circle with its center at (3, I) and a radius of 5.
In Eq. (L), removing parenthesesand combining terms gives x ' + y '  Z h x Z k y+ ( h , + k 2 r r ) : 0
Q)
Equation (2) is of the form )e+y'*Dx+Ey*F:0
(3) 2h, 2k, : : : where D E and F h2* k2 f. Equation (3) is called the general form of an equation of a circle, whereas Eq. (1) is called the centerrsdiusform. Becauseevery circle has a center and a radius, its equation can be put in the centerradius form, and hence into the general form. We now consider the question of whether or not the graph of every equation of the form x'+y,*Dx*Ey*F:0 is a circle. To determine this, we shall attempt to write this equation in the centeriadius form. We rewrite the equation as x'+y,*Dx*EyF and completethe squaresof the terms in parenthesesby adding]nDzand iE' on both sides, thus giving us
F + +Dz* iE, (x' * Dxt +D')+ (y' * Ey++E'z) or, €guivalently, (x + LD)2+ (y + +E)2: i(D' + Ez 4F)
( 4)
Equation (a) is in the form of Eq. (1) if and only if
+(D,+E24F):rz We now consider three cases,namely, (D'+ E2 4F) as positive, zero, and negative. Case7: (D, + E2 4F) > 0. Then r': *(D' + E2 4F), and so Eq. (a) is an equation of a circle having a radius equal to it/FE= 4F and its center at (*D,+E).
1.6 THE CIRCLE
C a s e2 : D z + E z 4 F : 0 . Equation ( ) is then of the form (x + tD)z + (Y * iE)z :0
(5)
 L2D and Becausethe only real values of r and y satisfying Eq. (5) are v +E). Comparing Eq. (5) with y:+E, the graph is the point (+D, and r  0. Thus, this point can be Eq. (1),we see that h:+D, k:iE, called a pointcircle. Case3: (Dt + E2 4F) < 0. Then Eq. (4) has a negative number on the right side and the sum of the squaresof two real numbers on the left side. There are no real values of r and y that satisfy such an equation; consequently,we say the graph is the empty set. Before stating the results of these three casesas a theorem, we observe that an equation of the form Axz*Ay'+Dx*Ey*
F:0
w h e r eA + 0
(6)
can be written in the form of Eq. (3) by dividing by A, thereby obtaining F r c ' + y ' +Dt r *, *E= A,U +i:0 Equation (6) is a special case of the general equation of the second degree: Axz * Bxy * Cy' t Dx * Ey * F : 0 in which the coefficients of x2 and yz are equal and which has no xy term. We have, then, the following theorem. 1.6.3 Theorem
The graph of any seconddegree equation in R2 in r and U, in which the coefficients of x2 and y2 are equal and in which there is no xy term, is either a circle, a pointcircle, or the empty set. o rLLUSrRArroN 2: The equation 2x2*2y'*LZxBy+31:0 is of the form (6), and therefore its graph is either a circle, a pointcircle, or the empty set. If the equation is put in the form of Eq. (1), we have
x'+y't6x4yF#:o ( x ' + 6 x )+ ( y '  4 y ) :  T ( x ' * 6 x* 9 ) + ( y '  4 y + 4 ) :  # + (r*3)2 + (Y2)':E Therefore,the graphis the empty set.
9+ 4
REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
ExAMPLE2: Find an equation of the circle through the three points A(4,5), B(3, 2), and
c(1,,4).
soLUrIoN:
The general form of an equation of the circle is
x'+y,tDxfEy*F:0 Because the threepoints A, B,and C must lie on the circle,the coordinates of these points must satisfy the equation. So we have
L6+25+4D+5E+F:0 9+ 4+3D2E*F:0 1+16+ D4E*F:0 or, €guivalently, 4D+ 5E+F:47 3D2E+F:L3 D  4 E* F :  1 7 Solving these three equations simultaneously, we get D:7
E:5
F:44
Thus, an equation of the circle is
x'+y'*7x5y44:0
In the following example we have a line which is tangent to a circle. The definition of the tangent line to a general curve at a specific point is given in Sec. 3.L. However, for a circle we use the definition from plane geometry which states that a tangent line at a point P on the circle is the line intersecting the circle at only the point P.
nxeuprr 3: Find an equation of the circle with its center at the point C(1,6) and tangent to the line / having the equation xv1:0.
soLUrIoN: See Fig. L.6.2. Given that h: L and k: 6, if we find r, we can obtain an equation of the circle by using the centerradius form. Let I, be the line through C and the point P, which is the point of tangency of line I with the circle.
r: lPCl Hence, we must find the coordinates of P. We do this by finding an equation of /, and then finding the point of intersection of /1 with /. Since 11 is along a diameter of the circle, and / is tangent to the circle , I, is perpendicular to l. Because the slope of / is L, the slope of 11is L. Therefore, using the pointslope form of an equation of a line, we obtain as an equation of I,
y51(r1)
1,6THE CIRCLE
or, equivalently,
x+Y7:0 Solving this equation simultaneouslywith the given equation of l, namely, X A L:0 we get x: 4 and y: 3. Thus, P is the point (4, 3). Therefore,
,:lPCl:ffi or, equivalently,
r: !18 So, an equation of the circle is
(xl)'+
(Y6)':
(\ft)'
or, equivalently,
F i g u r e1 . 6 . 2
x ' + y '  2 x  1 2 y* L 9 : 0
1.6 / )Exercises In Exercises 1 through 4, find an equation of the circle with center at C and radius r. Write the equation in both the centerradius forrr and the gmeral form. ' , . c ( Q ,0 ) , r : 8 r . c 1 4 ,  3 ) ,r : S 3. C(5, 12), f :3
4. C(l,l),r:2
In Exercises5 through 10, find an equation of the circle satisfying the given condition$. 5. Center is at (1, 2\ and through the point (3, 1). 5. Center is at (2,5) and tangent to the line x:7. is at (3, 5) and tangent to the line l}x t 5y  4:0. C"n /Z) w "r 8. Through the three points (2, 8) , (7, 3), and (2,0). 9. Tangent to the line 3x * y * 2:0 at (L, 1) and through the point (3, 5). 10. Tangent to the line 3x * 4y  16:0 at (4,1.) and with a radius of 5. (Two possiblecircles.) In ExercisesL1 through 14, find the center and radius of each circle, and draw a sketch of the graph. ll. f*y'6x8y*9:0 13. 3f * 3y' * 4y  7 :0
12.2f*2y'2x*2y*7:0 1 4 .* * y ' 
10rlOy+25:0
In ExercisesL5 through 20, determine whether the graph is a circle, a pointcircle, or the empty set. 16. 4* l 4y' * 24x  4y t L : 0 1 5. 12* y' 2x * LO y* L9:0 1 7 .f * y ' 
10r* 6y*36:0
1 8 .x 2 * y ' * 2 x  4 y
*5:0
48
REALNUMBERS, INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS
19. *+ f {f.x+36v719:0 20. W 1 9 f I 6 x  6 y + 5  0 21. Find an equation of the comrnon chord_of the two cirdes f*
 6y  12:0 and f* 2yrg:o. 3F+rlt l+gr(rrrrvr: If the coordinates of a Point gatisfy two different equations, then thjcoordinates also satisfy the difierence of the two equatione.)
22. Find the points of intersection of the two cirdes in Exercise 21. i\
Find an equation of the line which is tangent to the circle .d + y,  4t + 6y  12: O at the point (5, 1). 24. Findanequationofeachofthetwolineshavingslopejgwhicharetangenttothecirclef*l*2xWg:0. I
25. From the origin, dbrda of the cirde t' * I * 4r: 0 are drawn. Pr€ve that the set of midpoinb of these chords is a circle. 26. Prove analytically that a line from th€ center of any circle biEecting any drord is perpendicular to the chord. 27. Prcve analytically that an angle inscribed in a semicircle is a right angle. 28. Given the line y: flr + b tangent to the circle rP * y, = r2, find an equation involving m, b, and t
1.7 FUNCTIONS AND We intuitively consider y to be a function of r if there is some rule by THEIR GRAPHS which a unique value is assignedto y by a correspondingvalue of x. Familiar examples of such relationships are given by equations such as y:2f+5
(1)
Y:tPg
(2)
and It is not necessary that r and y be related by an equation in order for a functional relationship to exist between them, as shown in the following illustration. o ILLUsTRATIoN1,: lf y is the number of cents in the postage of a domestic first class letter, and if r is the number of ounces in the weight of the letter, then y is a function of x. For this functional relationship, there is no equation involving r and /; however, the relationship between r and y may be given by means of a table, such as Table 1.7.L. . T a b l e1 . 7 . 1
r: number of ounces in the weight of the letter
x 0 a n d r  2 > 0 ; o r x < } a n d x  2 < 0 .
two cases holds:
CaseT: r>0andrZ>0. That is, x>0
and x>2
Both inequalities hold if x > 2, which is the interval [2, +*;. Case2: x
ifr1
[1, and g(x):lZx Ll
ifr 0 suchthat r 0 there existsaD ) 0 suchthat
ls'=21 I I l <e
v3
c
Now
w h e n e v e r0 ( l t  z l
0 and rais any positive integer, or if L < 0 and n ts a positive odd integer. A proof of this theorem is given in Sec.2.6, and as is the casewith the preceding theorem, we apply it when necessary.
F
LIMITSAND CONTINUIW
o ILLUSTRATToN 6: From the result of Illustration 5 and Limit theorem 1.0, it follows that
lim e4
J?#
L
Following are some examplesillustrating the application of the above _ theorems. To indicate the limit theorem being ,s"d, we use the abbreviation "L.T." followed by the theorem number; for example, ,'L.T. 2,, refers to Limit theorem 2. EXAMPLE L:
lT
Find
(* +7x5)
SOLUTION:
lim (f *7x
5) : lim f * lim 7xlim 5 .1'3
and, when applicable, indicate the limit theorems which are being used.
(1.r.s)
IB
: lim x . Iim r* t3
."3
lim 7 . limr
lim 5
:rS
r_g
3'3+7.35
tB
(L.T.6)
G.T.3andL.T.2)
:9*2L5
It is important, at this point, to realize that the limit in Example 1 was evaluated by direct application of the theorems on limits. For the f u n c t i o n/ d e f i n e db y f ( x ) : x 2 i r x  S , w e s e et h a t f ( g ) : g z + 7 . g  5  25, which is the same as lim (x' + 7x  5). It is not always true that we havet_i1/(r) : f (a) (seeExample4). In ExampleL, rim (x) :/(3) f cause the function / is continuous at x:3. continuous functions in Sec. 2.5.
be
we discuss the meaning of
rxatvrpr,n 2: Find
tFzxc+3
VT+s
and when applicable indicate the limit theorems being used.
(1.T.10) lim (r3 * 2x t 3)
(L.r.e)
lim 12 * lim 5 t2
u2
(L.r.s)
2.2 THEOREMSON LIMITSOF FUNCTIONS
imx)3+lim2'limr*lim3
(L.T.7 and L.T. 8) (L.T. 3 and L.T. 2)
8+4+3
EXAMPLE 3:
Find
27 r. x3 llm^ xB x J
and, when applicable, indicate the limit theoremsbeing used'
solurroN: Here we have a more difficult problem since Limit theorem 9  3) because lim (r  3)  g. cannot be applied to the quotient (rt  27)l(x However, factoring the numerator, we obtain xs : : 27 
(x  3) (x'z* 3x f 9)
x3
x3
This quotient is (r2*3x+9) if x*3 (sinceif x* 3 we can divide the numerator and denominator by (r  3)). When evaluatingItT [(xt  27)l(x  3)], we are consideringvaluesof r close to 3, but not equal to 3. Therefore, it is possible to divide the numerator and denominator by (x 3). The solution to this problem takes the following form: xs_27
..
ttfr3:itTT :
lT
(r3)(f+3x*9) (x' i 3x + 9)
: lim
: (lim r)2 + 1.8
dividing numerator and denominator by (,  3) since x * 3
(1.r.4) (L.T. 8 and L.T. 1)
I3
:32 * L8
(L.r.3)
:27
Note that in Example 3 (rt  27)l(r  3) is not defined when x:3, but lim [(rt  27)l(x  3)l exists and is equal to 27.
NY
80
LIMITSAND CONTINUITY
ExAMPLE4: Given that / is the function defined by
'f[(cx ):
(* 3
li
ifx*4 ifx:4
soLUrIoN:
when
evaluating lim f (*), we are considering values o f x
closeto 4 but not equalto 4. Tf;il, we have lim /(r) : lim (x  3)
find lim f (x).
:t
(1.r.1)
I n E x a m p l e4 l ; r y f @ ) : t
b u t f ( 4 ) : 5 ' t h e r e f o r e ,t T f @ ) + f ( ) .
This is an example of a function which is discontinuousat x:4. In terms of geometry, this means that there is a break in the graph of the function at the point where x: 4 (seeFig. 2.2.1).The graprr or ttre function consistsof the isolatedpoint (4,5) and the straight line whose equa_ tion is y : x  3, with the point (4, L) deleted. There are two limits that we need in order to prove Limit theorems 9 and 10 in Sec.2.6. They are given in the following two theorems.
Figure2.2.1
2.2.11Theorem If. a is any number except zero, ,.
1
llm:taX
1 a
pRooF: We need to consider two cases: a > 0 and a < 0. We prove the theorem if a > 0 and leave the proof for a 1 O as an exercise (see Exercise 30). so, if a is any positive number, from Definition 2.L.j, we must show that for any € ) 0 there exists a 6 > 0 such that
Ir
1l
liil <e lx al Now
w h e n e v e r0 < I r  o l < a
1 1 t l ___ l ta_  x l
lx al
lid I * I:loll* and because a ) 0, we have
: tx otL 11ll al alxl
lx
(6)
We proceed to find some upper bound for the fraction Llalxl. If we require the D for which we are looking to be less than or equal to La, then certainly whenever lx al < 6, we know l* ol < ia. The inequality l*ol1 (a) show that lim /(r) and lim /(r) both exist but ar€ not equal, and hencelim /(r) doesnot exist. r_l

Il+
t_l
(b) Show that lim g(x) and lim g(r) both exist but are not equal, and hence u lim g(r) mg I 1tl+
.ifil. E\rl. (g) Find formulastor ; } k J _ r . r r r q a v l r t . u r q r l ( r l I \ L t f(x) 7(G)"Frove (
\.
.1'r"r .X 7'{ * r l "I * f  j .  '1: , 'fr, ),:. '. i",,,t
, _ . * , ' , . ' , Ii .l "i 'f. i .: : that lim f (x) .g(r) existsby' sfiowing that tim /(x) . S(r) : Iim (x) . S(r). frl ,_r_
does not exist.
tl
'
l
Ir l +1 +
L5. Prove Theorem2.3.9. 17. Let rf \( 'x ) : l
(  l' i f x < 0 L 1 ifx>0
show that lim /(x) does not exist but that lim l/(r)l does exist. .z'0 rO
2.4 INFINITE LIMITS
Let f be the tunction defined by
i@'):& A sketch of the graph of this function is We investigate the function values of / when .r is close to 2. Letting
2.4 INFINITELIMITS
x approach 2 from the right, w€ have the values of f (x) given in Table 2.4.L. From this table we intuitively see that as r gets closer and closer T a b l e2 .4 .7
x
3
t T
3
t2
I
5
4
27
48
2l TO
300
20r T0o
30,000
2001 T0T6
3,000,000
to 2 through values greater than 2, f (*) increaseswithout bound; in other words, we can make f (x) as large as we please by taking r close enough to 2. To indicate that /(x) increases without bound as x approaches 2 through values greater than 2, we write
Figure2.4.1
?
l i m f t : r  *
.rz* (I
Z)
If we let r approach 2 from the left, we have the values of f (x) given in Table 2.4.2.'We intuitively see from this table that as r gets closer and T a b l e2 .4 .2
x
1
3
.)
B
4
3
t2
27
48
D
l9 10
300
199
T00.
30,000
1999
TIITT'
3,000,000
closer to 2, through values less than 2, f (x) increaseswithout bound; so we write f i m $ : *c c
,z \X
z)
Therefore/as )c approaches2 fuomeither the right or the left, f (x) increases without bound, and we write h m j T: *z \x
*oo
We have the following definition. 2.4.1 Definition
Let f be a function which is defined at every number in some open interval I containing a, exceptpossibly at the number a itself. As x approaches a, f (x) increaseswithout bound,which is written lim /(r) IA
: *co
(1)
LIMITSAND CONTINUITY
if foranynumberN > 0 thereexistsa 6 > 0 suchthat/(r) > Nwhenever
0 < l r  a l< D . In words, Definition 2.4.1states that we can make f (x) as large as we please (i.e., greater than any positive number N) by taking r sufficiently close to a. NorE: It should be stressed again (as in Sec. 1.1) that *m is not a real number; hence, when we write it does not have the same 2fO:*a, meaning ur l,ti f@) : L, where L is a real number. Equation (1) can be read as "the limit of f(x) as r approaches a is positive infinity." In such a case the limit does not exist, but the symbolism r'+@" indicates the behavior of the function values f(x) as x gets closer and closer to a. In an analogous manner, we can indicate the behavior of a function whose function values decrease without bound. To lead up to this, we consider the function g defined by the equation
3
/\
g\x):lJ2]l A sketchof the graph of this function is in Fig.2.a.2. The function valuesgiven by g(r):31(xZ)'are the negativesof the function valuesgiven by f (x)31(x2)'. So for the functiong as r approaches 2, either from the right or the left, g(r) decreaseswithout bound, and we write
Figure 2.4.2
3
#W:q) In general,we have the following definition. 2.4.2 Definition
Let / be a function which is defined at every number in some open interval / containing a, exceptpossibly at the number a itself.As x approaches a, f(x) decreases without bound,which is written
rim f(x)
( 2)
tA
if for any number N < 0 t h e r e e x i s t s a 6 ) 0 s u c h t h a t / ( x ) < N w h e n e v e r N when
2.4 INFINITELIMITS
91
a < 6. Similar definitions can be given if lim f(x):an lim /(x) _ @, and lim f (x) : a. "o* *o* suppose ,t"i n is the function defined by the equation
ever 0 < x
,2x h(x):
(3)
A sketch of the graph of this function is in Fig.2.4.3. By referring to Figs. 2.4.I, 2.4.2, and 2.4.3, note the difference in the behavior of the function whose graph is sketched in Fig. 2.4.3 from the functions of the other two figures. We see that
2x
lrfll rt X
Figure2.4.3
:
r,
L
and 2x lim ;: L 2vX
aco
That is, for the function defined by Eq. (3), as r approaches1 through values less than L, the function values decreasewithout bound, and as x approachesL through values greater than L, the function values increase without bound. Before giving some examples,we need two limit theorems involving "infinite" limits. 2.4.3 Limit theorem 11
If r is any positive integer, then (i) lim *: ,tO+
**
4
...\ ,. L [o if r is odd (tt, lt_TF: l+.o if r ts even pRooF: We prove part (i). The proof of part (ii) is analogous and is left as an exercise (see Exercise 19). We must show that for any N > 0 there existsa6)0suchthat 1 ;tN
whenever 0(r 0 and N ) 0, 1
t"aft
w h e n e v e r0 ( r < 6
or, equivalently, since r ) 0,
x
N
w h e n e v eor < l r  o l < o
(4)
Sincelim g(r)  c > 0,by taking e: tc in Definition 2.1.1,it follows that tA
2.4 INFINITELIMITS
93
there exists a 0r ) 0 such that lg(r)
 cl < trc whenever 0
. s@): *oo. (ii) if c I 0,l\f(x).
g(r) : o.
The theorem is valid if. "x > a" is replaced by . rr,r,usrRArrorq 3: q
l i m . . * f u : * o 5) r_s tx
r
r.
x*4
and lim *s x
. 7 4
Therefore, from Theorem 2.4.6(ii), we have S i. I ' x*Af 'iT LG3F x41:€'
96
LIMITSAND CONTINUITY 2.4.7 Theorem
If t;yf@)_oo
and limg(x):c,
where c is any constant except 0,
then (i ) i f c ) O,Li ^ f (x)' S (r) : oo. (i i ) i f c I 0,l t^ f (x)' g(r): T h e th e orem i s val i d i f " x+
* o.
a" i s repl aced by " )c+ a* " ot" x+
a . "
o ILLUsrRArroN 4: In Example 2(b) we showed
{44 ,llm. :
oo
;:; x2 Furthermore, ,. x3 llm  _ ;:;x + 2
t 4
Thus, from Theorem 2.4.7(ii), it follows that
.lfm . l$=7
l:;:;L x2
Exercises 2.4 ,,Q
In Exercises1 through"l.4, evaluatethe limit. 1. lim x2a+X
'
4
t+)
4.rimffi ,. t/s+*
/. rlm .r*o
X'
'\R 10. lim r4
X' 4
Dm(i+) 16. Prove that lim 17. Prove that lim
?
ffi: ) ffi:
*o by using Definition 2.4.1. @ by using Definition 2.4.2.
18. Use Definition 2.4.1,to prove that
jlils+'l :*,.
15rl
x  3 1 :
x*21
r€
o
AT A NUMBER OF A FUNCTION 2.5 CONTINUIry 19. Prove Theorem 2.4.3(ii).
20. Prove Theorem 2.4.4(1i).
21. Prove Theorem 2.4.4(iii).
22. Prove Theorem Z.a.a(iv\.
23. Prove Theorem 2.4.5.
24. Prove Theorem 2.4.6.
25. Prove Theorem 2.4.7.
2.5 CONTINUITY OF A FUNCTION AT A NUMBER v
In Sec.2.1 we considered the function / defined by the equation ,.,.\ (2x * 3) (x  1) I\x): x_.1 We noted that / is defined for all values of r except 1. A sketch of the graph consisting of all points on the line y :2x * 3 except (1, 5) is shown in Fig. 2.5.1.There is a break in the graph at the point (1,5), and we state at the number 1.. that the function f is discontinuous for instance, the function is defined for all If we define f(l):2, values of x, but there is still a break in the graph, and the function is still however, there is no break in discontinuous at L. If we define f(l):5, the graph, and the function / is said to be continuousat all values of x. We have the following definition.
2.5.1 Definition
The function / is said to be continuousat the number a if and only if the following three conditions are satisfied: (i) f (a) exists. (ii) /(x) exists. L'i (iii) lim f (x): f (o). If one or more of these three conditions fails to hold at a, the function f at a. is said to be discontinuous We now consider some illustrations of discontinuous functions. In each illustration we draw a sketch of the graph, determine the points where there is a break in the graph, and show which of the three conditions in Definition 2.5.1fails to hold at each discontinuity. L: Let / be defined as follows: . rLLUSrRArroN ?r. [(2x+3)(l1) ,1 f(x):j lz
ifx*t if x:'1.
A sketch of the graph of this function is given in Fig. 2.5.2.We see that there is a break in the graph at the point where x : 1, and so we investi
LIMITSAND CONTINU]TY
gate there the conditions of Definition 2.5.1.. f (1) :2; thercfore,condition (i) is satisfied. LtT fttl Itm f (*):
:5; therefore,condition (ii) is satisfied. 5, but f (L) :2; therefore,condition (iii) is not satisfied.
We conclude that f is discontinuous at L.
o
Note that if in Illustration I f (l) is defined to be 5, then lim /(r) : f (1) and / would be continuous at 1.. o rLLUsrRArroN2: Let f be defined bv 1
,f,\( xt \ : +x _ 2
Figure2.5.3
A sketch of the graph of f is given in Fig. 2.s.2. There is a break in the graph at the point where x : 2, and so we investigate there the conditions of Definition 2.5.1.. f (2) is not defined; therefore, condition (i) is not satisfied. We conclude that f is discontinuous at 2. o o rLLUsrRArroN 3: Let g be defined by
rJ
8(r): l[3
Figure2.5.4
if x # 2 ifx2
A sketch of the graph of g is shown in Fig. 2.5.4.Investigatingthe three conditions of Definition 2.5.1,at 2 we have the following. S(2):3; therefore,condition (i) is satisfied. : *mi therefore, condition (ii) is not liT g(x) : m, and ]i1 S(r) satisfied. Thus, g is discontinuousat 2. . . rLLUsrRArroN4: Let h be defined by 7., [3*x n\x):tg_,
if.x="]. ifj,<x
A sketchof the graph of h is shown in Fig. 2.5.5.Becausethere is a break in the graph at the point where x:1, we investigate the conditions of Definition 2.5.1at L. We have the following. h(t1 :4; therefore,condition (i) is satisfied. lim h(r) : lim (3 + r1 :4 t7lim h(r)  lim (3 x):2
a1"
Figure2.5.5
,tl+
AT A NUMBER 99 OF A FUNCTION 2.5 CONTINUIW Because limh(x)
+limh(r),
we conclude that lim h(x) does not exis
lrl+
tl
t'l
therefore, condition (ii) fails to hold at 1. Hence, h is discontinuous at 1. 5: LCt F bC dCfiNCdbY o ILLUSTRATION
F(r) Figure2.5.6
flrsl L2
ifx*3 i f .x : 3
A sketch of the graph of F is shown in Fig. 2.5.5. We investigate the three conditions of Definition 2.5.1at the point where x: 3. We have F(3) :2; therefore, condition (i) is satisfied.
nttl : 0. So lim F(r) exists and is 0; therefore, tttl : 0 and l,T_ l.T condition (ii) is satisfied. tim F(r) :0 but F(3) :2; therefore,condition (iii) is not satisfied. c'3
Therefore, F is discontinuous at 3.
o
It should be apparent that the geometric notion of a break in the graph at a certain point is synonymous with the analytic concept of a function being discontinuous at a certain value of the independent variable. After Illustration L we mentioned that it f(1) had been defined to be 5, then / would be continuous at 1. This illustrates the concept of.aremovable discontinuity. In general, suppose that f is a function which is discontinuous at the number a,but for which f(tl exists. Then either f(a) + Tq fOl or else /(a) does not exist. Such a discontinuity is called a refi movable discontinuity because if f wete redefined at a so that f(a) l;a f@), / becomes continuous at a.If. the discontinuity is not removable it is called an essentialdiscontinuita.
nxervrpr.El: In each of the Illustrations 1 through 5, determine if the discontinuity is removable or essential.
solurroN:
In Illustration 1 the function is discontinuous at 1 but lilf
(x):
5. By redefining f0 :5, we have lim /(x) : f (t), and so the discontirr nuity is removable. In Illustration 2, the function / is discontinuous at 2. lim /(r) does not r'2 exist; hence, the discontinuity is essential. In Illustration 3, lim g(r) does not exist, and so the discontinuity is I2 essential.
In Illustration 4, the function h is discontinuousbecauselim h(r) does tr not exist, and so again the discontinuity is essential. F(x):0, but F(3): 2, and so F is discontinuousat In IllustrationU, TT 3. However, if F(3) is redefined to be 0, then the function is continuous at 3, and so the discontinuity is removable.
lOO
LIMITSAND CONTINUITY
Exercises 2.5
y ,1
In Exercises 1 through 22, draw a sketch of the graph of the function; then by observing wher€ there are breals in the graph, detemrine the values of the independent variable at which the fun*ion is discontin;ous and show why Definition 2.5.1 is not satisfied at each discontinuitv.
4 l . f .( ,x )\ : t , fx+ 4 Lt
irx#4 i f .x : 4
ra_16 a.h@):fr
. r+3 7.h(x):Vfrj_6
I t+x 1 0 .H ( r ) : l 2  x l2x1
x2+x6 3. F(r) : x*3
6. f(x):
f2x2llx*12 x25x*4
e.f(x):l
t1, if x < 0 0 if x :0 I r if 0 < x
v24
8.g(r):;d
if x 0 such that lf(x)Ll <e
w h e n e v e r0 < 1 "  a l
for any € ) 0 there existsa 0 there exists a 6 > 0 such that lf(x)f(a)l <e
w h e n e v e r0 ( l x  o l
0 becausewhen x: a, statement(6)obviously holds. We have, then, the following definition of continuity of a function by using e and 6 notation.
2.6.4Dehnition
The function f is said to be continuous at the number a if / is defined on some open interval containing a and if for any € > 0 there exists a 6 > 0 such that lf(x)f(a)l <e
wheneverlx41 0 there exists a 6r ) 0 such that  (b) < ., whenever lS(t)  bl < D' (e) lf QQ))  f From statements (9) and (8), we conclude that for any er ) 0 there exists a 6z ) 0 such that ( l*  ol < D' IfGGD f(b)  < ., whenever 0 from which it follows that lim/(s(r)):/(b)
104
LIMITSAND CONTINUITY
or, equivalently,
) : /(tims(r)) fif G2
xt:2
3 . f ( x ) : l r  3 l ;x , : 3
a.fk):L*lx+2lr;lh:2
?t s.f(x):trl1
6.f(x):lbll; == 3
ifro
rr:0
xr:0 (r2
i.f
1
Find the values of c and, 8o that/'(1)
exists.
32. Suppose IVJ:
(f lot"+bt +,
if x 0, however small, there exists a number N ) 0 such that lf(*)Ll
<e
w h e n e v e rr ) N
166
TOPICSON LIMITS,CONTINUry,AND THE DERIVATIVE
NorE: When we write x + *m, it does not have the same meaning as, for instance, x + 1000. The symbolism "x >*a" indicates the behavior of the variable x. However, w€ can read Eq. (1) as "the limit of f(x) as x approaches positive infinity is L," bearing in mind this note. Now let us consider the same function and let r take on the values 0, L, 2, 3, 4, 5, 10, L00, 1000, and so on, allowing x to decrease through negative values without bound. Table 4.7.2 gives the corresponding function values of f (x). Table 4.7.2
0 1 ') v2
f(*):ft+I
2
3
4
5
8183250
5
10 tr%
10
100
1000
200 20,000 2,000,000 101 L0,001. 1,000,001
Observe that the function values are the same for the negative numbers as for the corresponding positive numbers. So we intuitively see that as x decreases without bound, f (x) approaches 2, and formally we say that for any e ) 0, however small, we can find a number N < 0 such that lf (x) 2l < , whenever x < N. Using the symbolism "x >@" to denote that the variable r is decreasing without bound, we write 2x2
1. jl*i+2
In general, we have the following definition.
4.1.2 Definition
Let f be a function which is defined at every number in some interval (*, a). The limit of f(x), as x decreasesraithoutbound,is L, written
f?):t ,t1T
Q)
if for any € ) 0, however small, there exists a number N < 0 such that Ll oo. We have the following additional theorem. 4.1.3 Limit theorem 1,3 If r is any positive integer, then
(t) j: o "t1T_
(ii),11a4:o
6" (2) as Limit a" is limit
4.1 LIMITSAT INFINITY
pRooF oE (i): To prove part (i), we must show that Definition 4.1,.1, holds tor f (x): tlx' and L: 0; that is, we must show that for any e ) 0 there exists a number N ) 0 such that lr
I
l+01 <e I lx'
w h e n e v e rr ) N
or, equivalently, 1
lrl"=Z
wheneverrlN
or, equivalently, since r ) 0, /1\rb
whenever r)N
ltl t(;/
In order for the above to hold, take N : (lle)rtr. Thus, we can concludethat rr  /l\ur l+  0l < e whenever x ) N, if N: l: I ter I lx' r This proves part (i). The proof of part (ii) is analogous and is left as an exercise (see Exercise 22).
rxlvrplr r. rm 'j+@
L:
Find
4x3 ;J= L*  J
and when applicable indicate the limit theorems being used.
sol,urroN: To use Limit theorem 1.3,we divide the numerator and the denominator by r, thus giving us 4x3
"11
,.
43lx
zrc+s:"t1t7+ijlim (4  3lx)

(L.r. e)
tlo
lim (2 + slx) tl@
lim 4_
t+6
Iim (3/r) t*o
j1t2+"lit $tx') lim 4
lim 3' lim (Ux\
lim 2 * lim 5 ' lim (Ux) J *@
_43.0 2+5.0 2
(L.r.4)
Jfjo
(L.r. 6)
t*@
(L.T.2 and L.T. 13)
168
TOPICS ON LIMITS, CONTINUITY, ANDTHEDERIVATIVE
EXAMPLE2: Find r.
2x2r*5
,'14 4,,*l and when applicable indicate the limit theorems being used.
soLUrIoN: To use Limit theorem 13, we divide the numerator and the denominator by the highest power of x occurring in either the numerator or denominator, which in this case is 13. So we have r:* 2x2x*5_
rj_ 2lx'l,lxz*Slx}
"'14 AxsL:"'1a@ lim (2lx  tlx2 + Slx")

t'@
(L.r.e)
(4 ttxs)
Ji*
lim (zlx)  lim (Ux') * lim (5/r') 04@
I+@
lim 4 
(1.r.5)
t+@
lim (Ux')
Q6
2 ' l i m (ux)  lim (Llf) * lim 5 . lim (Llx") I@
I@
4  lim (Ux')
!t6
(1.r.6)
fi@
2. 0 0 + 5 . 0
(L.T.2 and L. T. 13)
40
EXAMPLE
,.
riii:t
Find 
3x*4
rl.a V2*
 5
and when applicable indicate the limit theorems being used.
soLUrIoN:
We divide the numerator and the denominator of the fraction by r. In the denominator we let r: t/7 since we are considering only positive values of r.
r:_
3x*4
3+4lx
,:
"'11ffi,:,t1tffilw
lim G + alx) lim 1D=W
(L.r.e)
Ilq
Ilq
lim G + alx) t*a
lim
,(L.r. 10)
(2  5lx2)
f*m
]g3*_li1 @tx) .T*€
(1.r.4)
t*a
lim 3* lim 4. lim (Tlx) t*a
Vlim 2
Il@
,f,*co
lim 5. lim (tlfl
(L.r. 6)
t'ta
3+4.0
(L.T.2 andL.T.13)
4.1 LIMITSAT INFINITY 1 6 9
The function is the same as the one in Example 3; however, solurroN. x or, since we are considering negative values of x, in this case #: equivalen tly, t/P : r. Hence, in the first step when we divide numerin the denominator, and we ator and denominator bv x, we let x :!* have ,.
3xt4
,:
Ja6s:*'11
3*4lx
ffi
lim (3 + alx) t@
lim 1t/ffi) lim 3* lim (alx) Jf@
c@
@
The remaining steps are similar to those in the solirtion of Example 3, and in the denominator we obtain tfZ instead of t/2. Hence, the limit is
4trt.
"Infinite" limits at infinity can be considbred. There are formal definitions for each of the following. l i m /(r) : * m
l i m /(r) : *m
t@
tl@
l i m /(r) :  m
lim
/(x)
: @
iX@
fil@
For exampte, f@): *m if the function/is defined on some interval "lit (a, +*1 and if for any number N > 0 there exists an M > 0 such that f (x) > N whenever r > M. The other definitions are left as an exercise (seeExercise17). Find
EXAMpLE 5: ,2
lim*
rr*o X t
L
Dividing
solurroN; obtain
the numerator
Y2
= *rlim t.1 t w x ' lim _*: r  * c ox f L
and the denominatot
by x2, we
1
Evaluating the limit of the denominator, we have
l i m f t * +f ) : H m1X * r i m4 :o*o:o T
r+
\I
/
r+*
r+*
Therefore, the limit of the denominator is 0, and the denominator is approaching 0 through positive values. The limit of the numerator is 1, and so by Limit theorem lz(i) (2.4.4) it follows that ryz
lim _*: r*a
X+
L
*cc
170
TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE
EXAMPLE 6:
soLU'oN, ,rit#*litf,ffi
Find
,l r.m . 2 x  *
We consider the limits of the numerator and the denominator separately.
x*@ 5X + 5
/)\t
lim I'r+o
1l:
\I
/
lim /q* 4): f/
r+*
\I
lim:,+*
lim 1:01:1
X
r+cc
tim !+ ti* i": o * o: o f
. . r  + c oI
*+*
Therefore, we have the limit of a quotient in which the limit of the numerator is  L and the limit of the denominator,is 0, where the denominator is approaching 0 through positive values. By Limit theorem 1.2(iii) it follows that r.
llm r+€
2x xz
:fr, JX +
5
Exercises 4.1 In ExercisesL through'l'4, find the limits, and when applicable indicate the limit theorems being used. 4
a. r. u mr+a
2x*l JX 
,^ . _r .l l f l4's32J* 3 I
L
{FT4
*Zx*5
.,' X,i.T x s + x + 1
5. trm u**o y+4
,. 4f+2f5 '',11 w
to.
,{1 fffi
x x)
^ ,. 3xa7f +2 o' 2x4+ l "1T. 11. lim (\7F;  fF + 1)
9. lim ,3*@
12. lim [
\FT
*o
r s .n ^ ?5 "y * 34 ya
For the functions defined in Ex€rrises 15 and 16, prove that lim l(x) :1 by applying Definition 4.1.1; that is, for any e ) 0, show that there exists a number N > 0 such that lf (*l  1l < . whenever x ) N.
r s .f ( x ) : *
16.f(x):H
17. Give a definition for each of the following:
(a) lim /(x)  @; (b) lim f (x):
*o;
,f6
18. Prove tlut lim
(f 4):+€
(c) lim /(r) : oo f@
by showing that for anyN > 0 there exists anM > 0 such t}at (f 4) > N whenever
x> M. 19. Prove that lim (6  I  t') : 6 by applying the definition in Exercise 17(a).
l 4.2 HORIZONTALAND VERTICALASYMPTOTES
171
20. Prove part (i) of Limit theorem 12 (2.4.4) if "x + a" is replaced by "x + *a." 21. Prove that
,._ 8r*3 _n j1i 2' 1 = by showing that for any e ) 0 there exists a number N < 0 such that
 l 0, no matter how large, the line y: k will intersect the graph of f in two points; the distance of these two points from the line x: a gets smaller and smaller as k gets larger and larger. The line r : A is called a "vertical asymptote" of the graph of /. Following is the definition of a vertical
4.2.1 Definition
The line x: A is said to be a aerticalasymptoteof the graph of the function f if at least one of the following statementsis true: (i) lim f ( x ) : 4 a td+
(ii) lim 8'd+
(iii) lim trA
f(x):* f@)a*^.
(iv) lim f ( x ) :  a . fi'A
For the function defined by Eq. (1), both parts (i) and (iii) of the above definition are true. If g is the function defined by Figure 4.2.2
. q\ (. ,x/ ) : o
1
(X _
A),
172
TOPICSON LIMITS,CONTINUITY,AND THE DERIVATIVE
then both parts (ii) and (iv) are true, and the line r : A isa vertical asymptote of the graph of g. This is shown in Fig. 4.2.2. A "hotizontal asymptote" of a graph is a line parallel to the x axis. 4.2.2 Definition
The line A: b is said to be a horizontalasymptoteof the graph of the function f tf at least one of the following statementsis true:
(i) ,lit f @):u.
(ii) lim f (x):to. ExAMPLE L: Find the horizontal asymptotes of the graph of the function / defined by r/
r
I\x):ffi
I
and draw a sketch of the graph.
solurroN:
First we consider,lim
f (x), and we have
lirfe)J1*# To evaluate this limit we write x: t/x' (x > 0, because r > +co) and then divide the numerator and the denominator, under the radical sign, by xr. x t:llm ....=: r*co Vxz * 1
1.  \/P ltm .J; Vf +  :
lim ,Tl@
(Ltf) 1
l
Therefore,by Definition a.2.2(1), the line y:
L is a horizontal asymptote. Now we consider (x); in this case we write x  \/P because ,lim f if x +@, x < 0. So we have l./&
lim/(r):,tiaffi :limta
t I * Llxz
(uf) 1 Figure4.2.3
Accordingly,by Definition 4.2.2(ii), the line y  1 is a horizontal asymptote. A sketch of the graph is in Fig.4.2.2.
Exavprn 2: Find the vertical and horizontal asymptotes of the graph of the equation
solurroN:
y:t2
Solving the given equation for y, we obtain
4.2 HORIZONTALAND VERTICALASYMPTOTES
xy'  2y'  4x  0, and draw a
Equation (2) defines two functions:
sketch of the graph. U
v
: f ,(x)
where fi is defined by
f ,(x)
: *2
{
and
 ztlh A: fr(x) wheref2 is definedby fr(x) The graph of the given equation is composed of the graphs of the two functions f, and fr. The domains of the two functions consist of those > 0. By using the result of Example 8, values of x f.or which xl(xz) we see that the domain of /r and fz is )c:2, excluding 1.7, and Sec.
(*,0] u (2,+m). Figure 4.2.4
Now, consider /r necause
Ir2f,(x): ri^.21#:
*oo
by Definition 4.2.1,(i)the line x  2 is a vertical asymptote of the graph of ft'
2rl#:,t11 zyftO:z ,tit f,@):*t1t Figure4.2.5
so by Definition 4.2.2(1)the line y : 2 is a horizontal asymptoteof the graph of ft. similarly,li1; fr@):2. A sketch of the graph of f is shown in Fig. 4.2.4.
Hence, by Definition 4.2.1,(ii)the line x: graph of fz.
2 is a vertical asymptote of the
so by Definition a.2.2(1)the line y : 2 is a horizontal asymptoteof the graph of fr. 2. A sketchof the graph of fris shown in Fig.4.2.5. Also, 11* fr(x): The graph of the given equation is the union of the graphs of.ft and fz, and a sketchis shown in Fig. 4.2.6.
4.2 Exercises In Exercises I through 14, find the horizontal and vertical asymptotes of the graph of the function defined by the givm equation, and draw a sketch of the graPh.
l.f(x):#
2. f (x):;,
_a
3. g(x):
_?
G;zy
174
TOPICS ON LIMITS, CONTINUry, ANDTHEDERIVATIVE
4. F(x):
x2*8x*16 4x2
s .f ( x ) : # x _ 6 8./(r):;a
/ . h \ x ): 7 ; J
10.s(r):ffi6 /Ly2
6. G(r):
6x2*11.r10
y2
11.F(x):A;+
A.f(x):w#fr
13.tf(x)+r\, \/x"_2
In ExercisesL5 through 21, find the horizontal and vertical asymptotes of the graph of the given equation, and draw a sketch of the graph. 1 5 . 3 x y 2 x  4 y  3 : 0 '/..8. Zxyz* 4y,  Jx:0 21. x2y * 6xy  f + 2x * 9y * 3 : 0
4'3 ADDITIONAL THEOREMS ON LIMITS OF FUNCTIONS
4'3'1'Theorem
1 6 .Z x y* 4 x  3 y * 6 : 0 1 9 .( y '  1 ) ( x  3 ) : 6
17. x2y2 x2+ 4yr:0 20. xyzI 3y,  9r: 0
We now discuss five theorems which are needed to prove some rmportant theorems in later sections.After the statementof each theorem, a graphicalillustration is given. If liry /(x) existsand is positive, then there is an open interval containing a such that f (x) > 0 for every x # a in the interval. . rLLUsrRArroN1: Consider the function defined bv f
f(x) : 2 x  l A sketchof the graph of / is in Fig. 4.9.r. Becausetiq /(4 : L, and 1 > 0, according to Theorem 4.3.1 there is an open interval containing 3 such that /(r) ) 0 for every x * 3 in the interval. such an interval is (2, 4). Actually, any open interval (a,b)for which i = a< 3 and b > gwill do. o pRooFoF rHEoRa''4.3.7: Let L: tjT f@).By hypothesis,L > 0. Apply_ ing Definition z.r.r and taking e:!L, we know thereis a 6 > 0 such that
lf (*)  Ll < +f whenever 0 < l*  ol < a (1) Also, lf (x) LI < lL is equivalentto *L < f (x) L < tf (refer to Theorcm'/..2.2),which in turn is equivalent to
Figure4,3.1
tt M whenever 0 < l* ol < E. > Then tt /(r) exists and is equal to L, L M' lT The proof is left as an exercise(SeeExercise3)' 5: Figure 4.3.4 also illustrates Theorem 4.3.5.From the o rLLUSrRArroN  1l < i; and, becauseas figure we see that /(r) = I whenever 0 < lt (*, B) exceptat L, Theointerval pieviously stated, f is dehnedon the open o L > +' L, then rem 4.3.5statesthat if lim /(r) exists and is
4.3 Exercises 1. Prove Theorem 4.3.2.  cl ts sufficiently small, and € > 0, the following inequalities 2. prove Theorem 4.3.3. (rrrNr: Ftust show that when lr m u s t h o l d : L  e < f ( x ) < I * e , a n d L  e < h ( x )< L * ' e ' \ 3. Prove Theorem 4.3.5. and that l'(0) : 0. (IIrNr: UEeTheo4. Ler f be a function such that ll(r)  = I for all r. Prove that f is differentiable at 0 rem 4.3.3.)
ON 4.4 CONTINUITy INTERVAL AN
In Example 3 of Sec. 2.6 we showed that the function h, for which is continuous at every number in the oPen interval h(x): {4P,
178
TOPICSON LIMITS, CONTINUIry, AND THEDERIVATIVE
' 4'4'l Definition
ExAMPLE1: If f (x) : U (x  g) , on what open intervals is f continuous?
(2, 2). Because of this fact, we say that h is continuous on the open interval (2, 2).Following is the general definition of continuity on an open interval. A function is said to be continuoLtson an open interaalif and only if it is continuous at every number in the open interval.
solurroN: The function / is continuous at every number except3. Hence, by Definition 4.4.1,/ is continuous on every open interval which does not contain the number 3.
We refer again to the function hfor which h(x)  {E4.We know that h is continuous on the open interval (2,2). However, becauseh is not defined on any oPen interval containing either 2 or 2, we cannot consider lim h(r) or lim h(x).Hence, to discussthe question of the r2
r2
continuity of h on the closedinterval [2,21, we must extend.the concept continuity to include continuity at an endpoint of a closed interval. _of We do this by first defining righthandcontinuity and,lefthandcontinuity. 4'4'2 Definition
The function / is said to be continuous from the right at the numbera if. and only if the following three conditions are satisfied: (t) f (a) exists.
(ii)
exists. liT /(r)
(iii) lim f(x):f(o). 4.4.3 Definition
Th9 fungtion / is said to be continuous from the teft at the numbera if and. only if the following three conditions are satisfied: (i) f(a) exists.
Gt)I'T /(x) exists. (iii) lim f(x):f(a).
4.4.4 Definition
rxaupr.n 2: Prove that the function h, f.or which h(x): l/4 xr, is continuouson the closedinterval l2,21.
A function whose domain includes the closed interval la, bl is said to be continuous on fa,b] if and only if it is continuous on the open interval (a, b), as well as continuous from the right at a and,continuous from the left at b.
soLUTroN: The function h is continuouson the open interval (2,2) and
\/47:0: "ti1.
h(2)
4.4 CONTINUIW ON AN INTERVAL
tim GE
179
 o: h(2)
I2
Thus, by Definition 4.4.4,h is continuous on the closedinterval l2,2f . 4.4.5 Definition
nxavrprr 3: Given / is the function defined bY
f(x) : determine whether f is continuous or discontinuous on each of the following intervals:
(3,2),?3,21,[3,2),(3,21.
(i) A function whose domain includes the interval halfopen on the
right la, b) is said to be continuous on la, b) if and only if it is continuous on the open interval (a, b) and continuous from the rig}:rt at a. (ii) A function whose domain includes the interval halfopen on the left (a, bl is said to be continuouson (a,bl if and only if it is continuous on the oPen interval (a,b) and continuous from the left at b. solurroN: We first determine the domain of /. Th" domain of / is the is nonnegative. Thus, any set of all numbers for which (2x)l(3*x) of this fraction the denominator and numerator the which r for values of changes numerator The domain. the from excluded are signs have opposite 3. We : I : when sign changes denominator artd the 2, I sign when *ik" use of Table 4.4.L to determine when the fraction is positive, negative, zero,or undefined, from which we are able to determine the values of.x for which /(x) exists. The domain of / is then the interual halfopen on the left (3, 21. The function / is continuous on the oPen interval (3,z).It is continuous from the left at 2 because
t2'
l'_t{ffi:s:f(2) 3 because However, f i" not continuous from the right at b
,ri1.rl'#:*rc We concludethat f is continuous on (3, 2] and discontinuouson [3,2] and [3, 2). Table4.4.7
2 x
3*x
2 x 3+x
f(x)
x f (x) for all I in this interval. Figures 4.5.1,and 4.s.2 each show a sketch of a portion of the graph of a function having a relative maximum value at c.
x
4.5.2 Definition
Figure4.5.2
The function / is said to have a relatioe minimum aalue at c if there exists an open interval containing c, on which is defined, such that (c) = (x) / f f for all r in this interval. Figures 4.5.3 and 4.5.4 each show a sketch of a portion of the graph of a function having a relative minimum value at c.
ac
Figure4.5.3
Figure 4.5.4
If the function / has either a relative maximum or a relative minimum value at c, then / is said to have a relatiue extremum at c. (The plurals of maximum and minimum are maxima and minima; the plural of extremum is extrema.) The following theorem enables us to locate the possible values of c for which there is a relative extremum.
4.5 MAXIMUMAND MINIMUMVALUESOF A FUNCTION
183
4.5.3 Theorem lf f(x) exists for all values of x in the open interval (a,b) and if / has a relative extremum at c, where a < c 1 b, then It f' (c) exists,f' (c) : 0. pRooF: The proof will be given for the casewhen / has a relative minimum value at c. tf f '(c) exists,from formula (5) of Sec.3.3 we have
(1)
f,(,):tjTfff Because / has a relative minimum existsaD)0suchthat
value at c, by Definition 4.5.2, there
w h e n e v e r(o l x  c l < 6
f(x)f(c) =o
If r is approachingc from the right, x  c ) 0, and therefore f(c) = o f(x) xc
whenevero ( r  c ( D
By Theorem 4.3.5,if the limit exists, ,.
fk)f(c)
lrrll>U nc+
)C

 n
(2)
C
Similarly, if x is approaching c from the left, x  c ( 0, and therefore f(x) f(c) o xc
whenever6 < x c < o
so that by Theorem 4.3.4,if the limit exists, Ilm
;c
f(x)  f(c) x
c
Sv
(3)
Becausef '(c) exists, the limits in inequalities (2) and (3) must be '(c). So from (2) we have equal, and both must be equal to f
f'(c) =o
(4)
and from (3),
f'(c)'o
(5)
Becauseboth (4) and (5) are taken to be true, we conclude that
f ' ( c ): o which was to be proved. The proof for the case when / has a relative maximum value at c is I similar and is left as an exercise (see Exercise 37). The geometrical interpretation of Theorem 4.5.3 is that if / has a rela
184
TOPICS ON LIMITS, CONTINUITY, ANDTHEDERIVATIVE
tive extremum at c and if f ' (c) exists,then the graph of y : /(x) must have a horizontal tangent line at the point where x: c. If / is a differentiable function, then the only possible values of x for which f canhave a relative extremum are thosefor which f ' (x): 0. However, f ' (x) can be equal to zero for a specific value of x, and yet f may not have a relative extremum there, as shown in the following illustration. 1: Consider the function f defined bv r rLLUSrRArroN f(x):(r1;' A sketch of the graph of this function is shown in Fig. 4.5.5. ,(x): f 3 ( x  L ) 2 ,a n d s o f ' ( 1 ) : 0 . H o w e v e r f, ( x ) < 0 , i f x .   ' 1 ,a, n d / ( r ) j 0 , if. x > 1,.So / does not have a relative extremum at L. o A function / may have a relative extremum at a number and f, may fail to exist there. This situation is shown in Illustration 2.
Figure4.5.5
o rLLUSrRArroN2: Let the function f be defined as follows:
f(x): {';_: lf; .=i
x
A sketch of the graph of this function is shown in Fig. 4.5.6.The function / has a relative maximum value at 3. The derivativ" irorr, the left at 3 is given by f (3) : 2, and the derivative from the right at 3 is given by f *(3)  1. Therefore,we concludethat /'(3) does notlxist o Illustration 2 demonstrateswhy the condition ,'f ,(c) exists,,must be included in the hypothesis of Theorem 4.5.3. In summary, then, if. a function / is defined at a number c, anecessary condition fot f to have a relative extremum there is that either ,(c) :'O f or f '(c) does not exist. But we have noted that this condition is not sufficient.
Figure4.5.6
4.5.4 Definition
If c is a number in the domain of the function and if either , (c): f f f'(c) does not exist, then c is called a critical number of f.
0 or
Because of this definition and the previous discussion, we can conclude that a necessary condition for a function to have a relative extremum at a number c is for c to be a critical number.
ExAMPLE1: Find the critical numbers of the function / defined by f (x)  x4ts+ 4xrt3.
SoLUTIoN:f,(x)!*,,,+t*_,,":t*_,,'1x*):w '(x): 0 when r_ L, and f '(x) doesnot existwhen t: 0. Both1 f and 0 are in the domain of f therefore, the critical numbers of f are L and 0.
AND MINIMUM VALUES 4.5 MAXIMUM OF A FUNCTION 185 We are frequently concerned with a function which is defined on a given interval, and we wish to find the largest or smallest function value on the interual. These intervals can be either closed, open, or closed at one end and open at the other. The greatest function value on an interval is called the "absolute maximum vallJe," and the smallest function value on an interval is called the "absolute minimum value." Following are the precise definitions. 4.5.5 Definition
The function / is said to have an absolute maximum aalue on an interaal if there is some number c in the interval such that /( c) > f (x) tor all r in the interval. In such a case, f(c) is the absolute maximum value of / on the interval.
4.5.6 Definition
The function / is said to have an absoluteminimum aalue on an interaal if there is some number c in the interval such that /(c) = f (x) for all r in the interual. In such a case, f(c) is the absolute minimum value of / on the interval. An absolute extremum of. a function on an interval is either an absolute maximum value or an absolute minimum value of the function on the interval. A function may or may not have an absolute extremum on a given interval. In each of the following illustrations, a function and an interval are given, and we determine the absolute extrema of the function on the interval if there are any. o rLLUsrRArroN 3: Suppose / is the function defined by
Figure4.5.7
f(x) 2x A sketchof the graphof / on ll, 4) is shownin Fig.4.5.7.The function/ has an absoluteminimum valueof 2 on 11,4).Thereis no absolutemaxilim /(r) : 8, but /(r) is alwayslessthan mum valueof / on lL, 4),because 8 on the given interval. o rLLUsrRArIoN 4: Consider the function / defined by
f(x):* A sketch of the graph of f on (3,2f is shown in Fig. 4.5.8.The function / has an absolute maximum value of 0 on (3' 21. There is no absolute minimum value of / on (3,21, because lim /(r) 9,but, f(x) is always t'Bo greater than 9 on the given interval. o TLLUSTRATIoN 5: The function / defined by Figure4.5.8
f(x) :
L
x2
186
TOPICSON LIMITS,CONTINUITY,AND THE DERIVATIVE
has neither an absolute maximum value nor an absolute minimum value on (1, 1). A sketch of the graph of / on (1, 1) is shown in Fig. 4.s.9. Observe that ,t1T. f
@) : a
and l_tT_fOl:
*oo
o
o rLLUSrRArroN 6: Let f be the function defined by
f(x): {X,*_rr* *,
ifx 0  *a) ro and less than a suitable positive 6. Also, 1ry can be Ir1/tr) made greater than any positive number by taking (x  3) > 0 and less than a suitable positive D. We may speak of an absolute extremum of a function when no interval is specified. In such a case we are referring to an absolute extremum of the function on the entire domain of the function.
F i g u r e4 . 5 . 1 1
4.5.7 Definition
/(c) is said to be the absolute maximum aalue of the function f if c is in the domain of f and it f (c) > f (*) for all varues of r in the domain of f.
4.5 MAXIMUMAND MINIMUMVALUESOF A FUNCTION 187
4.5.8 Definition
/(c) is said to be the absoluteminimumaalueof the function f if c is in the domain of / and if f (c) = f (x) for all values of r in the domain of /. . rLLUsrRArIoN 8: The graph of the function / defined by f(x):x24x*8 is a parabola, and a sketch is shown in Fig. 4.5.12. The lowest point of the parabola is at (2, 4), and the parabola oPens upward. The function has an absolute minimum value of 4 at 2. There is no absolute maxio mum value of /. Referring back to Illustrations 38, we see that the only case in which there is both an absolute maximum function value and an absolute minimum function value is in Illustration 5, where the function is continuous on the closed interval l5,4]. In the other illustrations, either we do not have a closed interval or we do not have a continuous function. If a function is continuous on a closed interval, there is a theorem, called the extremeoalue theorem, whrch assures us that the function has both an absolute maximum value and an absolute minimum value on the interval. The proof of this theorem is beyond the scope of this book, but we can state it without proof. You are referred to an advanced calculus text for the
F i g u r e4 . 5 . 1 2
proof.
4.5.9 Theorem ExtremeVglue Theorem
If the function / is continuous on the closed interval la, bf , then / has an absolute maximum value and an absolute minimum value on la, blAn absolute extremum of a function continuous on a closed interval must be either a relative extremum or a function value at an endpoint of the interval. Because a necessary condition for a function to have a relative extremum at a number c is for c to be a critical number, we can determine the absolute maximum value and the absolute minimum value of a continuous function f on a closed interval la,bl by the following procedure: (1) (2) (3) .
rxeuplr 2:
Given
f(x):x3+x2x*l find the absolute extrema of f on
[_2,+1.
find the function values at the critical numbers of.f on la, bl; find the values of f(a) and f(b); the largest of the values from steps (1) and (2) is the absolute maximum value, and the smallest of the values from steps (L) and (2) is the absolute minimum value.
sor,urroN: Because/ is continuous on l2, +1,the extremevaluetheorem applies. To find the critical numbers of f *e first find /': '(x) :3x2 * 2x 'l' f ' will f (x) exists for all real numbers, and so the only critical numbers of /
188
TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE
be the valuesof x for which f '(x):0. ( 3 x  1 ) ( x* 1 ) : s
Settingf ' (x)  0, we have
from which we obtain and
The critical numbers of f are1 and *, and each of these numbers is in the given closed interval l2, +l.We find the function values at the critical numbers and at the endpoints of the interval, which are given in Table 4.5.7. The absolute maximum value of f on l2, +l is therefore 2, which occurs atL, and the absoluteminimum value of /on l2,+l is1, which occurs at the left endpoint 2. A sketch of the graph of this function on [2, +l is shown in Fig. 4.5.73.
I
Table4.5.1 F i g u r e4 . 5 . 1 3
EXAMPLE 3:
Given
f (x) : (x  21rr" find the absolute extrema of f on [1,5].
solurroN: applies.
Because / is continuous on Il, sl, the extremevalue theorem
rf ' (\ "x/ ) : *3J( x  21u a There is no value of x for which f'(x):0. However,becausef,(x) does not exist at 2, we conclude that 2 is a critical number of , so that the f absolute extrema occur either at 2 ot at one of the endpoints of the interval. The function values at these numbers are given in tabte 4.s.2. From the table we conclude that the absolute minimum value of / ot [L5] is 0, occurring at2, and the absolutemaximum value of /on [1,5J is179, occurringat 5A sketchof the graph of this function on [1,5] is shown in Fig. 4.5.14. Table4.5.2
Exercises4.5 In ExercisesI through 10, find the critical numbers of the given function. 1. f (x) : x3I7x2  5x 2. f (x) :2x3  2x2 1.6xf 1.
3. f(x) : xa * 4x3 2x2 12x
INVOLVINGAN ABSOLUTEEXTREMUM ON A CLOSEDINTERVAL 189 4.6 APPLICATIONS
a. f@) : v7t3+ x4tl 3xrt3 7. f (x): 1 0 .f t x ) :
(*  A)ztz
6. f(x) : xa* 1L13* 34* + 75x 2 x
5.f(x)x6t512rrts 8. f (x\:
(r'  3f * 4;tte
e.flx):74
r*1 f _Sx+4
In Exercises 11 through 24, find the absolute extrema of the given function on the given interval' if there are any' and find the values of r at which the absolute extrema occur, Draw a eketch of the graph of the function on the interval. ,l
1,2.f (x) : x2 2x l 4; (*, **)
73,f (x):; ?L3J
1'a. f k) :;; [2,3)
15.f (x) : \/t4;
[3, +1
16. f(x):fi,
4 17. f (x) : ^\,,; 12,51 J)'
18./(r) : \84';
e2,2)
1 9 f. ( x ) : l x  4 l + r ; ( 0 , 6 )
r]1,'f (x)
 4  3x;(1, 2f 1
\x
[2
]
irx* s[; t3,s] ifr:5J
20. f (x) : 14 12l;(co, 1m;
z r .f ( x ) : l r  5 L2
23.f (x) : )c [rn; (1,3)
2 a .f @ ) : U ( x ) U ( x  1 ) ; (  1 , 1 )
?3,2)
22f(x):{r *'' ;i:=_l};e, t
In Exercises25 through 36, find the absolute maximum value and the absolute minimum value of the given function on the indicated intervul by the method used in Examples2 and 3 of this section. Draw a sketch of the graph of the function on the interval. 26. f (x): rs * 3x2 9x; l4, 4l 2 5.f (x): 13* 5x  4; [  3, 1 ] 'l'6; 28. f (x) : x4 8x2+ 16; l1, 4l 27. f (x) : x4 8x2+ l4, 0l 30. f (x) : 74 8xzI 16; l3 ' 2l 2 9 .f ( x ) : x 4 8 x 2* 1 6 ; 1 0 , 3 f
31.f (x): f* ,, l1, 2)
tl6 32.f (x): fi; ls,2l
3 3 .f ( x ) : ( r * 1 . ) 2 t 3 ; l  2 , " 1 . 1
3 a .f @ )  L 
35f (x): {'Jt lf;'=; :: t},r3,3r
(x31't'' l5,41
36f (x): {rl [; I ?]; ;l:=:=oa];r0,o1
37. Prove Theorem 4.5.3 for the case when f has a relative maximum value at c.
4.S APPLICATIONS INVOLVING AN ABSOLUTE EXTREMUM ON A CLOSED INTERVAL
We consider some problerns in whidr the solution is air absolute extremum oI a function on a dosed interval. Use is made of the extremevalue theorem, which assures us that both an absolute maximum value and an absolute minimum value of a function exist on a closed interval if the function is continuous on that closed interval. The procedure is illustrated by some examples.
TOPICSON LIMITS, CONTINUITY, AND THEDERIVATIVE
L: A cardboard box EXAMPLE manufacturer wishes to make open boxes from pieces of cardboard 12 in. square by cutting equal squares from the four corners and turning up the sides. Find the length of the side of the square to be cut out in order to obtain a box of the largest possible volume.
i ft N
I N ri
rn.
r in. (L2 zr) in.
solurroN: Let x:
the number of inches in the length of the side of the square to be cut out; V: the number of cubic inches in the volume of the box. The number of inches in the dimensions of the box are then r, (12  2x), and (12 2r). Figurc 4.6.t representsa given piece of cardboard, and Fig. 4.6.2representsthe box. The volume of the box is the product of the three dimensions, and so V is a function of x, and we write V(x) : xc(Lz 2x)(12  2x) (1) lf. x : 0, l/ :0, and if.x: 6, V :0. The value of r that we wish to find is in the closedinterval l0,6l.BecauseV is continuous on the closedinterval 10, 61,it follows from the extremevaluetheorem that V has an absolute maximum value on this interval. We also know that this absolute maximum value of V must occur either at a critical number or at an endpoint of the interval. To find the critical numbers of V, we findV, (r), and then find the values of x for which either V' (x) : 0 or V, (x) does not exist. From Eq. (1), we obtain V (x) : LMx  48x2* 4xg Thus, V' (x) : lM  96x * l,xz
Figure4.6.1
V' (x) exists for all values of r. Setting V, (r)  0, we have l2(*8r*12):g from which we obtain x: 6 and x:2
ktrz
2r) in.
Figure 4.6.2
The critical numbers of V are2 and 5, both of which are in the closed interval [0, 6]. The absolute maximum value of v on [0, 6] must occur at either a critical number or at an endpoint of the interval. BecauseV(0)  g and v(6) : 0, while v(2) : 128,we concludethat the absolutemaximum value of V on [0, 6] is L28,occurring at 2. Therefore, the largest possible volume is l2g in.B, and this is obtained when the length of the side of the square cut out is 2 in. We should emphasize that in Example 1 the existence of an absolute maximum value of I/ is guaranteed by the extremevalue theorem. In the following example, the existence of an absolute minimum value is guaranteed by the same theorem.
nxaprprr 2: An island is at point 4,6 miles offshore from
S O L U T I O N : Refer to Fig. 4.6.3.Let P be the point on the beach where the man lands. Therefore, the man rows from A to P and walks from p to c.
4.6 APPLICATIONSINVOLVINGAN ABSOLUTEEXTREMUMON A CLOSED INTERVAL
the nearest point B on a straight beach. A store is at point C,7 miles down the beach from B. If a man can row at the rate of 4 mi/hr and walk at the rate of 5 mi/hr, where should he land in order to go from the island to the store in the least possible time?
191
the number of miles in the distance from B to P. the number of hours in the time it takes the man to make the trip from A to C. Then T: the number of hours in the time to go from A to P plus the number of hours in the time to go from P to C. Because time is obtained by dividing distance by rate, we have Let r: T:
: _5 lAPl , , 4
lPCl
(2)
From Fig. 4.6.3,we seethat triangle ABP. Therefore,
is the length of the hypotenuse of right
lAPl: \/F+ 35 We also see from the figure that lPe  : 7  x. So from Eq. (2) T can be expressedas a function of x, and we have 6 miles
\456,7x
T(x):+?
Becausethe distance from B to C is 7 miles and becauseP can be any point on the line segment BC, we know that r is in the closed interual Figure4.6.3
n. 10,
We wish to find the value of x for which T has an absolute minimum value on [0, Zl. Because T is a continuous function of x on [0, 71,we know that such a value exists. The critical numbers of T are found by first computing T (x):
? "\*'' (r):LL 4\/i, +56 5 T' (x) existsfor all valuesof x. Setting T'(r)  0 and solving for x, we have (3)
'zi::2':,:*' 5x:4\m
36
x2:64 r:
+8
The number 8 is an extraneous root of Eq. (3), and 8 is not in the interval l0,7l.Therefore, there are no critical numbers of T in [0, 7]. The absolute minimum value of T on 10,7) must therefore occur at an endpoint of the interval. Computing T(0) and T(7), we get
T(0): i3 and T(7): +\65
192
TOPICSON LIMITS, AND THEDERIVATIVE CONTINUIW,
Since +\65 < H, the absolute minimum value of T on [0, 7] is +\/85, occurring when x:7. Therefore, in order for the man to go from the island to the store in the least possible time, he should row directly there and do no walking. EXAMPLE3: A rectangular field is to be fenced off along the bank of a river and no fence is required along the river. If the material for the fence costs $4 per running foot for the two ends and $6 per running foot for the side parallel to the river, find the dimensions of the field of largest possible area that can be enclosed with $1800 worth of fence.
soLUrIoN: Let r: the number of feet in the length of an end of the field; a: the number of feet in the length of the side parallel to the river; A: the number of square feet in the area of the field. Hence, A: )cA
(4)
Since the cost of the material for each end is g4 per running foot and the length of an end is x ft, the total cost for the fence for each end is 4r dollars. Similarly, the total cost of the fence for the third side is 5y dollars. We have, then, 4x * 4x * 6y :1800
(s)
To expressA in terms of a single variable, we solve Eq. (5) for y in terms of x and substitute this value into Eq. (4), yielding A as a function of x, and A ( x ) : r ( 3 0 0 t x ) (6) If y  0, x: 225,and if x: 0, A :300. Becauseboth r and y must be nonnegative, the value of x that will make A an absolute maximum is in the closed interval 10, 2251.BecauseA is continuous on the closed interval 10,2251,we conclude from the extremevaluetheorem that A has an absolute maximum value on this interval. From Eq,.(6),we havei A ( x ) : 3 0 0 r t x ' Hence, A ' ( x ) : 3 0 0 t x BecauseA'(x) exists for all x, the critical numbers of A will be found bv settingA'(x):0, which gives x:'l,t2t The only critical number of A is ILz+, which is in the closed interval 10,2251.Thus, the absolute maximum value of A must occur at either 0, tl2+, or 225.Because,4(0) :0 and A(225):0, while A(1,1,2t):16,875, we conclude that the absolute maximum value of A on [0, 22Slis L6,BT occurring when x: LL2i and y: L50 (obtained from Eq. (5) by substi
INVOLVINGAN ABSOLUTEEXTREMUMON A CLOSEDINTERVAL 193 4.6 APPLICATIONS
ExAMPLE4: In the planning of a coffee shop it is estimated that if there are places for 40 to 80 people, the weekly profit will be $8 per place. However, if the seating capacity is above 80 places, the weekly profit on each place will be decreasedbY 4 cents times the number of Placesabove 80. What should be the seating capacity in order to Yield the greatestweekly profit?
SoLUrroN: Let r: the number of placesin the seating capacity; P: the number of dollars in the total weekly profit. The value of P depends uPon r. and it is obtained by multiplying r by the number of dollars in the profit per place. When 40 < r < 80, $8 is the profit per place, and so P :8x. However, when r ) 80, the numthus giving ber of dollars in the profit per place is [80.04(r80)], P  r[8  0.04(r  80)] : 11.20x 004x2.so we have if40 > that in order r, 0. Show f to occur at a + o(i t), where r is in [0, b] and tl < na b\/F=e' must be satisfied: (0, inequality following b) the nuurber in the open interval 9 . Find the dimensions of the rightcircular cylinder of greatest lateral sur{ace area that can be inscribed in a sPherc with a radius of 6 in. 10. Find the dimeneions of the rightcircular cylinder of greatest volume that can be inscribed in a sphere with a radius of 6 in. point on the 1 1 .Given the cirrJe haviirg the equation f + 4 : 9, 6tt4 ,", the shortest distance from the Point (4' 5) to a the circle' on (4, 5) to a from the longist distance Point circle, and (b) the Point more than 8OOitems are Produced each week. The Profit item if not each profit of on a $20 12. A manufacturer can make to have declreases2 C€nts per item over 800. How many items shpuld the manufacturer produce eadl week in order the grcatest Profit? 13. A schoolsponsored trip will cost each student $15 if not morc than 150 student8 make the lriP; however, the cost Per student wiil be reduced Qcents for each student in excessof 150. How many students should make the trip in order fur the school to receive the Large8tSross income? L4. Solve Exerciee 13 iI the reduction Per student in excess of 150 i5 7 cent8. 15. A pdvate club charges annual memhrship dues of $lfi).per qrember less 50 cents for each member over 600 and plus S0'cents for eactr mJmber less than 600. How many members would give the club the most revenue from annual dues? 15. Suppoae a weight is to be held 10 ft below a horizontal line AB by a wire in the shape of a Y If the Points A and B art 8 ft lpa*, wttai is the shortest total length of wire that can be used? 12. ' A piece of wire 10 ft long ie cut into two pieces. One piece i8 bent into the sh4pe of a circle and th€ other into the shaPe i; shouli the wire be cui eo that (a)the combined area of the two fiSures i8 as small as possible and ;;;;;;;(b) the combined area of the two figues is as large as Possible? lg, Solve Exercise 17 if one piece of wire is bent into the shape of an equilateral tdangle and the other Piece is bent into the ehape of a square.
4.7 ROLLE,S THEOREM AND TIIE MEANVALUE THEOREM .
Letl be a function which is continuous on the closed interval [4, b], differ€ntiable on the open interval (a, b), and such thatf(a):0 andl(b):0. The French mathematician Michel Rolle (16521779) proved that if a
196
TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE
x
function / satisfies these conditions, there is at least one number c between a and b for which f '(c) : 0. Let us see what this means geometrically. Figure 4.7.1,shows a sketch of the graph of a function / that satisfies the conditions in the preceding ParagtaPh. We intuitively see that there is at least one point on the curve between the points (a,0) and (b,0) where the tangent line is parallelto the r axis; that is, the slope of the tangent line is zero. This situation is illustrated in Fig. 4.7.7 at the point P. So the abscissa of P would be the c such thaLf ' (c): 0. The function, whose graph is sketched in Fig. 4.7,1, not only is differentiable on the open interval (a, b) but also is differentiable at the end
It is necessary, however, that the function be continuous at the endpoints of the interval to guarantee a horizontal tangent line at an interior point' Figure 4.7.3 shows a sketch of the graph of a function that is continuous on the interval [a, b) but discontinuous at b; the function is differentiable on the oPen interval (a, b), and the function values are zero at both a and b. However, there is no point at which the graph has a horizontal tangent line. We now state and prove Rolle,s theorem.
Figure 4.7.3
4,7.'1,Theorem Rolle's Theorem
Let f be a function such that (i) it is continuous on the closed interval la, bl; (ii) it is differentiable on the open interval (a, b); (iii) f ( a ): f ( b ): 0 .
Then there is a number c in the open interval (a, b) such that
f ' ( c ): o PRooF: We consider two cases. Case1: f (x) :0 for all r in la, bl. Then f '(x): 0 for all r in (a, b); therefore, any number between a and b can be taken f.or c. Case2: /(r) is not zero for some value of r in the open interval (a,b). Because / is continuous on the closed interval fa, bl, we know by _ Theorem 4.5.9 that f has an absolute maximum value on la, bl and an a b s o l u te mi ni mum val ue on l a, bl . B y hypothesi s, (r):f(b):0. Fur _ f thermore , f (x) is not zero f.or some x in (a, b). Hence, we can conclude that / will have either a positive absolute maximum value at some c, in (a, b) or a negative absolute minimum value at some c, in (a, b), or both. Thus,
THEOREM 197 ANDTHEMEANVALUE THEOREM 4.7 ROLLE'S tot c: Cr, or c: cz as the case may be, we have an absolute extremum at an interior point of the interval la, bl.Therefore, the absolute extrernum ' f (c) is also a relative extremum , and because f (c) exists by hypothesis, it follows from Theorem 4.5.3 that f' (c) :0. This proves the theorem. I It should be noted that there may be more than one number in the open interval (a, b) for which the derivative of / is zero. This is illustrated geometricalty in Fig. 4.7.4, where there is a horizontal tangent line at the ' point where X: Crand also at the point where *: ,r, so that both f (ct) :0 '(tr) :0. a n df The converse of Rolle's theorem is not true. That is, we cannot conc l u d e th a t i f a functi on/i s such thatf' (c) 0, w i th a 1c ( b, then th e conditions (i), (ii), and (iii) must hold. (See Exercise 40.)
Figure 4.7.4
EXAMPLE
Given
f(x):4x39x verify that conditions (i), (ii), and (iii) of the hyoothesis of Rolle's theorem are satisfied for each of the following intervals: [8, 01, [0,3r],and [8, B].Then find a suitable value for c in each of these intervals for which
f
' ( c ): o '
sor,urroN: f'(x):12*9; f'(x) existsfor all valuesof x, and so / is differentiable on (oo, 1m) and thereforecontinuous on (*, +*). Conditions (i) and (ii) of Rolle's theorem thus hold on any interval. To determine on which intervals condition (iii) holds, we find the values of x fot which f (x)  0. Settingf (x) :0, we have Ax(f  *) :0 which gives us  _
rL
_3
2
_ 'r
3 2
and b:0, we see that Rolle's theorem holds on [8, 0]. Taking a:E Rolle's theorem holds on [0, srland I9, gl. Similarly, To find the suitable values for c, we set /'(r)  0 and get  9:0 12x2 which gives us
x  +\E
and *: *15
tfi. In the Therefore, in the interv al lar,0l a suitable choice for c is gl, possiare two there the interval interval lO, we take c: +\/3.In l9, *i bilities for c: either +\/g or t{5. We apply Rolle's theorem to prove one of the most important theorems in calculusthat known as the meanaalue theorem (or law of the mean). The meanvalue theorem is used to prove many theorems of both differential and integral calculus. You should become thoroughly familiar with the content of this theorem.
4.7.2 Theorem Let f be a function such that
MeanValue Theorem
(i) it is continuous on the closed interval la, bl; (ii) it is differentiable on the oPen interval (a,b).
198
TOPICS ON LIMITS, CONTINUIry, ANDTHEDERIVATIVE Then there is a number c in the open interval (a,b) such that
f(u)  f(')
' f'(c)
0a
r r n l_ f @ )
f'(c):tff
Refer to Fig. 4.7.5. By taking the r axis along the line segment AB,we observethat the meanvalue theorem is a generalization of Rolle's theorem, which is used in its proof. p,(b, f(b)) pRooF: An equation of the line through A and B in Fig. 4.7.s is
yf(a):fffk_a) or, equivalently,
f ( b ) f ( a ) ( x a )+ f ( a ) Y:'ff Figure4.7.5
Now if F(r) measuresthe verticaldistancebetweena point (x,/(r)) on lhe graph of the function / and the corresponding point on the secant line through A and B, then
F ( r ) :f ( x )  f % P k  a )  f ( o )
(1)
We show that this function F satisfies the three conditions of the hypothesis of Rolle's theorem.
But
(ulfu) F'(x): f '(x)f
THEOREM199 ANDTHEMEANVALUE THEOREM 4.7ROLLE'S Thus
F'(c) 1'G) f(u)r=fJ'l Therefore, there is a number c rn (a, b) such that
o:f,(c)_fW or, equivalently, r,,.\
f(b)f(a)
!f l\ c ,t :    
b_a
which was to be proved.
2: Given nxarurPr,n f(x):x35x23x verify that the hlryothesis of the meanvalue theorem is satisfied for a: 1 and b : 3. Then find all numbers c in the oPen interval (1,3) such that
f'(c\:fW
solurroN: Because/ is a polynomial function, / is continuous and differentiable for all values of r. Therefore,the hypothesis of the meanvalue theorem is satisfied for any a and b.  Lor  3 f' (x) :3xz z and (3) : 22 f f (t) : Hence,
_27 _(7):_l.o 2
31
' Settingf (c): 1.0, we obtain 3c210c3:L0 3c210c*7:0 (3c7)(c1):0 which gives us c:&
and c:t
Because1.is not in the open interval (1,3), the only possible value for c is 6.
nxavrpr.n3: Given
solurroN:
A sketch of the graph of f is shown in Fig. 4.7.6.
f (x) : xzrs draw a sketch of the graPh of f . Show that there is no number c in the open interval (2,2) such that
Figure4.7.6
TOPICS ON LIMITS, CONTINUlry, ANDTHEDERIVATIVE
Ic ,\/c^)\:f (22=) l f (  2 ) Which condition of the hypothesis of the meanvalue theorem fails to hold for f when a: 2 a n db : 2 ?
f (2) f(2)
4u3_ 4u3
2 (2)
4 0
There is no number c for which 2l3crts: e. The function / is continuous on the closedinterval l2,21; however, is not differentiable on the open interval (2,2) because / f,bl does not exist. Therefore, condition (ii) of the hypothesis of the meanvalue theorem fails to hold for f when a: 2 and,b : 2.
Exercises 4.7 In Exercises1 through 4, verify that conditions (i), (ii), and (iii) of the hypothesis of Rolle's theorem are satisfied by the given function on the indicated interval' Then find a suitable value for i that satisfies the conclusion of Rolle,s theorem. r.f(x)*4x*3; : [1,3] 2.
f(x)
x3 2xz x * 2; lt, Zl
3. f(x) : x3 2xz x * 2; l1, 2f a. f Q): .r3 1.6x;l4,01 5' For the function f defined bv f(r) : Ltt l 12*  x 3, determine three sets of values for 4 and D so that conditions (i), (ii)' and (iii) of the hypotheiis of Rolle's *t r*L rina a suttabt".,,ar." to" ir, of tlr" th*" op.r. intervale (a, b) for which /, (c) : 0. "",itii"llir",' " """t In 6 thtough 13, verify that the hypothesis of the meanvalue theorem ,Exercises is satisfied for the given function on the indicated interval. Then find a ;uitable value for c that sati"ri"" ti" *".ro.ior, of tle meanvalue theorcm.
6.f(x):xs**x;l2,tl
7 . f (x ) : xz l 2x  l ; [0, 1]
9. f (x) = x2t3; [0, 1]
1,0.f (x) : ffifi' ];
12. f(x):ff,
1,J. f(x):ffi,?r,4f
12,61
l6, 8l
8.f(x)tcr+!_y[*,s] 1,r.f (x): \ETz; 14,61
In Ex€rcises 14 through 23, (a) draw a sketch olthe_graph of the given function on the indicated interval; (b) test the three 6:T {itt) .oj 1ty hypothesis of _nole,itheorem ind determine which conditions are satisfied and rdhich, fi:l'l: !i)1 iia, if any, are not satisfied; and (c) if the three conditions in part (b) are satisfied, detenni i ne a point at which there is a hori
zontal tangent line.
M. f (x) : y3t4 2xtt4. 10, 4l
15. f(x) : y4t!  3xrt3. [0, 3]
16. f(x):'#;F+,sl
if x=21. 1 7 .f ( x ) : [ x + l3'71 3
1,8. r(x): {t::X il,;:l.}; ?2, +l
lz * irz 0, the function values are increasing as r increases; and when /,(r) < 0, the function values are decreasing as .r increases. We have the iollowing theorem. .
5.1.3 Theorem
Let the function f be continuous on the closed interval la, bl and differentiable on the open interval (a, b): (i) { f'@) > 0 for all x n (a, b), then f is increasing on [a, b]; (ii) if l'(r) < 0 for all x in (a, b), then / is decreasingon [a, b]. pRooF oF (i): Let r, and r, be any two numbers in [a, b] such that r, < :r. Then / is continuous on [rr, rr] and differentiable on (xr, rr). From the meanvalue theorem it follows that there is some number c in (xr, x2) such that
1' 1r'':f(x,)

f(x')
Because:, a rr, tn"^ rz rr )0. Atso, f,(c) > 0 by hypothesis. Theretore f(x)  f(rr) > 0, and so f(xr) > f(xr). We have shown, then, that f(xr) < f(h) whenever x, < xr, where x., and r, are any numbers in the interval [4, b]. Therefore, by Definition 5.1.1, it follows ti.rat is increasing / on fa, bl. The proof of part (ii) is similar and is left as an exercise(see Exer_ cise 34). I An immediate application of Theorem 5.1.1,is in the Droof of what is known as the firstderiaatioetestfor relafiaeextremaof a funcuon. 5.1'4 Theorem Let the function f be continuous at all points of the opm interval (a, b) FitstDerioatioe Test Extrena containing the number c, and suppose that lor Relatioe l' exists at'aii points of (a, b) except possibly at c: (1) if f' (x) > 0 for all values of r in some open intewal having c as its right endpoint, and if f'(x) < O for all values of r in iome open interval having c as its left endpoint, then f has a relative maximum value at c;
TEST ANDTHEFIRSTDERIVATIVE FUNCTIONS AND DECREASING 5 . 1 INCREASING
207
(ii) if f '(x) < 0 for all values of r in some oPen interval having c as its right endpoint, and if f ' (x) ) 0 for all values of x in some open interval having c as its left endpoint, then /"has a relative minimum value at c.
F i g u r e5 . 1 . 2
F i g u r e5 . 1 . 3
pRooF or (i): Let (d, c) be the interval having c as its right endpoint for which /'(r) > 0 for all r in the interval. It follows from Theorem 5.1.3(i) that / is increasing on ld, c). Let (c, e) be the interval having c as '(x) < 0 for all x in the interval. By Theorem its left endpoint for which f 5.1.3(ii) / is decreasing on lc, ef. Because/is increasing onld, c], it follows from DefinitionS.l..L that if x, is tnld, c] and xr* c, then f(xt) < f (c) .Also, because / is decreasing on fc, ef , it follows from Definition1.'1..2 that if x, is in [c, e] and xr* c, then f(c) > f(rr). Therefore, from Definition 4.5.L, / has a relative maximum value at c. The proof of part (ii) is similar to the proof of part (i), and it is left I as an exercise (see Exercise 35). The firstderivative test for relative extrema essentially states that if '(x) changes algebraic sign from positive to is continuous at c and f / negative as r increases through the number c, then / has a relative maxi'(x) changes algebraic sign from negative to posimum value at c; and tf f tive as r increases through c, then / has a relative minimum value at c. Figures 5.1.2 and 5.1.3 illustrate parts (i) and (ii), respectively, of Theorem 5.1,.4when /'(c) exists. Figure 5.1,.4shows a sketch of the graph '(c) of a function / that has a relative maximum value at a numbet c,but f d o e s n o te x i s t; how ever,f' (x) ) 0w hen x l c,andf' (r) < 0w hen x ) c . In Fig. 5.1.5, we show a sketch of the graph of a function / for which c is (r) f ' ( c ) l > 0 o r , e q u i v a l e n t l yf '(x) '(c)l < 0 or, equivalently, 0, and it follows from (1) that lf f < f'(x) f'(r). '(r) > 0, and if But becausef '(c) : 0, we concludethat if x is in l' , f '(r) '(r) < 0. Therefore,f changesalgebraicsign from positive r is in 1" , f to negativeas r increasesthrough c, and so, by Theorem5.L.4,/hasa relative maximum value at c. The proof of part (ii) is similar and is left as an exercise (see ExerI cise L8).
212
ADDITIONAL APPLICATIONS OF THEDERIVATIVE
EXAMPLE
Given
SOLUTION:
f(x):xa*$x34x2 find the relative maxima and the relative minima of f by applying the secondderivativetest.
f'(x) :4xs * 4xz 8x f"(x):12x2*8r8 Settingf '(x):0, we have
4 x ( x+ 2 )( x  1 ) which gives
x:L
ff:0
Thus, the critical numbers of f are2,0, and L. We determine whether or not there is a relative extremum at any of these critical numbers by finding the sign of the second derivative there. The results are summ arized. in Table 5.2.1. T a b l e5 .2 .1
f'(x)
+
0
0
0
/ has a relative maximum value
0
/ has a relative minimum value
_5
x
f" (*)
f(x)
3
T
Conclusion t' has a relative minimum
value
rf f " (c) : 0, as well as f '(r) :0, nothing can be concluded regarding a relative extremum of f at c. The following three illustrations lustify thii statement. .IL L U ST R A TToN7: If f(x) x4, then f ' (x):4x3 and ,,(x):' !,2x2. Thus, f '(0), and ''(0) all have the value zero. By applying the firstderivative f (0), f f test we see that / has a relative minimum value at 0. A sketch of the graph of / is shown in Fig. 5.2.1. o
Figure5.2.1
Figure5.2.2
o IL L U S TR A TION 2: If 8(x):xn, then g' (r) :4x3 and g,,(r):  12x2. Hence, g(0) : g'(0) : g"(0) : 0. In this case,g has a rerative maximum value at 0, as can be seen by applying the firstderivative test. A sketch of the graph of g is shown in Fig. 5.2.2. o . rLLUsrRArroN 3: If h(x) : r3, then h'(*) : 3x2 and,h" (*) : 6x; so h(0) : h'(0) : h" (0): 0. The function h does not have a relative extremum at 0 because if r < 0, h(x) < h(0); and if x ) o, h(x) > h(0). A sketch of the graph of h rs shown in Fig. 5.2.9. o
Figure5.2,3
In Illustrations 7,2, and 3 we have examples of three functions, each of which has zero for its second derivative at a number for which its first derivative is zero; yet one function has a relative minimum value at that
5.3 ADDITIONALPROBLEMSINVOLVINGABSOLUTEEXTREMA
213
number, another function has a relative maximum value at that number, and the third function has neither a relative maximum value nor a relative minimum value at that number.
5.2 Exercises In Exercises 1 through 14, find the relative o(trema of the given function by using the secondderivative test, if it can be applied. If the secondderivative test cannot be applied, us€ the firstderivative test. 2.s@):13sx*6 5. f(x) : (x  4)2 8 . f(* ) : x (x  1)'
1.f(x):3x22x*l a. h@) :2xs  %c2+ 27 7. G(x) : ( r  3) a 1 0.f(x):
x\/R
13.F ( r ) : 6 x r t 3  x z t l
11'.f(x)  lx6rtz+ xrtz
3.f(x):4x3+3f *'8r 6. G(x): (x * 2)3 9. h(x): xfr$ Ov2
1 2 .S U ) : ; * ;
 4)' M. G(x)  xszts(a
15.G i v e n f ( x ) : x 3 + s r x * 5 , p r o v e t h a t ( a ) i f r ) 0 , / h a s n o r e l a t i v e e x t r e m a ; ( b ) i f r /  0 , / h a s b o t h a r e l a t i v e m a x i m u m value and a relative minimum value. 16. Givenf(r):x'1.r+ k, where r > 0 and r + 1, prove that (a) if 0 < r< 1,/hasa relativemaximumvalue at 1; (b) if r > 1.,f ha8 a relative minimwn valu€ at l. 77. Given f(x) : f * rr1, prove that regardless of the value of r, f has a relative qrinimum value and no rclative maximurn value. 18. Prove Theorem 5.2.1(ii).
5.3 ADDITIONAL
The extremevalue thmrem (Iheorem 4.5.9) guarantees an absolute maximum value and an absolute minimum value for a function which is continuous on a dosed interval. We now consider some functions defned on intervals for which the extremevalue theorem does not apply and which may or may not have absolute extrema.
Given
solurroN: f is continuous on the interval [0, 6) becausethe only discontinuity of f is at 6, which is not in the interval. t, t ^"\_zx(ffi _ xz L2x* 27 _ (r 3) (r 9)
PROBLEMS INVOLVING ABSOLUTE EXTREMA
EXAMpLE l,:
f(x\:
t\re,
yz_27 x_5
find the absolute extrema of f on the interval [0, 5) if there are any.
I\x)
@:
1r_5y::ffiy
' '(r) : 0 when x: 3 or 9; so f (*) exists for all values of r in 10,6), and f the only critical number of / in the interval [0, 5) is 3. The firstderivative test is applied to determine if f has a relative extremum at 3. The results are summarizedin Table 5.3.1. Because/ has a relative maximum value at 3, and f is increasing on the interval [0, 3) and decreasingon the interval (3,6), we conclude that
214
ADDITIONALAPPLICATIONS OF THE DERIVATIVE
on [0, 6) f has an absolute maximum value at 3, and it is f(3), which is 6. Noting that lim f (x) : m, w€ conclude that there is no absolute mini
mum value of / on [0, 6). T a b l e5 .3 . 1
Conclusion
EXAMPLE
Given
f ( x \ : r r(X., *
t vrz
*
6)2
find the absolute extrema of f on (0, +m; if there are any.
0 d
c
o
(6)
To prove that TQ ) 0, we show that both of the factors on the right side of Eq. (5) have the same sign. rf (x  c) > 0, then x ) c. And because d is between r and c, then d > c; therefore, from inequality (6), lf'(d)f_'(c)l >0. If (rc) 0 if x) 0, the graph is concave upward at points on the graph immediately to the left of (0, 0) and at points immediately to the right of (0, 0). Consequently, (0, 0) is not a point of inflection. In Illustration 1 of Section 5.2 we showed that this function / has a relative minimum value at zero. Furthermore, the graph is concave upward at the point (0, 0) (see F i g . 5 .2 .1). o The graph of a function may have a point of inflection at a point, and the second derivative may fail to exist there, as shown in the next illustration. If / is the function defined by f (x) : tr1l3,then * $yl tl f' (x ) : !x6zt? and f" (x) :
. rLLUSrRerrox 4:
r ,rr6,r* rc /f\
/l
Figure5.4.8
I
4
(r) < 0. f " ( 0 ) d o e s n o t e x i s t ;b u t i f x < 0 , f " ( x ) > 0 , a n d t f x ) 0 , f " Hence, / has a point of inflection at (0, 0). A sketch of the graph of this function is shown in Fig. 5.4.8. Note that for this function /'(0) also fails to exist. The tangentline to the graph at(0,0) is the y axis. . In drawing a sketch of a graph having points of inflection, it is helpful to draw a segment of the tangent line at a point of inflection. Such a tangent line is called an inflectional tangent.
5.4 CONCAVITYAND POINTSOF INFLECTION
EXAMPLE 1: For the function in Example1 of Sec.5.1, find the points of inflection of the graph of the function, and determine where the graph is concaveupward and where it is concave downward.
SOLUTION:
f(x):x36x2*9x*1' f ' ( x ) : 3 x 2 r 2 x* 9 f " ( x ): 5 x  L 2 f " (x) existsfor all valuesof x; so the only possiblepoint of inflection is where f " (x): 0, which occurs at x:2. To determine whether there is a point of inflection at x:2, we must check to see if f " (x) changes sign; at the same time, we determine the concavity of the graph for the respective intervals. The results are summanzed in Table 5.4.1. Table5.4,1
f (x) oo
l
7.f(x):(x))rrs (*2 :ff 0 r t x < c ; f ' ( x )
< 0 ifr > c;f"(x) < 0itx < c;f"(r)
c.
1 6 .f ' ( x ) > 0 i t x < c ; f ' ( x ) > 0 i f r > c ; f " ( x ) > 0 i t x < c ; f , , ( t ) < 0 i f r > c . 7 7 .f " ( c ) : o rf t ( c ) : o ; 1 " ( x ) > 0 i f r < c ; 1 " ( x ) > o i I x > c . 1 8 .f ' ( c ) : 0 ; f ' ( x ) > 0 i f r < c;f"(x) > 0ifr>c. 79. f' (c) 0; f' (x) < 0 if r < c; f" (x) > 0 iI x > c.
m . f " ( c ): 0 ;f ' ( c ): * ; 1 " ( r ) > 0 i f r < c ; f , , ( x ) < o i t r > c . 21.l'(c) doesnot exist;f" (r\ >0itr < c;f,(x\ > 0 ifr > c. 22. l'(c) doesnot exisf /" (c) doesnot eist f" (x) < 0 ifr < c; f,'(r) > 0 ifr > c. 23. lim f'(x)  +co,lim l'(t) : o; f" (x\ > 0 ifr < c;f"(x) < 0 ifr > c. 24. lim f'(x):*@j
limf' (r):
*;f"
(x) > 0 if r < c;f" (x) > 0 if x > c.
E. Draw a Bketchof the graph of a functionf for which l(x), f'(r), andf,,(r) exist and are positive for all r. 26. II f(t\ :3f + xlxl, prove that l" (0) does not exist but the graph of f is concaveupward everyrvhere. 27. Prove Theorem 5.4.3(ii). 28. SuPPos€ that I is a function for which l" (x) exists for all values of r in some open interval I and that at a number c in I, f" (r) : O and'l"' (c) €xists and is not zero. Prove that the point (c, l(c) ) is a point of inflection oI the graph of /. (HrNr; The proof is similar to the prcof of the secondderivative test (Theorem 5.2.1.).)
TO DRAWING A SKETCHOF THE GRAPHOF A FUNCTION 5.5 APPLICATIONS
5.5 APPLICATIONS TO DRAWING A SKETCH OF THE GRAPH OF A FUNCTION
nxanaptnL: Given f(x):x33**3 find: the relative extrema of f the points of inflection of the graph of f the intervals on which f is increasing;the intervals on which / is decreasing;where the graph is concaveupward; where the graph is concavedownward; and the slope of any inflectional tangent. Draw a sketch of the graph.
We now apply the discussionsin Secs.5.'1.,5.2,and 5.4 to drawing a sketch of the graph of a function. If we are given f (x) and wish to draw a sketchof the graph of.f ,we proceedas follows. First, findf' (r) and f"(x). Then the critical numbers of f arc the values of r in the domain of f for which either f '(x) does not exist or f ' (r):0. Next, apply the firstderivative test (Theorem 5.1.4)or the secondderivative test (Theorem 5.2.1)to determine whether we have at a critical number arelativemaximum value, a relative minimum v3lue, or neither. To determine the intervals on which '(x) is positive; to de/ is increasing, we find the values.of.x for which f termine the intervals on which / is decreasing,we find the values of r for which f '(x) is negative. In determining the intervals on which / is monotonic, we also check the critical numbers at which / does not have a relative extremum. The values of r forwhich f"(*):0 or f"(x) does not exist give us the possible points of inflection, and we check to seeif f " (x) changessign at eachof these values of r to determine whether we actually have a point of inflectioh. The values of r for which f " (x) is positive and those for which f" (r) is negative will give us points at which the graph is concaveupward and points at which the graph is concavedownward. It is also helpful to find the slope of each inflectional tangent. It is suggestedthat all the information so obtained be incorporated into a table, as illustrated in the following examples.
S e t t i n gf ' ( x ) : 0 , w e o b sol,urroN f' (r)3x25x; f" (r):5x5. tain r : 0 and x: 2. Setting f " (x) : 0, we obtain x: t. In making the table,we considerthe points at which x:0, x:L, and r:2, and the intervals excluding these values of.x: oo N. However, the marginal cost may decrease for some values of x; hence, for these values of x, C" (x) < 0, and therefore the graph of the total cost function will be concave downward for these values of x. The following example involving a cubic cost function illustrates the case in which the concavity of the graph of the total cost function changes.
234
ADDITIONALAPPLICATIONSOF THE DERIVATIVE
2: Draw a sketch of ExAMPLE the graph of the total cost function C for which C(r):x36x2*
13r*1'
Determine where the graph is concaveupward and where it is concavedownward. Find any points of inflection and an equation of any inflectional tangent. Draw a segment of the inflectional tangent.
SOLUTION:
C'(x) :3xz  12x+ 13 C"(x):6xt2 C'(x) can be written as 3(r  2)2 + 1. Hence, C'(x) is never zero. C" (x): 0 when r: 2. To determine the concavity of the graph for the intervals (0, 2) and (2, am) and if the graph has a point of inflection at x:2, we use the resultssummarizedin Table 5.6.1. T a b l e5 . 6 . 1
C(x)
C'(r)
C" (x)
Conclusion
0<x 0 if and only if R'(r) > c, (x); therefore, the profit is increasing if and only if the marginal revenue is greater than the marginal cost. Let us determine what level of production is necessary to obtain the greatestprofit. S has a relative maximum function value at a number x for which S'(r) : 0 and S" (x) ( 0. From Eqs. ( ) and (5)
5.6 AN APPLICATION OF THE DERIVATIVE IN ECONOMICS
(r, R(r))
Figure5.6.7
4: Supposethat the ExAMPLE demand equation for a certain commodityis P:4  0.0002r, where x is the number of units produced each week and p dollars is the price of each unit. The number of dollars in the total cost of producing x units is 600 + 3x. If the weekly profit is to be as large as possible, find: (a) the number of units that should be produced each week; (b) the price of each unit; (c) the weekly profit.
we observe that this will occur at a value of x for which the marginal revenue equals the marginal cost and R" (r) < C" (x). The values of r will be restricted to a closed interval, in which the left endpoint is 0 (since r = 0) and the right endpoint (the largest permissible value of x) is determined from the demand equation. At each of the endpoints there will be no profit, and so the absolute maximum value of 5 occurs at a value of r where S has a relative maximum value. To illustrate this geometrically, refer to Fig. 5.6.7, where both the x total revenue curve and the total cost curye are drawn on the same set of axes. The vertical distance between the two curves for a particular value of r is 5(r), which gives the profit corresponding to that value of x. When this vertical distance is largest, S(r) is an absolute maximum. This is the distance an in the figure where A and B are the points on the two curves where the tangent lines are parallel, and hence C'(x)  R'(r).
sor,urroN: The price function is given by P (r)  4  0.0002x,and r is in the closed interval [0, 20,000]becauser and P(r) must be nonnegative. R(r) : xP(x), and so R(x) :4x  0.000212 r in [0, 20,000]
(6)
The cost function is given by C ( x ) : 6 0 0* 3 r
(7)
If S(r) dollarsis the profit, S(r): S(r):r0.0002f
600
R(r)  C(x), and so r in [0,20,000]
From Eqs. (6) and (7) we obtain R'(r)  4 0.0004x c'(r) : 3 Also, R"(r) : 0.0004 and C"(x):0. 4  0.0004x: 3
EquatingR'(r) and C'(r) we get
x:2500 BecauseR" (r) is always less than C" (x) , S" (x): R" (x)  C" (x) < 0, and so we conclude that 5(2500) is an absolute maximum value. (Note that S(x) is negative for r:0 and r:20,000.) P(2500):3.50 and s(2500):650. Therefore, to have the greatest weekly profit, 2500 units should be produced each week to be sold at $3.50each for a total profit of $550.
238
ADDITIONALAPPLICATIONS OF THE DERIVATIVE
ExAMPLE 5: Solve Example 4 if a tax of 20 cents is levied by the govemment on the monopolist for each unit produced.
soLUrIoN: With the added tax, the total cost function is now given by
C(x):(500*3r) *0.20x and so the marginal cost function is given by C'(x):3.20 EquatingR'(r) and C'(r), we get 4  0.0004x:3.20
As in Example4 it follows that when x: 2000,S has an absolute maximum value on [0, 20,000]. From P(x)  4 0.0002r and S(x) : O.gx  0.0002x2  600,we have P(2000): 3.60and S(2000): 200. Therefore, if the tax of 20 cents per unit is levied, only 2000units should be produced each week to be sold at $3.60 in order to attain a maximum total weekly profit of $200. It is interesting to note that in comparing the results of Examples 4 and 5, the entire 20 cents rise should not be passedon to the consumer to achieve the greatestweekly profit. That is, it is most profitable to raise the unit price by only L0 cents. The economic significance of this result is that consumers are sensitive to price changes, which prohibits the monopolist from passing on the tax completely to the consumer. A further note of interest in the two examplesis how the fixed costs of a company do not affect the determination of the number of units to be produced or the unit price such that maximum profit is obtained. The fixed costof a company is the cost that does not change as the company's output changes.Regardlessof whether anything is produced, the fixed cost must be met. In Examples 4 and 5, becauseC(r) : d00+ 3r and 600 + 3.2x, respectively, the fixed cost is $600. If the 600 in the expressions for C(r) is replacedby any constantk, C'(x) is not affected;hence, the value of x for which the marginal cost equals the marginal revenue is not affected by any such change. Of course, a change in fixed cost affects the unit cost and hence the actual profit; however, if a company is to have the greatestprofit possible, a change in its fixed cost will not affect the number of units to be produced nor the price per unit.
Exercises5.6 1. The number of dollars in the total cost of manufacturing .x watches in a certain plant is given by C(r) :1500 + 30r + 20/r' Find (a) the matginal cost function, (b) the marginal cost when x: 40, and (c) the cost of manufacturing the fortyfirst watch. 2. lf C(r) dollars is the total cost of manufacturing r toys and C(r) : 110 + 4r * 0.02f, find (a) the marginal cost function, (b) the marginal cost when r: 10, and (c) the cost of manufaduring the eleventh toy.
5.6 AN APPLICATION OF THE DERIVATIVE IN ECONOMICS 239
3. Suppose a liquid is produced by a certain dremical process and the total cost function C is given by C(x) :514rrr, where C(r) dollars is the total cost of producing t gallons of the liquid. Find (a) the marginal cost when 16 gal are prcduced and (b) the number of gallons produced when the marginal cost is 40 cents per gal. 4. The number of dollaB in the total cost of prcducing x units of a certain commodity is C(r) : & + 3x + 9{2t. Fi^d, (a) the marginal cost when 50 units are produced and (b) the nurnber of units produced when the marginal cost is $4.50. 5. T'he number of dollars in the total cost of prcducing r units of a commodity is C(r) : ae14r * * Find the function grving (a) the average cost, (b) the marginal cost, and (c) the marginal average cost (d) Find the absolute ninimum average unit cost. (e) Draw sketches of the total cost, average cost, and marginal cost curves on the same Bet of axes. Verify that the average cost and marginal cost are equal when the average cost has it8 least value.  6t + 4, find (a) the average cost 6. II C(x) dollars is the total cost of producing t units of a commodity and C(x):3f function, (b) the marginal cost function, arld (c) the marginal average cost function. (d) What is the range of C? (e) Find the absolute minimum average unit cost. (f) Draw sketches of the total cost, average cost, and marginal cost curves on the same set of axes. Verify that the average cost and marginal cost are equal when the average cost has its least value. 7. The total cost function C is given by C(x) = lat  2a2+ 5x + 2. (a) Determine the range of C. (b) Find the marginal co9t function, (c) Find the interval on whidr the martinal cost function is decreaaing and the interval on which it is incr€asing. (d) Draw a sketch of the graph of the total cost function; detemdne where the graph is concave upward and where it is concave downward, and find the points of inflection and an equation of any inflectional tangent. 8. If C(r) dollars b the total cost of producing t units of a commodity and C(r) = 2f  8r + 18, find (a) the domain and range of C, (b) the average cost function, (c) the absolute minimum average unit cost, and (d) the marginal cost tunction. (e) Draw sketches of the total cost, average cost, and marginal cost curves on the same set of axe8. 9. The fixed overhead expense of a manufactuer of childrcnls toys is $4fi) per week, and other costs amount to $3 fot each toy ploduced. Find (a) the total cost function, (b) the average cost function, and (c) the marginal cost function. (d) Show thaf there is no absolute minimum average unit cost. (e) What is the smallest number of toys that mu8t be Produced so that the average cost per toy is le8s than $3.42?(f) Draw sketches of the graphs of the functions in (a), (b), and (c) on the 6ame Bet of axes. 10. The number of hundreds of dollars in the total cost of producing 100xradios per day in a certain factory is C (x) : 4x + 5' Find (a) the average cost function, (b) the marginal cost function, and (c) the martinal average cost function. (d) Show that there i8 no absolute minimum average urlit cost. (e) What is the smallest number of radios that the factory must produce in a day so that the averagecost per radio is lessthan $7?(f) Draw sketchesof the total cost, averageco6t, and martinal cost curves on the same set of axes. 11. If the demand equation for a particular commodlty is 3r I 4p : 12, find (a) the price function, O) the total r€venue function, and (c) the marginal revenue function, Draw sketches of the demand, total revenue, and marginal revenue curves on the same set of axes. Verify that the marginal revenue curve intersects the t axis at the Point whose abscissa is the value of r for whidr the total levenue iB greate8t and that the demand curve interEects the t axis at the Point whose absci66a is twice that. 12. The derrand equation for a particular commodity is pf + ?  18:0 wherc p dollars is the price per unit when 100r units are demanded. Find (a) the price function, (b) the total revenue function, and (c) the marginal revenue function. (d) Find the absolute maximum total revenue. 13. Follow the instructiona of Exercise 12 if the demand equation is t' + Pt  36: 0' 14. Let R(x) dollars be the total revenue obtained when r rrnits of a comrnodity are demanded and R(r):2+31i1, where I is in the closed interval [2, 17]. Find (a) the demand equation, O) the Price function, and (c) the marginal revenue function. (d) Find the absolute maxtnum total levenue. (e) Draw skekhes of the demand, total revenue, and marginal revenue curves on the Sameset of axe8. 1.5.The demand equation for a certain commodity is r + p: 14, where r is the numbet of units Produced daily andp is the number of hundreds of dollars in the price of each unii. The number of hundre& of dollars in the total cost of Pro
240
ADDITIONALAPPLICATIONS OF THE DERIVATIVE
ducing x units is given by C(x) : *  2x I 2, and r is in the closed interval [1, 1a]. (a) Find the profit function and drav, a sketch of its graph. (b) On a set of axes different from that in (a), draw sketches of the total revenue and total cost curves and show the Seometrical interpretation of the profit function. (c) Find the maximum daily profit. (d) Find the marginal tevenue and rrarginal cost functions. (e) Draw sketches of the graphs of the marginal revinue and margi nal cost functions on the same set of axes and show that they intersect at the point for which the value of r makes the prcfit a maximum, 16. Follow the instructions of Exercise15 if the demand equation is f + p :32 and C(x) :5r. 17' The demand equation for a ce*ain commodity is p : (x  8),, and the total cost function is given by C(r) : l8t  f , where C(r) dollars is the total cost when x units are purchased. (a) Detemine the permissible vaiues of r. (b) Find marginal revenue and marginal co6t functions. (c) Find the value oI r which yields the maximum piofit. (d) Draw $e sketches of the marginal revenue and marginal cost functions on the same set of axes. 18. A monoPotst d€termines that if C(r) cents is the total cost of producing x units of a certain commodity, then C(r) : 25r + 20/000. The demand equation is x + sOp: 5000, where x units are demanded each week when the unit Drice tj.P If the weekly profit is to be maximized, find (a) the numbe! of units that should be produced each week, "".b: O) the price of each unit, and (c) the weekly profit. 19. solve Exercise 18 if the govemment levies a tax on the monopolist of 10 cents per unit produced. 20, Solve Exercise18 if the govemment imposes a 10% tax based upon the consume/s price, 21. For the monopoligt of Exerciee 18, determine the amount of tax that shoutd be levied by the government on each unit prcduced in order fo. the tax revenue received by the government to be maxirnized. 22' Find the maximum tax revenue that can be received by the government if an additive tax for each unit produced is leyied on a monopolist for which th_edeaTd equation is xllp:75, where.' units are demanded when p dollars is the price of one unit, and c(r) :3x r 1e0,where c(r) dollars is the total cost of producint r units. 23. The demand equation for a certain commodity pmduced by a monopolist is p : a  6r, ,ne total cost, C(r) dollars, of Prcducing :r units is detennined by C1r; : r a dx, where a, b, c, aid, d arc positive constants. "rrd If the government levies a tax on_the monopolist of f dollars per unit produced, show that in order ior the monopolist to nJximize his profits he should pass on to the consumer only onehalf of the tax; that is, he should increase his unit price bv it dollars.
ReaiewExercises (Chnpter5) 1. Find the shortest distance from the point (*, 0) to the curve y:
t[.
2. Prove that among all the rectangleshaving a given perimeter, the square has the greatestarea. 3. Prove that among all the rectangles of a given area, the square has the least perimeter. 4. The demand in a cerlain market for a particular kind of breakfastcereal is given by the demand equation p, + 25p 2000: 0, where P cents is the price of one box and x thousandso{ boxes is the quantity demandedper week. If the current Price of the cerealis 40 cents per box and the pdce per box is increasing at th€ rate of 0.2 cent each week, find the rate of change in the demand. 5. Two Particles start theit motion at the same time. one particle is moving along a horizontal line and rts equation of motion i6 t = t:  2t, where r ft is the directed distance of the particle from the o;igin at t sec.The otherparticle is moving along a vedical line that intersectsthe horizontal line at the odgin, and its equation of motion is y : p  2, where y ft is the directed distance of the partide from the odgin at t sec. Find when the directed distance Letween the two particles is least, and their velocities at that time.
REVIEWEXERCISES 241 the intervals on which f In Exercises 5 through 9, find: the relative extremi of f the points of inflection of the traph of t glaph i6 concave downwhete the upward; graph is concave the tf," i.tlrvals on which I is decreasing witere i'i""r""ri"g graph' of the Draw a sketch tangmt. ward; the slopeof any inflectional 6.f(x):(x*2)at3
8. /(r) : x\/FT
7.f(x): (r 4)'(x+2)' 9 . f ( x ) : ( r  3 ) s r*r t
10. If I\x):
x*L xr+1
gxaPh. prove that the graph of I ha8 three points of inflection that are collinear. DrarY a Eketch of the 11. If l(r) : 1111,show that the graph of I has a Point oI inflection at the origin' and only xo where r is a positive integer. Prove that the gnph of I has a point oJ inflection at th€ origin if at 0. value relative minimum has a is wen, , > 1. Furthirmore show that if I / Jd int"g". "nd (f * ar)!, where p is a rational numb er and'p * 0, prove that if P < ]the graph ofl has two Point8 of inflectg. fif(x\: ti6n and if p > * the graph of f has no Pointg of inflection' what can you condude, if an''ll1ing, about (a) the L4. Suppose the glaph of a function has a point of inflection at l: c. of continuity (c) c; the of at f" at c? continuity of / ai c; (b) the continuity /' P is the numbef of millions 15. Suppose c is the number of millions of dollars in the caPitalization of a certain colPoration,  0.004c'. r its capitalization is incleasing at the rate of : P 0.05c profit, and annual of dollars in the corpofation's profit if its current caPitalization is (a) $3 million $400,000per year, find the rate of chanie of the corporation's annual and (b) $6 million. when r aPartmentsare rented and 16. A property development company rents each apartment at p dollars per month is zero? revenue marginal How many apartments must be rented before the p :2(;\/ffi=E !2. Let fe\: ;; f;
about a rightcircular cylinder L7 . Find the dimensions of the rightcircular cone of leastvolume that can be circumscribed of radius r in. and altitude h in. radius to the measure of the altitude for a 1g. A tent is to be in the shape of a cone. Find the mtio of the measure of the tent of given volume to require the least material' :l18' whete P dollars is the Price per unit when 19. The demand equation for a particular commodity is (P + 4)(r + 3) the marginal revenue function' lg0r units are demanded. Find (a) the price functi;n, 16) the total rwenue function, (c) and margiral revenue revenue, total the demand, of (d) Find the absolute maxrsrum iotal revenue. (e) Draw sketches cuw$ on the same set of axes. $ and the total cost function is given by c(r) : 2r 1' 20. The demand equation for a certain commodity i8 p + 2V?J: (c) Find the functions. cost marginal permissible values of r. iul ritta the marginal revenue and t"loJ""rr""'ae value of r which yields the maximum profit' 21. The demand equation for a certain cpmmodity is l16px:70s  2. 1.03r* 18 ' 10'3f 6f unit, and I > 100. The number of where r is the number oI units produced weekly and p dollars is the price of each ' : 10k1' Find the number of units * 11 24 dolLars in the avenge cost of producing each unit is given by QQ) "\7for the weekly unit in oldet price of each Profit to be marirnized. that should be produced each week and the
242
ADDITIONALAPPLICATIONS OF THE DEBIVATIVE
22. When 1.000rboxes of a c€rtain*ind of material are produced, thenurnber of dollars in the total cost of production is givenbv c(r)  135r1i3 + ft50.Find (a)the marginicost when8000br""r ifi'r,*uu. ot uor"" produced whm the marginal cost "i"pau""a?ffi Grer thousand boxes) is g20. 23' A ladder is to rcach over a fence ' ft high to a watl ur ft behind the fence. Find the length of the shortest ladder that may be used. 24. Dtaw a sketch of the graph of.a function on the interval I in eachcase: (a) I is the open interval (0, 2) ard/is f con_ tinuous on r' At 1, I has a relative maximum value but /'(1) does not exirt. 1iy i i, th" to, zl. rrru ror,""io"eJiiiu*"i
i',iio ua,,"Jt7;;f 0.6ilil; Hll,Ti,""tiffi?ffifrTff""j,1'.but theabsorute
olenintervar (0.z),
25. (a) It t'@):3lrl + alx  tl, prove that/has an absolute minimum value of 3. (b) If g(r) =alll + 3lr _ 11,prove that 8 ha8 an absolute minimum value o_f3. (c) rl h(tc): alxl + tp  il, where d > 0 and > b o, plove that , has an abso_ lute minimum value that is the smalter of the t*,o ,rr.t"""'o f. "r,i 26' rl f(x): lxl" ' lr  tlb, where c and b are positive rational numbers, prove that has a relative maximum / value of L'bbl@.+ bl+b.
The differential and antidifferentiation
THE DIFFERENTIAL AND ANTIDIFFERENTIATION
6.1 THE DIFFERENTIAL
Supposethat the function / is defined by Y:f(x) Then, when f ,(x) exists, A/ '(x) lim /f \  ar;o A'r
( 1)
where Ly:f(x+ Lx)  f(x). From (1),itfollows thatfor any€ ) 0there existsa6>0suchthat
l +  f ' '( x ) 1 I ..
lAr
w h e n e v e( r ol A r l< 6
which is equivalent to
l\l,f'
(x) Arl lA;: i:1
PROOF:
n  F ( 1 ) ] + [ r ( 3 ) F ( 2 ) ] + ' tZ/ t r t i l  F ( i  1 ) l : [ F ( 1 ) F ( 0 ) ] + [ F ( 2 )
i:1
+ l F ( n  1 )  F ( n  2 ) l + l F ( n )  F ( n  1 )l
THE DEFINITEINTEGRAL
:  F ( 0 ) + [ F ( 1 )_ r ( r ) ] + [ F ( 2 )_ r 1 z ; 1 + . . . + l F ( n  1 )  F ( n 1 ) l + F ( n ) :F(0) +0+0+ . . . + 0+ F(n) : F(n)  F(0)
I
The following formulas, which are numbered for future reference, are also useful. n
7.'l.,.5Formula L
. n(n*71 ,:T
i:r n
7.1.6 Formula 2
. r_b n ( n * 1 ) ( 2 n + 1 )
i:r n
7.'1,.7Formula 3
,, r  f l '(4n+ 7 ) '
i:r 'n
7.'1,.8Formula 4 i:1
n(n r t) (6nBt 9n2+ n  1) ,+_ 30
Formulas 1 through 4 can be proved with or without using mathematical induction. The next illustration shows how Formula L can be proved without using mathematical induction. The proof of Form ula 1,by mathematical induction is left as an exercise (see Exercise 11). . ILLUSTRATToN 4: We prove Formula 1. n
) l:1,+2+3+
. . . t (nt) *n
and n
)i:n*(n1)+
(n2)+..
+2+t
,it*" add thesetwo equationsterm by term, the left side is n
2>i ""0
.;;he
right side are n terms, each having the value (n + 1). Hence,
n
2 > i : ( n + 1 ) + ( n * L ) + ( n+ 1 ) + ' i:r
:n(n*t) Therefore,
S\ : , _ n ( n,)+ t )
z.,l i=l
'
'+(n+I)
nterms
7.1 THE SIGMA NOTATION
EXAMrLE1: Prove Formula 2bY mathematical induction.
solurroN:
We wish to Prove that
*,, _n(n+L)(Zn+I)
>r:
6
'L.The left sideis then2 ,': l'. When ,rrr,ln" formulais verifiedfor n: '2'3li6 l. Theren : L , t h er i g h ts i d ei s [ 1 ( 1+ 1 ) ( 2+ D ) t e  ( 1 fore, the formula is true when n: 1. Now we assume that the formula is true for n: k, where k is any positive integer; and with this assumption we wish to prove that the formula is also true for n:k + 1. If the formula is true for tt: k, we have k
S .// i , : i:1
Wh e n n : k+l
f
k(k + 1,)(2k+ 1,) 6
(2)
k * L , w e have
i r : 1 2 * 2 2 + 9 2 + . . . + k 2+ ( k + 1 ) '
i:l k
:)i2+(k+1)' + L) (k 1), (by applyingEq. (2)) _ k(k + L)_(zk + + 6
_ k ( k+ L )( 2 k+ L ) + 6 ( k+ L ) ' z 6 _ ( k + 1 )[ k ( 2 k+ 1 ) + 6 ( k+ 1 ) ] _ (k+L)(zk'z+7k+6) 5 _ (k+1)(k+2)(2k+3) 6  ( / c +1 ) [ ( k + 1 ) + 1 ] [ 2 ( k +1 ) + 1 ] 6 Therefore, the formula is true f.ot n: k * L. We have proved that the formula holds f.or n: 1, and we have also proved that when the formula holds f.or n: k, the formula also holds for n: k * 1. Therefore, it follows that the formula holds when n is any positive integer.
A proof of Formula 2, without using mathematical induction, is left as an exercise. The proofs of Formulas 3 and 4 are also left as exercises (see Exercises L2 to 76).
THE DEFINITE INTEGRAL
ExAMpLE 2: n
Evaluate
soLUTroN: From Property 4, where F(i) : 4i,1t follows that n
(4' 4i') 4  4 0 > i:r
(4'  4i')
t
.{J
i:r
:4nI
nxaivrPrn3:
Evaluate
SOLUTION:
n
i(3i 2) > i:r
n
r(3t 2) > i:r
(3i2  2i)
by using Properties 14 and Formulas 14.
(3i')
:3
+> (2i)
i' Z2 i > i:7
(by Property 3)
(by Property2) n(n+ 7) 2 (by Formulas2 and 1 )
2nBl3nzIn2n22n
2 2n3*n2n
Exercises7.1 In Exercises 1 through 8, find the given sum. 6"
1 .> ( s i  2 ) i=l
t $
2.>(i+1.)' i:r
2
KFz
i . + , (  t ; u * r ) , k F
3
U ,D,''
3.
it
?
i1,
s1
6. z '  l + i 2
3 b 8 " '. y t I ,rork+3
9. Prove Property 1 (7.1,.1,). 10. Prove Property 3 (7.1.3). 11. Prove Formula 7 (7.1.5)by mathematicalinduction. L2. Prove Formula 2 (7.1.6\without using mathematicalinduction. (HrNr: iB (i  7)":3i '  3i + L, so that nn
)
t ; ' ( i  1 ) ' l : )
tgt, 3t+ 1)
On the left side of the above equation, use Property 4; on the right side, use Properties '1,,Z, and 3 and Formula 1.)
7.2 AREA 281 13. Prove Formula 3 (7.1.7)without using mathematicalinduction. (HrNr: ia (i  qa:4p  rO * 4i 1, and use a method similar to the one for Exercise12.) 14. Prove Formula 4 (7.1.8)without using mathematical induction (seehints for Exercises12 and 13 above). 15. Prove Formula 3 (7,1.4 by mathematical induction. 16, Prove Formula 4 (7.1.8)by mathematicalinduction. In Exercises17 through 25, evaluatethe indicated sum by using Properties 1 thlough 4 and Formulas 1 through 4.
t
20
1 8 .> 3 i ( i 2 + 2 )
2zi(i1,)
i:r n
n
20. > (2ktzk)
19.>(10i*'10,)
k:r
40 22.>3/T+r \Dll
'f:,Lk k+L) ,r.''flff=+l ,t
24iz(i
i:r n
 z\
24.>\i(L+i2) il lr
n
25. > [(3o  3o)' (3t't+ 3kr)2]
f
/i\21r12
2 6 . P r o v e :) l t  {  }  \nl J ,?nL
:2}
n
? = ,[ ' 
(*)']''' *,
Prope rty4 toshowthat
,.i;Use
) F ( n ) F ( 1 ) r ( 0 ) > ttt, + 1) F(i 1)l : F(nr1'+
Lli:1
(b) Prove that n
t
,1J
t(t+1)'(i'
1 ) ' l: ( n * 1 , ) 3 * n 3  L
i:r
(c) Prove that nnn
)
t t ; + 1 ) '  ( i  1 ) ' l : 2 ( 0 i '+ 2 ) : 2 n* 6 )
i2
i:r
(d) Using the results of parts (b) and (c), prove Formula 2. 28. Use the method of Exercise 27 to prove Formula 3. n )t'
n
29. If X:El  , pr ov e n
t ha t )
(x ,  X)' :2
i:r
7.2 AREA
n
i=l
r.
r,'  X 2 * , iL
We use the word mensure extensively throughout the book. A measure refers to a number (no units are included). For example, if the area of a triangle is L0 irt.z, we say that the measure of the area of the triangle is 10. You probably have an intuitive idea of what is meant by the measure of the area of certain geometrical figures; it is a number that in some way
THE DEFINITEINTEGRAL
Figure7.2.1
Figure7.2.2
gives the size of the region enclosed by the figure. The area of a rectangle is the product of its length and width, and the area of a triangle is half the product of the lengths of the base and the altitude. The area of a polygon can be defined as the sum of the areas of triangles into which it is decomposed, and it can be proved that the area thus obtained is independent of how the polygon is decomposed into triangles (see Fig. 7.2.1). However, how do we define the measure of the area of a region in a plane if the region is bounded by a curve? Are we even certain that such a region has an area? Let us consider a region R in the plane as shown in Fig. 7.2.2. The region R is bounded by the r axis, the lines x: a and x: b, and the curve having the equation y  f(x), where f is a function continuous on the closed interval [a,b]. For simplicity, we take f (x) > 0 for all x in [a,bl.We wish to assign a number A to be the measure of the area of R. We use a limiting process similar to the one used in defining the area of a circle: The area of a circle is defined as the limit of the areas of inscribed regular polygons as the number of sides increases without bound. We realize intuitively that, whatever number is chosen to represent A, that number must be at least as great as the measure of the area of any polygonal region contained in R, and must be no greater than the measure of the area of any polygonal region containing R. We first define a polygonal region contained in R. Divide the closed interval [a, b] into resubintervals. For simplicity, we shall now take each of these subintervals as being of equal length, for instance, Ar. Therefore, Ar: (b  a)ln. Denote the endpoints of these subintervals by xo, xt, x 2 , . . , x n  r ,r r , w h e r € f f o : f l , x t : a i L x , . . . , x i : A * i L , x , . . , xnr: a l (n  t) A,x,xn: b. Let the ith subinterval be denoted by lxrr, xrf. Because / is.continuous on the closed interval la, bf, it is continuous on each closed subinterval. By the extremevalue theorem (4.5.9), there is a number in each subinterval for which / has an absolute minimum value. In the lth subinterval, let this number be ci, so that /(c) is the absolute minimum value of f on the subinterval fx,r, r1]. Consider n rectangles, each having a width Ax units and an altitude f (r,) units (see Fig. 7.2.3). Let the sum of the areas of these n rcctangles be given by S, square units; then
S":f(cr) Ax*f(rr) Ax*'
'+f(c,) Ax.r
* f(c") Lx
or, with the sigma notation, n
sn:\
f (c) Ax
(1)
The summation on the right side of Eq. (L) gives the sum of the measuresof the areasof n inscribed rectangles.Thus, however we define A, rt must be such that A= S,
7.2 AREA
Figure7.2.3
Figure7.2.4
In Fig. 7.2.3 the shaded region has an area of 5n square units. Now, let n increase. Specifically, multiply nby 2; then the number of rectangles is doubled, and the width of each rectangle is halved. This is illustrated in Fir.7.2.4, showing twice as many rectangles as Fig. 7.2.9. By comparing the two figures, notice that the shaded region in Fig. 7.2.4 appears to
THE DEFINITEINTEGRAL
approximate the region R more nearly than that of Fig.7.2.3. So the sum of the measures of the areas of the rectangles in Fig. 7.2.4 is closer to the number we wish to represent the measure of the area of R. As n increases, the values of s, found from Eq. (1) increase, and successive values of 5, differ from each other by amounts that become arbitrarily small. This is proved in advanced calculus by a theorem which states that if / is continuous on fa, b], then as n increases without bound, the value of S, given by (1) approaches a limit. It is this limit that we take as the definition of the measure of the area of region R. 7.2.'l' Definition
Suppose that the function/ is continuous on the closed interval la,bl,with f (x) > 0 for all x in fa, b), and that R is the region bounded by the curve y  f(x), the x axis, and the lines x: a and r: b. Divide the interval la,bI into n subintervals, each of length Ax: (b  a) ln, and denote the ith subinterval by lxtt, xil. Then if f(c) is the absolute minimum function value on the ith subinterval, the measure of the area of region R is given by
rim ) /(r,) ar
A*+€b {*1
(2)
.
Equation (2) means that, for any e ) 0, there is a number N ) 0 such that
ol l>f,',,a,x . .
whenever rz ) N
and n is a positive integer. We could take circumscribed rectangles instead of inscribed rectangles. In this case, we take as the measures of the altitudes of the rectangles the absolute maximum value of / on each subinterval. The existence of an absolute maximum value of / on each subinterval is guaranteed by the extremevalue theorem (4.5.9). The corresponding sums of the measures of the areas of the circumscribed rectangles are at least as great as the measure of the area of the region R, and it can be shown that the limit of these sums as n increases without bound is exactly the same as the limit of the sum of the measures of the areas of the inscribed rectangles. This is also proved in advanced calculus. Thus, we could define the measure of the area of the region R by
A:
n
lim 2 f @) Lx
(3)
wheref (";.;"'.]1" urrotutemaximumvatueof / on lx,,, x,l. The measure of the altitude of the rectangle in the lth subinterval actually can be taken as the function value of any number in that subinterval, and the limit of the sum of the measures of the areas of the rectangles is the same no matter what numbers are selected. This is also proved in advanced calculus, and later in this chapter we extend the definition of the measure of the area of a region to be the limit of such a sum.
7.2 AREA EXAMPLEL: Find the area of the region bounded by the curve A : x', the r axis, and the line x: 3 by taking inscribed rectangles.
solurroN: Figure 7.2.5 shows the region and the ith inscribed rectangle. We apply Definition 7.2.L. Divide the closed interval [0, 3] into n subintervals, each of length A,x: xs: 0, xr: Lx, xz: 2 Ax, . . ., xi: i Lx, . . r x n r: (n  t) A,x,xn: 3. ^

rJ,,t,
30 nn
3
The function value /(x)  xz and because / is increasing on [0, 3], the absolute minimum value of f on the ith subinterval lxrr, r;] is f (xrr). Therefore, from Eq. (2) n
A:
lim D fk,r) Lx
Because ::,::;
( 4)
1) Ax and /(x) : ,sz,we have
f (x,,) [(t 1) Ar]'z Therefore, nn
) f (x,') Ar: > (t 1)'z(Ar)g 8". ;; : ztn,ro tnut: n n s ( t  L ) ' Ln:,2 7 n A r : / ( r ' ' ) ) > sL i:r i:r
:4[$;,2$ no
t+$
r), ']
,?:, ,?:, f4 and using Formulas 2 and L and Property L from Sec.7.'1., we get
._27 ln(n+t)(2n+L) ,: 2fQ'')Ax:lirt 27
_2.ry!*"1
2n3*3n2*n6n26n*6n
n3
9 2
2n23n*L ___v
Then, from Eq. (4), we have
(,1.#) :*(2_0+0) 9 Therefore, the area of the region is 9 square units.
THE DEFINITEINTEGRAL
2: Find the areaof the ExAMPLE region in ExampleL by taking circumscribed rectangles.
solurroN: we take as the measureof the altitude of the lth rectangle the absolute maximum value of / on the lth subintervaI lxi4, 11] which is f (x,). From Eq. (3), we have n
A: lim 2 f k,) tx Because [!
r'ir,thenf (x,): (i a,x)z,and so
n
n
Ar: 2f(',) i:r
]
L/ i:1
)7n
i2(A,x)S:
+n"s? i , l = l
27 l n ( n+ 7 )( 2 n+ 1 ) ' l n3
L6l
2n2*3n*L _____7Therefore, from Eq. (5), we obtain q
A:
Itm n+6
,
.(r*q+4) \
/r
n"/
( a s in Example1)
uxaupru 3: Find the area of the trapezoid which is the region bounded by the line 2x * y : 8, the x axis, and the lines x : 1 and x: 3. Take inscribed rectangles.
soLUTroN: The region and the ith inscribed rectangle are shown in Fig. 7.2.6.The closed interval [1, 3] is divided into n subintervals, each of length A,x; xs: L , xr: "1.* Lx, xz: I * 2 A,x, xi:t*i Lx,
1.t (n  L) L,x,xn 31
Solving the equation of the line for y, we obtainy:2x * B. Therefore, f (x): 2x* 8, and because / is decreasing on [1,3], the absolute minimum value of f on the ith subinterval lxi_r, ri] is f (rr). Because ri: 1 + i Ax a nd f(* ) : 2x* 8, then f(x,) : 2(L+ i A x) * 8 : 6  zi Ax. Fr om Eq,. (2), we have
f(x) A,x
lim fl*@
: lim i tu 2i a,x)ax n+@ i:l F i g u r e7 . 2 , 6
:
n
lim )
2+@
i:l n
i:r
I0 Ax  Zi(A,x)'z1
,,G)',1 I'G)
7.2 AREA
:
lim lllo
287
l+yiE'l
we have Using Property 1,(7.1.1)and Formula l, (7.1,.5),
: lim ftA) nl  \
h*@
8 Therefore,the area is 8 square units. Using the formula from plane geometry for the area of a trapezoid,A:ih(b1*b2), where h,br, andb, are, respectively,the number of units in the lengths of the altitude and the two bases,we get A: +(2)(6+ 2): 8, which agreeswith our result.
7,2 Exercises In Exercises 1 through 14, use the method oI this section to find the area of the given region; use inscribed or circumscribed rectantles as indicated. For each exercise, draw a figure showing the redon and the ith rectangle. The region bounded by y: r', the x axis, and the line r = 2; inscribed rectangles, J. 2. The region of Exercise 1; cirormscribed rectangles. ..\ The re6on boun dedby y :2x,
1 and r : 4; circumscribed rectangles.
tte r axis, and the lines r:
4. The region of Exercise 3; insclibed rectangles. 5l The region above the r axis and to the right of the line r:1 ''' y=4f, inscribed rectangles.
bounded by the r axis, the line r= 1, and the curve
6, The region of Exercise 5; circumsoibed rectangles. 7i The region lying to the left of the line : = 1 bounded by the curve and lines of Exercise 5; circumscribed rectangles. 8. The region of Exercise 7; inscribed rectangles. 9. The region bounded by y : 3/, dre r axis, and the line r:
1; inscribed rectangles.
10. The region of Exercise 9i circumscribed rectantles. 11. The region bounded by y:
ri, the r axis, and the lines r:
L and r:
2; inscribed rectangles.
12. The region of Exercise lli circumscribed rectantles. li.lnre region bou nded by y: ' rectangles.
mx, with tt > 0, the r axis, and the lines l:
a and r:
b, with b ) a > 0; circumscribed
14. The region of Exercbe 13; inscribed rectangles. 15. Use the rrethod of this s€ction to find the ar€a of an isosceles trapezoid whoee bases have measr:res b. and D"and whose altitude has measure l.
284
INTEGRAL THE DEFINITE
4 tox:4lorm 76. The Snph of y:4l.zland the r axis hom x: area of this triangle.
a triarrgle. Use the method of this section to lind the
In Exercises 17 throtgb 22, tind the area of the region by taking as the measure of the altitude of the ith rectangle l(nr), where z1 is the midpoint of the ith subinterval. (rrrwr: n1: !(4 a 4 1).) 17. The region of Example 1..
18. The region of Exercise L.
19. The region of Exercise 3.
20. The region of Exercise 5.
21. The region of Exercise 7.
22. The region of Exercise 9.
7.3 THE DEFINITE INTEGRAL
In the preceding section, the measure of the area of a region was defined as the following limit: n
lim ) f(c1) A,x
n++6
(1)
i:l
To lead up to this definition, we divided the closed interval fa, bl into subintervals of equal length and then took ci as the point in the ith subinterval for which / has an absolute minimum value. We also restricted the function values f (x) to be nonnegative on fa,bl and further required f tobe continuous on la, bl. To define the definite integral, we need to consider a new kind of limiting process/ of which the limit given in (1) is a special case. Let / be a function defined on the closed interval [a, b].Divide this interval into n subintervals by choosing any (n  1.)intermediate points between a andb. Let t xnr be the intermediate points so that ro(xr1xr1
1 xn_r I
x,
The points x6,xr, xz, . r xnrrx?rarenot necessarilyequidistant.Let Arr be the length of the first subinterval so that Arr : xr  xil let Arr be the length of the secondsubinterval so that Azr  x2 xr; and so forth, so that the length of the lth subinterval is A1r, and a.g: xi A set of all such subintervals of the interval [a, b] is called apartition of the interval [a, b].Let A be such a partition. Figure 7.3.1 illustrates one such partition A, of la, bl. I a:ro
I


I
x7
)t2
13
x4
I xn_L
l.>r In:o
F i g u re7 .3 .1 The partition A contains n subintervals. One of these subintervals is longest; however, there may be more than one such subinterval. The length of the longest subinterval of the partition A, called the norm of the partition, is denoted by llAll. Choose a point in each subinterval of the partition A: Let fr be the point chosen in [xo, xr] so that ro < €, < xr. Let f, be the point chosen in
7.3 THE DEFINITEINTEGRAL
v
lxr, xrf so that xr s tz = x2, afld so forth, so that fi is the point chosenin lxur,x], and xrr 3 tt  x.. Form the sum
0
' 'tf({") A'"x
f ( ( , )A , r * f ( ( , )L , x t ' ' ' + f ( ( , ) L p t ' \
or n
> f (€,) Lix \
Su.nl sum is called a Riemannsum,named for the mathematicianGeorg Friedrich Bernhard Riemann (18261.866). o ILLUSTRATIoN 1: Suppose/(x) :1.0  x2, with i = x < 3. We will find the Riemann sum for the function f on [i, e1 for the partition A: xo:i, x.t:1, xz:li, xs:Lt, xq:2*, xs:3, andf1: t, tz:Lt, ts:11, €q:2,
t": 21.
Figure 7.3.2 shows a sketch of the graph of / on l*, Z1 and the five rectangles,the measuresof whose areasare the terms of the Riemannsum.
5
5
2f tt,l L&: f(€,)A,r * f(il A,rx*f(€') A'r * f(t^) Anr*/((') A'r
 1)+ f@(L* r+) : fE)(l +)+ fG)0+ + f (2)(2+1*)+ f (+)(3 2+) : (e*)(+)+ (8+)(+)+ (6€)(+)+ (5)(+)+ (2+)(+) : 18#
.
Thenorm of A is the lengthof the longestsubinterval.Hence,llAll:*.
o
xot,
Figure7.3.2
xrtrl
xr\xn
l, t, !.
€rx
Because the function values f (x) are not restricted to nonnegative values, some of the /(fi) could be negative. In such a case/ the geometric interpretation of the Riemann sum would be the sum of the measures of the areas of the rectangles lying above the r axis plus the negatives of the measures of the areas of the rectangles lying below the r axis. This situation is illustrated in Fig.7.3.3. Here
v
a\
y : f(x)
J
li
IC
Arl
.i,
la'
dt, \
tn
t, *rl i4 trv
xott \
€z O
x2
,{N
A^ 'l
: i:f: :tt
As i/
\
i
{o xo
.xs
f,l f' I ,/,,
t1
b .{rc
;\ Aa
t F i g u r e7 . 3 . 3
fto r,s
A,I
T I
t
1
rto
THE DEFINITEINTEGRAL
j
ftg,l Lix: Ar* A2_ As_A4 Au* Au* A7 As An Aro
because f(€"), fG), fG), f$r), f$n), and /(f,o) are negative numbers. We are now in a position to definewhat is meantby a function / being "integrable" on the closedinterval la, bl.
7.3.1 Definition
Let f be a function whose domain includesthe closedinterval [a, b]. Then / is said to be integrableon fa, bl if there is a number L satisfyingthe condition that, for every e ) 0, there exists a 6 ) 0 such that I Lix tl ' ' fte,l l) ln
for every partition A for which llAll < 6, and for any fi in the closed interval f x r _ r x, 1 l i, : 1 , 2 , . . . , n .
In words, Definition7.3.L states that, for a given function / defined on the closed interval la, bl, we can make the values of the Riemann sums as close to L as we please by taking the norms llAllof all partitions A of [a, b] sufficiently small for all possible choices of the numbers f1 for which xir s €i = xi.If Definition 7.3.1 holds, we write n
(2)
lim ) /(f,) A,ix: L llAll0 t:r
The above limiting process is different from that discussed in Chapter the number L in (2) exists if for every e ) 0 there 2. From Definition 7.3.'I.., existsaD)0suchthat
lf rtr,l Lix rlI . . lr:r' for every partition A for which llAll< 6, and for any f, in the closedinterval , n. f x r  r ,x i l ,i : L , 2 , In Definition 2.1.1we had the following: (3)
lim/(r):1 if for every e > 0 there exists a 6 > 0 such that lf(x) Ll < e
w h e n e v e r 0< l r  a l
< 6
In limiting process(2), for a particular 6 > 0 there are infinitely many partitions A having norm l[ll < 6. This is analogousto the fact that in limiting process (3), for a given 6 > 0 there are infinitely many values of x for which 0 < lr  al < 6. However, in limiting process(2) , for eachpartition
7.3 THE DEFINITE INTEGRAL 291 A there are infinitely many choices of ti.It is in this respect that the two limiting processes differ. In Chapter 2 (Theorem 2.1,.2) we showed that if the number L in limiting process (3) exists, it is unique. In a similar manner we can show that if there is a number L satisfying Definition 7.3.1, then it is unique. Now we can define the "definite integral,."
7.9.2 Definition
If /is a function defined on the closed interval [a,bf, then the definite integral of / from a to b, denoted bV Il f @) dx, is given by
u,' fl'f;w'n*t' ,tiil,:/tut
(4)
if the limit exists. Note that the statement"the function / is integrableon the closedinterval la, bf" is synonymouswith the statement"the definite integral of / from atob exists." In the notation for the definite integral I! f @) dx, f (x) is called the integrand,a is called the lower limit, and b is called the upper limit. T}l.e symbol
I
J
is called an integral sign. The integral sign resembles a capital 5, which is appropriate because the definite integral is the limit of a sum. It is the same symbol we used in Chapter 6 to indicate the operation of antidifferentiation. The reason for the common symbol is that a theorem (7.6.2), called the fundamental theorem of the calculus, enables us to evaluate a definite integral by finding an antiderivative (also called an indefinite integral). A question that now arises is as follows: Under what conditions does a number L satisfying Definition 7.3.Lexis that is, under what conditions is a function f integrable? An answer to this question is given by the following theorem. 7.3.3 Theorem
If a function / is continuous on the closed interval lq, bf, then/ is integ ra b l e o n l a , b l . The proof of this theorem is beyond the scope of this book and is given in advanced calculus texts. The condition that / is continuous on la,bl,while being sufficient to guarantee that/is integrable on la,bl,is not a necessary condition for the existence of [l f @) dx. That is, if / is continuous on la,bT,then Theorem 7.3.3 assures us that Il f G) dx exists; however,
THE DEFINITEINTEGRAL
it is possible for the integral to exist even if the function is discontinuous at somenumbers in la, b]. The following examplegives a function which is discontinuous and yet integrable on a closed interval.
EXAMPLE1: defined by
Let f be the function
,,, [o ifx*o l \ x ) : I "Ltr r i f r _ o Letfa, bl be any interval such that a 1 0 < b. Show that/is discontinuous on la, bl and yet integrable on [a, bl.
solurroN:
Becauselim f (x) :0
+ f (0),/is discontinuous at 0 and hence
discontinuous on [a, b]. To prove that / is integrable on la, bl we show that Definition 7.3.7 is satisfied. Consider the Riemann sum n
) /(f,) l,' If none of the numbers (r, €r, zero.Supposethat $: 0. Then
is zero, then the Riemann sum is
n
. > /(f,) Aix: I L,x
,r, "u:,iur.ur"
{rl s llall lf, rr*,, Hence,
If, f rc,l
Lic_ol . .
whenever llAll< e
Comparingthe abovewith DefinitionT.3.'1, where D: e and L:0, that f is integrable on la, bl.
we see
In Definition7.3.2, the closed interval [a, b] is given, and so we assume Ihat a < b. To consider the definite integral of a function from a ta b / when o s b, or when fl: b, we have the following definitions. 7.3.4 Definition
I f a > . , b, then rb
J"f(x) dx:it tf f( x) 7.3.5 Definition
dx [: f(x)
dx exists.
It f Q) dx 0 if f (a) exists. At the beginning of this section, we stated that the limit used in Definition 7.2.1to define the measure of the area of a region is a special caseof
7.3 THE DEFINITEINTEGRAL
the limit used in Definition7.3.2 to define the definite integral. In the discussion of area, the intervalfa, b] was divided into n subintervals of equal length. Such a partition of the interval [a, bj is called a regular partition.If Ar is the length of each subinterval in a regular partition, then each Lix: Ar, and the norm of the partition is Ar. Making these substitutions in Eq. (4) , we have fbn
I f tr>dx: Alim ) f ((,) Lx Ja r ,o i:l Furthermore, ba n and
Lx So from Eq. (6) lim lL
(8)
t@
and from (7),because b > a and Ar approaches zero through positive valu e s (b e c a u s eA x > 0), lim n:
*oo
(e)
Ar0
From limits (8) and (9), we concludethat Ar +0 is equivalent to n >+@
(10)
Thus, we have from Eq. (5) and statement(10), fbn
)i : l f (€,)Lx I f t*l dx: nlim Ja _*o
(11)
It should be remembered that fi can be any point in the ith subinterval lxit, xrl. In applications of the definite integral, regular partitions are often used; therefore, formulas (5) and (L1) are especially important. Comparing the limit used in Definition 7.2.L, which gives the measure of the area of a region, with the limit on the right side of Eq. (11), we have in the first case,
lim ) f(c) Lx
(t2)
n+@ i:l
where/(ci) is the absolute minimum function value on lxit, x1).In the sec
THE DEFINITEINTEGRAL
ond case, we have n
rim ) f(€,) Lx n*q it
(13)
where (1 is any number in [r1r,r,]. Becausethe function / is continuous on fa,bl, by Theorem 7.9.9, Il f @ dx exists;therefore,this definite integral is the limit of all Riemann sums of / on la, bl including those in (12) and (13).Becauseof this, we redefine the area of a region in a more general way. 7.3.6 Definition f(x)
Letthe function/becontinuousonfa, bl and f(x) > 0forall rin la,bl.Let R be the region bounded by the curve y  f (x), the r axis, and the lines x: fr and r : b. Then the measure of the area of region R is given by
o'* P,rtr'lo":I)f(x) dx ,,1il,
F i g u r e7 . 3 . 4
ExAMpLE2: Find the exact value of the definite integral f3
I x2dx
Jt
Interpret the result geometrically.
The above definition states that if f (x) > 0 for all r in Ia, bl, the definite integral [! f (x) dx can be interpreted geometrically as the measure of the area of the region R shown in Fig. 7.i.4. Equation (11) can be used to find the exact value of a definite integral as illustrated in the following example.
soLUrIoN: Consider a regular partition of the closedinterval [1, 3] into n subintervals.Then A^x: 2ln. If we choose ti as the right endpoint of each subinterval, we have
€ r : L + 1 , € z : 1 * r G ) , t s : L+3G),
€i:1*'(1), tn:1
*"(t
Becausef (x): x',
f c,): (r* +)' : (! j4)' Therefore,by using Eq. (11) and applying properties and formulas from Sec.7.'1,, we get
I' *' dx:lri) :
(+)'t"
)n
lim h*a
+s n"?
1:l
(n' t 4ni l 4i2)
7.3 THE DEFINITEINTEGRAL
:
lim ll
*6
#l*z
'l., t 4n
295
n
n
i+4
i'] i:r
i:r
:Ii hl"'"
,r,+ n ' n ( n * L ) , T4l n ( n+ t ) ( 2 n + 1 ) l Z
:,li #l**2nst2n2.Wl
:.llils*f*8n2 ;+ L 2n + I Jn'
:,Illo*f* 8g , 4i ,
4f g"')
:6*O+8+0+0 _83 (1, L)
Figure7.3.5
We interpret the result geometrically. Because xz = 0 for all x in the x axis, and the lines [1,3], the region bounded by the curve A:x', : 3 has an area of 83 square units. The region is shown x: 1 and x in Fig. 7.3.5.
Exercises7.3 In Exercises1 through 6, iind the Riemannsum for the function on the interval, using the given partition A and the given valuesof $. Draw a sketchof the graph of the function on the given interval, and show the rectanglesthe measureof whose areasare the terms of lhe Riemannsum. (SeeIllustration 1 and Fig. 7,3,2.) 1. f (x) : f , O s r = 3; for A: xo: 0, xr: tr,xz: 7*,rs : 2i, xe: 3; : L ta : 1' &: lt, €n= 2i 2.f(x):f,g 0 there exists a 6 > 0 such that ln r + s(f,)la# (M+ N)l. . t/(f,) l> lir I t
forall partitions A for which llAll< 6 and for any 4i in lxr_r,xil. Because
M:,Jim t fG,) o,* and IY: fim XS(f,) a,t l l a l l o t : t llalloi:r it follows that for any e ) 0 there exist a 6, ) 0 and a 6z ) 0 such that
.*z and r(ri)Aixrl.f  Air utl r,*,, l> l7:r'  lp, for all partitions A for which llAll< 61 and llAll< 6r, and for any fi in fxrr, ri]. Therefore,if 5 : min(6r, 6r), then for any e ) 0
Air,l*l> r(ri)AirNl.f ' 2+i.:, l>rte,l le' I l?r"" 
(1)
for all partitions A for which llAll< 6 and for any fi in lxi_r, xif. By the triangle inequalit!, we have \ /n \l Aix *) * (=>,t(fi) Air t)l l(p fre,l l/n
= fG,)Licrl * lI
,^, s(fi) Air Nl
From inequalities (L) and (2), we have lln
m
\
I
lx* ) s(f,)t,r)  (M+N)l . . l() ftr'l
(3)
From Property 3 (7.1,.3) of the sigma notation, we have nnn
> /(4') A,r* ) s((,) Lix: > t/(f,) + s(f,)l Lix ," ;; ,"bstitutin ,: n"^(4) e)0 ln
(4)
into'O, *" are able to concludethat for any I
+s(f,)lL& (M+N)l.. l}_,ffrc; for all partitions A for which llAll< 6, where 6: min(6r, 5r) and for any€i
OF THE DEFINITEINTEGRAL 7.4 PROPERTIES r2l. This proves
in lxia,
that f * g is integrable
299
on [a, bl and that
dx: ,o, dx* [' su)a* f rru>+ s(r)I Theorem 7.4.4 can be extended to any number of functions. That is, if the functions fr, fr, . . , fn are all integrable on la, b), then (ft+ fr* ' ' ' + f,) ,s integrable on [a,bf and rb IJ a lf,(x) + f"(x) + . . . + f"(x)l dx
: lo f,(x)dx* fuf,(*) dx* ' ' ' + [' f,(x) dx Jct Ja Jo'"
The plus sign in the statementof Theorem7.4.4canbe replacedby a 'l'. minus sign as a result of applying Theorem 7.4.3,where k: 7.4.5 Theorem If the function / is integrableon the closedintervals la, bf,la, c], andlc, bf,
f rr.tdx:["r,,,tdx* ['
fr*>o*
wherea g(x) for all x in la, bl. We wish to prove that Il lf G)  s?)l dx = 0. fb
I lf (x) s(r)l d x : Ja
fbn
I h(x) dx:
Ja
lim ). ft(f,) Air
l l a l l' o F r
(8)
Assume that
(e)
l i m L,l) h ( { , ) A i x : L ( 0 llAll0 i=1
Then by Definition 7.3.1, with e:L,
,(fi) Air .l . t l> li:l I t ? ' l
there exists a 6 > 0 such that
whenever llAll< 6
But because n
ln h(t) A,* L = l t h(t) Aix Ll > i:r lt li:l
from inequality (10)we ha VC n
tt(il A,ix L <  L 2 i:r
wheneverllAll< D
(10;
OF THE DEFINITEINTEGRAL 7.4 PROPERTIES
n
h(il A;r ( 0
wheneverllAll< S
(1 1 )
i:r
But statement(11) is impossible becauseevery h((1) is nonnegative and every L,g 7 0; thus, we have a contradiction to our assumption (9). Therefore, (9) is false, and lim
). ft(f,) A1x > 0
(J2)
LJ
llall*o t:t
From (8) and (L2),we have fb IJ a t/tt)  s(r)I dx> 0
Hence,
["ro [' sal
and so
f(x) g(r)
x
dx> f sroa' f tal
. rLLUSrRArroN 3: Figure 7.4.3 gives a geometric interpretation of Theorem 7.4.8 when f(x) > g(x) > 0 for all r in la,bl. Il f @ dx gives the measure of the area of the region bounded by the curye y: f (x)'the x axis, and the lines x: A and r : b. Ig g@) dxgives the measure of the area of the region bounded by the curve y : 8(x), the r axis, and the lines x: a and r : b.In the figure we see that the first area is greater than the second area.
Figure7.4.3
7.4.9 Theorem
Supposethat the function / is continuous on the closedinterval la, bl.If m and M are, respectively, the absolute minimum and absolute maximum function values of f on la, bl so that mf(x)=M
m ( b a )
forax=b
=r
f(x) dx < M(b a)
pRooF: Because / is continuous on la, bl, the extremevalue theorem (4.5.9) guarantees the existence of m and M.
THE DEFINITE INTEGRAL
By Theorem 7.4.7 fb
J,
* d x : m ( b a )
(13)
, dx: M(b s)
(14)
and fb
J"
Because/ is continuous on fa, bl, itfollows from Theorem 7.3.j that/ is integrable on fa, bl. Then because f (*) > m for all r in [a, b], we have from Theorem 7.4.8 fb
fb
I f ( * ) d x = lJ n m d x Ja' which from (13) gives fb
I f' @ d x > m ( b  a ) Ja
(15)
Similarly, becauseM = f (x) for all .r in la, bf, it follows from Theorem7.4.8 that
f,,,dx>[)roo, which from (14) gives
M(b_a)=f) fato
(16)
Combining inequalities (15) and (16) we have
v: f@)
Figure7.4.4
nxevrprn1: Apply Theorem7.4.9 to find a smallestand a largest possiblevalue of r4 r l 2 ( r '  6 x 2 * 9 x + 1 . ) d x
J
m(ba)=['fildx=
M(b  a)
o rLLUSrRArroN4: A geometric interpretation of Theorem 7.4.9 is given in Fig.7.4.4, where f (x) > 0 for all x in la, b). The integral t!, (x) dx gives the f measure of the area of the region bounded by the curve y : (x),the r axis, f and the lines x: A and x : b. This area is greater than that of the rectangle whose dimensions are m and (b  a) and less than that of the rectangle whose dimensions are M and (b  a). o
solurroN: Referring to Example 1, Sec.5.1, we see that has a relative / minimum value of 1 at x:3 and a relative maximum value of 5 at x:"1.. and f $):5. Hence, the absolute minimum value of / on 1i, 41isI, f e):# and the absolute maximum value is 5. Taking m: '!. and M : 5 in Theorem 7.4,.9,we have
L(4+) = f (rr 6xz*9x*t) dx=5(ai) I rtz
OF THE DEFINITEINTEGRAL 7.4 PROPERTIES
Use the results of ExamPleL, Sec.5.1.
and so
t=l
r4
(x'6x2* 9x*t)
dx
I'
*, dx but
f'
* a* =
Ii
o dx. Do not evaluatethe definiteintegrals.
306
THE DEFINITEINTEGRAL
20. If / is continuous on la, b], prove that
l['lfro*l=I:lf(x)ldx
(Hrur:
(x)l = f (x) s l/(x)l.)
7.5 THE MEANVALUE Before stating and proving the meanvalue theorem for integrals, w€ THEOREM FOR INTEGRALS discuss an important theorem about a function that is continuous on a closedinterval. It is called the intermediateaalue theorem,and we need to use it to prove the meanvalue theorem for integrals. 7.5.1 Theorem I nt erm edint eV ahte Tlrcor ent
v f(b) k f(a)
F i g u r e7 . 5 . 1
y : f(x)
If the function / is continuous on the closedinterval [a, bl andif f (a) + f (b), for any number k between (a) and (b) there exists a number c f fhen f between a and b such that f (c) : 16. The proof of this theorem is beyond the scope of this book; it can be found in an advancedcalculustext. However, *" dir..rss the geometricinterpretation of the theorem. In Fig.7.s.1, (0,k) is any pointon the y axis between the points (0,f (a)) and (0,f (b)).TheoremT.s.j.statesthat the line y : k must intersect the curve whose equation is y : (x) at the point f (c'k), where c lies between a and b. Figure 7.5.1.shows ihir it,t"rsection. Note that for some values of k there may be more than one possible value f.ot c. The theorem statesthat there is always at least or" lrulue of.c, but that it is not necessarilyunique. Figure 7.5.2 showsthree possible values of c (cr, cr, and c.r)for a particular k. Theorem 7.s.j,statesthat if the function / is continuous on a closed interval [a, bf, then / assumesevery value between f (a) andf (b) as.r assumesall values between a and.b.The importanceof the continuity of /on la,bl is demonstratedin the following illustration. . ILLUSTRATTON 1: Consider the function I defined by tf \('x \/ : [ x ,  1 lx"
ifo<x 0 for all values of x in la, bl. Then tl fG) dx is the measure of the area of the region bounded by the curve whose equation is y : (x), the f x axis, and the lines x: a and x: b (see Fig.7.s.il. Theorem7.s.2 states that there is a number X in la, b] such that the area of the recta ngle AEFB of height /(X) units and width (b  a) units is equal to the area of the r€Bion ADCB. . The value of X is not necessarily unique. The theorem does not provide a method for finding X, but it states that a value of X exists, and
7.5 THE MEANVALUETHEOREMFOR INTEGRALS 309
this is used to prove other theorems. In some particular cases we can find the value of X, as is illustrated in the following example.
2: Find the value of X EXAMPLE such that f3
I 'f ( * ) d x  l ( x ) ( 3  1 ) Jr if f (x) : )c2.Use the result of Example 2 in Sec.7.3.
solurroN: In Example2 of Sec.7.3, we obtained r3 I x 2d x : 8 3
JL
Therefore,we wish to find X such that ' (2) :zfL /(x) that is, X2: # Therefore, X: _+_ +\/99 We reject +\E
since it is not in the interval [1, 31,and we have
f3
dx fG\E)(3  1) I f *m
(or Ax  0), we have, if the limit exists,
nfb
I /(4')a'
l!r_ e!_ ba ,_;*
:
 to
1tg)ax = ba
This leads to the following definition. 7 '5'3 Definition
If the function / is integrable on the closed interval la, bf, the aaerageaalue of / on [a, b] is rb
f{r) a*
J, F, EXAMPLE3: Find the average value of the function / defined by f (*) : x2 otr the interval [1,,3].
SoLUrroN: In Example 2, Sec.
we obtained
r3 I x2dx: +q
Jr
So if A.V. is the average value of f on 11,gl, we have _
236 
31
4^
_rJ
3
An important application of the averagevalue of a function occursin physics and engineering in connection with the conceptof center of mass. This is discussedin the next chaDter. In economics,Definition 2.5.1can be used to find an averagetotal cost or an averagetotal revenue. o rLLusrRArroN4: If the total cost function C is given by C(r): #, where C(r) thousands of dollars is the total cost of producing 100r units of a certain commodity, then the number of thousandsof dollars in the average total cost, when the number of units produced takes on all values from 1O0to 300, is given by the averagevalue of C on [1, 3]. From Example 3, this is rra:4.33. Hence, the averagetotal cost is $4333. o
Exercises7.5 ln Exercises 1 through 8, a function / and a closed interval [a, D] are given. Determine if the intermediate_value theorem holds lor the given value of ft. If the theorem holds, find a number c sucihthat c) : lc. If the theorem does not hold, give the f( reason. Draw a sketch of the curve and the line v : k.
THEOREMOF THE CALCULUS 7.6 THE FUNDAMENTAL
311
1. f(x):2 + x  *t la,bl: [0,3]; k: t 2. f({: f a5r6;la,bf : l1,2l; k: a 3. f (x) : tfi=P, lq, b1: la.5,37;k: 3 a. fk):vffi=F; fa,bf : f0,81;k: B 4
s f {x\ : ;}1; (1*r
6 .f @ : 1 ;  ; (*+
?.f@ : lf _;
.
,la, bl : [3, 1]; k: t it4 =x =?1
ii.,
 1 l; Io'bl:[4'r],k=L
itZt 0 and l(1) < 1 consider the in [0, 1] sudr thatl(c):c. trr.b"" " : x ar\d apply the intermediatevalue theotem to 8 on [0, 1] .) function g for which g@) l(x\ 21..lf f is continuous on [a, b] and Jl lG) dx:o,
7.6 Tp.E FUNDAMENTAL THEOREM OF THE CALCULUS
prove that there is at least one number c in [c, b] such thatl(c):0.
Historically the basic conceptsof the definite integral were used by the ancient Greeks,principally Archimedes (287212B.c.),more than 2000years ago,which was many yearsbefore the differential calculuswas discovered.
312
INTEGRAL THE DEFINITE
In the seventeenth century, almost simultaneously but working independently, Newton and Leibniz showed how the calculus could be used to find the area of a region bounded by u curve or a set of curves, by evaluating a definite integral by antidifferentiation. The procedure involves what is known as the t'undamentaltheorem of the calculus.Before we state and prove this important theorem, w€ discuss definite integrals having a variable upper limit, and a preliminary theorem. Let f be a function continuous on the closed interval la, bl. Then the value of the definite integral ft f Q) dx depends only on the function f and the numbers a and b, and not on the symbol r, which is used here as the independent variable. In Example 2,Sec.7.3,we found the value of .frtx2dx to be 83. Any other symbol instead of x could have been used; for example,
l'
,' or:
u2du: r2dr 8a 1," l,'
If / is continuous on the closed interval fn, bl, then by Theorem 7.3.3 and the definition of the definite integrcI, [! f (t) dt exists. We previously stated that if the definite integral exists, it is a unique number. If r is a number rnfa, b], then/is continuous onfa, x] because it is continuous on [a, bl.Consequently, If f ft) df exists and is a unique number whose value depends on x. Therefore, I{ fU) dt defines a function F having as its domain all numbers in the closed interval [a, b] and whose function value at any number x in la, bl is given by
_r"f(t) dt
F(x) _
f(t)
F i g u r e7 . 6 . 1
7.6.1 Theorem
J,
(1)
As a notational observation, if the limits of the definite integral are variables, different symbols are uged for these limits and for the independent variable in the integrand. H8nce, in Eq. (1), because x is the upper limit, we use the letter f as the independent variable in the integ.ur,d. If, in Eq. (1),f (t) > 0 for all values of f in la, bl, then the function value F (x) can be interpreted geometrically as the measure of the area of the region bounded by the curve whose equation i" y: f (t), the f axis, and t h e l i n e s t : a a n d f : x . ( S e e F i g . 7 . 6 . 1 ,N. )o t e t h a t F ( a ) : I t fU) dt,which by Definition 7.3.5 equals 0. We now state and prove an important theorem giving the derivative of a function F defined as a definite integral having a variable upper limit. Let the function / be continuous on the closed interval la, bl and let x be any number in la, bl.If F is the function defined by
F(x):f ru>0, then
F'(x): f(x)
THEOREM OF THECALCULUS 313 7.6THEFUNDAMENTAL (If x: x:b,
a, the derivative in (2) may be a derivative from the right, and if the derivative in (2) may be a derivative from the left.)
pRooF: Consider two numbers 11 and (r1 + Ar) in [a, b]. Then
F(r,):
dt f' fG)
and
F(x,* Ax)
ft) at f'*^* f
so that
F(x,* Ar) F(r,):
(t) dtdt f' f U) f'*'" f
(3)
By Theorem7.4.6, f e:r+tt
f Jrr+Lx
f rt
dt: or* J" fG\ J., f(t) J,
fQ)dt
or, equivalently, rrr+ir
f n+ Lr
lrt
fo dt J, fG)dt: J*, f(t)dt
J,
(4)
Substituting.from Eq. (4) into (3), we get F(xrf,ax) F(x,) :
f,'*o"
f(t) dt
(s)
By the meanvalue theorem for integrals (7.5.2),there is some number the closed interval bounded by x, and (r, + Ax) such that f xr+L.r
J.,
f (t) dt: 1(X)Ar
From Eqs. (5) and (6), we obtain F ( x 1* A x )  F ( r ' ) : / ( X ) A r or, if we divide by Lx, F(x'*Ax)F(x'):f(X) r \/ Ar Taking the limit as Ax approaches zerol we have ,.
F(rr * A{)  F(xr) :
,n";;'ide
lim f (X)
(7)
or,* ,r, i, r' (',i"il'u"r"rine lim f (x), recallthatX is in
314
THE DEFINITEINTEGRAL
the closedinterval bounded by .rr and x, * Lx, and because
: ll.T r,: r, and llq (r, * Ar) r, it follows from the squeeze theorem (4.3.3) that continuous at .x1,w€ have lim /(X) :
l:l*:xt.Because/is
lim f (X) : f (xt); thus, from Eq.
(7) we get F'(xr):f(xr)
(8)
If the function/ is not defined for values of x less than abut is continuous from the right at a, then in the above argumenl, if. xr: a rn Eq. (T), A,x must approach 0 from the right. Hence, the left side of Eq. (g) will be Fi(rr). Similarly, if f is not defined for values of x greater than b but is continuous from the left at b, then if xr: b inEq. (7), A,x must approach 0 from the left. Hence, we have F'_(xr) on the left side of Eq. (8). Because r, is any number in la, bl, Eq.(8) states what we wished to Prove. I Theorem 7.6.L states that the definite integral t{ f(t) dt,withvariable upper limit r, is an antiderivative of f.
7.6.2 Theorem F t r n d n m e n t aT l heorem of the Cnlutlus
Let the function / be continuous on the closed interval la, bl and let g be a function such that
g'Q):f(x)
o)
for all x in fa, bl. Then fb
I f U )d t : s f t )  s ( a ) Ja' (If x: x:b,
a, the derivative in (9) may be a derivative from the right , and if the derivative in (9) may be a derivative from the left.)
pRooF: If is continuous at all numbers in / [a, bf, we know from Theorem 7.6.t that the definite integral Ii f (t) dt, with variable upper limit r, defines a function F whose derivative on la, bl is Becauseby hypothesis /. 8'(x) : f (x), it follows from Theorem 6.3.3 that fr
8 ( r )  l fJ < tldt+k a
(10)
where k is some constant. Letting x: b and r : a, successively, in Eq. (10), we get
:T:
8@)
f(t) dt + k
(11)
THEOREMOF THE CALCULUS 315 7.6 THE FUNDAMENTAL
and (12)
s(a)f,urdt+k From Eqs. (11) and (L2), we obtain
8(b)s@\:["fUl
ot
dt f fu)
But, by DefinitionT.3.5, f Al dt:0, and so we have [" fb s(b) s@): J, f U)at
which is what we wished we Prove. If f is not defined for values of x greater than b but is continuous from the left at b, the derivative in (9) is a derivative from the left, and we have g@) : F'(b),from which (11) fotlows. Similarly, if f is not defined for values of x less than a but is continuous from the right at a, then the derivative in (9) is a derivative from the right, and we have 7n(a):F'*(a), from I which (12) follows. We are now in a position to find the exact value of a definite integral by applying Theorcm 7.6.2.In applying the theorem, we denote lD ts(b) stu)l by s(r)l,
o rLLUSrRArroN 1: We apply the fundamental theorem of the calculus to evaluate r3 I x2dx Jr Here, f(x):
x2. An antiderivative of.x2 is ixt. From this we choose .Y3
g(x): ?
Therefore,from Theorem 7.6.2,we get
["*'dr:+l':t* 3l'
Jr*
3
:83 Compare this result with that of Example 2, Sec. 7.3. Because of the connection between definite integrals and antiderivatives, we used the integral sign J for the notation J/(r) dx for an an
316
THE DEFINITEINTEGRAL
tiderivative. We now dispense with the terminology of antiderivatives and antidifferentiation and begin to call I f Q) dx the indefiniteintegralof "t' of x, dx." The processof evaluating an indefinite integral or a definite integral is called integration. The difference between an indefinite integral and a definite integral should be emphasized. The indefinite integral Ifk) dx is defined as a function g such that its derivative D*[g(x)]: f (x). However, the definite integral Il f Q) dx is a number whose value dependson the function f and, the numbers a andb, and it is defined as the limit of a Riemann sum. The definition of the definite integral makes no reference to differentiation. The general indefinite integral involves an arbitrary constant; for instance, f*3
I x2dx:++C JJ This arbitrary constantC is calleda constantof integration.lnapplying the fundamental theorem to evaluate a definite integral, we do not need.to include the arbitrary constantC in the expressionforg(r) becausethe fundamental theorem permits us to select any antiderivative, including the one for which C:0. Evaluate
SOLUTION:
 6xz* ex+ 1,)itx Ii, o'
(xt 
6x2* 9x * 1,) dx:
[:,,
xsdx6[^rxzdx
*t
lnn,
:t6.{*, +*,1:, : ( 6 4  1 , 2 8 + 7 2 + 4 ) ( #  * + g + * )
Evaluate
SOLUTION:
(r4l3 + Axrts)
dx: *xzrl, + 4 . t*n,tlt_,
:++3 (++3)
(x6ars * n*rrs) dx
_ 76
sor,urroN t
:)\
f2
J"
Zx't/xt + '1,dx : 3
rz
J
lFn
(ixz) dx
7.6 THE FUNDAMENTAL THEOREMOF THE CALCULUS
317
: 6 ( 8 * L 1 t r z # ( 0 * L } s r z
: + ( 2 7 L ) _
104 I
f_
ExavrprE 4: r3
Evaluate
I xt/t+x dx
Jo
solurroN: u:
To evaluate the indefinite integral l.f,+x
u2:'!,*x
J
xVt I x dx we let dx2udu
x:tt2"!.
Substituting, we have r_l
I xt/t + x dx:  (u' t)u(2udu) JJ 2
r  (un uz) du
J
: E u '  f i u Bt C : ? ( 1 * x ) s t z ? ( 1 * x ) s t za . , Therefore, the definite integral f3 13 I x \ / 1 ,* x d x : 9 ( 1 * x ) s r z 3 ( 1 * x ) s r zI
Jo
lo
: ?@);tz z(4Y,  !"(1)5t2 + ?(t1r,,
:+!+E+3 _
116 T5
Another method for evaluating the definite integral in Example 4 involves changing the limits of the definite integral to values of a. The procedure is shown in the following illustration. Often this second method is shorter and its justification follows immediately from Theorems 6.3.10 and 7.6.2. o TLLUSTRATToN 2: Because u: l.'ffix, : and when .r 3, u :2. Thus, we have f3
I xt/t + x dx:2
Jo
12
 (un u2)du
Jr
12
:?u5  &ut I
It
:91+q?+e _
116 15
we see that when ff:
0, u:
L;
318
THE DEFINITEINTEGRAL
EXAMPLE 5'
Evaluate
l x + 2 1d x
soLUrIoN: If we let f (x): lr * 21,insteadof finding an antiderivativeof / directly, we write f(x) as ifx>2 ifx 0. Prove that F is a constant function by showint that F,(.r):0.
(nrnr: Use Theorem
7.6.1afterwriting the given integlal as the difference of two integrals.) 32' Make up an examPleof a discontinuous function fo{ which the meanvaluetheorem for integrals (a) does not hold, and (b) does hold. 33. Let/be continuous on [a,D] ^na f{t) di * 0. Show that for any numbert in (0,1) there is a nurnber cin (a, b) such Jfu o' that f.
tb
I f(t) dt:k I t'() dt 1"r*i",cor,"ia".*" ,""or." F for whichF(x): [" f(t) d, / [' ruldr andapplythe intermediatevarue theorem.) J"' / J"'
Applications of the definite integral
APPLICATIONS OF THE DEFINITEINTEGRAL
8.1 AREA OF A REGION IN A PLANE
From Definition 7.3.6, if f is a function continuous on the closed interval lo, bl and if f (x) = 0 for all r in la, bl, then the number of square units in the area of the region bounded by the curve y : f (x), the r axis, and the l i n e s x : a and x: b i s n
lim )  f(f,) Ax 0 t:l
llAll
which is equal to the definite integral ft f Q) dx. suppose that f (x) < 0 for all x in la, bl. Then eachf (il is a negative number, and so we define the number of square units in the area of the regionboundedby y : f(x), the r axis,and the lines)c: aand x : b to be n
lim >. tf(f') I Ax
llAll'0 i=l
which equals
 fb f(x) dx Ja'
nxavrprE 1: Find the areaof the region bounded by the cuwe y : f  4x, the r axis, and the lines x:Landr:3.
soLUrIoN: The region, together with a rectangular element of area, is s h o w n i n Fi g. 8.1.1. If one takes a regular partition of the interval [1, 3], the width of each rectangle is Ar. Because x2 4x { 0 on [1,3], the altitude of the lth rectangle is  (€r'  a€r) : 4€i  trt.Hence, the sum of the measures of the areas of n rectangles is given by 7Z
i=1
The measure of the desired area is given by the limit of this sum as Al approaches 0; so if A square units is the area of the region, we have n
hm ) Lx:0
(afi  t,') Lx
i:l
(4x  xz) dx :2)c2 (€i, €i' atr)
*rt
 T ,.)
F i g u r e8 . 1 . 1
Thus, the area of the region is ]2 square units. rxavrprn 2: Find the area of the region bounded by the curve y : xs 2x2 5x * 5, the r axis, and the lines x: L andx:2.
solurroN: The region is shown in Fig. g.l.z.Let (x) : )cB 2x2 sx * 6. f Because f (x) > 0 when x is in the closed interval [1, 1] and (r) = o f when x is in the closed interval [1, 21, we separate the region into two parts. Let At be the number of square units in the area of the region when
8 . 1 A R E A O F A R E G I O NI N A P L A N E
t
of square units in the area of the t ,,'l,f risin J t, dtlnd let A, be the number region w he e n l x: i si i n lr, zl. Then n
( f Ar l'"1)i =l1 f ( 'f,)
A1
litl :
( € t, f ( t i ) )
rII'r If_ l' J' l
rr 1( 't ' l
(r)t ddx a x3_ 2x25x*6)
dx
F(r,
f(x)
 2x2 5x * 6) dx
If A square units is the area of the entire region, then Ar* A,
A:
:
F i g u r e8 . 1 . 2
12 (x' 2x25x * 5) dx G'  2xz 5x f 6) dxJ, J_,
, lr l_  3x' Ex'* U*1 2  3x" Ex' * U* ), L*"' liro l_r: [ ( * z '  E + o ) ( ++ ?  B  5 ) ]  l ( 4  a f  1 0+ 1 2 ) ( + ?  B + a ) l
: Gi, fGt))
rt
t
(?9) The area of the region is thereforc lft
square units.
Now consider two functions/ and g continuous on the closed interval la, bf and such that f (x) > 8(x) for all x in la, bl. We wish to find the area of the region bounded by the two curvesy : f (x) and! :8Q) and the two lines x: a and x : b. Such a situation is shown in Fig. 8.1'3. Take a regular partition of the interval la, bl, with each subinterval having a length of Ar. In each subinterval choose a point f1. Consider the rectangle having altitude lf (€,)  S(6;) ] units and width Ar units. Such a rectangle is shown in Fig. 8.1.3. There are n such rectangles, one associated with each subinterval. The sum of the measures of the areas of these n rectangles is given by the following Riemann sum: (f i, g(ti))
F i g u r e8 . 1 . 3
n s(fi)l Ax > t/(f')
fni":tiiu*ann
sum is an approximation to what we intuitively think of as
A P P L I C A T I O NOSF T H E D E F I N I T E INTEGRAL
the number representing the "measure of the area" of the region. The larger the value of nor, equivalently, the smaller the value of Axthe better is this approximation. If A square units is the area of the region, we define n,
A : l i m > t / ( f , )* s ( f , ) l A x
(1)
Because f andg are continuouson la, bl, so alsois (f  g); therefore, the limit in Eq. (1) exists and is equal to the definite integral rb
J, tf G) nxavrru 3: Find the area of the region bounded by the curyes A:x2 andy:x2*4x.
y:
8@)
(2,4)
G' , f ( ti )
 g(x)ldx
soLUrIoN:
To find the points of intersection of the two curves, we solve the equations simultaneously and obtain the points (0, 0) and (2,4). The region is shown in Fig. 8.I.4. Let /(r) : )c2 * 4x, and g (x) : 12. Therefore, in the interval l0, Zl the curve y : f (x) is above the curve y : g(x). We draw a vertical rectangular element of area,having altitude lf (€r)  s((o)] units and width Ar units. The measure of the area of this rectangle then is given by Ax. The sum of the measures of the areas of n such lf((r) S({i)] rectangles is given by the Riemann sum n
lf (€,) s(f,)I Ax
y : f(x) i:1
If A square units is the area of the region, then n
lim .// \
(f;,9(fr))
Ar0
f:1
lf G,) s(fi)I Ar
and the limit of the Riemann sum is a definite integral. Hence,
lf(x)  s(x)l dx F i g u r e8 . 1 . 4
*t' Jo
l(x'+ 4x) xzldx
:I:
(2x'*
:
4x) dx
[ l,x'.u'f',
:_++g_0 _ 58
The area of the region is $ square units.
8 . 1 A R E A O F A R E G I O NI N A P L A N E
EXAMPLE4r Find the area of the region bounded bY the Parabola : x  5. : Az 2x 2 andthe line y
Z) and (9, 4) . The The two curves intersect at the points (3, solurroN: region is shown in Fig. 8.1.5. Theequationy2:2x2isequivalenttothetwoequations
and y\Ei=T
y:\Ez :
, : (9,41
, :
g(xl
ltkl
fr\x)
F i g u r e8 . 1 . 5
u : g(r) v :frkl \e, 4,
Gi, f Jtil) Gt, f r(Eil),
with the first equation giving the upper half of the parabola ald thr sectLx 2 and ond equation ginittg the bottom half. If we let /r(x): :  \Ex  2, the equation of the top half of the parabola is.A : f t(x) , f ,(x) and the equation of the bottom half of the parabola is U: fr(x)' If we let gG) : x  5, the equation of the line is y : S (r) . In Fig. 8.1.6 we see two vertical rectangular elements of area. Each rectangleLas the upper base on the curve A : f r(r). Becausethe base of the frG)] first rectangle is on the curve y: fr(x), the altitude is [/t({,) : y its curve is on the rectangle second the base of 8(r), the units. Because using by problem this solve to wish If we units. gG)l altitude is l/,(f,) vertical rectangular elements of area, we must divide the region into two separate regions, for instance R1 and R2,where R, is the region bounded by and u:fr(x) and the l i ne x:3, and w here R 2 i s th e th e c u rv e s y :fi (r) region bounded by the curyes A : f r(x) and y : 8@) and the line r: 3 (see Fig. 8.1'.7). If A, it the number of square units in the area of region R1, w€ have
Ar: rim ) lf'Go)  frG,)l Ar i:l
(€i, g({i ))
AeO
_t' lf ,(x) _ J, _ f ' l \ E  z J'
(€', /r(€i ))
F i g u r e8 . 1 . 6
fz@)l dx
:, It t E x  Z
+t/zxzlax ax
l3 : I (Zx Z)srzI
lr
:+L y : g(x) y : f ,(x')
It Az is the number of square units in the area of region R2, w€ have n
Az: lim )
:J; :T:
l f ' ( € , ) s ( f , ) l A x
[/'(r)  g(x)l dx
Y: f 2Gl
t\E2
F i g u r e8 . 1 . 7
 (x s)) dx
A P P L I C A T I O NOSF T H E D E F I N I T E INTEGRAL
:
 z)'z +x'+ sxf'" [+Qx
:[g++451t*8+ts1 Hence At * Az: # + # : 18. Therefore, the area of the entire region is 1g square units. nxevrprE 5: Find the area of the region in Example 4by taking horizontal rectangular elements of area.
(3,  2l
soLUrIoN: Figure 8.1.8 illustrates the region with a horizontal rectangular element of areaIf in the equations of the parabola and the line we solve for x,we have x:iQ'+2)
* 2) and r(y) : y + s, the equation of the parabola r : tr(y) Letting QQ) :t(y' : QQ) and the equation of the line as tr: L(y). maybe written as x r*e \9, 4l consider the closed interval l2,4] on the y axis and take a regullr partition of this interval, each subinterval will have a length of Ly.In the ith t r ( { i ) ,{ i ) subinterval lyir, ai], choose a point f;. Then the lenglh of the ith rectangular element is [r(1,) d(fr)] units and the width is ay units. The measure of the area of the region can be approximated by ihe Riemann sum: n
F i g u r e8 . 1 . 8
and x:y*S
i:r
[r(f,)  6G)l Lv
If A square units is the area of the region, then n
A: lim > tl(f,)  d(f,)l Ly A.tl*0 i:1 Because). and Q arc continuouson l2,4], so arsois (rd), limit of the Riemann sum is a definite integral:
o:I^,[r(y) 6U)] dv
: I:,l ( y+ 5 ) i ( y ' + 2 ) l d v
t ,
:+
l n (y' + 2y + 8) dy
J_,
[ *y"* v'+ 8yf_,
: + [  g s + 1 6 + 3 2 ) ( 8+ 4  1 , 6 ) l :L8
and the
8.1A R E AOF A R E GION IN A P LA N E 32 9 Comparing the solutions in Examples 4 and 5, we see that in the first casewe have two definite integrals to evaluate, whereas in the second case we have only one. In general, if possible, the rectangular elements of area should be constructed so that a single definite integral is obtained. The following example illustrates a situation where two definite integrals are necessary. EXAMPLE6: Find the area of the region bounded by the two curves A: xB 6X2* 8r and !: x2 4x.
€i , f(ti))
solurroN: The points of intersection of the two curves are (0, 0), (3, and (4,0). The region is shown in Fig. 8.1.9.
Let f (r) : xB 6x2* 8x and g(x) : )c2 4x. In the interval [0, 3] the curvey: f (r) is abovethe curve y: g(r), and in the interval [3, 4] the curvey: g(x) is abovethe curvey: f (r). So the region must be divided into two separateregionsRt and R2,where Rt is the region bounded by the two curves in the interval [0, 3] and R2is the region bounded by the two curvesin the interval 13, 41. Letting A, squareunits be the areaof R1and A, squareunits be the area of Rr, we have n
y:8@) y : f(x)
Ar:
lim
n
Az:lim Ar0
R2
( f ; , g ( f; ) )
€i,f(ti))
3, 3)
(i,s((;))
F i g u r e8 . 1 . 9
lf (€,) s(fr)I Ax
AJr 0 i : 1
)
[ S ( f , ) f ( € , ) ] A x
i:l
so that
A, * Or:
6xzI 8x)  (x' ax)l dx It [(x'+ [^ l@'  4x)  (r'  6x2+ 8x)l dx
:f
Js
(x,  Txz* I2x) dx + f^ (x, t Txz I2x) dx J3
 u*]^, :lr*^  &x,+ 6r]',+ * Tx" f+r 4T5rs727 _ 6
7L
Therefore, the required area is ff square units.
8.1 Exercises h Exercfues 1 though 20, find the area of the region bounded by the given curves. In each Problem do the following: (a) Draw a figure showing the region and a rectangular element of area; (b) expressthe area of the region as the limit of a Rigmann sum; (c) find the limit in part (b) by evaluating a definite ihtegral by the fundamental theorem of the calculus. t J .x. 2 :  y ; y   4 4. yt  x; x: 2; x: 4
330
APPLICATIONS OF THE DEFINITE INTEGRAL
3. tt + y I 4: 0; / : 8. Take the elemmts of area perpendicular to the y axis. 4, The same region as in Exercise 3. Take the elements of area parallel to the y axis. 'l 5. ir:2y2.x=O,y:2 d.y":4t;r:O;y:2 ,).y:2i".y:7 8.y:f;y:r' 9. y2:y1;a:g 1 0 y : y z ; f : 1 6  , 1 1 .V : { r ; y : *
t 2 .x : 4  y 2 ; x : 4  4 y l\. ys: f;gy 4 4:g 1 4 .r y z : y z 7 ; x : 7 t y = l ; y : 4 1 5 .x : y 2  ) ; y : g  y z 1 6 .t 6 : y z  y ; x : y  y z : : lV. y 2rt 3g 9x; y S 23 tr tB. 3y : as 2rz  t1x; y : I  ng _ ,r, * *
b. y:lxl,y:Sr,x=r,x:t
2 0y. : l x + 1 1 * l x l , y : 0 , x :  2 , x  _ 3
Find by integration the area of the triangle having vertices at (S, l), (1, 3), and (1,_2). T. 22. Find by integation the area of the triangle having vertice8 at (3,4), (2,0), and (0, 1). 23. Findthe areaof the region bounded by the curvc r1 f +2ry y2:Oandthelirrer:4. (HrNr: Solvethe cubic equation for y in terms of r, and express y as two functions of r,) 24. Find the area of the region bounded by the three curves y:
f , x: yB, and x + y:2. 2 6 . F i n d t h e a r e ao f t h e r e g i o n b o u n d e d b y t h e t h r e e c u r v e s y : * , y : g  y r , a n d . 4 r  y + 1 2 : 0 . 26. Find the_areaof the region above the panbola * : 4py and inside the triangle formed by the r axis and the lines y : x * 8P and Y : a 1gp. 27. (a) Find the area of the region bounded by the curves y2 : Apx and f: of the area in part (a) with respect to p when p :3.
4py. (b) Find the rate of change of the measure
28' Find the rate of drange of the measurc of the area of Exercise 25 with respect to p when p: *. 29. If d square unit6 i8 the area of the region bounded by the parabola y, : 4r and the line : y flr (z > 0), find the rate of change of A with respect to t, 30. Detersdne n so that the region above the line y : 1nx and below the parabola y = 2x _ .f has an area of 36 square units.
8.2 VOLUME OF A SOLID OF REVOLUTION: CIRCULARDISK AND CIRCUTAR.RING METHODS
F i g u r e8 . 2 . 1
We must first define what is meant by the "volume" of asolid of revolution; that is, we wish to assign a number,for example V,to what we intuitively think of as the measure of the volume of such a solid. We define the measure of the volume of a rightcircular cylinder by mzh, where r and h are, respectively, the number of units in the base radius and altitude. Consider first the case where the axis of revolution is a boundary of
C:I R C U L A R  D I SAKN D C I R C U L A R  R I NM GE T H O D S 3 3 1 8 . 2 V O L U M EO F A S O L I DO F R E V O L U T I O N
the region that is revolved. Let the function / be continuous on the closed i n te rv a l l a ,b l a nd assume that f (x) > 0 foral l x i n l a,bl . Let R be the region bounded bythe curve y: f (r), the x axis, and the lines x: A and x: b. Let S be the solid of revolution obtained by revolving the region R about the r axis. We proceed to find a suitable definition for the number V which gives the measure of the volume of S. Let A be a partition of the closed interval la, bl given by a:xo1xtlxr(
1 xnt I
xr:
b
. T h e n w e h a v e n subi nterval sof the form l * rr,r;], w herei :7,2, ,rt , any Lrx: xixiy Choose being length of the ith subinterval with the point fi, with xir s ti s xi, in each subinterval and draw the rectangles having widths A1r units and altitudes f (€r) units. Figure 8.2.3 shows the region R together with the ith rectangle. When the ith rectangle is revolved about the x axis, we obtain a circular disk in the form of a rightcircular cylinder whose base radius is given by f ((r)units and whose altitude is given by A'ixunits, as shown in Fig.8.2.4. The measure of the volurne of this circular disk, which we denote by L,1V,is given by
(t; 'f(tt))
L,V nlf (il)' L,x
F i g u r e8 . 2 . 3
(1)
Because there are n rectangles, n curclrlardisks are obtained in this way, and the sum of the measures of the volumes of these n circular disks is given by
fi:r o,r: fi:r rrlf(€,)l'L,x
(2)
which is a Riemann sum. The Riemann sum given in Eq. (2) is an approximation to what we intuitively think of as the number of cubic units in the volume of the solid of revolution. The smaller we take the norm llAllof the partition, the larger will be n, and the closer this approximation will be to the number V we wish to assign to the measure of the volume. We therefore define V to be the limit of the Riemann sum in Eq. (2) as llAllapproaches zero. This limit exists because f i" continuous on la, bf, which is true because / is continuous there. We have, then, the following definition.
F i g u r e8 . 2 . 4
8.2.1,Definition
Let the function / be continuous on the closed interval lo, bl, and assume that f (r) > 0 for all r in la, bl. If S is the solid of revolution obtained by revolving about the r axis the region bounded by the curve y : f (x), the x axis, and the lines x : a and x : b, and if V is the number of cubic units in the volume of S, then
v
: ' u)f' dx f. ,tfG,)l'A,x f ,,t;il,
(3)
APPLICATIONS OF THE DEFINITEINTEGRAL
v
A similar definition applies when both the axis of revolution and a boundary of the revolved region is the y axis or any line parallel to either the r axis or the y axis.
(tr,€i'
o TLLUSTRATIoN 1: We find the volume of the solid of revolution generated when the region bounded by the curve U : xr, the x axis, and the lines x:"1. and x:2 is revolved about the r axis. Refer to Fig. B.z.s, showing the region and a rectangular element of area. Figure 8.2.6 shows an element of volume and the solid of revolution. The measure of the volume of the circular disk is given by L1V: n(€r')'
Lix:
rr(f A,ix
Then Figure8.2.5
v: .Jimf ntf t* llAlloi:t :,
r2 I x4dx
Jr
: rr(*rt)l' lr
_
37_ 5ta
Therefore, the volume of the solid of revolution is #a
cubic units.
o
Now suPPose that the axis of revolution is not a boundary of the region being revolved. Let f and 8 be two continuous functions on the
we wish to find a value for V. Let A be a partition of the interval la, bl, given by A:XolXrlXrl
Figure8.2.6
1 xn_, I xr:
fu
and the lth subinterval fr,r, ri] has length Lix: xr xr_.tIn the lth subinterval, choose any €i, with xir s €i = xi.Consider the n rectangular elements of area for the region R. See Fig.8.2.7 illustrating the region and the ith rectangle, and Fig. 8.2.8 showing the solid of revolution. When the ith rectangle is revolved about the x axis, a circular ring (or "washe{'), as shown in Fig. 8.2.9, is obtained. The number giving the difference of the measures of the areas of the two circular regions is (rlf ((,)l'rlg(f,)1,) and the thickness is A;x units. Therefore, the measure of the volume of the circular ring is given by
L v : r ( l f ( f , ) l '  [ s ( { , ) ] 'A ) ir The sum of the measures of the volumes of the n circular rings formed by
METHODS AND CIRCULARRING CIRCULARDISK 8.2 VOLUMEOF A SOLID OF REVOLUTION:
f(€t) ti, f(t;))
f(x)
s(x)
F i g u r e8 , 2 . 9
Figure8.2.8
Figure8.2.7
revolving the n rectangular elements of area about the r axis is
)
i=L
(4)
l,Y i:l
The number of cubic units in the volume, then, is defined to be the limit of the Riemann sum in Eq. (4) as llAllapproacheszero.The limit exists since , f and g2 are continuous on la, b) because f and I are continuous there. 8.2.2 Definition
Let the functions f and I be continuous on the closedinterval la, bl, and assumethatf (x) > g(x) > 0 forall x in la,bf. Then if Vcubic units isthe volume of the solid of revolution generatedby revolving about the r axis and A: gk) and the lines the region bounded by the curves y:/(r) x : n a n dx : b ,
As before, a similar definition applies when the axis of revolution the y axis or any line parallel to either the r axis or the y axis.
Exavrpr.r l.: Find the volume of the solid generated by revolving about the r axis the region bounded by the parabola : x * 3. A : xcz* L and the line y
solurroN: The points of intersection are (1., 2) and (2,5). Figure 8.2.10 shows the region and a rectangular element of area. An element of volume and the solid of revolution are shown in Fig. 8.2.1'L.
andg(r) T a k i n gf ( x ) : x * 3 the volume of the circular ring is
n(lf G)l'
[ s ( 1 ' ) ] ' )A i x
* L, we find that the measure of
334
APPLICATIONS OF THE DEFINITEINTEGRAL
ti , fGi)
Gi , s ft))
Fi g ur e 8. 2. 10
F i g u r e8 . 2 . 1 1
If V cubic units is the volume of the solid, then n
v : lim ) '( lf G)l'  [s(f )],) Aix iIr llallo
:7r If2 (lf(x)l'Jt f2
: T I
Jr
ls@)fz)dx
[ ( r ' + 3 ) '  ( x ' + t ) ' z 1d x
f2
:rT I lxnf Jr
+6x*sldx
: nlEx' *x3* gr'+ grlt L
lr
: rl(+
S + L z+ 1 6 ) ( t + + + 3  8 ) l
_rL7_ 5tl
Therefore, the volume of the solid of revolution
is LFr
cubic units.
AND CIRCULARRING METHODS CIRCULARDISK 8.2 VOLUMEOF A SOLID OF REVOLUTION:
ExAMPLE 2r Find the volume of the solid generatedby revolving about the line x: 4 the region bounded by the two parabolas )c:yyz andx:A'3.
solurroN: The curyes intersectat the points (2, 1.) and (+,il. The region and a rectangular element of area are shown in Fig. P.2.12.Figure 8.2.13shows the solid of revolution as well as an elementof volume, which is a circular ring.
F i g u r e8 . 2 . 1 3
Let F(y)  y  y2 and G(y) = y'  3. The number of cubic units in the volume of the circular ring is L1V: n(14 + F(f')1,  14+ G(f')l') Lua
(G(6r),fi)
Thus,
v
F i g u r e8 . 2 . 1 2
n
+ F(€,)f' 14+ G(€)]') L,y lim ). l l A l l  0 i : 1 "([4
,
f3l2 y')'  (4* y'g)'l dY J  r [ ( 4 + y
n
I ?2y'9y'*8y+L5) dy Jr
l3l2
13tz  3Yt* 4Y'* tu,l, : n ll+rn _ 
875* 32,.
The volume of the solid of revolution is then Y*n
cubic units.
8.2 Exercises In Exercises 1. through 8, find the volume of the solid of revolution when the given region of Fig. 8.2.1,4is revolved about the indicated line. An equation of the curve in the figure is y2 : *2.
2. OAC about the line AC
l. OAC about the r axis 3. OAC about the line BC
'
4. OAC about the y axis
5. OBC about the y axis
6. OBC about the line BC
7. OBC about the line AC
8. OBC about the x axis
F i g u r e8 . 2 . 1 4
336
APPLICATIONS OF THE DEFINITEINTEGRAL
9 . Find
the volume of the sphere generated by revolving the circle whose equation is 12 * A2: 12 about a diameter. 10.Find by integration the volurne of a rightcircular cone of altitude /z units and base radius a units.
1 1 .Find
the volume of the solid generated by revolving about the x axis the region bounded by the curve : x3 and the A linesy:0andx:2.
12. Find the volume of the solid generated by revolving the region in Exercise L1,about the y axis. 13. Find the volume of the solid generated by revolving about the line x:
4
the region bounded by that line and the
p a r a b o l ax : 4 i 6 y  2 y 2 .
14. Find the volume of the solid generated by revolving the region bounded by the curves U2: 4x andy: r about the x axis. 1 5. Find the volume of the solid generated by revolving the region of Exercise 14 about the line x: 4. 16. An oil tank in the shape of a sphere has a diameter of 50 ft. How much oil does the tank contain if the depth of the oil is 25 ft?
1 7. A paraboloid of revolution is obtained by revolving the parab ola y2 : 4px about the r axis. Find the volume bounded by a paraboloid of revolution and a plane perpendicular to its axis if the plane is 10 in. from the vertex, and if the plane section of intersection is a circle having a radius of 5 in.
18. Find the volume of the solid generated by revolving about the x axis the region bounded by the loop of the curve whose equation is 2y2 : x(*  4).
19. Find the volume of the solid generated when the region bounded by one loop of the curve whose equation is xzyz: (x'
9) (1  x2) is revolved about the x axis.
20. The region bounded by a pentagon having vertices at (4,4), (2,0), (0, 8), (2,0), and (4,4) is revolved about the x axis. Find the volume of the solid generated.
8.3 VOLUME OF A SOLID OF REVOLUTION: CYLINDRICALSHELL METHOD
Figure8.3.'l
In the preceding section we found the volume of a solid of revolution by taking the rectangular elements of area perpendicular to the axis of revolution, and the element of volume was either a circular disk or a circular ring. If a rectangular element of area is parallel to the axis of revolution, then when this element of area is revolved about the axis of revolution, a cylindrical shell is obtained. A cylindrical shell is a solid contained between two rylinders having the same center and axis. Such a cylindrical shell is shown in Fig. 8.3.1. If the cylindrical shell has an inner radius r, units, outer radius r, units, and altitude h units, then its volume V cubic units is given by V: nrzzh Trrzh (l) Let R be the region bounded by the curye y : f (x), the r axis, and the lines x: a and r : b, where / is continuous on the closed interval la, bl and f(x) > 0 for all x in la, b]; furtherrnore, assume that a > 0. such a region is shown in Fig. 8.3.2. If R is revolved about the y axis, a solid of revolution S is generated. Such a solid is shown in Fig. 8.3.3. To find the volume of S when the rectangular elements of area are taken parallel to the y axis, we proceed in the following manner. Let A be a partition of the closed interval [a, b] given by A:Xo 0) and the line r : a is revolved about r: a, 18. Find the volume of the solid genented by revolving the region bounded by the curve li3 * yrl": a2i' about the y axis. 19. Find the volume of the solid generatedby revolving about the y axis the region outside the cuwe y : f and between the Lines y : 2a  1 ?n\d.y : x + 2. 20. Through a spherical shaped solid of radius 6 in., a hole of radius 2 in. is bored, and the axis of the hole is a diameter of the sphere. Find the volume of the part of the solid that remains. 21. A hole of radius 2\6 in. is bored through the center of a spherical shaped solid of radius 4 in. Find the volume of the portion of the solid cut out. 22. Find the volume of the solid generatedif the region bounded by the parabola y, :4pr about the y axis.
and the line r:
p is revolved
PLANESECTIONS 341 8.4 VOLUME OF A SOLIDHAVINGKNOWNPARALLEL
8.4 VOLUME OF A SOLID Let S be a solid. By a plane sectionof S is meant a plane region formed by HAVING KNOWN PARALLEL the intersection of a plane with S. In Sec.8.2 we learned how to find the PLANE SECTIONS volume of a solid of revolution for which all plane sections perpendicular to the axis of revolution are circular. We now generalize this method to find the volume of a solid for which it is possible to express the area of any plane section perpendicular to a fixed line in terms of the perpendicular distance of the plane section from a fixed point. We first define what we mean when we say that a solid is a "cylinder." 8.4.1. Definition
A solid is a right cylinder if it is bounded by two congruent plane regions R1 and R2 lying in parallel planes and by a lateral surface generated by u line segment, having its endpoints on the boundaries of R1 and Rz, which moves so that it is always PerPendicular to the planes of Rt and R2.
F i g u r e8 . 4 . 1
Figure8.4.2
Figure8.4.3
Figure 8.4.1.illustrates a right cylinder. The heighf of the cylinder is the perpendicular distance between the planes of Rr and R2, and the base is either R1 or R2. If the base of the right cylinder is a region enclosed by a circle, we have a rightcircular cylinder (see Fig. 8.a.4; if the base is a region enclosed by u rectangle, we have a rectangular parallelepiped (see F i g . 8 .a .3 ). If the area of the base of a right cylinder is A square units and the height is h units, then by definition we let the volume of the right cylinder be measured by the product of A and h. As stated above, we are considering solids for which the area of any plane section that is perpendicular to a fixed line is a function of the perpendicular distance of the plane section from a fixed point. The solid 5 of Fig. 8.4.4 is such a solid, and it lies between the planes perpendicular to the r axis at a and b. We represent the number of square units in the area of the plane section of S in the plane perpendicular to the r axis at x by A(x), where A is continuous on la, bl. Let A be a partition of the closed interval la, bl given by a:xo/xr1xzl A(€ t) Figure 8.4.4
'1xn:6
. , Wehave, then, n subintervals of the form lxrr, ri], wherei:1,2, any Choose x^. xiLrx: being n,wittr the length of the ith subinterval
APPLICATIONS OF THE DEFINITEINTEGRAL
number f1, with xit s €i a xi, in each subinterval and construct the right cylindersof heights A;x units and plane section areasA(tr) square units. The volume of the lth cylinder is A(f,) A;x cubic units. We obtain n right cylinders,and the sum of the measuresof the volumes of thesen cvlinders is given by the Riemann sum n
A(() ap
(1)
i:r
This Riemann sum is an approximation of what we intuitively think of as the measure of the volume of S, and the smaller we take the norm llAllof the partition A, the larger will be n, and the closer this approximation will be to the number we wish to assign to the measure of the volume. We have, then, the following definition. 8.4.2 Definition
Let S be a solid such that S lies between planes drawn perpendicular to the x axis at a and b. If the measure of the area of the plane section of S drawn perpendicular to the x axis at r is given by A(x) , wherc A is continuous on la, bl, then the measure of the volume of S is given by
lim 2 a(f,) Aix: [' A ( x ) d x .0 i=l J(l
(2)
lllll
Note that Definitions 8.2.7 and 8.2.2 are special cases of Definition 8.4.2. If in Eq,. (2) we take A (x): nl f (x)12,w e have E q. (3) of S e c.g. Z. lf w e t a k e A ( x ) : n ( l f ( x ) l '  [S(x)]'), *" have Eq. (S) of Sec. 8.2.
nxervrpr.r 1: If the base of a solid is a circle with a radius of r units and if all plane sections perpendicular to a fixed diameter of the base are squares, find the volume of the solid.
solurroN: Take the circle in the xy plane, the center at the origin, and the fixed diameter along the r axis. Therefore, an equation of the circle is 12 I y' : r2. Figure 8.4.5 shows the solid and an element of volume, which is a right cylinder of altitude A1r units and with an area of the base given by l2f (il]2 square units, where /(r) is obtained by solving the equation of thpirc,fe j?, v and setting y : f (x). This computation give f (x) : Yr"  xz. Therefore, if V cubic units is the volume of the solid, "we have
lim tt llall0
n
lzf((,)l' Lix
i:r
(r'
: +l'"*
x 2 )d x
8.4 VOLUMEOF A SOLID HAVINGKNOWN PARALLELPLANESECTIONS
F i g u r e8 . 4 . 5
rxeuPrn 2:
A wedge is cut from a rightcircular cylinder with a radius of r in. bY two Planes, one perpendicular to the axis of the cylinder and the other intersecting the first at an angle of measurement 60oalong a diameter of the circular plane section. Find the volume of the wedge.
solurroN: The wedge is shown in Fig. 8.4.6. The xy plane is taken as the plane perpendicular to the axis of the cylinder, and the origin is at the An equation of the circular plane section is then point of plrp"ndicularity. : 12.Every plane section of the wedge perpendicular to the r axis is * * y, a right triangle. An element of volume is,a right cylinder having altitude (€r)12 in.2, where /(r) is obA,rin., and irea of the base given bV i$lf for y and setting y : f (x), circle tained by solving the equation of the in.3 is the volume of the if V Therefore, \fr'=T. thereby giving f (x) wedge, n_
V:
U,  €rr)L,x
lim )iVe .0,:r
llall
_
:+\/g
fr

Jr
(r,  x2)dx I
 i*"1'_, : +\/5lr'* : fi\/3r3 Hence, the volume of the wedge is 3 tEr3 in'3
Figure8.4.6
U4
APPLICATIONS OF THEDEFINITE INTEGBAL
Exercises 8.4 l
The baseof a solid is a circlehawing a radius of / units. Find the volume of the solid if all plane sectionsperpendicular to a fixed diameter of the base are equilateral triangles,
The base of a solid is a circle with a radius of r units, and all pLanesections perpendicular to a fixed diameter of the base are isoscelesright triangles haYing the hypotenuse in the plane of the base. iind the volume of the solid. 3 . Solve Exercise2 if the isoscelesdght t angles have one leg in the plane of the base. 4. Find the volume of a right pyramid having a height of t units and a square base of side n units. 5 . Find the volume of the tebahedmn havint 3 mutually perpendicular faces and three mutually perpendicular edges whose lengths have measuresa, b, and c. 6 . The base of a solid is a cirde with a radius of 4 in. , and each plane section perpendicular to a fixed diameter of the base is an isosceles triangle having an altitude of 10 in. and a chord of the cirde as a base. Find the volume of the solid. 7. The base of a solid is a circle with a radius of 9 in., and each plane section perpendicular to a fixed diameter of the base is a square having a chord of the circle as a diagonal. Find the volume of the solid. 8. Two rightcircular cylinde{s, each having a radius of / units, have axes that intersect at right angles. Find the volume of the solid common to the two cylinders. 9 . A wedge is cut ftom a solid in the shapeof a rightcircular cylinder with a radius of r in. by a the plane through a diameter of the base and inclined to the Plane of the base at an angle of measurement 45". Fi;d volume" of the wedge. 10. A wedge is cut from a solid in the shaPeof a rightcircular cone having a base radius of 5 ft and an altitude of 20 ft by two half planes through the axis of the cone, The argle between the two planeshas a measurement of 30.. Find the volume of the wedge cut out.
8'5 WORK
The "work" done by u force acting on an object is defined in physics as "force times displacement." For example, suppose that an otiect is moving to the right along the r axis from a point a to apoint b, and.a constant force of F lb is acting on the object in the direction of motion. Then if the displacement is measured in feet, (b  a) is the number of feet in the displacement. And if W is the number of footpounds of work done by the force, W is defiped by (1) o tttustRArroN '1.: lI W is the number of footpounds in the work necessary to lift a 70lb weight to a height of 3 ft, then W:70
' 3:270
.
In this section we consider the work done by a variable force, which is a function of the position of the object on which the force is acting. We wish to define what is meant by the term "work" in such a case. suppose that /(x) , where /is continuous on la, bl, is the number of units in the force acting in the direction of motion on an object as it moves to the right along the x axis from point a to pointb. Let A be a partition of
8.5 WORK
the closedinterval la, bf : 1 xnr I xr:
a:xolxrlxzl
b
The ith subinterval is [x11, x]; and if xi, is close to xi, the force is almost constant in this subinterval. If we assume that the force is constant in the ith subinterval and if fi is any point such that ri, s fi s ri, then if AiW is the number of units of work done on the object as it moves from the point xi1 Io the point xi, from formula (1) we have A ,W: f (€ ,) (x, xi r) Replacin E )ci xir by A;x, w€ have
A1W:f (t) L'x and n
AiW:2fG,) L,x
i:1
(2)
i:r
The smaller we take the norrn of the partition A, the larger n will be and the closer the Riemann sum in Eq. (2) will be to what we intuitively think of as the measure of the total work done. We therefore define the measure of the total work as the limit of the Riemann sum in Eq. (2). 8.5.1 Definition
Let the function / be continuous on the closed interval la, bl and /(r) be the number of units in the force acting on an object at the point / on the r axis. Then if W units is the work done by the force as the obfect moves from a to b, W is given by (3)
In the following examplewe use Hooke'slaw, which statesthat if a spring is stretchedx in. beyond its natural length, it is pulled back with a force equal to kx lb, where k is a constant depending on the wire used. nxalvrpn L: A sPring has a natural length of 14 in. If a force of 5 lb is required to keep the sPring stretched 2rrt., how much work is done in stretching the sPring from its natural length to a length of 18 in.?
solurroN: Place the spring along the r axis with the origin at the point where the stretching starts (see Fig. 8.5.1). 14in.# T
8 .5 .1 F i g u re Let r: the number of inches the spring is stretched; : the number of pounds in the force acting on the spring r in. f (x) beyond its natural length.
APPLICATIONS OF THE DEFINITEINTEGRAL
Then, by Hooke's law, f (x) : kr. Because/(2) : 5, we have . 5:k'2
k:E Therefore,
f(x):Ex If W: the number of inchpounds of work done in stretching the spring from its natural length of 'J,4in. to a length of 1g in., we have
w: li3 f. f G,)o,* llall0 t:I
f(x) dx Exdx
: T x z 1I4
JO
:20 Therefore, the work done in stretching the spring is 20 in.lb. EXAMPLE2: A water tank in the form of an inverted rightcircular cone is 20 ft across the top and 15 ft deep. If the surface of the water is 5 ft below the top of the tank, find the work done in pumping the water to the top of the tank.
solurroN: Referto Fig. 8.5.2.The positive r axis is chosenin the downward direction becausethe motion is vertical. Take the origin at the top of the tank. We considera partition of the closedinterval [5, 15] on the r ixis and let fi be any point in the ith subinterval l*rr, r;]. An elementof volume is a circular disk having thickness A,ixft and radiu s (il ft, where the f function / is determined by an equation of the line through the points (0, 10) and (L5, 0) in the form y  f (r). The number of cubic feet in the volume of this elementis given by A^iV: nlf (&)1, Lrx.If w is the number of pounds in the weight of L fts of water, then the number of pounds in the weight of this element is wnlf(t)1, A;r, which is the force required to pump the element to the top of the tank. If xi, is closeto xi, then the distance through which this element moves is approximately& ft.Thus, if A'rWftlb is the work done in pumping the element to the iop of the tank, AiW is approximately(wrrlf G)l' Lrx) . fii so if w is the number of footpounds of work done,
w:
. Ji3 f .df G,)1, €i Lix
llall0 f:t
: *n Figure8.5.2
fr5
(x)lzxdx J, lf
To determine f (x), we find an equation of the line through the points
8.5 WORK
(15, 0) and (0, 10) by using the interceptform:
*15 + + : 10 1
or yBx*Lo
T h e r e f o r e , f ( x ) :  & x * L 0 ,a n d
w  *n f" ?3x * \o)zxdx Js rr5 : rpir  (6rt  #x't Js
toox) dx
:'u)n lt*n  Xqr3* 5012 1ts rJt : + (10,000nar) Therefore,the work doneis 1,0,000rrw19 ftlb.
8.5 Exercises 1. A spring has a natural length of 8 in. If a force of 20lb stretchesthe spring i in., find the work done in stretching the spring from 8 in. to 11 in. 2. A spring has a natural length of 10 in.. and a 30lb force stretches it to 11iin. Find the work done in stretching the sprint from 10 in. to 12 in. Then find the work done in stretchint the sPdng from 12 in. to 14 in. 3. A spring has a natural length of 6 in. A 12,000lb force compressesthe spring to 5* in Find the work done in compressing it from 6 in. to 5 in. Hooke's law holds for compression as well as for extension. 4. A spring has a natural length o{ 6 in. A 12001b force compresses it to 5} in. Find the wotk done in compressing it from 6 in. to 4* in. 5. A swimming pool full of water is in the form of a rectangularparallelepiped5 ft deep, 15 ft wide, and 25 ft long. Find the work required to pump the water in the pool up to a level I ft above the surface of the pool. 6. A trough full of water is 10 ft long, and its cross section is in the shape of an isosceles triangle 2 ft wide acrossthe top and 2 ft high. How much work is done in pumping all the water out of the trcugh over the top? 7. A hemispherical tank with a radius of 5 ft is filed with water to a depth of 4 ft. Find the work done in pumping the water to the top of the tank. 8. A rightcircular cylindrical tank with a depth of 12 ft and a radius of 4 ft is half tull of oit weighing 60 lb/ft3. Find the wotk done in pumping the oil to a height 6 ft above the tank. 9. A cable 200 ft lont and weighing 4 lb/ft is hanging vertically down a well. If a weight of 100Ib is susp€nded from the lower end of the cable, find the work done in pulling the cable and weight to the top of the well, 10. A bucket weighing 20lb containing 60 lb of sand is attached to the lower end of a 100 ft long chain that weighs 10lb and is hanging in a deep well. Find the work done in raising the bucket to the top of the well, 11. SolveExercise10 if the sand is leaking out of the bucket at a constantrate and has all leakedout iust as soon as the bucket i8 at the toD of the well.
348
APPLICATIONS OF THE DEFINITEINTEGRAL
12. As a water tank is being raised, water spills out at a constant rate of 2 fC per foot of rise, If the tank originally contained 1000 fe of water, find the work done in raising the tank 20 ft. 13. A tank in the form of a rectangular parallelepiped 6 ft deep,4 ft wide, and 12 ft long is fuJl of oil weighing 50lb/fi3. When onethird of the work necessary to pump the oil to the top of the tank has been done, find by how much the sudace of the oil is lowered. 14. A cylindrical tank 10 ft high and 5 ft in radius is standing on a platform 50 ft high. Find the depth of the water whm onehalf of the work required to fill the tank faom the Found level through a pipe in the bottom has been done. 15. A one horsepower motor can do 550 ftlb of work per second. If a 0.1.hp motor is used to pump water from a full tank in theshapeofarectangularparallelepiped2ftdeep,2ftwide,and6ftlonttoapoint5ftabovethetopofthetank,how long will it take? 16. A meteorite is a miles from the center of the earth and falls to the surface of the earth. The folce of sraviw is inverselv proportional to the square of the distance of a body from the center of the earth. Find the work doie by gravity if the weight of the meteorite is r, lb at the sudace of the earth. Let R rriles be the Edius of the earth,
8'6 LIQUID PRESSURE Another application of the definite integral in physics is to find the force causedby liquid pressureon a plate submergedin the liquid or on a side of a containerholding the liquid. First of all, supposethat a flat plate is insefted horizontally into a liquid in a container. The weight of the liquid exertsa force on the plate. The force per square unit of area exerted by the liquid on the plate is called the pressureof the liquid. Let w be the number of pounds in the weight of one cubic foot of the liquid and h be the number of feet in the depth of a point below the surfaceof the liqui d. If p is the number of pounds per square foot of pressure exertedby the liquid at the point, then 'u:oi"'rne
(1)
number of square feet in the area of a flat plate that is submerged horizontally in the liquid, and F is the number of pounds in the force causedby liquid pressure acting on the upper face of the plate, then F:PA
( 2)
Substituting from formula (1) into (2) gives us F : whA
(3)
Note that formula (1) states that the size of the container is immaterial so far as liquid pressure is concerned. For example, at a depth of 5 ft in a swimming pool filled with salt water the pressure is the same as at a depth of 5 ft in the Pacific Ocean, assuming the density of the water is the same. Now suPPose that the plate is submerged vertically in the liquid. Then at points on the plate at different depths the pressure, computed from formula (1), will be different and will be greater at the bottom of the plate than at the top. We now proceed to define the force caused by liquid Pressure when the plate is submerged vertically in the liquid. We use
8.6 LIQUIDPRESSUBE 349 Pascal's principle: At any point in a liquid, the pressure is the same in all directions. In Fig. 8.6.1 let ABCD be the region bounded by the x axis, the lines x: a and r : b, andthe curve y: f (x), where the function /is continuous and/(r) > 0 on the closed interval fa, bl. Choose the coordinate axes so they axis lies along the line of the surface of the liquid. Take the r axis vertical with the positive direction downward. The length of the plate at a depth r ft is given by f (x) ft. Let A be a partition of the closed interval la, bl which divides the interval into n subintervals. Choose a point f; in the ith subinterval, with xrr 3 €i = xr. Draw n hofizontal rectangles. The ith rectangle has a length of f (€r) ft and a width of L,ix ft (see Fig. 8.6.1). If we rotate each rectangular element through an angle of 90o,each element becomes a plate submerged in the liquid at a depth of f6 ft below the surface of the liquid and perpendicular to the region ABCD. Then the force on the rth rectangular element is given by wtrf (t) L# lb. An approximation to F, the number of pounds in the total force on the vertical plate, is given by
F i g u r e8 . 6 . 1
n
wt,fGi) Air 2 i:r
(4)
which is a Riemann sum. The smaller we take llAll,the larger n willbe and the closer the approximation given by (4) will be to what we wish to be the measure of the total force. We have, then, the following definition. 8'6'1 Definition
Suppose that a flat plate is submerged vertically in a liquid of weight ut pounds per cubic unit. The length of the plate at a depth of r units below the surface of the liquid is /(x) units, where / is continuous on the closed interval la, bl and f(x) > 0 on la, bl. Then F, the number of pounds of force causedby liquid pressureon the plate, is given by
t:
Air: wxf(x)dx [: 2*rOrri) ,,Til,
(s)
Exevpr.E 1: A trough having a soLUrIoN: Figure 8.5.2 illustrates one end of the trough together with a trapezoidal cross section is full of rectangularelement of area. An equation of line AB it y :g  *r. Let water. If the trapezoid is 3 ft wide f (x) : B *x.If we rotate the rectangularelement through 90', th" force on at the top,2 ft wide at the bottom, the element is given by 2w(1fG) Lfi lb. If F is the number of pounds in and 2 ft deep, find the total force the total force on the side of the trough, owing to liquid pressure on one n end of the trough. F  lim zwt,f (t) L#
2
H
llAll0 f:t
APPLICATIONS OF THE DEFINITEINTEGRAL
:2us :Ztll
[, oo) d x
I,.*
 ix) dx
lrr'# ' ] , 2 , O )A ( 2 , l ) Figure8.6.2
Taking w : 62.5, we find that the total force is 291,.2lb.
Exavprr 2: The ends of a trough are semicircular regions, each with a radius of 2 ft. Find the force caused by liquid pressure on one end if the trough is full of water.
soLUrIoN: Figure 8.5.3 shows one end of the trough together with a rectangular element of area. An equation of the semicircle is xz * yr:4. Solving for y gives y : t[=7, and so let f (x) : {E 7. The force on the rectangular element is given by 2w(j(fr) Arr. so if F pounds is the total force on the side of the trough,
F: ,,limf ,*g,f(f,) Air llall.0 t:l :2ut Ir2 xf(x) dx Jo
r2 :2'ut I xt/E=P
dx
JO
: &w(4 xr)r,rl' JO
Figure8.6.3
Therefore, the total force is ffut lb.
8.6 Exercises 1. A platein the shaPeof a rectangle is submerged vertically in a tank of water, with the upper edge lying in the surface. If "ia" the width of the plate is 10 ft and the depth is 8 ft, find the {orce due to liquid presiure ott otr" of the plate. 2. A square plate of side 4 ft is submerged ve ically in a tank of water and its center is 2 ft below the surface.Find the force due to liquid ptessure on one side of the plate. 3. Solve Exercise 2 if the center of the plate is 4 ft below the surface. 4. A Platein the shapeof an isoscelesright triantle is submergedvertically in a tank of water, with one let lying in the surface.The legs are each 6 ft lont, Find the force due to liquid pressureon one side of the plate. 5. A rectantular tank fuIl of water is 2 ft wide and 18 in. deep, Find the force due to liquid pressure on one end of the tank. 6. The,endsof a trough are equilateraltriangles having sides with lengtls of 2 ft. If the water in the trough is 1 ft deep, find the force due to liquid prrssure on one end.
8.7 CENTEROF MASSOF A ROD
351
7. The face of a dam adiacent to the water is vertical, and its shape is in the form of an isosceles triangle 250 ft wide across the top and 100 ft high in the center. If the water is 10 ft deep in the center, find the total force on the dam due to liquid Pressure. 8. An oil tank is in the shape of a rightcircular cylinder 4 ft in diameter, and its axis is horizontal. If the tank is half full of oil weighing 50 lb/fe, find the total force on one end due to liquid Pressure. 9. The faceof the gate of a dam is in the shape of an isoscelestriangle 4 ft wide at the top and 3 ft high. If the upper edge of the face of the gate is 15 ft below the surface of the water, find the total force due to liquid pressure on the gate. 10. The face of a gate of a dam is vertical and in the shape of an isoscelestrapezoid 3 ft wide at the top, 4 ft wide at the bottom, and 3 ft high. If the upper base is 20 ft below the surface of the water, find the total force due to liquid pressure on the gate. 11. The face of a dam adjacentto the water is inclined at an angle of 30ofrom the vertical. Tlie shape of the face is a rectangle of width 50 ft and slant height 30 ft. If the dam is full of water, find the total force due to liquid pressure on the face. 12. Solve Exercise11 if the face of the dam is an isoscelestrapezoid 120ft wide at the top, 80 ft wide at the bottom, and with a slant height of.40 ft13. The bottom of a swimming pool is an inclined plane. The pool is 2 ft deep at one end and 8 ft deep at the other. If the width of the pool is 25 ft and the length is 40 ft, find the total force due to liquid pressure on the bottom. 14. If the end of a water tank is in the shapeof a rectangle and the tank is full, show that the measure of the force due to liq' uid pressureon the end is the product of the measureof the areaof the end and the measureof the force at the geometrical center.
8.7 CENTER OF MASS OF A ROD
In Sec. 7.5 we leamed that if the function / is continuous on the closed interval la,bf', the averagevalue of f on fa, b] is given by
I rat dk b: a
An important application of the average value of a function occurs in physics in connection with the concept of centerof mass. To arrive at a definition of "mass," consider a particle that is set into motion along an axis by a force of F lb exerted on the particle. So long as the force is acting on the particle, the velocity of the particle is increasing; that is, the particle has an acceleration.The ratio of the.force to the acceleration is.constant regardless of the nagnitude of the force, and this constant ratio is called the massof the particle. . rLLUsrRArror L: If the acceleration of a certain particle is 10 ft/sec2 when the force is 30Ib, the mass of the particle is 30 lb :  3lb 10 ftlsecz 1 ftlsec2 Thus, for every 1 ft/secz of acceleration,a force of 3 lb must be exerted on . the particle. When the unit of force is 1 lb and the unit of acceleration is7 ftlsecz,
,.w
352
APPLICATIONS OF THE DEFINITEINTEGRAL
the unit of mass is called one slug. That is, L slug is the mass of a particle whose acceleration is 1. ftlsec2 when the magnitude of the force on the particle is 1 lb. Hence, the particle of Illustration 1 that has an acceleration of 10 ftlsec2 when the force is 30 lb has a mass of 3 slugs. From physics, if W lb is the weight of an object having a mass of m slugs, and g ftlsec2 is the constant of acceleration due to gravity, then W: mB Consider now a horizontal rod, of negligible weight and thickness, placed on the r axis. On the rod is a system of n particles located at points x 1 , x 2 , .. . , l n . T h e i t h p a r t i d e ( i : 1 , 2 , . ,n) is atadirecteddistance 4 ft from the origin and its mass is mi slugs. See Fig.8.7.1. The n
number of slugs in the total mass of the system ir )
/ni. We define the
i :l
F m2
lTl s
xs
O
x4
x,r
{
t
x,2
x
F i g ur e 8 . 7. 1
momentof massof the ith particle with respect to the origin as mixi slugft. The moment of mass for the system is defined as the sum of the moments .all of mass of the particles. Hence, If M, slugft is the moment of mass of the system with respect to the origin, then
Now we wish to find a point f such that if the total mass of the system were concentrated there, its moment of mass with respect to the origin would be equal to the moment of mass of the system with respect to the origin. Then f must satisfy the equation nn
t> mi:2 i:l
flltXt
i :l
and so
(1)
The point r is called the centerof massof the system, and it is the point where the system will balance. The position of the center of mass is indepehd'entof the position of the origin; that is, the location of ttre center of mass relative to the positions of the particles does not change when the origin is changed.
8,7 CENTEROF MASS OF A ROD
ExAMPLE L: Given four particles of masses 2, 3,1,, and 5 slugs located on the r axis at the Points having coordinates 5, 2, 3, and 4, respectively, where distance measurement is in feet, find the center of mass of this system.
solurroN: lf 7 is the coordinate of the center of mass, we have from formula (1) 7
fi, Thus, the center of mass is # tt to the left of the origin.
The preceding discussion is now extended to a rigid horizontal rod having a continuously distributed mass.The rod is said tobehomogeneous if its mass is directly proportional to its length. In other words, if the segment of the rod whose length is Arr ft has a mass of L1mslugs, and'A,1m: k Air, then the rod is homogeneous. The number k is a constant and k slugs/ft is called the.linear dcnsity of the rod. Lim = p((i)Lir
il*],
,
,,_r€i
;.
O
tfi
,,
L
Figwe 8.7.2
Suppose that we have a nonhomogeneous rod, in which case the linear density varies along the rod. Let L ft be the length of the rod, and place the rod on the r axis so the left endpoint of the rod is at the origin and the right endpoint is at L. See Fi9.8.7.2. The linear density at any point r on the rod is p(r) slugs/ft, where p is continuous on [0, L]. To find the total mass of the rod we consider a partition A of the closed interval [0, L] into z subintervals. The fth subinterval is [rat, ra], and its length is A4x ft. If f6 is any point in lxp1, xrf, ffir approximation to the mass of the part of the rod contained in the ith subinterval is Lam slags, where Arnt: p(t) L* The number of slugs in the total mass of the rod is approximated by nn
s A ; t n  t pG) Lfc
LJ"Ll i:l
(2)
i:l
The smaller we take the norm of the partition A, the doser the Riemann sum in Eq. (2) will be to what we intuitively think of as the measure of the mass of the rod, and so we define the measure of the mass as the limit of the Riemann sum in Eq. (2). 8.7.1 Definition
A rod of length L ft has its left endpoint at the origin. If the number of slugs per foot in the linear density at a point x ft from the origin is p(x), where p is continuous on 10,L), then the total massof the rod is M slugs,
APPLICATIONS OF THE DEFINITEINTEGRAL
where
p(x) dx
ExAMPrn 2: The density at any point of a rod 4 ftlong varies directly as the distance from the point to an external point in the line of the rod and 2 ft from an end, where the density is 5 slugs/ft. Find the total mass of the rod.
solurroN: Figure 8.7.3shows the rod placed on the r axis. If p(r) is the number of slugs per foot in the density of the rod at the point r ft from the end having the greater density, then P(x):c(6x) where c is the constant of proportionality. Becausep(4):5, we have 5:2c or c:*. Hence,p(x):E(6x). Therefore,if M slugs is the total mass of the rod, we have from Definition 8.7.1.
M : , l i p f Z < r f , )A , x llAll0 i=l E(6 x) dx
40 The total mass of the rod is therefore 40 slugs. 6'* ti x
ox.,(E>,46 F i g u r e8 . 7 . 3
we now proceed to define the center of mass of the rod of Definition 8.7.L. However, first we must define the moment of mass of the rod with respect to the origin. ti
L rm : p(t;)Aix
t
,,
,r_, \; Figure8.7.4
As before, place the rod on the r axis with the left endpoint at the origin and the right endpoint at L. see Fig. 8.7.4.Let A be a partition of [0, L] into n subintervals, with the lth subinterval lxo_r, xolhaving length Aix ft. If f' is any point in lx;r, ri], an approximation to the moment of mass
I
8.7 CENTEROF MASS OF A ROD
355
with respect to the origin of the part of the rod contained in the ith subinterval is & Lim slugft, where Liftr: pG) L*. The number of slugfeet in the moment of mass of the entire rod is approximated by n
n
i:l
i:l
(4)
PGt) Lix
The smaller we take the norm of the partition A, the closer the Riemann sum in Eq. (4) will be to what we intuitively think of as the measure of the moment of mass of the rod with respect to the origin. We have, then, the followi.g definition. 8.7.2 Definition
A rod of length L ft has its left endpoint at the origin and the number of slugs per foot in the linear density at a point x ft from the origin is p (x) , where p is continuous on [0, L]. The moment of mass of the rod with respect to the origin rs Mo slugft, where
(s) The center of massof the rod is at the point r such that if M slugs is the total mass of the rod, iM : Mo. Thus, from Eqs. (3) and (5) we get
(6)
EXAMPLE 3:
Find the center of mass for the rod in Example 2.
soLUrIoN: In Example 2, we found M:4A. E(5  x), we have
r
Using Eq. (6) with p(r) :
Ex(6 x) dx 40
Theref.ore, the center of mass is at I ft from the end having the greater density. 2: If a rod is of uniform density k slugs/ft, where k is a constant, then from formula (6) we have
O ILLUSTRATION
kLz 2L
kx2 2 L
kx
A T
kL2
0
Thus, the center of mass is at the center of the rod, as is to be expected.o
356
APPLICATIONS OF THE DEFINITEINTEGRAL
8.7 Exercises In Exercises1 through 4, a system of particles is located on the r axis. The number of slugs in the mass of each particle and the coordinate of its position are given. Distance is measured in feet. Find the center of mass of each svstem. l. mr:5 at 2; m2: 6 at 3; ms: 4 at 5; mq: 3 at 8. 2. mr:2
a t  4 ; m z : 8 a l  7 ; m s : 4 a t 2 ;m q , : 2 a t 3 . 3. mt: 2 at 3; frrz: 4 at 2; ms: 20 at 4; mn: 70 at 6; ms: 30 at 9. 4. m, : 5 at 7; ttrz: 3 al 2; m": 5 at 0i ffiE: ! at 2; m5: 8 at 10. In Exercises5 through 9, find the total mass of the given rod and the center of mass. 5 . The length of a rod is 9 in. and the linear density of the rod at a point r in. from one end is (4r * 1) slugs/in. 6 . The length of a rod is 3 ft, and the linear density of the rod at a point r ft from one end is (5 * 2x) slugslft. 7. The length of a rod is 10 ft and the measure of the linear density at a point is a linear function of the measureof the distance of the point from the left end of the rod. The linear density at the left end is 2 slugs/ft and at the right end is 3 slugs/ft. 8 . A rod is 10 ft long, and the measure of the linear density at a point is a linear function of the measureof the distance from the center of the rod. The linear density at each end of the rod is 5 slugs/ft and at the center the linear density is 3i slugs/ft. 9 . The measureof the linear density at a point of a rod varies directly as the third power of the measureof the distance of the point from one end. The length of the rod is 4 ft and the linear density is 2 slugs/ft at the center.
10.A rod is 6 ft long and its mass is 24 slugs. If the measure of the linear density at any point of the rod varies directly as the square of the distance of the point from one end, find the largest value of the linear density. L L . The length of a rod is L ft and the center of mass of the rod is at the point *L ft from the left end. If the measureof the linear density at a point is proportional to a power of the measure of the distance of the point from the left end and the linear density at the right end is 20 slugs/ft, find the linear density at a point x ft from the left end. Assume the massis measured in slugs.
t2. The total mass of a rod of length L ft is M slugs and the measure of the linear density at a point r ft from the left end is
proportional to the measureof the distance of the point from the right end. Show that the linear density at a point on the rod r ft from the left end is 2M(L  x) lLz slugs/ft.
8.8 CENTER OF MASS OF A PLANE REGION
Let the massesof z particleslocated at the points (xr,yr), (x",yr) , . . . , (xn, An)in the ry pline be measuredin slugs by *r,;r, . . . , mn, and consider the problem of finding the center of mass of this system. we may imagine the partides being supported by a sheet of negligible weight and negligible thickness and may assumethat each particle has its position at exactly one point. The center of mass is the point where the sheet will balance. To determine the center of mass, we must find two averages:f, which is the average value for the abscissasof the n points, and y', the averagevalue for the ordinates of the z points. we first define the moment of mass of a system of particles with respect to an axis. ' If a particle at a distance d ft from an axis has a mass of.m slugs,then if
8.8 CENTEROF MASS OF A PLANEREGION
M, slugft is the moment of mass of the particle with respect to the axis, Mt:
(1)
md
If the ith particle, having trraissftti slugs, is located at the point (xi, y) ,Its distance from the y axis is xrft; thus, from formula (1), the moment of massof this particle with respectto the y axis ismixl slugft. Similarly, the moment of mass of the particle with respectto the r axis is miylslugftThe moment of the system of n particles with respectto the y axis is Mu slugft, where (2) and the moment of the system with respect to the x axis Ls Mr slugft, where (3) The total mass of the system Ls M slugs, where (4)
The center of mass of the system is at the point (f , Y), where
i,:i'.u.llilliillffilr::,i
:'....it.*.. ...':,..
(s)
(6)
The point (i, !) canbe interpreted as the point such that, if the total mass M slugs of the system were concentrated there, its moment of mass with respectto the y axis,Mn slugft, would be determin edby Mu : Mi , and its moment of mass with respect to the I axis, M" slugft, would be determined by Mr: My. SOLUTION: ExAMPLEL: Find the centerof massof the four particleshavi^g Mo: masses 2, 6, 4, and 1.slugslocated at the points (5, 2) , (2, L) , (0,3), and (4,L), respectively. M, '. 
M
4
t f t i x r  2 ( 5 ) + 6 (  2 ) + 4 ( 0 ) + L( 4 ) : 2
i:7 4
\l
.at i:t
 s4
Z,r i:t
rniAt2(2) + 6(1)+ 4(3)+ L(1) :13 lfii:2 + 6 + 4 + L : 13
358
A P P L I C A T I O NO SF T H E D E F I N I T EI N T E G R A L
Therefore,
W:a M13JM13
andyW
1 3
1
The center of mass is at (#, L ).
0i, f0))
xirTi
\/
L;x
F i g u r e8 . 8 . 1
f(r)
consider now a thin sheet of continuously distributed mass, for example, a piece of paper or a flat strip of tin. we regard such sheetsas being two dimensional and call such a plane region a lamina.In this sectionwe confine our discussion to homogeneous laminae, that is, laminae having constant area density. Laminae of variable area density are considered in Chapter 2L in connection with applications of multiple integrals. Let a homogeneous lamina of areaA ff have a mass of M slugs. Then if the constant area density is k slugs/ff , M: kA. rf the homogeneous lamina is a rectangle,its center of mass is defined to be at the center of the rectangle. we use this definition to define the center of mass of a more general homogeneous lamina. Let L be the homogeneous lamina whose constant area density is k slugs/ff, and which is bounded by the curve y : (x), the r axis, and the f lines x : a and x : b. The function / is continuous on the closed interval l!' bl, andf (x) > 0 for all r in fa,bf . see Fig. 8.8.1.Let A be a partition of the interval la, bl inton subintervals.The ith subinterval is [4_r, 4] and A6x: xi  xir The midpoint of [r11, xi] is yi.Associatedwith eachsubinterval is a rectangular lamina whose width, altitude, and area densitv are given by Luxft, f (yt) tt, and k slugs/ft2, respectively, and whose centlr of mass is at the point (7,, if 0)). The area of the rectangularlamina is f (yr) L,,xft"; hence, kf (y) Ag slugs is its mass. Consequently,if L,iMo slugft is the moment of mass of this rectangularelemeniwith respeciti the y axis, LtMo: yftf(y)
A1x
The sum of the measures of the moments of mass of n such rectangular laminae with respect to the y axis is given by the Riemann sum n
kv,f(y,) aic i:l
If M, slugft is the moment of mass of the lamina L with respectto the y axis, we define
Similarly, if AiM, slugft is the moment of mass of the ith rectangular lamina with respect to the x axrs,
LoM":if(y)kf(y)
Aux
andthe sumof the measuresof the momentsof massof tt suchrectangular laminaewith respectto the r axis is given by the Riemannsum
> +klf(v,)f'L,x
(8)
Thus, if M* slugft is the moment of massof the laminaL with resPect to the r axis, w€ define
' ... . ., ;,.
+*ur+*+il*#*t* i+ffi
r,,
(e)
The mass of the lth rectangular lamina is kf (y) Alx slugs, and so the sum of the measuresof the massesof n rectangular laminae is given by n
)
kf (y) L1x
(10)
the point (x, y) , we detine
which by using formulas (7) , (9) , and (10) gives b
b
k furr*l dx Ja
and
y
fb
k I f(x) dx Ja
Dividing both the nu merator and denominator by k, we get
*
rb  *f(*) dx
ttl
rb IJ a f@) dx
(11)
and
lf (x)l'
L
f(x) dx
(12)
360
A P P L I C A T I O NO SF T H E D E F I N I T EI N T E G R A L
In formulas (11) and (12) the denominator is the number of square units in the area of the region, and so we have expressed a physical problem in terms of a geometric one. That is, r and ! can be considered as the averageabscissaand the averageordinate, respectively, of a geometric region. In such a case,x and y depend only on the region, not on the mass of the lamina. so we refer to the center of mass of a plane region instead of to the center of mass of a homogeneous lamina. In such a case,we call the center of mass the centroidof the region. Instead of moments of mass, we consider moments of the region. we define the moments of the plane region in the above discussion with respect to the r axis and the y axis by the following:
If. (I, y) is the centroid of the plane region and tf.M, andMu are defined as above,
ExAMPLE2: Find the centroid of the first quadrant region bounded by the curye y' : 4x, the x axis, and the lines N  L and x: 4.
soLUrIoN: Let f (x) : 2x1t2. The equation of the curve is then y : /(x). In Fig. 8.8.2, the region is shown together with the lth rectangular element. The centroid of the rectangle is at (7;, *f @).). The areaA square units of the region is given by
A
n
lim 2 f0) l l a l l o i : r
Lic
l,^
f(x) dx
r
0i, f 0))
2xrt2dx
Ti Lix
F i g u r e8 . 8 . 2
We now computeMo and Mr. Mo:
lim f,r,f' (yr) Aix
l l a l l o , ? : r
xf(x) dx
re;:
8.8 CENTEROF MASS OF A PLANEREGION
361
x(2xrrz1dx
t24 5

M*
 li* Zif1,) f(v,) Ln llalloi:1 : * :t
f4
J, lf(x)f' f4
Jr
dx
4xdx
 xrln lr
:L5 Hence,
+_Ma rw A
_+
Zsg
_93 35
and M
r, rA: r 2 84
15
:E:
45
Therefore,the centroid is at the point (89,#).
ExAMPr,n3: Find the centroid of the region bounded bY the curyes y:x2 andy2x*3.
solurroN: The points of intersection of the two curves are (1,1) and (3, 9). The region is shown in Fig. 8.8.3, together with the ith rectangular element. Let f (x) : f and g(r) : 2x * 3. The centroid of the ith rectangular elementis at the point (71,ilf 0) + S(yr)l ) where 7; is the midpoint of the ith subinterval lxrr, xr]. The measure of the area of the region is given by : tim [s(y,)f(y,)] Lrx
i
llalloi:l
:f
f3 Jt
ts(r)f(x)ldx
f3
 
Jr
u
3
lZx+3xzldx
APPLICATIONS OF THE DEFINITEINTEGRAL
v : f@lt y : s(*) (3,9)
(v t, g(y ,)
Q,,tlfrl + s(y,)l) 0i, f0))
(1,1)
Figure 8.8.3
We now compute Mu and M*. n
lim ) y,lSQ)  f(y,)l Lix
Mo:
l l a l l s , 7 : 1
: f xkk)  f(x)l dx Jr : f x l 2 x + 3 x r l d x :t;'3
M,: lim >+ls4) + f (y)I tsQo) f (y)l 4x s ,:1 llallfg
:u
+ (x)]ts(r), f (x)l dx J_,[s(r) f
1 i
I lZ, + 3 * x'flzx + 3  xrf dx Jr
l
r ,:E l
t/^\tr^/
fB
fg Jt
:#
t
l+xr*I2x+9xnf dx
8 , 8 C E N T E RO F M A S S O F A P L A N ER E G I O N
363
Therefore,
+:#1, 3 .n
, : 5i # andy : TM: E
_L7
Hence,the centroidis at the point (1, +). If a plane region has an axis of symmetry, the centroid of the region lies on the axis of symmetry. This is now stated and proved as a theorem. 8.8.1 Theorem
f(v,)) f0t))
Lix
Aix
If the plane region R has the line L as an axis of symmetry, the centroid of R lies on L. pRooF: Choose the coordinate axes so that L is on the y axis and the origin is in the region R. Figure 8.8.4 illustrates an example of this situation. In the figure, R is the region CDE, C is the point (a,0), E is the point (a,0) , and an equationof the curve CDE is y : f (x). Consider a partition of the interval 10, af .Let y5be the midpoint of the ith subinterval lxor, xrl. The moment with resPectto the y axis of the rectangular element having an altitude /(7,) and a width Air is yo[f1) Alr]. Becauseof symmetry, for a similar partition of the interval la, 0f there is a corresPondingelement having as its moment with respect to the y axis ytf 0) A1r.The sum of thesetwo moments is 0; theref.ore,Mo: 0. Because i : Mul A, we concludethat f : 0. Thus, the centroid I of the region R lies on the y axis, which is what was to be proved.
F i g u r e8 . 8 . 4
By applying the preceding theorem, we can simplify the problem of finding the centroid of a plane region that can be divided into regions having axes of symmetry.
ExAMPLE4: Find the centroid of the region bounded bY the semiand the r axis. circle y  \tr4
soLUTIoN: The region is shown in Fig. 8.8.5. Because the y axLSis an axis of symmetry, w€ conclude that the centroid lies on the y axLS;so .r : 0. The moment of the region with respect to the x axis is given by
M*: lim >+l\fyf'L,x i:1 ;;alls
 2 . + f G  x ' )d x JO
F i g u r e8 . 8 . 5
The area of the region .ts 2n square units; so
v
+8 2rr
3rr
364
A P P L I C A T I O NO SF T H E D E F I N I T EINTEGRAL
There is a useful relation between the force causedby liquid pressure on a plane region and the location of the centroid of the region. As in sec. 8.6, let ABCD be the region bounded by the r axis, the lines x: a artd x:b, and the cur''ey:f(x), where/is continuousand f(x) > 0 on the closed interval la,bl.The region ABCD can be consideredas a vertical plate immersed in a liquid having weight ar pounds per cubic unit of the liquid (see Fig. 8.8.6).If F lb is the force owing to liquid pressureon the vertical plate,
r:,li3
f w{f((1)a1x
llAll0 i:t
or, equivalently, Figure 8.8.6
xf(x) dx
F_
(13)
If i is the abscissaof the centroid of the region ABCD, then i : Mo lA. BecauseMn: Il xf(x) dx, we have
I:
xf(x) dx A
and so
I:
xf (x) dx  iA
(14)
Substituting from Eq. (14) into (13), *e obtain F wIA
(ls)
Formula (15) states that the total force owing to liquid pressure against a vertical plane region is the sameas it would be if the region were horizontal at a depth f units below the surface of a liquid. . ILLUsrRArroN1: Consider a trough full of water having asends semicircular regions each with a radius of.2 ft. Using the result of Example4, we find that the centroid of the region is at a depth of Sljn ft. Therefore,using formula (15), we see that if F lb is the force on one end of the trough, _8t6 F:tu.*.2r:;. This agreeswith the result found in Example2 of Sec.8.6.
.
For various simple plane regions, the centroid may be found in a table. when both the areaof the region and the centroid of the region may be obtained directly, formqla (15) is easy to apply and is used in such casesby engineers to find the force caused by liquid pressure.
K 8.8 CENTEROF MASS OF A PLANEREGION
Exercises 8.8 l,2,andSslugsandlocatedatthepoints (1,3),(2'1), 1. Findthecenterof massof thethreeparticleshavingmassesof 1), (3, respectivelY' and 2) ' 2. Find the center of mass of the four particles having masses of.2, 3,3, and 4 slugs and located at the points (1, (1, 3), (0, 5), and (2,l) , respectivelY. prove that the centroid of three particles, having equal masses,in a plane lies at the point of intersection of the medians 3. of the triangle having as vertices the points at which the partides are located. In Exercises4 through 11, find the centroid of the region with the indicated boundaries the Y axis. 4. The parabolax:2y  y' aurrd rP and the r axis.
5. The parabolay:4
6. The parabola A2: 4x, the y axis, and the line y:
4.
and the line Y:4. 8. The lines y : 2x * l, x * Y : 7, and x: 8. 9. The curyes A : f and A : 4x in the first quadrant. 7. The parabola!:
f
10. The curves A: * and'A: x3. 4artd.y:2xf. 1 1 . T h e c u r v e sy : f 12. prove that the distance from the centroid of a triangle to any side of the triangle is equal to onethird the length of the altitude to that side. 13. If the centroid of the region bounded by the parabola!':4px value of a.
and the line r:
a is to be at the point (p, 0), find the
14. Solve Exercise4 of Sec.8.6 by using formula (15) of this section. 15. Solve Exercise5 of Sec.8.5 by using formula (15) of this section. 16. The face of a dam adjacent to the water is vertical and is in the shapeof an isoscelestrapezoid 90 ft wide at the top, 50 ft wide at the bottom, and 20 ft high. Use forrrula (15) of this section to find the total force due to liquid Pressureon the face of the dam. 17. Find the moment about the lower base of the trapezoid of the force in Exercise L5. L8. Solve Exercise 6 of Sec. 8.5 by using formula (15) of this section. L9. Find the center of mass of the lamina bounded by the parabola 2y'  18  3r and the y axis if the area density at any point (x,y) is \ffi slugs/ftz.
fr(x)
20. Solve ExerciseL9 if the area density at any point (x, y) is r slugs/fP. 2 L . Let R be the regionboundedby the curvesy: fr(r) and y: fr(x) (seeFig. 8.8.7).lf A is the measureof the areaof R and if y is the ordinate of the centroid of R, prove that the measureof the volume,V, of the solid of revolution obtained by revolving R about the r axis is given by V  2nyA
fr(x)
F i g u r e8 . 8 . 7
Stating this formula in words we have: If a plane regionis revoloeil abouta line in its plane that doesnot cut the region,then the measureof the aolumeof the solid
366
OF THE DEFINITE INTEGRAL APPLICATIONS of rettolutiongenetateilis equalto the productof the measureof the areaof the regionand the measureof the distancetra,eled by the centroidof the region. The above statement is known as the theoremot' Pappusfor volumes of solids of revolution.
22. Us_ethe theorem of Pappus to find the volume of the torus (doughnutshaped) generatedby revolving a circle with a radius of r units about a line in its plane at a distance of b units from its center, whereb ) t.
23. Use the theorem of Pappus to find the centroid of the region bounded by a semicircle and its diameter. 24. Use the theorem of Pappus to find the volume of a sphere with a radius of r units. 25. LetR be the region bounded by the semicircle V: moment of R with respect to the line y : 4.
t/rt=
land
the x axis. Use the theorem of pappus to find the
26. If R is the region of Exercise25, use the theorem of Pappus to find the volume of the solid of revolution generated by revolving R about the line x  y : r. (nrrrn: use the result of Exercise24 in sec. 5.3.) l'
8.9 CENTER OF MASS OF A SOLID OF REVOLUTION
v:f@) 0i,f0))
xitTi)cib Lix
Figure8.9.1
To tind the center of mass of a solid, in general we must make use of multiple integration. This procedure is taken up in chapter 21 as an application of multiple integrals. However, if the shape of the solid is thai of a solid of revolution, and its volume density is constant, we find the center of mass by a method similar to the one used to obtain the center of massof a homogeneous lamina. Following is the procedure for finding the center of mass of a homogeneous solid of revolution, with the assumption that the center of mass is on the axis of revolution. We first set up a threedimensional coordinate system. The r and.y axesare taken as in two dimensions, and the third axis, the z axis, is taken perpendicular to them at the origin. A point in three dimensions is then given by (x, y , z) . The plane containing the r and y axesis called the ry plane, and the xz plane and the yz plane are defined similarly. suppose that the r axis is the axis of revolution. Then under the assumption that the center of mass lies on the axis of revolution, the y andz coordinates of the center of mass are each zero, and so it is only necessary to find the r coordinate, which we call i. To find r we make use of thb moment of the solid of revolution with respect to the yz plane. Let / be a function that is continuous on the closed interval la , bl , and assumethat/(r) > 0 for all r in la, bf. R is the region bounded by the cufite y : f(x), the r axis, and the lines r : a and x : b;S is the homogeneous solid of revolution whose volume density is k slugs/ft', where k is a constant, and which is generated by revolving the region R about the r axis. Take a partition A of the closed interval la, bf , and denote the ith subintervalby lxnr,4] (with i: 7,2, . . . ,n ). Let 7i be the midpoint of Lxrr, xtf.Form n rectangleshaving altitudes of f (y) ft and baseswhose width is A;r ft. Refer to Fig. 8.9.1, showing the region R and the ith rectangle.If each of the n rectanglesis revolved about the r axis, n circular disks are generated. The ith rectangle generates a circular disk having a radius of f(y) ft and a thicknessof Lp ft; its volume is rrff(y)fz Aaxff ,
OF MASSOF A SOLIDOF REVOLUTION 8.9 CENTER
and its mass is knlf (y)1, Ap slugs.Figure8.9.2shows the solid of revolution S and the ith circular disk' The center of mass of the circular disk lies on the axis of revolution at the center of the disk (yi, 0, 0). The moment of mass of the disk with respect to the yz plarte is then LMo" slugft where LrMo,: Tt(knlf (yr)l' Lrx) The sum of the measuresof the moments of mass of the n circular disks with resPectto the yz plane is given by the Riemann sum
Zy,t
nlf(y)fz L$
(1)
The number of slugsfeet in the moment of massof s with respectto theyz plane, denoted by Mu", is then defined to be the limit of the Riemann sum in (1.)as llAllapproacheszero; so we have
Figure8.9.2
The volume , V ft3, of S was, defined in Sec.8.2 b v fb
v  lim t nlf (yi)12A i r  r T I t f @ ) f ' d x Ja Fr llallo
The mass,M slugs, of solid S is defined by
We define the center of mass of S as the point (r, 0, 0) such that
+ffi
+i=_qi#+++$ffffi Substituting from Eqs. (2) and (4) into (5), we get
xlf(x)l' dx lf(x)l' dx
368
APPLICATIONS OF THE DEFINITEINTEGRAL
From Eq. (6), we see that the center of mass of a homogeneous solid of revolution depends only on the shape of the solid, not onits substance. Therefore, as for a homogeneouslamina, we refer to the center of mass as the centroid of the solid of revolution. when we have a homogeneous solid of revolution, instead of the moment of mass we consider the moment of the solid. The moment, Maz, of the solid S with respectto theyz plane is given by (7)
Thus, if (I, 0, 0) is the centroid of S, from Eqs. (3), (d) , and,(7) we have
ffi ExAMPLE1: Find the centroid of the solid of revolution generated by revolving about the r axis the region bounded by the curve y : x2, the x axis,and the line x=3.
SOLUTION:
The region and a rectangular element are shown in Fig. 8.9.3.
The solid of revolution and an element of volume are shown in Fig . 8.g.4.
f(x) :
x2
Mor:
lim 2 y,nlf (y,)f' L,x
n
Ilallo f:r
xlf(x)f' dx
v=f@)
x5 dx
_ryn f(v))
v ,,Til, rilf(vt)f'Nx } f3
I tf?)f'dx Jo _1r[t*nA* Jo Figure8.9.3
:#n Therefore,
, =T n  5 x  TM : 4, ,q_ 2 Therefore, the centroid is at the point (8, O, 0 ) .
F i g u r e8 . 9 . 4
8.9 CENTEROF MASS OF A SOLID OF REVOLUTION
0i, f0))
Figure8.9.6
Figure8.9.5
The cenkoid of a solid of revolution also can be found by the cylindricalshell method. Let R be the region_bou_nded by the curve > 0 on la, bl, the r axis, and the i : f @),where / is continuous and /(r) of revolution generated by solid the be Let s and, x:b. r:a iirr", of S is then at the point centroid y axis. The the about R revolving (0, , , O).lf the rectangular elements are taken parallel to the y axis, then the Llement of volume is a cylindrical shell. Let the ith rectangle have_a width of L1x xi  xi_1,u.'.dlit 7abe the midpoint of the interval lxir, xtf. The centroid of the rylindrical shell obtained by revolving this rectangle about the y axis is at the center of the cylindrical shell, which is the point (0,tf Q), 0). Figure 8.9.5shows the region R and a rectangular element of area,and Fig.8.9.5 shows the cylindrical shell. The moment, Mrz, of s with respect to the xz Plarreis given bY
lf V cubic units is the volume of S,
v
lim f ,nr,f(y,) \x
Ilall o tr
xf(x) dx
i::;
APPLICATIONS OF THE DEFINITEINTEGRAL I
EXAMPLE 2: Use the cylindricalshell method to find the centroid of the solid of revolution generated by revolving the region in Example L about the y axrs.
solurroN: Figure 8.9.7 shows th e region and a rectangular element of area. The solid of revolution and a cylindricalshell element of volume are shown in Fig. 8.9.8.
M,,: ,lip 2 +ffvt)2nyi(yt)A$ llAll0it f3
v:f@) (3,9)
=rr I xlf (x)f' dx JO 7Tft"d* JO
=Tn V  lim f, rorrf (y,) Lix llalloFi r3 ,2n I xf(x) dx
0 i, f0 i))
JO
:2tr ft *, d* JO
:#n Q,,tftv,l)
Therefore, 2!3n
M y_f:#_3 aix
Figure8.9.7
Hence, the centroid is at the point
( 0 ,3 , 0 ) .
Figure8.9.8
8.9 CENTEROF MASS OF A SOLID OF REVOLUTION 371
ExAMPLE3: Solve Example L by the cylindricalshell method.
Figure 8.9.9 shows the region and a rectangular element of solurroN: area, and Fig. 8.9.10 shows the cylindricalshell element of volume obtained by reiolving the rectangle about the r axis. The centroid of the
,7.'n),0,0)' The cylindricalshellis it its center,i"hi.h is the point (+(3 + (f, 0,0)' is at centroidof the solid of revolution
v (3,9)
M,,
 6l  lim i +(3+ 6,1 2nyo(3
Lil
l l a l l *o 7 : t
fs : n l" y(3+ frl(3  {vl dy JO fs
:n
 (gvyz) dy
JO
(tE , v,)
(*tt + tfr) ,y)
Ai
v
:#n
AiA
(3, Yi )
Finding V by the cylindricalshell method, we have
v
A it
lim $ 2nyo(3 ,6)
Lra
l l a l l o ? t fs
2n I V(3  y't') dY JO
#n F i g u r e8 . 9 . 9
_
M.,"
*v2 t::
x
5
and the centroid is at (8,0,0). The result agreeswith that of Examplet.
t
8.9 Exercises revolving the given region about the inIn Exercises1 through 16, findthe centroid of the solid of revolution generatedby dicated line. elements perpendicular to the 1. The region bounded by y : 4x f and the x axis, about the y axis. Take the rectangular axis of revolution. 2. Same as Exercise 1, but take the rectangular elements parallel to the axis of revolution.
972
APPLICATIONS OF THE DEFINITEINTEGRAL
3. The region bounded by x * 2y :2, the x axis, and the y axis, about the r axis. Take the rectangular elementsperpendicular to the axis of revolution.
Same as Exercise3, but take the rectangular elements parallel to the axis of revolution. The region bounded by y' : r3 and x  4, about the ff axis. Take the rectangular elements perpendicular to the axis of revolution.
Same as Exercise5, but take the rectangularelementsparallel to the axis of revolution. The region bounded by y : x3,x  z, and the r axis, about the lin e x: 2. Take the rectangular elements perpendicular to the axis of revolution. 8. Same as Exercise7, but take the rectangular elements parallel to the axis of revolution. 9. The region bounded by xny: l, y :1, and y :4, about the y axis. 10. The region in Exercise9, about the r axis. 11. The region bounded by the lines y : x, ! : 2x, and.x * y : 6, aboutthe y axis. 12' The region bounded by the portion of the circle r3 t y' :4 in the first quadrant, the portion of the line 2r  y : 4 in the fourth quadrant, and the y axis, about the y axis. L3. The region bounded by y: rP and !: x + 2, aboutthe line y:4. 1.4.The region bounded by y": 4x and :16  4x, about the r axis. !2 L5. The region bounded by y : \/W , the x axis, and the line r: p, aboul the line r: p. 15. The region of Exercise 15, about the line y:2p. L7. Find the centroid of the rightcircular cone of altitude ft units and radius r units. L8. Find the center of mass of the solid of revolution of Exercise3 if the measure of the volume density of the solid at any point is equal to a constant k times the measure of the distance of the point from tn" y, pi*". 19. Find the center of mass of the solid of revolution of Exercise5 if the measure of the volume density of the solid at any point is equal to a constant k times the measure of the distance of the point from the yz p?ane. 20. that a cylindrical hole with a radius of r units is bored through a solid wooden hemisphere luppose of radius 2r units, so that the axis of the cylinder is the same as the axis of ttre hemifrhere. Find the centroii of the solid remaining.
8.L0 LENGTH oF ARc OF A PLANE CURVE
Let the function f be continuous on the closed interval la, bl. Consider the graph of the equati on y : f (x) of which a sketch is shown in Fig. 9 . 1 0L. .
b
: f(x)
F i g u r e8 . 1 0 . 1
8 . 1 0 L E N G T HO F A R C O F A P L A N EC U R V E
373
The portion of the curve from the point A(a, f (a)) to the point B(b, f(b)) is called an arc. We wish to assigna number to what we intuitively think of as the length of such an arc. If the arc is a line segmentfrom the point (xr, yr) to the point (rr, Ur), we know from the formula for the distance between two points that its length is given We use this formula for defining the length of av ffi. an arc in general. Recallfrom geometry that the circumference of a circle is defined as the limit of the perimeters of regular polygons inscribed in the circle. For other curves we proceed in a similar way. Let A be a partition of the closed interval la, b) fotmed by dividing the interval into r subintervals by choosing any (n  1) intermediate numbers between a and b. Let X0: a, and xn:b, and let xt , Xz, Xs, 1 xnr I xr. Xnr be the intermediate numbers so that Xo I Xr I xz I
Then the fth subinterval is [ri1, x1]; and its length, denoted by Air, is . . . ,2. Thenif llAllisthenormofthepartition x1 xilrwherei:!,2,3, = A, each Ap llAll. u
8.10.2 Figure Associatedwith eachpoint (4, 0) on the r axis is a point Prktu f k)) on the curve. Draw a line segmentfrom eachpoint Pil to the next point P, as shown in Fig.8.10.2 The length of the line segmentfrom Pil to Pi is denoted UV lFrFr I and is given by the distance formula lPtt.tr,l:
xrr)' +
(1)
A*)'
The sum of the lengths of the line segments is
l p a l + l p  l p+r l p r 4 l+ . . . + l n ,  p , l + " '
+ lPrrPnl
which can be written in sigma notation as
lP,rPil
(2)
i:l
It seems plausible that if n is sufficiently large, the sum in (2) will be "close to" what we would intuitively think of as the length of the arc AB. So we define the length of the arc as the limit of the sum in (2) as the norrn
374
APPLICATIONS OF THEDEFINITE INTEGRAL
of A approacheszero, in which casen increaseswithout bound. we have, then, the following definition. 8.L0.L Definition
If the function / is continuous on the closed interval la, bl and,if there exists a number L having the following property: for any e ) 0 there is a 6>0suchthat lr. _  l ) l l P , _ , P ,  L l < e
l  " ' ' I
li:l
I
for every partition A of the interual la, bf forwhich llAll< 6, then we write n
t: ,lip > lF;El llAll0 i:r
(3)
and L is called the length of the arc of fhe curve y : (x) from the point f A(a, f(a)) to the point B(b, f(U11. If the limit in Eq. (3) exists,the arc is said tobe rectifiable. we derive a formula for finding the length L of an arc that is rectifiable. The derivation requires that the derivative of be continuous on f la,bf; such a function is said to be "smooth,, on lo,bl. 8.1.0.2Definition
A function / is said tobe smootfton an interval I if f , is continuous on I. (xi,yi)
Uit
:
LiU
(xir,Ait)
Figure8.10.3
Refer now to Figure 8.10.3. If. pi_r has coordinates (4_1, yi_1) and.p, has coordinates (xv y), then the length of the chord n*rr, is given by formula (1). Lettin g h  xrr
:
A# and Ui  Ai_r 
lPrrPrl:
AtA, we have
(4)
or/ equivalently,becauseAg # 0, lPrtP,l:
(Air)
(s)
CURVE 375 OFARCOFA PLANE 8.10LENGTH Becausewe required that f ' be continuous on [4, b], the hypothesis of the meanvalue theorem (4.7.2) is satisfied by f , and so there is a number z1in the open intervd (xrr, r1) such that ' (6) f (xt) f (xt') : f (z) (x, xrr) BecauseL$: f(x) * f (xrt) and A6r x1 ri1, from Eq. (6) we have
+r!_:f, (zr) Lrx
(7)
Substituting from Eq. (7) into (5), we get (8)
lmlffiLix where x*r 1 zi 1 xi. For each i from L to n there is an equation of the form (8). nn
(e) i :l
i:l
Taking the limit on both sides of Eq. (9) as llAllapproadres zeto,we obtain nn
lim ). lP;F;l : lim ) Vt * lf'(z)12 LP llAll0 i:l
(10)
llAll0 t=l
if this limit exists. To show that the limit on the right side of Eq. (10) exists, let g be the function defined by
s@): \ET(@F Because/' is continuous on [4, bl, g is continuous on la, bf. Therefore, . ,n, b e c a u sxe* r 1 ? , 1 l x y f o t i : L , 2 , nn
\ s@)t* li ) \trTTfGJP Ap: ,lim llAll0 l=1
llAll0 i=l
or, equivalently, (11) ll
B e c a u sge( x ) : f f i ,
f r o mE q . ( 1 1 )w e h a v e
rimtffiLrx:f'rydx \ o'J 11611*of:r
(12) Jo
Substituting from Eq. (L2) into (10),we get
rim>Finl : ['ry Jo
'=r llallo
dx
(13)
376
A P P L I C A T I O NO SF T H E D E F I N I T EI N T E G R A L
Then from Eqs. (3) and (13) we obtain rb
Ll ffidx Ja In this way, we have proved the followirg 8.L0.3 Theorem
(14) theorem.
If the functi on f and its derivativ e ' are continuous on the closed interval f fa, bf , then the length of arc of the curve y: f (x) from the point (a, (a)) f to the point (b, f (b)) is given by
we also have the following theorem, which gives the length of the arc of a curve when r is expressedas a function of.y. g.10.4 Theorem
If the function F and its derivative F' are continuous on the closed interval lr'lf ,thenthelglgat of arcof thecurvex:F(y) fromthepoint (F(c),c) to the point (F(d) , d) is given by
The proof of rheorem 8.10.4is identical with that of rheorem g.10.3; here we interchange x and y as well as the functions f and F, The definite integral obtained when applying Theorem g.10.3 or Theorem 8.10.4is often difficult to evaluate.Becauseour techniques of integration have so far been limited to integration of powers, we are further restricted in finding equations of curves for which we can evaluate the resulting definite integrals to find the length of an arc. EXAMPLE1.:Find the length of the arc of the curye y : x2t3from the p o i n t ( 1 , 1 ) t o (8, 4) by using Theorem 8.10.3.
'(x): solurroN: See Fig.8.10.4. Because f ( x ) : ) c 2 t 3f, Theorem 8.L0.3we have
&Xrtt. From
v v: To evalu ate this definite integral, let u  gxztT* 4; then du:6x1t W h e n x : l , u  L 3 ; when x  8, Lt 40. Therefore,
F i g u r e8 , 1 0 . 4
uLtz du
dx.
8 . 1 0 L E N G T HO F A R C O F A P L A N EC U R V E
ExAMPrn 2: Find the length of the arc in Example L by using Theorem 8.L0.4.
soLUTroN: Because y  x2t3and r Lettin g F (y)  At'', we have
377
0, we solve for x and obtain x  y3t2
F ' ( Y ) E Y u z Then, from Theorem 8.'1.0.4,we have fq
L  Vr+*y dy JT
rftMdv ,
v
J,
 l f
14
rg  3 (a + 9y)3/2
lr
L
 ;7 (40tr'  133t2)
: 7.6
is continuous on fa,bl, then the definite integral If f dt is a function of r; and it gives the length of the arc tf \trTV@P of the cuwey:f(x) fromthe point(a,f(a)) to thepoint (x,f(x)),where r is any number in the closed interval la, bf .Let s denote the length of this arc; thus, s is a function of x, and we have ar
s(x):f
Vl+tf'G)12dt
Ja
we have From Theorem7.6.'1.,
s'(r): ttt+TfTdP or, becauses'1r;: dsldxandf'(x): dyldx,
Multiplying by dx, we obtain dsSimilarly, if we are talking about the length of arc of the curve r:80)
(15)
378
APPLICATIONS OF THE DEFINITEINTEGRAL
f r o m ( S ( c ), c ) t o ( S Q ) , y ) , w e h a v e
ds
dy
(16)
Squaring on both sides of Eq. (15) gives
ffiffi
ffilF $#rl
(r7)
From Eq. (r7) we get the geometric interpretation of ds, which is shown in Fig. 8.10.5.
Q@ * Ar, V * tV)
Figure8.10.5
In Fig. 8.10.5, line T is tangent to the cur.rtey: f (x) at the
pointP. lFMl= Ax: dx; lMSl: Ay;lMRl: dV;lpRl:}r; ih" l"ngthof a r cP Q : 4 r .
Exercises 8.10 1. Find the length of the arc of the curve 9yz : 4rs from the origin to the point (g, 2\/t) . 2. Find the length of the arc of the curve *: (2V + 3)s from (1, t) ro (Z\/V, D. 3. Find the length of the arc of the curve 8y : * * 2x2 fuom the point where r: 1 to the point where r: 2. 4. Use Theorem 8.10.3to find the length of the arc of the curve ys:gf from the point (1,2) to the point (27,7g). 5. Solve Exercise4 by using Theorem 8.10.4. $ Find the entire length of the arc of the curve x2ts+ y2t3: 1 from the point where x : * to the point where r: 1. E Find the length of the arc of the curve y: *(* *21srz from the point where r:0 to the pointwhere r:3. 8. Find the length of the curve 6ry: yo * 3 from the point where y: I to the point where y: !.
BEVIEWEXERCISES 379 1 in the first quadrant from r : [a to x: a. 10. Find the length of the curve 9yz : 4(t * rP)a in the first quadrant from x : 0 to r :2\/2. 11. Find the length of the curve 9y' : x(x  3)' in the first quadrant from x: L to x:3. 12. Find the length of the curve 9y': x'(2x * 3) in the secondquadrant from r: 1 to r: 0. 9. Find the length of the cuwe (xla)2t} t (ylb)'t":
(Chapter8) ReoiewExercises 1. Find the area of the region bounded by the loop of the curve y2: *(4
x).
2. The region in the first quadrant bounded by the curyes r : y2 and,x : ya is revolved about'the y axis Find the volume of the solid generated. 3. The base of a solid is the region bounded by the parabola !2:8x and the line r: every plane section perpendicular to the axis of the base is a square.
8. Find the volume of the solid if
4. A container has the sameshape and dimensions as a solid of revolution formed by revolving about the y axis the region in the first quadrant bounded by the parabola *: 4py, the y axis, and the line y : p. If the container is full of water, find the work done in pumping all the water up to a point 3p ft above the top of the container. 5. The sur{aceof a tank is the sameas that of a paraboloid of revolution which is obtained by revolving the parabola y : f about the r axis. The vertex of the parabola is at the bottom of the tank and the tank is 35 ft high. If the tank is filled with water to a depth of.20 ft, find the work done in pumping all of the water out over the top. 6. A plate in the shape of a region bounded by the parabola * : 6y and the line 2y: 3 is Placed in a water tank with its vertex downward and the line in the surfaceof the water. Find the force due to liquid Pressureon one side of the plate' 7. Find the area of the region bounded by the curves y: lx 11,V: f  2x, the y axis, and the line r: 2. 8. Find the area of the region bounded by the curves y: lxl + lr 1l and y: x + t, : lx  21, the r axis, and the 9. Find the volume of the solid generated by revolving the region bounded by the curve y lines l: I and r: 4 about the r axis. 10. Find the length of arc of the curve ayz: f from the origin to (4a,8a). 11. Find the length of the curve 6y': x(x  2)2 fuom (2, 0) to (8' A\tr) ' (2,2),(2'4)'and(1,2). 12. Findthecenterofmassofthefourparticleshavingequalmasseslocatedatthepoints(3,0), l) ' and (5, 2). Find 13. Tfuee particles having masses5, 2, andS slugs are located, respectively, at the points (1, 3), (2, mass. the center of 14. The length of a rod is 8 in. and the linear density of the rod at a point r in. from the left end is 216 + 1 slugs/in. Find the total mass of the rod and the center of mass. 15. Find the centroid of the region bounded by the parabola !2: x and the line y : x  2. 16. Find the centroid of the region bounded by the loop of the curve !2: *  x3. 12. The length of a rod is 4 ft and the linear density of the rod at a point r ft from the left end is (3r * 1) slugs/ft. Find the total mass of the rod and the center of mass. 18. Give an example to show that the centroid of a plane region is not necessarily a point within the region. 19. Find the volume of the solid generatedby revolving about the lin e y : 'I..the region bounded by the line 2y : r * 3 and outside the curves!2 I x:0 and.y2 4x:0. 20. Use integration to find the volume of a segment of a sphere if the sphere has a radius of r units and the altitude of the segment is h units.
380
APPLICATIONS OF THE DEFINITE INTEGRAL
2 L . A church steepleis 30 ft high and every horizontal plane section is a square having sides of length onetenth of the distance of the plane section from the top of the steeple. Find the volume of the steeple. 22. A trough full of water is 5 ft long, and its cross section is in the shape of a semicircle with a diameter oI 2It at the top. How much work is required to pump the water out over the top? 23. A water tank is in the shapeof a hemisphere surmounted by a rightcircular cylinder. The radius of both the hemisphere and the cylinder is 4 ft and the altitude of the cylinder is 8 ft. If the tank is full of water, find the work necessaryto empty the tank by pumping it through an outlet at the top of the tank. 24. A force of 500lb is required to compress a spring whose natural length is 10 in. to a length of 9 in. Find the work done to compress the spring to a length of 8 in. 25. A cable is 20 ft long and weighs 2lblft, and is hanging vertically from the top of a pole. Find the work done in raising the entire cable to the top of the pole. 26. The work necessaryto stretch a spring from 9 in. to L0 in. is I times the work necessaryto stretch it from 8 in. to 9 in. What is the natural length of the spring? 27. Find the length of the curve 9x2tt+ Ayzrz 36 in the second quadrant from r: 1 to r: *. 28. A semicircular plate with a radius of 3 ft is submerged vertically in a tank of water, with its diameter lying in the surface. Use formula (15) of Sec.8.8 to find the force due to liquid pressure on one side of the plate. 29. A cylindrical tank is half full of gasoline weighing 42lblfts.If the axis is horizontal and the diameter is 6 ft, find the force on an end due to liquid pressure. 30. Find the centroid of the region bounded above by the parabola4*:369y
andbelow by the r axis. 3 1 . Use the theorem of Pappus to find the volume of a rightcircular cone with a base radius of r units and an altitude of h units.
32. Theregionboundedbythe parabolal: of the solid of revolution formed.
4y,thex axis,andthe liner:4
is revolvedaboutthey axis.Findthe centroid
33. Find the center of mass of the region of Exercise30 if the measure of the area density at any point (x, y) is l[  V 34. Find the center of mass of the solid of revolution of Exercise32 if the measureof the volume density of the solid at any point is equal to the measure of the distance of the point from the rz plane.
Logarithmlc and rl functions
FUNCTIONS AND EXPONENTIAL LOGARITHMIC
9.I THE NATURAL LOGARITHMIC FUNCTION
The definition of the logarithmic function that you encountered in algebra is based on exponents. The laws of logarithms are then proved from corresponding laws of exponents. One such law of exponents is Ac . Aa:
gc*u
(1)
If the exponents, r and ! t altl positive integers and a is any real number, (1) follows from the definition of a positive integer exponent and mathematical induction. If the exponents are allowed to be any integers, either positive, negative, ot zeto, and,a # 0, then (1) will hold if a zero exponent and a negative integer exponent are defined by ao: I and
n>0 If the exponents are rational numbers and a > 0, then Eq. (1) holds when a^tn is defined bv antn: (%)* It is not quite so simple to define a'when r is an ilrational number. For example, what is meant by 4fi? The definition of the logarithmic function, as given in elementary algebra, is based on the assumption that aa exists if a is any positive number and x is any real number. This definition states that the equation 4t:
N
where a is any positive number except 1 and N is any positive number, can be solved for r, and r is uniquely determined by x: loio N In elementary algebra,logarithms are used mainly as an aid to computation, and for such purposes the number a (called the base)is taken as 10. The following laws of logarithms are proved from the laws of exponents: Law 1 logoMN:logo M * logoN .M : loga M log" N Law 2 tog, f Law 3 logo1:0 Law 4 logoM:nlogoM Law 5 logoa:l In this chapter we define the logarithmic function by using calculus and prove the laws of logarithms by means of this definition. Then the exponential function is defined in terms of the logarithmic function. This definition enables us to define a' when r is any real number and a * 0.
LOGARITHMIC 9.1THENATURAL FUNCTION383 The laws of exponents will then be proved if the exponent is any real number. Let us recall the formula 
+n+l
*C I t"dt:*= n+ L
n#t
J
Consider the function This formula does not hold when n:1. defined by the equation A: tr, where f is positive. A sketch of the graph of this equation is shown in Fig. 9.1.1. LetR be the region bounded above by the curve y : Uf, below by the f axis, on the left by the line t : l, and on the right by the line f : r, where r is greater than 1. This region R is shown in Fig. 9.1.1.The measureof the areaof R is a function of r; call it A(x) and define it as a definite integral by
F i g u r e9 . 1 . 1
(2)
A(x)
Now consider the integral in (2) if 0 < r ( 1. From Definition 7.3.4, we have
f +dt:t:0, Then the integral I: Glt) df represents the measure of the area of the region bounded above by the curve y: Ut, below by the t axis, on the left by the line t:x, and on the right by the line f :1. So the integral li $lt) df is then the negative of the measure of the area of the region shown in Fig.9.1.2. rf.x:L, the integral li Glt) df becomesIl Olt) dt, which equals 0 by Definition7.3.5.In this casethe left and right boundariesof the region are the same and the measure of the area is 0. Thus, we see that the integral t{ Glt) dt for x } 0 can be interpreted s in terms of the measure of the area of a region. Its value depends on r and is therefore a function of r. This integral is used to define the "nafural logarithmic function."
Figure 9.1.2
9.1.1 Definition
The natural logarithmicfunction is the function defined by r>0 The domain of the natural logarithmic function is the set of all tive numbers. We read ln x as "the natural logarithm of x."
The natural logarithmic function is differentiable becauseby applying Theorem 7.5.1we have D"(ln r)
384
FUNCTIONS ANDEXPONENTIAL LOGARITHMIC Therefore,
D"(ln i:1
(3)
From formula (3) and the chain rule,7f.u is a differentiable function of x, and u(x) > 0, then
*#*g##;r{f*', .,
(4)
ffi'D.ua EXAMPLE 1:
Given
yln(3r25x+8)
fu__  r dx
tind dyldx. ExAMPw 2:
solurroN: Applying formula (4), we get
Given
y : l n [ ( 4 x '+ 3 ) ( 2 x 1 ) ]
SOLUTION:
D"U
find D *U.
3x2 6x + 8
'(6x6)
6x5
3x2 6x + 8
Applying formula (4) , we get 7
(4x' + 3) (2x 1)
[8r (2x  1) + 2(4xz+ 3) ]
24x28x+6 (4x'+ 3) (2x 1) SOLUTION:
dy
From formula (4) , we have 1
(x+1) )c
It should be emphasizedthat when using formula (4), u(x) must be positive; that is, a number in the domain of the derivative must be in the domain of the given natural logarithmic function. o rLLUsrRArroxL: In ExampleL, the domain of the given natural logarithmic function is the set of all real numbers because3x2 6x * 8 > 0 for all r. This can be seen from the fact that the parabola having equation  6x * 8 has its vertex at (1,5) and opensupward. Hence,(6x  6)l A :3f (3x' 6r * 8) is the derivative for all values of x. In Example 2, because (4x2t 3)(2x 1) > 0 only when x > i, the domain of the given natural logarithmic function is the interval (*, +).
FUNCTION 9.1 THE NATURALLOGARITHMIC
Therefore, it is understood that fraction (5) x>i. Becausexl(x * 1) > 0 when either x 11 the natural logarithmic function in Example and so 1.lx(xt 1) is the derivative if either x
0, the domain of 3 is (oo, 1) U (0, **), 1 or r ) 0. o
We show that the natural logarithmic function obeys the laws of logarithms as stated earlier. However, first we need a preliminary theorem, which we state and prove. positive numbers, then
lf a
9.1.2 Theorem
1
idt PRooF: In the integral ["u (Llt) dt, make the substitution t: aur then u : 1, and when t  ab, u  b. Therefore, we dtadu. When t:a, have
f'lo':I:# @
du)
: ['Lau u Jr
1,+"
,
I
F i g u r e9 . 1. 3
o TLLUsTRATIoN2: In terms of geometrY, Theorem 9."1..2states that the measure of the area of the region shown in Fig. 9.1.3 is the same as the measure of the area of the region shown in Fig.9."1..4. o If we take x '/. in Definition 9.1.1.,we have
l n 1 : [ ' I ta , Jr
The right side of the above is zero by Definition t
7.3.5. So we have (6)
lnL0
Equation (5) corresponds to Law 3 of logarithms, given earlier. The following three theorems give some properties of the natural logarithmic function.
F i g u r e9 . 1. 4
9.L.3 Theorem
If a and b arc any Positive numbers, then ln (ab):ln
a+ lnb
386
FUNCTIONS LOGARITHMICANDEXPONENTIAL PRooF:From Definition 9.1.1,
rn(ab):f"oIL at Jr
which, from Theorem7.4.6, gives
tn(ab): I:+ at+l"' I at Applying Theorem 9.1.2 to the second integral on the right side of the above, we obtain
tn(ab):f+at+lolat t t Jr
Jt
and so from Definition 9.1.1 we have ln(ab\ :ltt 9.1.4 Theorem
n * ln b
I
lL a and b are any positive numbers, then lnl:ln
a lnb
pRooF: Becauseo:
(alb) . b, we have
ln4:hf+.b) \al
Applying Theorem 9.1.3 to the right side of the above equation, we get
l na : t n l + n a ubtractingln b on both sidesof the aboveequation,w e obtain ln4:halnb b
9.1.5 Theorem
rf. n is any positive number and r is any rational number, then lna':rlna pRooF:
From formula (4), if. r is any rational number and r ) 0, we have 1
D"(lnr):*.rrrr!* and D,(rlnr):!*
9.1 THE NATURALLOGARITHMIC FUNCTION
Therefore, D'(ln x') : D,(r ln x) From the above equation the derivatives of ln x'and rln x are equal, and so it follows from Theorem 5.3.3 that there is a constant K such that lnx":rlnx*K
forallr>O
(7\
To determine K, substitute 1 for r in Eq. (7) and we have ln 1": r ln 1* K But, by Eq. (5), ln 1 : 0; hence,K : 0. ReplacingK by0 in Eq. (7),we get lrtx':rlnx
forallr>0
from which it follows that if x:
fl, where a is any positive number, then
lna':rlna
I
Note that Theorems 9.1.3,9.1.4,and 9.1.5are the same properties as the laws of logarithms 1.,2, and 4, respectively, given earlier. . rLLUsrRArroN3: In Example 2, if Theorem 9.1.3is applied before finding the derivative, we have y : ln(Axz + 3) * ln (2x  1)
(8)
Now the domain of the function defined by the above equation is the interval (4, aoo), which is the same as the domain of the given function. From Eq. (8) we get L'tU :
8x2 Ef+ gT E=T
and combining the fractions gives 8x(2x 1,)* 2(4x2* 3) un t A_.:_@ which is the same as the first line of the solution of Example 2.
.
o rLLUSrnerroN 4: If we apply Theorem 9.1,.4before finding the derivative in Example 3, we have
lnY:ln
x
ln(r+1)
(e)
ln r is defined only when x ) 0, and ln(x * 1) is defined only when x > 1. Therefore, the domain of the function defined by Eq. (9) is the interval (0, 1o). But the domain of the function given in Example 3 consists of the two intervals (, 1) and (0, +;. Finding D*y from Eq. (9), we have
x+"1,
388
FUNCTIONS LOGARITHMICAND EXPONENTIAL
but remember here that r must be greater than 0, whereas in the solution of Example 3 values of r less than 1 are also included. o Illustration 4 shows the care that must be taken when applying Theorems 9.1..3,9.L.4,and 9.1.5to natural logarithmic functions. ExAMPLE4: y:ln(2x find D *A.
Given  l)t
solurroN:
From Theorem 9.1.5we have y :ln(2x  l)t: 3 ln(2x  1)
Note that ln(2x  1)3 and 3ln(2x  1) both have the samedomain, x ) E. Applying formula (4) gives DrA:3 .
1
,rr1
. '_ r
6
2x_ 
In the di scutssron ror ) n tth ] Iflt tr follo,1 find this by.Utsin S rg fo: rla ((4), )ffn nu fon (l l' D,(ln lrll)) ' D)*"'(ln 1
ve need to make use of D,(ln lrl). $o is substitutedfor l*l,and so we ha{re
tfr,2)
w* r ' D " (({x' , ) .f
1 {x' x x2
Hence, D"(ln lxl):,
1
(10)
From formula (10) and the chain rule, if u is a differentiable function of x, we get (11) The followirg example logarithmic function, given plify the work involved in volving products, quotients, soLUrIoN:
illustrates how the properties oT the natural in Theorems 9.'1,.3,9."1,.4, and 9.'1..5,can simdifferentiating complicated expressions inand powers.
From the given equation
' f?E' = l:, lf'FTl lvl:lr(r*D\mllr+[ml
FUNCTION 9.1 THE NATURALLOGARITHMIC
Taking the natural logarithm and applying the properties of logarithms, we obtain
l r ,l y l: * l n l r + 1 l l n l r * 2 l  + l n l r + 3 1 Differentiating on both sides implicitly with respect to r and applying formula (11), we get 1, ^
I'rY :
I
11
1
2 / 1 rr + 1 \ T Ix 5  +2m
Multiplying on both sides by Ur we obtain DrA: y
2 ( x+ 2 ) ( r + 3 )  6 ( r * 1 ) ( r + 3 )  3 ( r * 1 ) ( x+ 2 )
5(x+1)(r+2)(r+3)
Replacing y by its given value, we obtain D rA equal to Z x z* 1 0 r + 1 2  6 x z  2 4 x  1 8
3x29x6
6(x+1)(r+2)(r+3) 7x2*23x12 Dra:
6 ( x + 1 ) 2 t 3 ( *x 2 ) ' ( r * 3 ) s r z
(r2) we have, for n any real number,
(13)
Evaluate
SOLUTION:
f
xzdx
I f 3x2dx
Jm:3Jffi:3'^ EXAMprn 7: Evaluate
I:#'.
1r
+C lr3+11
sor,urroN: Because(x' * 2) l(x + 1) is an improper fraction, we divide the numerator by the denominator and obtain x2*2
4,
'+1IxI+I+1
3
390
LOGARITHMICAND EXPONENTIAL FUNCTIONS
Therefore,
Q1.#) dx
I#dx:l:
l:
:txrx*3lnlr+11 22+3ln33ln 3In33
L
0
3In3
The answer in the above example also can be written asln27 because, by Theorem 9.'1,.5,3In 3 : ln 33. ExAMPr,r 8: Evaluate
SOLUTION:
f hr ,  ax Jx
LCt LI :
then du : dxlx; so we have
ln
I+dx1"d tt I " r +
C: + ( h r )z+ C
Exercises 9.1 In Exercises1 through 10, differentiate the given function and simplify the result. L f (x): ln(l + 4x') 2 .f ( x ) : l n \ M \r '\$; \E3. f(x): ln 4. S(r) : ln(ln r) 5. h(x) : ln (x2ln x) 7.f(x):ln
6.f(x):ln/# 8. f (x): f4nT xln(r* \ffi)U.G(r)
lr3+11
O F ( r ) \ E T
h(1 + lm)
In Exercises11 through 16, findD"y by logarithmic differentiation. xlffi
4A
LZ.y:ffi
ffi 14.1t: \' (x * l)zrs L6. y  (5x  4) (x, + 3) (3rt  5) In Exercises L7 and L8, fin d D ry by implicit differentiation.
(b. m!*xyl
W, lnxy*x*yz In Exercises 19 through
x
26, evaluatethe indefinite
integral.
\R
9.2 THEGRAPHOF THENATURALLOGARITHMICFUNCTION
21Jt+x L n x
fdx
J szx
\q'
ntlffi
5  nt z,\A + ' f g + 2 * a* J
In Exercises27 through 30, evaluatethe definite integral. 2l dx 27. [5 Jt i25
29. 31.P r o v e t h a t l n r n : n ' l n x b y f i r s t s h o w i n g t h a t l n t n a n d n ' l n x d i f f e r b y a c o n s t a n t a n d t h e n f i n d t h e c o n s t a n t b y taking r:
1. ln r ) 0 and 1 ln r
32. Prove that x7
Ux < 0 for all r > 0 and x * l,thus establishingthe inequality
r1.hr(xl x 1. (nrnr: Letf(x):x1lnrandg(x):tlnxllxanddeterminethesigns forallr>0andr# g'(r) on the intervals (0, 1) and (1, +o;.;
off'(x) and
33. Use the result of Exercise32 to prove lt* to
h(l*
r) : t
1C
which 34. Establish the limit of Exercise33 by using the definition of the derivative to find F'(0) for the function F for F(r) : ln(l + r). rt rc (Jlt) dt by using Theorem 7.4.8.) (ln r)/r : 0. (mNr: First prove ttrat ol \/i) o, = 35. prove J. J. "lif_ J 1
J l
g.Z TlflIE GRAPH OF THE NATURAL LOGARITHMIC FUNCTION
To draw a sketch of the graph of the natural logarithmic function, we must first consider some Properties of this function' Let / be the function defined by
f(r):lnt:
J[ r' It o ,
r>o
The domain of / is the set of all positive numbers. The function / is differentiable for all r in its domain, and
f'(x):+
(1)
Because/ is differentiable for all x > 0, / is continuous for all r > 0. From (1) weconclude thatf'(r) > 0 forall x) 0, andtherefore/is anincreasing function.
392
FUNCTIONS AND EXPONENTIAL LOGARITHMIC 1
f"(*):*
Q)
From (2) it follows that f" (r) < 0 when x > 0. Therefore, the graph of y : f (x) is concavedownward at every point. Because/(1) : ln L : 0, the r intercept of the graph is 1. f(2):ln
L t
To determine an approximate numerical value for ln 2, the definite integral l[lt) df is interpreted as the number of square units in the area of the region shown in Fig.9.2.1. From Fig. 9.2.1 we see that ln 2 is between the measuresof the areas of the rectangles, each having a base of length 1 unit and altitudes of lengths * unit and 1 unit; that is, 0.5< ln2 < 1
F i g ur e 9 . 2 . 1
(3)
An inequality can be obtained analytically, using Theorem 7.4.g,by proceedingas follows.Letf (t) :Llt andg(t) : *. Then/(t) > g(t) for all f in [1,2]. Becausef and g are continuous on 11.,21,they are integrable on f7,2f, and we have from Theorem 7.4.8
f+
dt>
or, equivalently,
l n2 = +
(4)
Similarly, it f (t) : l,lt and h(t'S: t, then h(t) = f (t) for all t in [1, 2]. Becauseh and / are continuous on l'1,,21,they are integrable there; and again using Theorem 7.4.8, we obtain
or/ equivalently, 1 > ln?
(s)
Combining (4) and (5), we get 0.5 2 or a 12, the straight line ;r: a intersects the graph in two points. In Fig. 9.8.2 we have a sketch of the graph of Eq. (2). Here we see that any vertical line intersects the graph in only one point. Equation (2) does not define x as a function of y because for each value of y gteatet than r., two values of r are determined. In some cases,an equation involving x and y will define both y as a function of r and x as a function oly, as shown in the following illustration. o TLLUSTRATToN1.: Consider
Figure9.3.2
Q)
the equation
Y : )c3+'l'
(3)
If we let f be the function defined by
f(x):f+r
(4)
then Eq. (3) can be written as y : f(x),and y is a function of.x. lf.we solve Eq. (3) for r, we obtain
xw
(5)
and if we let g be the function defined by
sQ)w then Eq. (5) may be written as x:
(6) g Q) , and r is a function of y.
o
The functions / and g defined by Eqs. (4) and (6) are called inuerse functions.wealso say that the function g is the inoerseof the function/and that/ is the inaerseof g. The notation f 1 denotes the inverse of function /. Note that in using 1. to denote the inverse of a function, it should not be confused with the exponent 1. we have the following formal definition of the inverse of a function.
9.3 THE INVERSEOF A FUNCTION
9.3.1 Definition
If the function / is the set of ordered pairs (x, y), and if there is a function fr such that : ::.::...::j::l::::.:::i::i.r':....r......:..:,.:,,::::r,:::i::ii:..:r': ,: ': . . . . : : .r.i: .r i::::..::..:1..: .1 . :....!..r. ,:. , . :. .,:.::: i: ::. .:i ir
i:::::::!:l:l:ii::::::i::i::i::rli::iil:li;:l;::tji:::::irj::!:::iil]j':::::ii::i::::S:::ii::::
.:.... ::::..:.:
i......:::.......
:. i
:::::i:::;:::i:::i
(7)
then/1, which is the set of ordered pairs (y, r), is called the inaerseof the function f , and /and fr arecalled inaersefunctions. The domain of /1 is the range of f , and the range of /1 is the domain of /. Eliminating y between the equations in (7) , we obtain
x  f '(f (x)),
(8)
and eliminating r between the same pair of equations, w€ get
v:f(f'(v))
(e)
So from Eqs. (8) and (9) we seethat if the inverse of the function/ is the function /t, then the inverse of the function /1 is the function /. Becausethe functions / and g defined by Eqs. (4) and (5) are inverse functions, fr can be written in place of g. We have from (a) and (6) (10) and f'(x):9i1 f(x):f+l Sketches of the graphs of functions / and f defined in (10) are shown in Figs. 9.3.3and 9.3.4,respectively. We observe that in Fig. 9.3.3the function / is continuous and increasing on its entire domain. The definition of an increasing function (Definition 5.1.1)is satisfiedby f for all valuesof r in its domain. We alsoobserve in Fig. 9.3.3 that a horizontal line intersects the graph of / in only one point. We intuitively suspect that if a function is continuous and increasing, then a horizontal line will intersect the graph of the function in only one point, and so the function will have an inverse. The following theorem verifies this fact.
Figure9.3.3
Figure9.3.4
9.3.2 Theorem
Supposethat the function / is continuous and increasing on the closedintewalla, bl. Then (i) / has an inverse /1, which is defined on If @), f (b)l; (ii) /' is increasineon If @), f (b)l; (iii) /1 is continuoason lf(a), f(b). PRooFor (i): Since/ is continuous on Ia, bl, then if k is any number such that f(a) < k < f(b), by the intermediatevalue theorem (7.5.1), there exists a number c in (a, b) such that /(c) : k. So if y is any number in the closed interval lf (a) , f (b)1, there is at least one number x in la, bl such that y : f (x). We wish to show that for each number y there corresponds only one number x. Supposeto a number y, in lf (a),l(b)l there correspond
398
LOGARITHMICAND EXPONENTIAL FUNCTIONS
two numbers x, and x, (x, # 12) in [a, b] such that !r: /(rr). Then, we must have f(x,):
f @r) and y, : (11)
f(xr)
Becausewe have assumedx1 * x2, either xr I xz ot x2 < xr. lt q 1 xr, because/ is increasing on la, bl, it follows that /(r1) < f(xr); this contradicts (11). lf. xr1r1, then f(x") < f(x), and,this also contradicts (11). ThelefoJe,our assumption that x, # xris false, and so to eachvalue of y in Lf(a), f (b)l there corresponds exactly one number x in [a, b] such ihat y : f (x). Therefore,/ has an inverse /r, which is defined for all numbers in ff(a), f(b). pRooFor (ii): To prove that /1 is increasing on lf(a), /(b)], we must show that y, and Az are two numbers in [/(a), /(b)] such that y1 I y2, if then fl(yr) < f,(yr). Because/1 is defined on lf(a), f(b), theie exist numbers r, and x2 in fa, bl such that y, : f (xr) ^dyr: f (xr). Therefore,
f
tQ) f'(f (x')) : xr
(12)
f
t(yr) 
(13)
and f
'(f
(xr)) : x,
If x2 < 11, then because / is increasing on fa,b), f (x) < (x1) or, f 'th"n equivalently, azl yt. But ur 0 for all r in (0, c),by Theorem5.l.3 (i), f is increasingon [0, c]; and so by Theorep !.1.!, /, has an inverse on [L, c2ll. The inverse function is given by
(2r) Similarly, because/2 is continuous on [c, 0] and becausef L@) < 0 for all x in (c,0), by Theorem5.1.3(ii), /, is decreasihgon [c, 0]. Hence, by Theorem 9.3.3,fzhas an inverse on [1, ,,  tl, and the inverse function is given by
fr'(y):\m Figure 9.3.5
(22)
Becausethe letter used to represent the independent variable is irrelevant as far as the function is concerned, r could be used instead of u in (21) and (22), and we write fr'(r)  \m
and r in tl , c2 1]
(23)
and
Qfu,a)
\ffi frt(r) 
o Rb, u)
and r in tl , cz 1]
(24) o
In Fig. 9.3.5 there are sketchesof the graphs of /, and its inverse fil, as defined in (19) and.(23), plotted on the same set of axes. In Fig. 9.3.6 there are sketchesof the graphs of f2 andits inverse/21, as defined in (20) and (24), plotted on the same set of axes. It appears intuitively true from Fig.9.3.5 that if the point Q@, o) is on the graph of fi, then the point R(o, u) is on the graph of fr'. From Fig. 9.3.6 it appears to be true that if the point Q@, a) is on the graph of fz, then the point R(o, u) is on the graph of fr'. In general, if Q is the point (u, a) and R is the point (2, z), the line
9.3 THE INVERSEOF A FUNCTION
segment QR is perpendicular to the line y: t and is bisected by it. We state that the point Q is a reflectionof the point R with respect to the line U : x, and the point R is a reflectionof the point Q with respect to the line : A : x.If r and y are interchanged in the equation y f(x), we obtain the x: (y) equation graph of the the and x: equation , f (y) is said to be a f refleciionof thegraph of the equati ony : f (x) with respectto the line y : 7 Becausethe equation x: f (y) is equivalent to the equation y : fr (x) , we conclude that the graph of the equation y: ft(x) is a reflection of the graph of the equationy : f (x) with respectto the line y : r. Therefore,if a function / haq an inverse /1, their graphs are reflections of eachother with respectto the line y: v. EXAMPLEL: Given the function / is defined by
f(x):# t find the inverse f if it exists. Draw a sketchof the graPh of f , and if 1r exists,draw a sketchof the graph of 1t on the sameset of axes as the graPh of f . v y 
f'(x) : /'U
v:fk)
so1urroN: f '(x) : Sl@ L)'. Because/is continuous for allvalues of r except7, andf' (r) < 0 if x *'l', itfollows fromTheorem9.3.3that/hasan inveise if r is any real number excePt1' To find/r,lety : f (x), and solve for x, thus giving x: ft(y).So we have 2x*3 u : T " x L
rYY:2x*3 x(y2):y*3 v*3
i:
='
a2
Thus,
f'(v):
y+3 y2
or, equivalently, v

f'(xl )
f'(x) The domain of fr is the set of all real numbers excePt2. Sketchesof the graphs of f and fL are shown in Fig. 9.3.7.
Figure 9.3.7
The following theorem bxpressesa relationship between the derivative of a function and the derivative of the inverse of the function if the function has an inverse. 9.3.4 Theorem
closedinSupposethat the function / is continuous and monotonic on the ' (r) * Ofor :f bltr'df on!a, (x).If differentiable tewal1a,bT,andlety /is any r in la, bl, then the derivative of the inverse function /1, defined by *:1r(A), is given by
402
LOGARITHMIC AND EXPONENTIAL FUNCTIONS
pRooF: Because is continuous and monotonic on / la, b], then by Theorems 9.3.2 and 9.3.3,/ has an inverse which is continuous and monot o n i co n [ f ( a ) , f ( b ) ] ( o r l f ( b ) , f ( a ) l i f f ( b ) < f ( a ) ) . If r is a number in la, bf, LetAr be an increment of.x, Ax # 0, such that x * Lx is also in [a, b]. Then the corresponding increment of.y is given by
Ly:f(x*Ar) f(x)
(25)
Ly * 0 sinceAr * 0 and/is monotonicon la,bf; that is, either onla,bl f(x+ Lx) < f(x) or f(x+ Lx) > f(x) If r is in [a,b] andy_: f (x),theny is in [/(a) , f (b)] (or lf (b),f (a)l). Also,if x t Ax is in [a, b], theny + Ly is in [/(a) ,f (b)] (or lf (D; @))) f because y + Ly: f(x* Ax)by (25).So x: fr(y)
e6\
)c+ Ar: f_r(y + Ly) It followsfrom (26) and (27) that Ar: ft(y + Ly) f10)
(27)
and
(28)
From the definition of a derivative,
+ Lv)  f '(y) Dox: lim f'Q substit"tiJ;"*
es) ^::(2s) into the aboveequatior,we set A,x
D o x : lim
;;; f(x*Ar) f(x)
and becauseAr + 0, Dox:
lim Ag 0
f(x*Ar) f(x)
(2e)
Lx Beforewe find the limit in (29) we show that under the hypothesis of this theorem "Lx> 0" is equivalent to ,'Ly + 0.,, First weshow that lim Ar: 0. From (28) we have As0
hm lf'!+ In at: As0
ArO
ty) f,(y)l
Because/r is continuous o n l f ( a ) , f ( b ) l ( o r l f ( b ) , f ( a ) l ) , t h el i m i to n the right side of the above equation is zero. So
ln Ar0
(30)
9.3 THE INVERSEOF A FUNCTION
that li$ AV:0. From (25) we have Now we demonstrate lim AY: lim lf (x + Lx)  f(x)l
Ar0
Ato
Because/ is continuous on Ia, bl, the limit on the right side of the above equation is zeto, and therefore we have
o Jn LY
(31)
From (30) and (31) it follows that Ar+0
ifandonlyif
(32)
Ly0
Thus, applying the limit theorem (regarding the limit of a quotient) to (29) and using (32), we have 1,
Dox:
Because/ is differentiable on la, bl, the,limit in the denominator of the above is/'(r) or, equivalently,DrA, and so we have
EXAMPLEZ: Show that Theorem g.3.Aholds for the function of Example L.
solurroN: lf y: f (x) : (2x + g)/(r  1), then
D,A:f'(*):C+ at all Because/,(r) exists if.x * 1,/is differentiable and hence continuous # l. x forall is_decreasing L,andsof * < ir, b x # l.F.irth"r*o"",f,(x) b] interval any for holds [a, 9.3.4 Theorem of esis Therefore, tt t ypott " 1. number the which does not contain In the solution of Example 1 we showed that x : f  r ( v\ )r :' Y + 3 y2 Computing Dux ftom this equation, we get
Dox:(f\'Q):G+ If in the above equation we let ,: 5
(2x + 3)/(r  1), we obtain
404
AND EXPONENTIAL FUNCTIONS LOGARITHMIC
EXAMPLE3: Given f is the function defined by f (x) : xs * x, determine if / has an inverse. If it does, find the derivative of the inverse function.
'(x) :3xz + 1. Therefore, ' (x) ) soLUrIoN: Because 0 for f (x) : x3* x, f f all real numbers, and so f is increasing on its entire domain. Because/ is also continuous on its entire domain, it follows from Theorem 9.3.2 that f has an inverse fr
L e t y  f ( x ) a n d t h e nx  f  ' U ) .
S o by Theorem9.3.4,
r\17
uax:W
g x z+ l
9.3 Exercises In ExercisesL through 8, find the inverse of the given function, if it exists, and determine its domain. Draw a sketch of the graph of the given function, and if the given function does not have an inverse, show that a horizontal line intersects the graph in more than one point. If the given function has an inverse, draw a sketch of the graph of the inverse function on the same set of axes as the graph of the given function.
1. f(x) : x3
2. f (x) : x2+ 5
3. f (x)
4 . f ( x ) ( x + 2 ) '
5. f (x)
6. f(x)
7. f (x)  2lxl+ x
8. f(x)
In Exercises9 through 14, perform each of the following steps: (a) Solve the equation for y interms of r and expressy as one or more functions of r; (b) for each of the functions obtained in (a) determine if the function has an inverse, and if it does, determine the domain of the inverse function; (c) use implicit differentiation to find D,y and.Duxand determine the values of r and y for which D,y and.Dux arereciprocals. 9. x2+A,:9 12.9y' 8130
10.x 2  4 Y '  t 6
11.
13.2x23xy+L0
1 , 4 . 2 x 2 * 2 y *L : 0
In Exercises 15 through 20, determine if the given function has an inverse, and if it does, determine the domain and range of the inverse function.
L 6 .f ( x )  ( x + 3 ) '
1 7f.( x ) x 2 + , r ) 0 x
19. f(x) : x5 * xB
20.f(x)
*x
FUNCTION 9.4 THE EXPONENTIAL
21. Let the function / be defined by f (x):
I
ttr=V
Jo
df. Determine if / has an inverse, and if it does, find the derivative of
the inverse function. : (r * 5)/(r + k) will be its own inverse' 22. Determine the value of the constant k so that the function d.efined by /(r) rfx < L fx it L < x < 9 23.Givenf (x)  l*' ifx>9 lzrtfr t Provethat / has an inversefunctionand find f (x) . 24. Given that the function / is continuous and increasing on the closed intervalla,bl.AssumingTheorem9'3'1(i) prove/1 is continuour i.o* the right atf(a) and,continuous from the left atl(b). 25. Prove Theorem 9.3.3 (ii).
Prove Theorem 9.3.3(i).
and (ii),
27. Prove Theorem 9.3.3 (iii).
Show that the formula of Theorem 9.3.4 can be written as
( f  ' ) 'v(' r ) :' (#f  t ( r ) ) f
29.use
the formula of Exercise 28 to show that
(f')"(r):ffi 9.4 THE EXPONENTIAL FUNCTION
9.A.1Definition
Becausethe natural logarithmic function is increasing on its entire domain, then by Theorem 9.3.2 it has an inverse which is also an increasing function. The inverse of the natural logarithmic function is called the "exponential function," which we now define' The exponentialfunction is the inverse of the natural logarithmic function' and it is defined by (1)
The exponential function "exp(r)" is read as "the value of the exponential function at r." 'Becausethe range of the natural logarithmic function is the set of all real numbers, the domain of the exponential function is the set of all real numbers. The range of the exponential function is the set of positive numbers because this is the domain of the natural logarithmic function' Becausethe natural logarithmic function and the exponential function are inverses of each other, it follows from Eqs. (8) and (9) of sec.9.3 that ln(exP(x)) : x
(2)
 x exP (ln r)
(3)
and Because 0 : ln L, we have exp 0  exP (ln 1)
406
AND EXPONENTIAL FUNCTIONS LOGARITHMIC
which from (3) gives us exP0:L We now theorems.
9.4.2 Theorem
(4) state some properties
of the exponential
function
AS
lf. a and b are any real numbers, then exP(a* b) : exp(a)' exp(b) PRooF: Let A:
exp(a), and so from Definition 9.4.1 we have
a:ln A Let B: exp(b), and from Definition 9.4.1.it follows that b:lnB From Theorem9.1.3 we have lnA+lnB:lnAB
(Z\
Substituting from (5) and (5) into (Z), we obtain a * b:ln AB So exp(a * b) : exp(ln AB)
(s)
From Eq. (3) it follows that the right side of Eq. (8) is AB, and so we have exp(a*b):An Replacing A and B by their values, we get exp(a* b): exp(a) . exp(b) 9.4.3 Theorem
r
lI a and b arc any real numbers, exP(a b) : exp(a) : exp(b) The proof is analogousto the proof of Theorem 9.4.2,where Theorem 9.1.3is replacedby Theorem 9.1,.4.rt is left as an exercise(seeExercise1).
9.4.4 Theorem
If a is any real number and r is any rational number, then exp(ra): [exp(a)]" pRooF: If in Eq. (g) x : lexp(a)1,,we have lexp (a)] " : exp{lnfexp(a)1,] Applying Theorem 9.1.5to the right side of the above equation, we obtain
FUNCTION 9.4 THE EXPONENTIAL
[exp@)]r : exP{rln[exP(a)] ] But from Eq. (2),ln[exp @)]  n, and therefore [exp(a)]':
T
exp(ra)
We now wish to define what is meant by a', where a is a positive number and r is an irrational number. To arrive at a reasonabledefinition, consider the casea,, where a ) 0 and r is a rational number. we have from Eq. (3)
(e)
sr : exp[ln (a')] But by Theorem9."!,.5,lna'
rln a; so from Eq. (9)
61r: exp(r ln a)
(10)
Becausethe right side of Eq. (10) has a meaning if r is any real number, we use it for our definition. 9.4.5 Definition
g.4.6 Theorem
lf a is any positive number and r is any real number, we define
lf. a is any positive number and r is any real number, lna':xlr':.a PRooF: From Definition 9.4.5, a' : exp(x ln a) Hence, from Definition 9.4.1, we have ln n": xln a
I
Following is the definition of one of the most important numbers in mathematics. 9.4.7 Definition
The number e is defined by the forimula e: exp L The letter "e,, was chosen because of the swiss mathematician and physicist Leonhard Euler (77071783)' number; that is, it cannot be exThe number e is a transcend.ental pressed as the root of any polynomial with integer coefficients. The number zr is another example of a transcendentalnumber. The proof that e is transcendentalwas first given inl873,by Charles Hermite, and its value
AND EXPONENTIAL FUNCTIONS LOGARITHMIC
can be expressed to any required degree of accuracy.In Chapter 16 we leam a method for doing this. The value of e to seven decimal places is 2.71,8281,8. 9.4.8 Theorem
ln e: L. PRooF' By Definition 9.4.7, eexpL Hence, ',
ln e  ln(exp 1)
(11)
Becausethe natural logarithmic function and the exponential function are inverse functions, it follows that ln(exP 1) : L
(r2)
Substituting from (L2) into (11), w€ have lneL 9.4.9 Theorem
exp (x) : st , for all values of x. PRooF' By Definition 9.4.5, st : exp (r ln e)
(13)
But by Theorem 9.4.8, ln e  1. Substituting this in (13), we obtain sr : exp (r)
From now on, we write e'in place of exp(r), and so from Definition 9.4.1we have
(r4) We now derive the formula for the derivative of the exponential function. Let
a:e' Then from (14) we have xlny
(ls)
Differentiatirg get
1:1
v
D"U
on both sides of (15) implicitly with respect to x, we
FUNCTION 9,4 THE EXPONENTIAL
409
So,
D"a y Replacing y by e*, we obtain ( 16,)
D r(e") : sr
If.u is adifferentiable function of.x,7t follows from (L6) and the chain rule that (17)
It follows that the derivative of the function defined by f(x):ke', where k is a constant, is itself. The only other function we have previously encountered that has this property is the constant function zero; actaally, this is the specialcaseof f(r) : ke' when k: 0. It can be proved that the : ke' most general function which is its own derivative is given by f(x) (seeExercise35). EXAMPLE
Given Y : srtrz find
+
dyI dx.
solurroN;
From (17)
: H:"ttz (i)
/grlrz
x3
From (I7) we obtain the following
indefinite
integration formula: (18)
EXAMptn 2:
l4a*
J\E
Find
Let
u  t E ; then du
r
2Jeudu 2e'+C 2en+C : Becausefrom (14), e' : A if and only if x : ln y, the graph of y t' i" : identical with the graph of tt: ln y. So we can obtain the graph of y et by interchanging the r and y axesin Fig' 9.22(seeFig' 9'a'1)' The gruph ol y : et carrbe obtained without referring to the graph of iogarithmic function. Because the range of the exponential natuial the set of all positive numbers, it follows that e' > 0 for all valis the function graph lies entirely above the r axis' D,A : e" ) 0 for all r; the r. So ues of so the function is increasing for all x. D r'y : e' ) 0 for all r; so the graph is concave upward at all Points.
410
FUNCTIONS AND EXPONENTIAL LOGARITHMIC
We have the followi^g
two limits:
(le)
lim stfm X
*@
and
e*: o
"tlq
(20)
The proofs of (19) and (20) are left as exercises(see Exercises44 and 45). To plot somespecificpoints use the table in the Appendix giving powers of e.
F i g u r e9 . 4 . 1
ExAMPr,u3: Find the area of the region bounded by the curye y : e', thecoordinate axes,and the line x: 2. Y:
The region is shown in Fig. 9.4.2.
solurroN:
tip
l lAal l o f , 1
T
er dx
e, (2, ezl
et
e2
e0
: e2 1 From a table of powers of e, we see that the value of e2to two decimal places is 7 .39. Therefore, the area of the region is approximately G.gg square units.
Ei Air
Figure 9.4.2
The conclusionsof Theorems9.4.2,9.4.3,and9.4.4are now restatedby using e" in place of exp(r). If. a and b are any real numbers, then ea+b: eo , eb eob: €ra:
ea + eb (sa)r
where r is any rational
number.
9,4 THE EXPONENTIAL FUNCTION
Writing Eqs. (2) , (3), and (4) by substituting have
411
er in place of exp (x) , we
ln(e") : )c 
,lnx
X
and eo:t
ExAMPrn 4: y :
t2r*ln
find D *A.
Given n
t2r*ln r :
soLUTroN '
t2r trnr :
e'" (x). So
y  xez' Therefore, D rU : s2r* 2xe2*
In Definition9.A.7, the number e was defined as the value of the exponential function at L; that is, e: exp L. To arrive at another way of defining e, we consider the natural logarithmic function f(x) :ln x '(x) :1/r; hence, We know that the derivative of / is given by f L. However, let us apply the definition of the derivative to find f,(t): /'(1). We have ' ): !f ' (\ L
:
r(1 l i m E !_L*)._lG)_ Lx
Aro
lim Ar0
_ lim A.r 0
Therefore, 1
lim ;* ln(1+ Ar) : 1 aro AI Replacing Ar by h, we have from the above equation and Theorem9'4.6 (27\ limln(l *h)un, ft0
Now, because the exponential function and the natural logarithmic function are inverse functions, we have lim (L * hlun: lim exp[ln(l * h)trn1 h0
hO
Q2)
AND EXPONENTIAL FUNCTIONS LOGARITHMIC
Becausethe exponential function is continuous and lim ln(l + h)unexists and equals L as shown in Eq. (21))rwe can apply H"orem right side of Eq. (22) and get
lg
2.5.5to the
(1+ h)trnexp h(l + h)',of: exp1 fm
Hence,
(23) Equation (23) is sometimes used to define e; however, to use this as a definition it is necessaryto prove that the limit exists. Let us consider the function F defined by F(h)(1 +h)un
(24)
and determine the function values for some values of h closeto zero. These values are given in Table 9.4.1. Table9.4.1 h
1 0.5 0.05 0.0L 0.001 0.0010.0L 0.05 0.5
F(h):(1 +hyrn
2 2.25 2.65 2.702.71,692.7196 2.73 2.79 4
Table9.4.1leadsus to suspectthat lim (1 + lr; ru ir probably a number that lies between 2.7169 and2.T191.As previously mentioned, in Chapter 16 we leam a method for finding the value of e to any desired number of decimal places.
Exercises 9,4 1. Prove Theorem9.4.g. A
1. 1
2. Draw a sketch of the graph of y :
3. Draw a sketch of the graph of A  el"l
es".
In Exercises4 throu gh 14, find dyl dx. 4. a:e5'
5. y :
g9tz
8' a:e"* \.trt. y: 1'4. y :
*lrlrna' gnlt/A+rz
I 9.4 THE EXPONENTIAL FUNCTION
413
In Exercises1.5through 1.8,find D*y by implicit differentiation.
r:.
: [F. e' * eu et+u
16. ye'" * xezu 1
tt[. y'e'* * xyt: I \/ 1.9through26, In Exercises
18. €u: ln(r3 + gy)
r
evaluate the indefinite
rl4't
19. 1 ezsr dx
20. l tzr+r dx
22. lr"r" 4*
23' J ,ks"Ydx
J
J f
,]t"'
integral.
J f
plr
?26. I#.
oZc
EY.J, Jl # d x ex+J
In Exercises27 through 30, evaluate the definite integral.
27. 29.
fr
J. f2
J,
?8. e2dx xea" dx
30.
In Exercises 31 and 32, find. the relative extrema of f , the points of inllection of the graph of.f , the intervals on which / is increasing, the intervals on which / is decreasing, where the gfaph is concave upward, where the graph is concave downward, and the slope of any inflectional tangent. Draw a sketch of the graph of /. 3 1 .f ( x ) : x e  c
&}f(*)e*'
d. fitta an equation of the tangent line to the curve !: e" that is perpendicular to the line 2r !:5. (0, 1) and (1, e). Cl fitra the area of the region bounded by the curve y: e' and the line through the points : 1. If every plane section per35. A solid has as its basethe region bounded by the curves ! : e' and A : e" and the line r pendicular to the r axis is a square find the volume of the solid. : 35. Prove that the most generalfunction that is equal to its derivative is given by /(r) : ke". (rrrNr: Let y /(x) and solve the differential equation dVldx: y.) id* lf p lblftz is the atmospheric pressure at a height of h ft above sea level, then p :2ll6eo'owo31t". Find the time rate of change of the atmospheric press.rr" outside of an airplane that is 5000 ft high and rising at the rate of 160 ftlsec. 3g. At a certain height the gaugeon an airplane indicates that the atmospheric pressrre is 1500lb/ff. Applying the formula of Exercise37, ipproximat by differentials how much higher the airplane must rise so that the pressure will be 1480 lblft2. Use differentials to find the aPProxg9. lllft is the length of an iron rod when f degreesis its temperature, then I : 60e0'00001'. l'0. imate increase in I when f increasesfrom 0 to (F.
A simple electric circuit containing no condensers,a resistanceof R ohms, and an inductance of L henrys has the electromotive force cut off when the current is 16amperes. The current dies down so that at f secthe current is i amperes and i:
loe 0 t h e r e e x i s t s a n N < 0 s u c h t h a t e s < e w h e n e v e r r ( N . 46. Draw a sketch of the graph of F if F(h) : (7 t h)tth.
9 q NTHFR FXPONFNTIAL AND I OGARITHMIC FTINCTIONS
From Definition 9.4.5, we have qx :
,"rln
u
(1)
The expression on the left side of Eq. (1) is called the " exponential function to the base a."
Definition
If a is any positive number and r is any real number; then the function / defined by f(x) : 6t is called the exponentialfunction to the basea. The exponential function to the base a satisfies the same Properties as the exponential function to tlEbase e. . TLLUSTRATTON 1.: If x and y are any real numbers and n is positive, then from Eq. (1) we have frr7a

gtlnarulna
_
grlna*ulna
:
g(t*u)ln
_
a
Or*U
From Illustration 7 we have the property ar aa 
,r*a
We also have the followirg At+fr.u:ara
(2) properties: (3)
9.5 OTHEREXPONENTIAL AND LOGARITHMIC FUNCTIONS 415 (a")u 
ata
(4)
(ab)' :
arbt
(s) (6)
ao: I
The proofs of (3) through (5) are left as exercises (see Exercises 1 to 4). To find the derivative of the exponential function to the base a, we set ac  ecrnaand apply the chain rule. We have At:
exlna
s c r n aD r ( x l n a )
Dr(a'):
: errno(lna) :4'ln
a
Hence, itu is a differentiable
function of r, (7)
o TLLUSTRATToN2: If y :
D*A:3"
Jr2, then from formula
ln 3(2x) z(ln 3)x3"'
(7) we have
o
From (7) we obtain the followi.g indefinite integration formula: (8)
Find dx
SOLUTION: I \4OT dX: I 103't2dx. Let Lt: Ex; then du :8 dx; thus JJ ? du: dx. We have then ff
dx: I to & du  1or"rz JJ :5 2 1' 0 " r  C lt:r L0
:
ExAMPrn 2: Draw sketches of the graPhs of y  2" andY :2* on the same set of axes. Find the area of the region bounded bY these two graphs and the line x: 2.
2 . t03Jft2
31,,L0
, C
The region is soLUrIoN: The required sketches are shown in Fig. 9.5.'1.. shaded in the figure. If A square units is the desired atea, we have A
lim i:r llallo
416
FUNCTIONS AND EXPONENTIAL LOGARITHMIC
(2  2") dx (2, 4)
(t i , zE')
2t
ln 2
2r
12
ln? Jo
\ ((i,zti'1
F i g u r e9 . 5 . 1
We can now define the "logarithmic function to the base a', if a is any positive number other than 1. 9.5.2 Definition
'1., If a is any positive number except the logarithmicfunction to the basea is the inverse of the exponential function to the base a; and we write
(e) The above is the definition of the logarithmic function to the base a usually given in elementary algebra; however, (9) has meaning for y any real number becauseaa hasbeen precisely defined. It should be noted,that 7fa: e, we have the logarithmic function to the base e, which is the natural logarithmic function. log" x is read as "the logarithm of.x to the base a.', The logarithmic function to the base a obeys the samelaws as the natural logarithmic function. We list them. logo Qy)  logo x + Iogo y logoQ:U):logoxlogoy logoL :0 logo xa : y logo x
(10) (11) (12) (13)
The proofs of (10) through (13) are left as exercises (see Exercises
5 to 8). A relationship between logarithms to the base a and natural logarithms follows easily. Let Y :logo x
9.5 OTHEREXPONENTIAL AND LOGARITHMIC FUNCTIONS 417
Then aa:)c ln aa: ln r ylnalnr Y
lnr l"a
Replacing y by logo x, we obtain
ll::r,,., ::rili.iiuiiiniii:ii.:i:i...iiirii:iitixi:liiiii,iiirli.,..ilillir::iiiffiiri ,,*fiffitff't :
r ' : : 1 r : : : : : r : : : r : : : t : : : : : . . : : r'
,tj : : : : . : : : : : : : : . : : :i :. :i : .: j: : : : . : : : :. . . :
(14)
.
Equation (14) sometimes is used as the definition of the logarithmic function to the base a. Becausethe natural logarithmic function is continuous at all x > 0, it follows from (14) that the logarithmic function to the base a is continuous at all r ) 0. If in (14) we take x: e,wehave
ot,
(ls) We now find the derivative of the logarithmic function to the base a. We differentiate on both sides of Eq. (14) with respect to r, and we obtain
D , ( l o g o x ) : + D , ( l nw lna
r)
D*(logox):#'+
(16)
Substituting from (15) into (16), we get e D r(logox) :lo8o x
(rz7
If u is a differentiable function of x, we have (18) Note that if in (18) we take n  e, we get
D*(rog,u) :
lo? t D*u u
FUNCTIONS LOGARITHMICAND EXPONENTIAL
418
or, equivalently, 1
D"(lnu):!D,u which is the formula we had previously for the derivative of the natural logarithmic function. EXAMPTT 3:
Y
Given
soLUrIoN: Using (11), w€ write
: logto x * L
y:
xz +'/',
l o g t o ( r + 1 )  l o g r o ( x '+ 1 )
From (18) we have
find dy ldx.
dy
e _Iogro e . )y _logn x+1,dx )c2+L'zL
Becauser" has been defined for any real number n, we can now prove the formula for finding the derivative of the power function if the exponent is any real number. 9.5.3 Theorem
rf n is any real number and the function / is defined by wherer)0 f(x):x" then f'(x) : ny"r PRooF: Lety: Y
rn. Then from (1) :
,nrnr
Thus, DrU 
,nrnt ,nln
ln r)
t
vn'L x  nxnr
I
Theorem 9.5.3 enablesus to find the derivative of a variable to a constant power. Previously in this section we leamed how to differentiate a
9.5 OTHER EXPONENTIALAND LOGARITHMICFUNCTIONS
4I9
constant to a variable power. We now consider the derivative of a function whose function value is a variable to a variable power. The method of Iogarithmic differentiationis used and is illustrated in the following example. EXAMPTr 4:
Given
y  x', find
dyldx.
solurroN: BecauseA: f , then lyl : lr'1. We take the natural logarithm on both sides of the equation, and we have ln lYl : ltt lr"l or, equivalent$,
ltt lYl: r ln lrl Differentiating on both sides of the above equation with respect to r,
+ * Lx
H:a}nl'l
+ 1)
 r'(ln lrl + 1)
9,5 Exercises In Exercises1 through 4, prove the given ProPerty if a is any positive. number and x and y are any real numbers. L.
'A*
+ Aa
3. (ab)t :
2. (a")u  ata
Asa
4 . a o: ' L
arbt 8, prove
5 through
In Exercises
the given
ProPerty
if a \s any positive number
and x and y arc any positive numbers.
5. logok : !) : logo x  logo y
5. logo(rV)  logo )c+ log" y 7. Iogo xu  y Log" x
8 . l o g oL : 0
'(x). In Exercises9 through24, find f , ' 9 rf.k \  3 5 '
i 1 0 .f ( x ) : $  B t logrox 1 ' 3 f. ( x ) : ' x
;r'
I?. f(x)
 (x3+3)2'*
l s .f ( x ) : @ 18. f (x) 
vtnx
21.f(x\:1ge' 24. f (x)  (ln x)tn r
i i
t"t. f(x)25t34cz
14. f ( x ) : l o g r o
#
r) ] 16. f (x)  logo[1o8o(log,
17. f (x) : logro[logto(r+ 1) ]
1 9 .f ( x ) : x 6
20. f ( x ) :
22.f(x):(x)""
23. f (x) : (4e"'1r"
xstz
FUNCTIONS AND EXPONENTIAL LOGARITHMIC In Exercises 25 through 32, evaluate the indefinite integral.
2s. *" a* [
27.
28. Iu**,(2xr+L)dx
'30.
I
a"e"dx
Ir"rnr(lnr+1)dx
31. *r,'3," dx I 33. Find an equation of the tangent line to the curve y24v /2x at (4,2). 34. A partide is moving along a straight line according to the equation of motion s  A2r' + B2ht,where s ft is the directed distance of the particle from the starting point at f sec. Prove that if a ft/sec2is the accelerationat f sec, then a is proportional to s. A company has leamed that when it initiates a new salescampaign the number of salesper day increases.However, the number of extra daily salesper day decreasesas the impact of the campaign wears off. For a specific campaign the company has determined that if S is the number of extra daily salesas a result of the campaign and r is the number of days that have elapsed since the campaign ended, then
s: 1000. 3'/2 Find the rate at which the extradaily salesis decreasingwhen (a) x: 4 and,(b) r:
10.
In Exercises 36 and 37,usedifferentialsto find an approximatevalueof the givenlogarithmandexpressthe answerto three decimal places. 36. logro 1..0L5
37. log,o 997
38. Draw sketchesof the graphs of.y : lo916r and y :ln
r on the same set of axes.
39. Given:f(x) :t@" * a"). Prove that f(b * c) * f(b  c) :zf(b)f(c). tl(). A particle moves along a straight line according to the equation of motion s : fll, where s ft is the directed distance of the partide from the starting point at f sec. Find the velocity and accelerationat 2 sec.
9.5 LAWS OF GROWTH AND DECAY
The laws of growth and decay provide applications of the exponential function in chemistry, physics, biology, and business. Such a situation would arise when the rate of change of the amount of a substancewith respect to time is proportional to the amount of the substancepresent at a given instant. This would occur in biology, where under certain circumstances the rate of grovWh of a culture of bacteria is proportional to the amount of bacteria present at any given instant. In a chemical reaction, it is often the casethat the rate of the reaction is proportional to the quantity of the substancepresen! for instance, it is known from experiments that the rate of decay of radium is proportional to the amount of radium present at a given instant. An application in business occurswhen interest is compounded continuously. In such cases,if the time is represented by f units, and A units represents the amount of the substancepresent at any time, then dA 7l:
Ktt
9.6 LAWS OF GROWTHAND DECAY
where k is a constant. If A increasesasf increases,then k > O,and we have lhe law of naturalgrowth.If A decreasesas f increases,then k ( 0, and we have the law of natural decny.In problems involving the law of natural decay, thehalf life of a substanceis the time required for half of it to decay. L: The rate of decay of. EXAMPLE is radium proportional to the amount present at any time. If 50 mg of radium are Presentnow, and its half life is 1.690years,how much radium will be Present100 years from now?
Let f : the number of years in the time from now; A : the number of milligrams of radium Present at f years. we have the initial conditions given in Table 9.6.1.The differential equation is
solurroN:
Separatingthe variables, we obtain
9.6.1 Table t
0
1590
100
A
60
30
Aro,
#kdt Integrating, we have
lnlAl:H+e lAl:ekt+eF'ekt Letting eE: C, we have lAl : Cro', and becauseA is nonnegative we can omit the absolutevalue bars around A, thereby giving us A:
Cekt
BecauseA:
50 when f : 0, we obtain 50 : C. So (1)
A  50ekt  'J..690, or we get 30 : 6l0e16e0k Because A  30 when t 0.5 :
gr6e0k
So ln 0.5 : 1,590k and
llr9:5 : 0.0004L0 k  1690 :o='929r 1690 Substituting this value of k into Eq. (1), we obtain A

50eo.o00410t
When t  L00, A
Aroo,and we have
Aroo:50eo'0410:57.5 Therefore, there will be 57.6mg of radium present 100 years from now.
422
FUNCTIONS AND EXPONENTIAL LOGARITHMIC
ExAMPtn 2: In a certain culture, the rate of growth of bacteria is proportional to the amount present. If there are 1000 bacteria present initially, and the amount doubles in L2 min, how long will it take before there will be 1.,000,000bacteria present?
The differential equation is the same as we had in Example above, the general solution is
Table9.6.2
t A
0
soLUrIoN: Let f : the number of minutes in the time from now; A: the number of bacteria present at f min. Even though by definition A is a nonnegative integer, we assumethat A can be any nonnegativenumber for A to be a continuous function of f. Table9.6.2 gives the initial conditions. The differential equation is
L2
T
L,000 2 ,000 L,000,000
hence, ds
A  Cekt
When t  0, A:
L000;hence,
1,000,which gives
A:1000ekt From the condition that A: erzk 2
2000when f :12, we obtain
and so k: #ln2:0.05776 Hence, we have A
1000eo.o5776t
Replacing t by T and A by 1,000,000,we get 1,000,000 1000e0'05776r e0.05776?:1000
0.057767: ln 1000 ln 1( ,: O ffi:179.6 Therefore, there will be 1,000,000bacteria present in t hr, 59 min, 36 sec. ExAMPrn 3:
Newton,s law of cooling states that the rate at which a body changes temperature is proportional to the difference between its temperature and that of the surrounding medium. If a body is in air of temperature 35oand the body cools
solurroN: Let f : the number of minutes in the time that has elapsed since the body started to cool; r: the number of degreesin the temperature of the body at f minutes. Table 9.6.3 gives the initial conditions. From Newton's law of cooling, we have dx
7t
k(x  35)
DECAY
from 120"to 60oin 40 min, find the temperatureof the body after 100 min.
Separating the variables, we obtain dx
ffikdt
T n b l e9 . 6 . 3
t
0
40
x
I20
60
kl dt
100 '
rroo
kt+ c x: cekt+ 35 Therefore,
When x85ekt+35
When t  40, x  60; and we obtain + 35 85e4ok
60:
40k ln #
n:l[il;'::'Lz) x
_ g5e0.0306t+ 35
:r f f r o o 8 5 e  3 ' o 6 1 3 5 : 3 9 Therefore,
ExAMPrn4: There are L00 gal of brine in a tank and the brine contains 70lb of dissolved salt. Fresh water runs into the tank at the rate of 3 gal/min, and the mixture, kePt uniform by stirring, runs out at the same rate. How many Pounds of salt are there in the tank at the end of 1.hr?
the temperature
of the body
is 39o after 100 min.
Let f : the number of minutes that have elapsed since the water started flowing into the tank; t: the number of pounds of salt in the tank at t min' Because1,00gal of brine are in the tank at all times, at f minutes the number of pounds of salt per gallon is r/100. Three gallons of the mixture run out of the tank each minute, and so the tank loses3(r/100) pounds of salt per minute. BecauseDp is the rate of change of x with respect to f, and r is decreasing as f increases,we have the differential equation
solurroN:
dt
100
we also have the initial conditions given in Table9.6.4. Separating the variables and integratingr w€ have
LOGARITHMIC AND EXPONENTIAL FUNCTIONS T a b l e9 . 6 . 4
t
0
60
x
70
reo
I+o.otIat ln lrl  0.03f + e x: Ce When
x:70, and so C:70. Letting
60 and x  xeo,we have
x6o: 70e1'8
:70(0.L553)  1'/,.57 So there are 11.57lb of salt in the tank after L hr.
The calculus is often very useful to the economist for evaluating certain business decisions.However, to use the calcuruswe must beconcemed with continuous functions. consider, for example, the following formula which givesA, the number of dollars in the amount after f years, ii P dollars is invested at a rate of 100i percent, compound.edz times per year: A : p ( t\ + 'Jm \l ^ '
(2)
Let us conceive of a situation in which the interest is continuously compounding; that is, consider formula (2), where we let the number of interest qerlo$s per year increasewithout bound. Then going to the limit in formula (2), we have
A: p mtim /r + a)'' _*@ m/ \
which can be written as
1)mtil* A : p t nl it m *@ l(t+ L\
m/
J
(3)
Now consider
tim /r + !\''n
z+o \
ml
Letting h : il m, we have mli : 1lh; and because',m >*o,, is equivalent tO ,rh + 0+r, we have
== tim /r + l\'" (1 * hlttn ' m/ nli$
n+o \
9.6 LAWS OF GROWTHAND DECAY
425
Hence, using Theorem 2.6.5, we have
.ri* [(t +h)*'')":
:
git
and so Eq. (3) becomes A  Peit
(4)
By letting t vary through the set of nonnegative real numbers, we seethat Eq. (4) expressesA as a continuous function of f. Anotherway of looking at the samesituation is to consider an investment of P dollars, which increasesat a rate proportional to its size. This is the law of natural growth. Then if A dollars is the amount at f years, we have
lnlAl:H+e A Cekt When t:
0, A:
P, and so C : P. Therefore,we have
A: pekt
(5)
Comparing Eq. (5) withEq. (4), we seethat they are the same if.k: i. So if an investment increasesat a rate proportional to its size, we say that the interest is compoundedcontinuously, and the interest rate is the constant of proportionality. . rLLUSrRArroN1: If P dollars is invested at a rate of 8vo Per year compounded continuously, and A dollars is the amount of the investment at f years, dA
ff: o'oee and A Peo.ost
o
If in Eq. (4) we take P : L,i:1, and t: l,we getA: e, which gives a justification for the economist's interpretation of the number "e" as the yield on an investment of one dollar for a year at an interest rate of 700Vo compounded continuously.
ii't .'ii
FUNCTIONS AND EXPONENTIAL LOGARITHMIC
EXAMPIE5: If $5000 is borrowed at an interest rate of L2Voper year, compounded continuously, and the loan is to be repaid in one payment at the end of a year, how much must the borrower repay? Also, find the effective rate of interest which is the rate that gives the same amount of interest compounded once a yeat.
solurroN: Letting A dollars be the amount to be repaid, and becauseP : 5000,i :0.12, and f : 1, we have from Eq. (4) d:
5gggro'tz : s000(1.1275) :5637.50
Hence, the borrower must rcpay $5637.50.Lettingl be the effective rate of interest, we have 5000(1*i) :5000e0'12 1*i:to'rz i:1.1275  L : 0.1275  12,7570
9;6 Exercises 1. Bacteriagrown in a certain culture increaseat a rate proportional to the amount present. If there are 1000bacteria present initially and the amount doubles in L hr, how many bacteria will there be in 3* hr? 2. In a certain culture where_the rate of growth of bacteria is proportional to the amount present, the number triples in 3 hr, and at the end of 12 hr there were 10 million bacteria. How many bacteria were present initially? 3' In a certain chemical reaction the rate of conversion of a substanceis proportional to the amount of the substancestill untransformed at that time. After 10 min onethird of the original amount of the substancehas been converted, and 20 g has been converted after 15 min. What was the original amount of the substance? 4. Sugar decomPosesin water at a rate proportional to the amount still unchanged. If there were 50 lb of sugar present initiallyandattheendof 5hrthisisreduced ro20 lb,howlongwillittaie u n t i l 9 0 V oo f t h e s u g a r i J d e l o m p o s e d l 5. The rate of natural increase of the population of a certain city is proportional to the population. If the population increasesfrom 40,000to 50,000in 40 years, when will the population be 80,000? 6. Using Newton's law of cooling (see Example 3), if a body in air at a temperature of 0ocools from 200oto 100' in 40 min; how many more minutes will it take for the body to cool to 50.? 7. Under the conditions of Exarnple3, alter how many minutes will the temperature of the body be 45'? 8. When a simple electric circuit, containing no condensersbut having inductance and resistance,has the electromotive force removed, the rate of decreaseof the current is proportional to the current. The current is i amperesf sec after the. cutoff, and i: 4Owhen t: 0. If the current dies down to L5 amperes in 0.01 sec, find i in terms of f. 9. If a thermometer is taken from a room in which the temperature is 75ointo the open, where the temperature is 35", and the reading of the thermometer is 55oafter 30 sec, (a) how long after the removal will the reading be Sb.f 1U)What is the thermometer reading 3 min after the removal? Use Newton's law of cooling (see Example 3). 10. Thirty Percent of a radioactive substancedisappears in 15 years. Find the half life of the substance.
REVIEWEXERCISES 427
11. If the half life of radium is 1690years, what percent of the amount present now will be remaining after (a) 100years and (b) 1000years? 12. A tank contains 200 gal of brine in which there are 3 lb of salt per gallon. It is desired to dilute this solution by adding brine containing t lb of salt per gallon, which flows into the tank at the rate of 4 gaUmin and runs out at the same rate. When will the tank contain 1+ lb of salt per gallon? 13. A tank contains 100 gal of fresh water and brine containing2lb of salt per gallon flows into the tank at the rate of 3 gal/min. If the mixture, kept uniform by stirring, flows out at the same rate, how many pounds of salt are there in the tank at the end of 30 min? 14. A loan of $100is repaid in one payment at the end of a year. If the interest rate is 8% compounded continuously, determine (a) the total amount repaid and (b) the effective rate of interest. s 15. If an amount of money invested doubles itself in L0 years at interest compounded continuously, how long will it take for the original amount to triple itself? 15. If the purchasing power of a dollar is decreasingat the rate of 8% annually, cornpounded continuously, how long will it take for the purchasing power to be 50 cents? Professor Willard Libby of University of Califomia at Los Angeles was awarded the Nobel prize in chemistry for discovering a method of determining the date of death of a onceliving object. Professor Libby made use of the fact that the tissue of a living organism is composed of two kinds of carbons, a radioactive carbon A and a stable carbon B, in which the ratio of the amount of A to the amount of B is approximately constant. When the organism dies, the law of natural decay applies to A If it is determined that the amount of A in a piece of charcoal is only 15Voof its original amount and the half life of A is 5500 years, when did the tree from which the charcoal came die?
(Chapter9) ReaiewExercises In Exercises1 through 8, differentiate the given function.
1 .f ( x ) : ( l nf ) 2
2 .f ( x ) : @ h
3 .f ( x ) :
a. f @)  105"
5. /(r)  ifttrr
6. f(x)ttt(
8. f(x):3"'" (7)t"' integral. the indefinite 14, evaluate 9 through Exercises In r aozc dx lo. I eu'*1xl)
logro
L*x tx
+xz)
z. f(x):
s. lfi*ax
r
nlffa,
BI#
(e" * at*) dx
In Exercises15 through L8, evaluate the definite integral.
ts.
fzfr
ln
17. J,fua* 18
16.
x2e8dx "l l 3
,
19. Find D*A if ye' * xea* x * y  0. '(x)2 0 . l f f ( x ) : l o g ( " ' y (+x 1 ) , f i n d f
Jo
kb *7)z dx
428
FUNCTIONS AND EXPONENTIAL LOGARITHMIC
2 1 . T h e l i n e a r d e n s i t y o fa r o d a t a p o i n t x f t f r o m o n e e n d i s U ( x * 1 ) center of mass of the rod. 22. Find an equation of the tangent line to the curve y : xsr af (2,2).
s l u g s / f tI.f t h e r o d i s 4 f t l o n g , f i n d t h e m a s s a n d
23. A particle is moving on a straight line wheres ft is the directed distance of the particle from the origin, o ftlsec is the velocityof the particle, and aftlse& is theaccelerationof the particle at f sec.If. a: et * etand o:1and s:2 when t: 0, find o and s in terms of l. 24.T'heareaoftheregign,boundedbythecurvey:es,thecoordinateaxes,andtheliner:b(b>0),isafunctionofb.lf / is this tunction, find /(b). Also find 1im /(b) . 25. The volume of the solid of revolution obtained by revolving the region in Exercise24 about the r axis is a function of b. If g is this function, find g(b). Also find la S(A). 25. The rate of natural increaseof the population of a certain city is proportional to the population. If the population doubles in 60 years and if the population in 1950was 60,000,estimate the population in the year 2000. 27. The rate of decay of a radioactive substanceis proportional to the amount present. If half of a given deposit of the substance disappears in 1900years, how long will it take for 95Voof the deposit to disappear? 28. Prove that if a rectangle is to have its base on the r axis and two of its vertices on the curve y : scz, then the rectangle will have the largest possible area if the two vertices are at the points of inflection of the graph. 29. Givenf (x) : ln lrl and r < 0. Show that / has an inverse function. If g is the inverse functioru find g(r) and the domain of g. In Exercises30 and 31, find the inverse of the given function if there is one, and determine its domain. Draw a sketch of the given function, and if the given function has an inverse, draw a sketch of the graph of the inverse function on the same set of axes.
3 0 .Jf '(\ x )  ! + 4 3 2x
3 2 . P r o v e t h a t i fr < 1 , L r r ( r .
(nrur: Letf(x):rlnrandshowthat/isdecreasingon(0,1)andfind/(l).)
33. When a gas undergoesan adiabatic (no gain or loss of heat) expansionor compression,then the rate of changeof the Pressurewith respect to the volume varies directly as the pressure and inversely as the volume. If the pressure is p lb/in.'zwhen the volume is u in.3, and the initial pressure and volume are polblin.z and a6 in.s, show thai pztk: p6ooi. U. ft W in.lb is the work done by a gas expanding against a piston in a cylinder and P lb/in., is the pressure of the gas when the volume of the gas is V in.3, show that if V, in.3 and.V"in.s are the initial and final volumes, respectively, then
W:I
fvz
PN
Jvr
35. Supposea piston comPressesa gas in a cylinder from an initial volume of 50 in.s to a volume of 40 in.3. If Boyle's law (Exercise L0 in Exercises3.9) holds, and the initial pressure is 50 lblin.z, find the work done by the piston. (Use the result of Exercise34.) 36. The charge of electricity on a spherical surfaceleaks off at a rate proportional to the charge. Initially, the charge of electricity was 8 coulombs and onefourth leaks off in 15 min. When will there be only 2 coulombs remaining? 37. How long will it take for an investment to double itself if interest is paid at the rate of 8% compounded continuously? 38. A tank contains 50 gal of salt water with 120lb of dissolved salt. Salt water with 3 lb of salt per gallon flows into the tank at the rate of 2 gallmin and the mixture, kept uniform by stirring, flows out at the same rate. How long will it be before there are 200 lb of salt in the tank?
REVIEWEXERCISES rt
39.Find I et"t dx if f is any real number. Jo
40. Provethat rf.x> 0, and f" ,ot dt:L,
(1 + h)rm.
then ti* x:lim
Jt
h_O
h_0
4 L . Prove that lim
1og,(l * r) : Logoe X
tO
(Norr:
Comparewith Exercise33 in Exercises9.1.)
42. Prove that li*a'1:loo cO
X
(nrnr: Lety:a'Tandexpress lim s(y).)
(a"l)lxas
a f u n c t i o no f y , s a y g ( y ) ' T h e n s h o w t h a t ! + 0 a s x + 0 '
andfind
43. Use the results of Exercises41 and 42 to Prove that
r i mU _ _Lb x'l
(HrNr:
X,
Write
xbL _to'n'L blnx )cL
.bLnx xl
Then let s b ln r and t:
x  L.)
44. Prove that
\
ean "l' : fl lim X
t0
eo'and find /'(0) by two methods.) cf(x) for allr where c is a constant, Prove that there is a con45. If the domain of /is the set of allreal numbers and f'(x): function 8 for which 8(x) : f(x)eco and find 8'(x)') stant k for which f(x)  ke"* for all r. (HrNr: Consider the (nrNr:
Let f (x) :
46. Prove that
Drn(lnx)(l)nr+ (nrxr:
Use mathematical induction' )
each integral' 47. DoExercise 17 in the Review Exercisesof chapter 7 by evaluating
Iiigonometric functions
1 0 . 1T H E S I N E A N D C O S I N EF U N C T I O N S 431
10.1.THE SINE AND COSINE FUNCTIONS
F i g u r e1 0 . 1 . 1
L0.L.LDefinition
In geometry an angleis defined as the union of two rays called the sides, having a cornmon endpoint called the vertex. Any angle is congruent to some angle having its vertex at the origin and one side, called the initial side,lyingon the positive side of the r axis. Such an angle is said to be in standardposition.Figure 10.1.1shows an angle AOB in standard position with AO as the initial side. The other side, OB, is called the terminal side. The angle AAB canbe formed by rotating the side OA to the side OB, arrd under such a rotation the point A moves along the circumference of a circle, having its center at O and radius IOA I to the point B. In the study of trigonometry dealing with problems involving angles of triangles, the measurement of an angle is usually given in degtees. However, in the calculuswe are concernedwith trigonometric functions of real numbers, and we use radian tneasureto define these functions. To define the radian measure of an angle we use the length of an arc of a circle. If such an arc is smaller than a semicircle, it can be considered the by Theorem $aph of a function having a continuous derivative; and so g.r0.z it has length. If the arc is a semicircle or larger, it has a length that is the sum of the lengths of arcs which are smaller than semicircles. Let AOB be an angle in standard position and 16Z'l :1. If s units is the length of the arc oi the circle traveled by point A as the initial side OA is rotated to the terminal side OB, the radianmeasure,f, of angle AOB is given by t:
s
if the rotation is counterclockwise
and t: s
if the rotation is clockwise
o rr.r,usrnerrow 1: By using the fact that the measure of the length of the unit circle's circumference is 2tr, we determine the radian measuresof the +T , ,\r ,  n%r , and angles in Fig. 10.1.2a,b, c, d, e, and f. They aretzt , *n , tn, respectively.
Ln 2
(d) 10.1.2 Figure In Definition 10.1.1 it is possible that there may be more than one complete revolution in the rotation of. OA'
TRIGONOMETRIC FUNCTIONS o ILLUSTRATIoN 2: Figure
10.1.3a shows such an angle whose radian
measure is +7r,and Fig. 10.1.3bshows one whose radian measure is *r. An angle formed by one complete revolution so that oA is coincident with oB has degree measure of 360 and radian measure of 2rr. Hence, there is the following correspondencebetween degreemeasureand radian measure (where the symbol  indicates that the given measurementsare for the same or congruent angles): 350"  2zr rad or, equivalently, 180' zr rad From this it follows that 1'  r#ozr rad and 180' 1 rad 
57"78'
From this correspondencethe measurement of an angre can be converted from one system of units to the other.
F i g u r e1 0 . 1 . 3
o ILLUSTRATToN 3:
L62"  162 . tl1n rad: r%.zrrad  T n5' 
1m" _ 75o n
Table 10.1.1gives the corresponding degreeand radian measuresof certain angles. Table10.1.1
' t
we now define ,n" ,r"" 10'1'2 Definition
; *n
;; tr
""0 "r;""
', &n
"; tr
;r;r'.;
Ezr
",jr
"""
*o
,5;
Eo
2n
,""
""0"".Suppose that f is a real number. Placean angle, having radian measure f, in standard position and let point P be at the interseciion of the terminal side of the.angle with the unit circle having its center at the origin. If p is the point (*, y),then the cosinefunctionlJaefinea Uy c o sf : r
10.1 THE SINE AND COSINEFUNCTIONS
and the sinefunction is defined by 0, 1)
ir/z)
F i g u r e1 0 . 1 . 4
sinf:/From the above definition it is seen that sin f and cos f are defined for any value of f. Hence, the domain of the sine and cosine functions is the set of all real numbers. The largest value either function may have is 1, and the smallest value is 1. It will be shown later that the sine and cosine functions are continuous everywhere, and from this it follows that the range of the two functions is [L, 1]' For certain values of f, the cosine and sine are easily obtained from a figure. From Fig. 10.1.4we see that cos 0: 1 and sin 0: 0, cos*zr: L\/, 0 and sin *z : L, cos rr:7 and sin 7r:0, and sin 1n: l!2, cosir: Table 10.1'.2gives these values and some cos *zr:0 and sin *rr:1. others that are frequently used.
+\n +\E 1 +\E +\n +\n +\E 1 1 +\E t{z z u  2 , & 1
1
n
r;
An equation of the unit circle having its center at the origin is x2  cos f and y : sin t, it follows that * y'  L. Because x (1) F i g u r e1 0 . 1 . 5
Note that coszf and sin2 f stand for (cos t)2 and (sin t)2. Equation (1) is called anidentity becauseit is valid for any real number f. Figures L0.1.5and 10.l..6show angleshaving a negative radian measure of lf and corresponding angles having a positive radian measure of t' From these figures we see that . '.i.anii,..., ..,'ffib{.Itli.''l#..'cod.,.*. , sin{*f } +',,usin:,f
(2)
These equations hold for any real number f becausethe points where the t) intersect terminal sides of the angles (having radian measures f and the unit circle have equal abscissasand ordinates that differ only in sign. Hence, Eqs. (2) are identities. From these equations it follows that the cosine function is even and the sine function is odd. From Definition 1.0.1.2the following identities are obtained. t
F i g u r e1 0 . 1 . 6
and
Bin:(f,*,2rrr) '= ein f
(3)
The property of cosineand sine statedby Eqs. (3) is called periodicity, which is now defined.
TRIGONOMETRIC FUNCTIONS
L0.L.3Definition
A function f ,t said to be periodicwith periodp * 0 if whenever x is in the domain of /, then x * p is also in the domain of f and
From the above definition and Eqs. (3) it is seen that the sine and cosine functions are periodic with period 2n; that is, whenever the value of the independent variable f is increasedby 2n, the value of each of the functions is repeated.It is becauseof the periodicity of the sine and cosine that these functions have important applications in physics and engineering in connection with periodically repetitive phenomena such as wave motion and vibrations. , we now proceed to derive a useful formula becauseother important identities can be obtained from it. The formula we derive is as fblows: (4)
P1 (cos a, sin a)
x
where a and b are any real numbers. Refer to Fig. 10.1.7 showing a unit circle with the points Q ( 1 , o ) , P r ( c o sa, sin a), Pr(cosb, srn b), Pr(cos(a * b), sin (a * b)), and Po(cosa, sin a)  Using the notatiot 8 to denote the measure of the length of arc from R to S, we have @, O t r r :n , @ r : b , f jP rPr: a, and
ffi:
lal: o.Because Gr:@+ffir,
w€have
C4b*a F i g u r e1 0 . 1 . 7
and because PnP, a
frr:&+OFr, +b
it fotlowsthat (5)
From Eqs. (5) and (5) we seethat @, :'pnFr;therefore, the length of the joining the p_ointsQand Ps is the same as the length of the chord :hg1d joining the points Pn and Pr. squaring the measuresof tfiese lengths, we have
loP;,:l4J'l
(7)
Using the distance formula, we get
IOF;!,: [cos(a + b)  1], + [sin(a * b)  0], : cosz(a f b)  2 cos(a+ b) + L * sin2(a+ b) and because cosz(a * b) + sinz (a * b) : 1,, the above may be written as
l84l':22cos(a+b) Again applying the distance formula, we have l@rp:
(cosbcos a)2+ (sin b+sin a),
(8)
I O . 1T H E S I N EA N D C O S I N EF U N C T I O N S 435 :
cosz b  2 cos a cos b + cosz a
* sin2 b * 2 sin a sinb*
sin2 a
and because coszb * sin2 b : 1.and cos2a * sinz a: l, the above becomes (9) cosa cosb * 2 sir.a sinb lF/,l':l2 Substituting from Eqs. (8) and (9) into (7),we obtain 2  2 cos(a+ b) : 2  2 cosn cos b * 2 sin a sin b from which follows cos(a * b) : cos a cosb sin a sin b which is formula (4). (a) by substiA formula for cos(a  b) can be obtained from formula tuting b
for b. Doing this, we have cos4 cos(b)  sin a sin(b) c o s( a * (  b ) ) :  sin b, we get Because cos(b)  cos b and sin (b) cssta *',b}"*
s'.n Gosl'li'* sin at'lbin h'
(10)
for all real numberc a NtdbLetting a:Lrr in formula (1'0),we have the equation,cos(Lnb): : sin *zr : L' it .or J"'"ol a + ,it Lrr sin b; and'becausecos *rz 0 and  b: c' b".orr,"s cos(*zr b) : sin b. Now if in this equation welet $n c and we obtain cos c: sin(*n c)' We have therefore then b :irrproved the following two identities.
*$r
(11)
Formulassimilarto(4)and(10)forthesineofthesumanddifference from those we have. A formula for the sine of of two numbers follow ""rity from (11) the sum of two numbers follo*, from (10) and (11)' We have b) s i n ( a* b ) : c o s ( i z  ( a * b ) ) : c o s ( ( * z t  a ) With (10) it follows that sin(a * b) : cos(tz a) cos b * sin(tzr a) sin b and from (L1) we get
$D' for all real numbers a and b. we To obtain a formula for the sine of the difference of two numbers have we write a b as a+ (b) and apply formula (12)' Doing this' sin(a * (b)) : sin a cos(b) * cos a sin(b) :sin b' we obtain and becausecos(b) : cos b and sin(b)
436
TRIGONOMETRIC FUNCTIONS
(13) :h is valid for all real numb ers a and b. Bylettingab f informulas (L2) and (4) , respectively,we get the (14)
(ls) Using the identity sin2 t + cos2 t 
I we may rewrite formula (15) as
cos2t:2coszt1,
(16)
or as c o s2 t : 1 ,  2
s i n 2f
(r7)
Replacingt by tt in formulas (15) and (lr) and performingsome algebraicmanipulations, we get, respectively, (18) and
(1e) (19) are identities because they hold for all
real numbers t. By subhacting the terms of formula (4) from the corresponding terms of formula (10) the following is obtained: sin a sin g:{[cos(a
* b) * cos(a b)]
e0)
By adding corresponding terms of formulas (4) and (10) we get cosd cos 6:glcos(a + b) + cos(a b)l
er) and by adding corresponding terms of formulas (12) and (13) we have sin a cos6:f fsin(a + b) + sin(a _ b)] ez) Bylettingc:a * b and d:ab informulas (22), (21),and (20),we obtain, respectively,
sinc+ sind2 sin rycos +
(zt1
c o sc + c o s d : 2 cos c + d cos c  d 2 T
(24)
and c o s c  c o s d :  n z
sln
c*4^r^cd 2
Sln
2
(2s)
10.1 THE SINE AND COSINEFUNCTIONS
If the terms of formula (13) are subtracted from the corresponding terms of formula (L2) and a * b and a  b arereplaced by c and il, tespectively, we have
P(x,v) Pt (rt , at)
s i n c  s i n d  2 c o s+ sz i2n
F i g u r e1 0 . 1. 8
v 7
P(
o
i
1
/ \
_L
4
(26)
You are asked to perform the indicated operations in the derivation of some of the above formulas in Exercises7 to 6. When referring to the trigonometric functions with a domain of angle measurements we use the notation f, to denote the measurement of an angle if its degree measure is 0. For example, 45ois the measurementof an angle whose degree measure is 45 or, equivalently, radian measure is *zr' Consider an angle of doin standard position on a rectangular cartesiancoordinate system. Chooseany point P, excluding the vertex, on the terminal side of the angle and let its abscissabe r, its ordinate be y, and lOf l : r. Referto Fig. 10.1.8.The ratios xly andy/r ateindependentof the choice of P becauseif the point Pr is chosen instead of P, we see by Fig. 10.1.8that xlr: xlrl and ylr: alrr Becausethe position of the terminal side depends on the angle, these two ratios are functions of the measurement of the angle, and we define cos 0o:x
rr
and
sin 0o:Y
(27)
Becauseany point P (other than the origin) may be chosen on the terminal side, we could choose the point for which / : 1, and this is the point where the terminal side intersects the unit circle * * y': 1 (see Fig. 10.1.9).Then cos gois the abscissaof the point and sin 0ois the ordinate of the point. This gives the analogy between the sine and cosine of real numbers and those of angle measurements.We have the next definition.
.9 Figure10.1
10.1.4 Definition
If a de'greesand r radians are measurements for the same angle, then cos co : cos r and sin ao: sin t
10.1. Exercises (10). 1. Derive formula (20) by subtracting the terms of formula (4) from the corresponding terms of formula 2. Derive formula (21) by a method similar to that suggestedin Exercise 1. 3. Derive formula (22) by a method similar to that suggestedin Exercise 1. 4. Derive forrrula (23) by using formula (22). 5. Derive formula QD by using formula (21). 6. Derive formula (25) by using formula (20). I. Derive a formula for sin 3t in terms of sin f by using formulas (72), (74), (15), and (1).
438
TRIGONOMETRIC FUNCTIONS
8 . Derive a formula for cos 3f in terms of cos f. Use a method similar to that suggested in Exercise7. 9 . Without using tables, find the value of (a) sin #r
and.(b) cos gz.
10. Without using tables, find the value of (a) sin *z and (b) cos *rr. 1 1 .Without using tables, find the value of (a) sin ttn and.(b) cos tzr. T2, Express each of the following in terms of sin f or sin(iz.t): (a) sin(Ezrt); (d) cos(Bzrf f).
(b) cos(gzr
(c) sin(Bzr+ t);
13. Express the function values of Exercise12 in terms of cos f or cos($z  f). t4. Expresseachof the following in terms of sin f or sin (tn  t): (a) sin(z  t); (b) cos(z  f); (c) sin(z + l); (d) cos(z * f). 15. Express the function values of Exercise 14 in terms of cos f or cos(*a.  f).
rc.Find all values of.t
for which (a) sin f :0,
and (b) cos f :0.
17. Find all values of f for which (a) sin f : 1, and (b) cos f : 1. 18. Find all values of f for which (a) sin f : 1, and (b) cost: 1. 19. Find all values of t for which (a) sin f : *, and (b) cos t: i. 20. Find all values of t for which (a) sin t: +\/r, and (b) cos f : Lt/2. 21. Suppose/ is a function which is periodic with period 2r , and,whose domain is the set of all real numbers prove that / is also periodic with period 22. 22. Prove that the function of Exercise 21 is periodic with period 2nn f.orevery integer n. (Hrr.rr: Use mathematical induction. ) 23. Prove that if f is defined by f (x)  )c [r]] , then ,t periodic. What is the smallest positive period of f? f
1.O.2DERIVATIVES OF THE SINE AND COSINE FUNCTIONS
Before the formula for the derivative of the sine function can be derived we need to know the value of ,.
sin f
1iT t
Letting f(t): (sin t)lt, we see that /(0) is not defined. However, we prove that lim /(f) exists and is equal to L. 10.2.LTheorem
lim tO
PRooF: We first assume that 0 < t < tn. Refel to Fig. 10.2.7,which shows the unit circle 12*yr:1 and the shaded sector BOP, where B is the point (1, 0) and p is the point (cos f, sin f). The area of a circular sector of radius r and central angle of radian measure f is determined by trzt; and so if s square units is the area of sector BOP,
s tt
OF THE SINE AND COSINE FUNCTIONS 10.2 DERIVATIVES
439
consider now the triangle BoP, and let K1 square units be the area of this triangle. Hence,
K r:t2l F l ' IOB I:j( sint) ' ( 1) :f sint The line through the points o(0,0) sin f/cos f; therefore, its equation is
( 2)
and P(cos f, sin f) has slope
sin f x u:cost " B (L, o)
F i g u r e1 0 . 2 . 1
This line intersectsthe line x:1 at the point (1, sin tlcos t), which is the point T in the figure. Letting K, square units be the area of right triangle BOT, we have
K,:+lBr . 16:+'#l' ! : +.#+
(3)
From Fig. 1'0.2.1'we see that K1
(4) , we get Substituting from Eqs. (1), (2), and (3) in inequality L
L
T s i nt < ; t . ; ' f f i
sinf
pos: Multiplying eachmember of the above inequality by Zlsin f , which is itive because0 < f < *n, we obtain r1
1'
"inT';o;7 Taking the reciProcal of each member of the above inequality and reversing the direction of the inequality signs, we get
(s)
si+f .1 c o st 1. From (17) and the chain rule, Lf.u is a differentiablefunction of r, we have
u) ffiD*u
ExAMPLE3: A picture 7 ft high is placed on a wall with its bas e 9 ft above the level of the eye of an observer. How far fuom the wall should the obsenrer stand in order for the angle subtended at his eye by the picture to be the greatest?
( 1 8)
sol,urroN: Let x ft be the distance of the observer from the wall,0 be the radian measure of the angle subtended at the observer's eye by the picture, o be the radian measureof the angle subtended at the observer's eye by the portion of the wall above his eye level and below the picture, and F: o + 0. Referto Fig. 10.8.1. We wish to find the value of r that will make d an absolute maximum. Becauser is in the interval (0, *), the absolute maximum value of 0 will
]
FUNCTIONS TRIGONOMETRIC
be a relative maximum value. We see from the figure that cotB+ and c o.t)oc: o B e c a u s e0
Take two cases:a > 0 and z < 0. du ^f 3.  :cos'z*C J \/l u, Is this formula equivalent to formula (1)? Why?
4'
fdu
Jr;7:
cotru * C
Is this formula equivalent to formula (2)? Why? du f 5. l;ffi:cscl
lul+c
Take two cases:z > 0 and z < 0. Is this formula equivalent to formula (3)? Why? In Exercises5 through 25, evaluate the indefinite integral.
7I#
i
,101 .L
13I#
14t4 J Iz,''.t f
L 6 .l L J\ffi
dx
^fdx
'
t
,89.Jffi a
,
22. [ = !d* J x2+x*L f
c o s zx
L7:l " J x2x*2
i&lJ,'ffi, 6
xdx
1 1Jt LV r 6  e t '
J xtF
\''"
i
S I x4+r6
. , . , .f  Q + x ) d x J V42xx2
3dx
z) a+l ' J G + D f f i 
In Exercis es 26 through 33, evaluate the definite integral. . . : 'f r 1 * x
27.' J o #d*
,
LtX'
30. f d x
J, x'z4x+13
4". Q,
r[L + (ln r)2]
\ 34.Find the area of the region bounded by the curve y : 8l (x2 * 4) , the x axis, the y axis, and the line x: 2. :i5. Find the abscissaof the centroid of the region of Exercise34. 36. Find the area of the region bounded by the curves x2: 4ay and y :8a31 (x2 * 4a2), 37. Find the circumference of the circle rP * y' : I by integration.
*/
t'i
FUNCTIONS TRIGONOMETRIC
488
38. A particle moving in a straight line is said to have simple harmonicmotion if the measure of its accelerationis always proportional to the measure of its displacement from a fixed point on the line and its accelerationand displacement are oppositely directed. So if at f sec,s ft is the directed distance of the particle from the origin and o ftlsec is the velocity of the particle, then a differential equation for simple harmonic motion is da  kzs dt
(7)
where k2 is the constant of proportionality and the minus sign indicates that the acceleration is opposite in direction from the displacement.Becausedoldt : (dulds)(dsldt) : o(dolds), Eq. (7) may be written as I'j
da E:
kzs
(8)
(a) Solve Eq. (8) for a to get Tr *k\m. Note: Take a2k2as the arbitrary constant of integration and justify this choice. (b) Letting a  dsldt in the solution of part (u), we obtain the differential equation i
o
parts. It II.2 INTEGRATION A method of integration that is quite useful is integrationby a product: of BY PARTS depends on the formula for the differential d(ua)uda*adu or, equivalently, udod(ua)adu
\
(1)
Integrating on both sides of (1), we have (2)
Formula (2) is called the formula for integrntionby pntts. This formula expressesthe integral [u do in terms of another integral, lo du. By a suitable choice of u and do, it may be easier to integrate the second integral than the first. When choosing the substitutions for u and do,we usually want da to be the most complicated factor of the integrand that can be integrated directly and u to be a function whose derivative is a simpler funciion. The method is shown by the following illustrations and examples. o ILLUSTRATTON
1.:
We wish to evaluate
r
I tanr x dx
J
Let u:
t a n  l r and da  dx. Then
TECHNIQUESOF INTEGRATION
494
So from (2) we have f
J
tanl x dx :
( t a n x ,) ( x *C , )
I(r*
xtanl x*crtanrx[+d*
C , )# n
rR''JR
r
dx
: x t a n  l x * C , t a n  r x  * l n 1 1+ x r l  C , t a n  r x * C , :xtanrr*ln (1,+x2)*C, I In Illustration L observe that the first constant of integration c1 does not appear in the final result. This is true in general, and we prove it as follows: Bv writing u * C, in place of a in formula 2, we have ff
J
u du: u(a* C,)
+ Cr)du J kt
:uo*cru[aducrIau JJ
:uu*cfl[oa"cfl :nn
[ o au J Therefore,it is not necessaryto write C1when finding a from do. EXAMPLE 1:
Find
f
I rln xdx
J
SOLUTION:Let Lt ln r and da : x dx. Then ,dxxz au: x2 ,v2
. h r d x  i t n xt I .vz
It+ _ 1;I^
'_. d( ,x
2
tnx
txz ln lx EXAMPLE 2: f
Find
f rsinxdx
J
xdx
tx' '+ C
SOLUTTON: Let Lt r andl d v : s i n r : d x Then dudx
a n d 7 )   cos S )c
Therefore, we have f
I r sin x dx:x J
f
c o sx )+ f c o sx;ddx :
ficos.
J
x + sinx+. C
BY PARTS 495 11.2INTEGRATION 2: IN
o ILLUSTRATION
Example 2, if instead of our choice s of u and du as
above, we let and
Lt: sin r
da:xdx
we get
and o:tf
du:cosxdx So we have f
J
f r sinx dx:i sin;
1f
J
f cosx dx
The integral on the right is more complicated than the integral with which we started, thereby indicating that these are not desirable choices for z o and ilo. It may happen that a particular integral may require repeated applications of integration by parts. This is illustrated in the following example.
EXAMPTn 3:
Find
r I xze*dx J
 x2 and da  et dx. Then soLUTIoN: Let u and 7):er
duZxdx We have, then, ff
I xre" dx: JJ
xzet  2  xe' dx
We now apply integration r,tx
by Parts to the integral on the right. Let
and d0:etdx
Then and u:e*
dildx So we obtain
ff  *t" dxxex
JJ
 e'dx
:xeter+C Therefore, f I x"" dx: )cze" zlxe" er + C]
J
: Nzen Zxe' * 2e'+ C
Another situation that sometimes occurs when using integration by parts is shown in ExamPle4.
I
TECHNIQUES OF INTEGRATION
Find
soLUTroN: Let u  e, and du  sin x dx. Then du:
sin x dx
e* dx
and
T) cos x
Therefore,
r I e" sin xdx
cos x dx
JJ
The integral on the right is similar to the first integral, exceptit has cosr in place of sin r. We apply integration by purts again by letting fr:et
and d0cosxdx
So
dile'dx
and osinx
Thus, we have ff
t * s i nx d x JI J
e'cosx* [e'sin x
 e* sinxdx]
Now we have on the right the same integral that we have on the left. so we add I e" sin x dx to both sides of the equation, thus giving us
r
2 l e ' s i nx d x J
Dividirg
e ' c o sx * e " s i n x * e
on both sides by 2, we obtain
r
I t " s i n x d x  + e " ( s i n x  c o sr ) + C J In applying integration by parts to a specific integral one pair of / chrrices fot u and du may work while another pair may nofl We saw this in Illustration 2, and another case occurs in Illustration 3. OI
LLUSrRarroN 3: In Example 4, rn the step where we have
fr
l t " s i nx d x :  e * c o s x *
JJ
l t * c o sx d x
i f v ve evaluate the integral on the right by letting
ucosff di: we get
and d0:e*dx
sin x dx and u : er
far
t" sin x dx Jl J ff
I t" sin x dx: JJ
e* cosx * (e" cosr * l t" sin x dx) I e* sin x dx
BYPARTS 11.2INTEGRATION
,497""
In Sec. 10.8 we noted that in order to integrate odd powers of the secantand cosecantwe use integration by parts. This processis illustrated in Example 5.
EXAMPLE 5:
soLUTroN: Let Lt secr and da : sec2x dx. Then du : sec r tan x dx and a  tan x
Find
f
I sect x dx J
Therefore, fr xdx=sec xtanxI sec3 JJ rf I secsx dx: secr tan xJJ
ffr I s e c tx d x : s e c JJJ
xtanx
f s e cx t a n z x d x  secr(seczx I) dx  s e c tx d x *
 s e cx d x
Adding I secsx dx to both sidet, yu get
r 2  t".t xdx:secrtanr*ln J
l s e cx * t a n x l+ 2 C
f
x d x  + s e c r t a n) c + * l n l s e cx + t a n r l + C lsecs
Integration by parts often is used when the integrand involves logarithms, inverse trigonometric functions, and products.
11..2 Exercises In Exercises L through t6, evaluate the indefinite
1.
f
Jrnxdx
u..! x3'dx I
t a n  lx d x lx r L0. sin(ln r) dx J 7.
integral.
, UJ
,6I#"
tA
? Icos2xdx g,/ 'i"'\x dx
3.
8. *' stn3x dx I
,.
''ir. I#. 14. /.r.'
6.
xdx Isecz I.'tnxdx I
e* cosx dx
L2. x'e" dx I x dx
1s.
/r".'xdx
TECHNIQUES OF INTEGRATION
In Exercises17 through 24, evaluatethe definite integral. t . [:'" sin 3r cosx dr Jo
F\. Jo ,j L'^
e*asin4xdx
cos2x itx {9 '' f _x2 Jn
'') f ,r,, a, $ Jo
,r. 1""'ox surr cotxcsc cscx r 4tr itx Jtu
zz. fn '"' ,.,
sec,t/i dx
,0. [""'' cos\/2x dx Jo
e. ,4'
r Jo
x sinr x ilx
/,$. fi"a the area of the region bounded by the cule y : In x, the r axis, and the line x: ez. ?$. fi"a the volume of the solid generated by revolving the region in Exercise25 about the r axis. {}. ni"a the volume of the solid generated by revolving the region in Exercise25 about the y axis. 28. Find the centroid of the region bounded by the cule y : er, the coordinate axes, and the line : r 3. 29' The linear density of a rod at a point r ft from one end is 2et slugs/ft. If the rod is 6 ft long, find the mass and center of mass of the rod. 30' Find the centroid of the solid of revolution obtained by revolving about the x axis the region bounded by the curve y : sin x, the x axis, and the line x: tr. 31' A board is in the shape of a region bounded by a straight line and one arch of the sine curve. If the board is submerged vertically in water so that the straight line is the lowei bound ary 2 ft below the surfaceof the water, find the force on the board due to liquid pressure. 32' A particle is rnoving along a straight line and s ft is the directed distance of the particle from the origin at t sec.rf o f t / s e ci s t h e v e l o c i t ya t f s e c , s : 0 w h e n f : 0 , a n d o . s : t s i n t , f i n d s i n t e r m so f f a n d a l s os w h e n t : t n . 33' A water tank full of water is inthe shape of the solid of revolution formed by rotating about the r axis the region bounded by the cuwe y: e", the coordinate axes, and the line r: 4. Find the work d6ne in pumping all the water to the top of the tank. Distance is measured in feet. Take the positive r axis vertically downward. 34. The face of a dam is in the shape of one arch of the curve y :100 cos z&ozrrand the surface of the water is at the top of the dam' Find the force due to liquid pressure on the face of the dam. Distance is measured in feet. 35' A manufacturer has discovered that if x hundreds_of units of a particular commodity are produced per week, the marginal iost is determined by x2'r' and the marginal revenue is d'etermined by s . zttz,wliere both the production cost thousands of dollars' If the weekly fixed costs amount to $2000,find the *a"i.r* *"etaf frorit ili:::T:ftTarein
11.3 INTEGRATION By TRIGONOMETRIC SUBSTITUTION
If the integrand contains an expression of the form \/ar=, \/FTE, or {F=V, where a } O,iiis often possible to perform the integration by making a trigonometric substitution which reiults in an inte_ gral involving trigonometric functions. we consider each form as a separate case. Case 1: The integrand contains an expression of the form \/F7A, a > 0. we introduce a new variable 0 by letting u: a sir, g, where 0 < 0 =+rif u > 0 and *rr < g < 0 iIu < 0. Ttrendu: a cos0 d0,and,
\/F=7 : \/r=}Gfr
0:{F6=R6
: a@0
Beggge'*n= 0 in,cos 0  0; hence,\re 0:cos 0, andwe have \/F=7: a cos0. Because sin 0: ula and.trr  0 = t,,it followsthat 0 : sinr(ula).
SUBSTITUTION 499 BY TRIGONOMETRIC 11.3 INTEGRATION
soLUrIoN:'Letr:3 sin d, where0 < 0 < tnif. x > 0 and*rr 0) and 11'3.2 (for x < 0). We see that cot9: V9  *lr. Therefore, t/g =7 f ..F
I
, t dx
\5=7
_,,__,x c  sinri*
F i g u r e1 1 . 3 . 2
Case 2: The integrand contains an expression of the form \817, a > 0. we introduce a new varidble 0 by letting u: a tan 0, where 0 < 0 < tn lf.u > 0 and Lo < e < O if u < }'Then du: a secz0 il9'and
\/F 1fr : GTV@o
: a{I+ffin:
a{@T
:sec0' andwehave B e c a u sei n < 0 < t n , s e c 0> 1 ; t h u s\ / s @ e in g: < izr, it followsthat < 0 ula and tan \/7 + fr: sec0. Because 0: tanr(ula).
EXAMPLE
2:
r  !x'+ JI
EVAIUATC
5 dx
solurroN: Letr: tEtan0,where0 < 0 < irif.x = 0andizr < 0 < 0if r < 0. Then dr: \6 se& 0 d0 and
\ffi
_m\6\@_\E
sec0
Therefore,
sec'ade 0 d0) dx:  \re secO(fi secz  \ffi / JJ Using the resultof Example5 of Sec.1L.2,we have  \m
J
dx:f
r " . 0 t a n0 + * t " P e ce + t a n g l + C
: TECHNIQUES OF INTEGRATION
r>0 F i g u r e11 . 3 . 3
F i g u r e11 . 3 . 4
we find sec0 from Figs.11.3.3(for x  0) and 1,1.3.4 (for r < 0), where tarr 0 : xlt6. We see that sec e: \/FTEIVE. Hence, ,5 ax: 2
tx \ffi tx \ffi
*Bh l\ffi *Bh (lffi
*xl Brr,\6 +c *r) +cl
Because\ffi*x)0,wedroptheabsolutevaluebars. case 3:_ The integrand contains an expression of the form t/F1v, a > 0. we introduce a new variable d by letting u: a sec0, where < 0 0 a andr < 0 a. ln Fig.11.3.5,u flt sec d: ula =1, and z < 0 0forall r in la, bh however, (1) holds for anyfunction which is continuous on la, bf.
f(x)
Figur1 e1.8.1 Then the integral I t' f @) dx is the measure of the area of the region bounded by the x axis, the lines x : a a n d x : x 1 r a n d the portion of the
11 . 8 T H E T R A P E Z O I D ARL U L E
curve from P0to P1.This integral may be approximated by the measure of the area of the trapezoid formed by the lines r : a, x: x1, P6P1,and the x axis. By a formula from geometry, the measureof the areaof 'this trapezoid is
iLf@,)+ f(x)l Lx
Similarly, the other integrals on the right side of (1) may be approximated by the measure of the area of a trapezoid. Using the symbol "" fot "is approximately equal to," we have then for the ith integral ftt
(*) ilx  tlf (x;,) + f (x,)l Lx J",_,f
(2)
So, using (2) for each of the integrals on the right side of (1) , we have fb
J,f{*) Thus,
d x = E l f ( x )+ / ( r ' ) l L x + i [ f @ ' ) + / ( r , )A] r * ' ' + tlf(x",) + f (x"')fLx+ tlf Q"r)* /(r")l Ar ,r,
Formula (3) is known as the trapezoidalrule.
ExAMPLE 1:
Compute
Pdx Joffi by using the trapezoidal rule with n: 6. Expressthe result to three decimal places.Check by finding the exactvalue of the definite integral.
Because la, bl  [0, 3] and
solurroN:
6, we have
,o:1:0.5 n6
Lx:b Therefore,
f'
d*
Jo LG*x2

+
+ 2f(xr)+ f (x)l [/(ro)+zf(x,)* 2f(xr)+ Zf(xr)+ 2f(xn)
where f (x)  Il(16 + x').The computationof the sum in brackets in the aboveis shown in Table11.8.1.So, [' j'
Jo 6fu:
0'25(0'5427)
I'J*
Q'l.607
Jorcfu:
Rounding the result off to three decimal places, we get
I'J'
Jorcfu:
Q"l'6!
We check by finding the exactvalue. We have It
d*
1 . _^1rlt
J of f i : a t a n ' Z l o
TECHNIQUEO S F INTEGRATION
: + tanl *
* tanl 0
 +(0) : +(0.6435)  Q.1,6A9 to four decimal places Table11.8.1
f (x,)
X1
0 1 2 3 4 5 6
0 0.5 L L.5 2 2.5 3
0.0625 0.0615 0.0588 0.0548 0.0500 0.0450 0.0400
1, 2 2 2 2 2 1,
I
o.o62si
0.L230 I 0.LL76 I 0."1.096 0.1000 0.0900 0.0400 I
 0.6427 io'r@')
i :0
I
To consider the accuracyof the approximation of a definite integral by the trapezoidal rule, we prove first that as Ar approaches zero and.n increases without bound, the limit of the approximation by the trapezoidal rule is the exact value of the definite integral. Let
T: i Lx[f(x) + zf(x) + . . . r 2f(x,_,)+ f(x")] Then
T :l f( xr ) + f( x) + . . . + f( x") l A^x+ilf( x) f( x,) l Lx or, equivalently, n
r: ) f(x) ax + ilf@)  f(b)l Ax i=1 Therefore, if. n >*o and At + 0, we have
lim ttf@)  f(b)) Lx ["] r: aao !* i 'f@1)ax* a*_o ?r
Aao
fb
: I f @ )d x + 0 Ja'
Thus, we can make the difference between Tand the value of the definite integral as small as we please by taking n sufficiently large (and consequently Ar sufficiently small). The following theorem, which is proved in advanced calculus, gives us a method for estimating the error obtained when using the trapez6idal rule. The error is denoted by e7.
11.8 THE TRAPEZOIDAL RULE
1,L.8.LTheorem
Let thre funct: Iu rcti10n f 1 be continuous on the closedinterval fa, bl, and f ' andf " )n both exisl ist or 1 [ a , bb l . If on fbb
tl f(' ( x )
€yT 
d: dl xx' T
,J/a
where3 TI'i is tl he aP rro ximate value of [f f (x) dx found by the trapezoidal th PPI rule, th en threre is3 Ssor then ( me number 4 in la, bl such that €y
EXAMPug 2: Find the bounds for the error in the result of Example L.
__
1 t2
,(b
at ) f '" (q) (Ar)'
(4)
soLUrIoN: We first find the absolute minimum and absolute maximum values of f" (x) on [0, 3]. f ( x ) : ( 1 0+ l )  t '(x) : 2x(16 * f1z f  2(LG+ f):2: 10*  g2)(L6t *1t f,,(x):B*(1.6 * f;s 6x(6xz  32)(1.6* *1+ * L2x(1'6* l; r f "' (x) : :24x(L5
xz)(L6* *1+
Becausef"'(x) > 0 for all x in the open interval (0, 3), then /"'is increasing on the open interval (0,3). Therefore, the absolute minimum value of f " on [0, 3] is f " (0), and the absolute maximum value of f " on [0, 3] is f " (3).
f"(il h
andf"\3)&
T a k i n g n  0 o n t h e right side of (4), w€ get
3 ( r\L _1 :r,.\ru)z
2048 Taking n 3 on the right side of (4), w€ have \L  3122 L2 \E2s )Z
11 45,000
Therefore, if e1 is the error in the result of Example L, we conclude
1L.8 Exercises In Exercises 1 through 14, compute the approximate value of the given definite integral by the trapezoidal rule for the indicated value of z. Express the result to three decimal places.In ExercisesI through 8, find the exact value of the definite integral and compare the result with the approximation.
TECHNIQUESOF INTEGRATION
526
1. ['4,n:S Jr )c'
4. 7.
r2
Jrx!4f fr
J,
z . f " Jt *lx ' , n : 8
3.
?rrdx 5 . J,ffi;n:5
6.
J,
dx;n8
sinr dx;n5
8.
fn
J,
ln(l * x') d x ;n  4 \R
dx;n6
tr cos x2 dx; n:
LL. [' ," dx; nJo
14.
fr
J.
4
5
g.
r2
Jo*"dx;n4 f3
J,
V l + x 2d x ; n  6
sin r dx; n [37rtz x Jnn
6
1 . 2f." + i " d x ; n  6 L*x Jo
VI + xsdx;n: 4
In Exercises15 through 20, find. the bounds for the error in the approximation of the indicated exercises. 15. Exercise1 15. Exercise2 12. Exercise3 18. Exercise6
19. Exercise 1.1
20. Exercise L0
21. The integral Io' e" dx is very important in mathematical statistics. It is called a "probability integral" and it cannot be evaluated exactly in terms of elementary functions. Use the trapezoidal rule with n :5 to find an approximate value and express the result to three decimal places. 22. The region bounded by the curve whose equation is y : gttz, the r axis, the y axis, and the line r: 2 is revolved about. the r axis. Find the volume of the solid of revolution generated. Approdmaie the definite integral by the trapezoidal rule to three decimal places, with z :5. 23' Show that the exact value of l& t/+  f dx ig z. Approximate the definite integral by the trapezoidal rule to three decimal places,with z:8, and compare the value so obtained with the exact value. 24. Show that the exactvalue of.t losdxl (x * 1) is ln 2. Approximate the definite integral by the trapezoidal rule with n : 6 to three decimal places, and compare the value so obtained with the exact value of ln2 as given in a table.
11'9 SIMPSoN'S RULE
1L.9.LTheorem
Another method for approximating the value of a definite integral is provided by Simpson'srule (sometimesreferred to as the parabotiirute). For a given partition of the dosed interval [a, bl, simpson,s rule rr,r"liy gives a better approximation than the trapezoidal rule. Howevei, simpson's rule requires more effort to apply. tn the trapezoidal rule, successive points on the graph of y: /(r) are connected by segments of straight lines, whereas in simpson's rule the points are conhectJaby ,"gments of parabolas. Before simpson's rule is developed, we statl and pnove a theorem which will be needed. If.Po(xs,Ao),Pr(xr, U), and pr(xr, Ar\ are three noncollinearpoints on the parabola having the equation y: A* + Bx * C, where Ao2 0, Ar > 0, Az > 0, xr : xo * h, and xz : xo * 2h, then the measure of the area of the region bounded by the parabola, the r axis, and the lines l: xo arrdx: x, is given by *h(yo*4yr*yr)
(1)
11 . 9 S I M P S O N 'R SU L E
( x o ,y o )
527
pRooF: The parabola whose equation is y : Ax2 * Bx * C has a vertical 'axis. Refer to Fig. 1'1..9.'1., which shows the region bounded by the parabola, the x axis,and the lines r : ro afld x: xz. BecausePo, Pr, and P, are points on the parabola, their coordinates satisfy the equation of the parabola. So when we replace xlby xs * h, and xrby xs*2h,wehave yr: Axoz* Bxsl C lr : A(xo + h)2 + B (16* h) + C : A(xs2I 2hxs* h2) + B (xo+ h) + C U z : A ( x o + 2 h ) z* B ( x s + 2 h ) + C : A ( x n z* 4 h x s +4 h 2 )* B ( x e+ 2 h ) + C Therefore, Ao* 4yr* yr
F i g u r e11. 9 . 1
A ( 6 x o ' * 1 2 h x 0+ t h ' ) + B ( 6 x o + 6 h ) + 6 C
(2)
Now if K square units is the area of the region, then K can be computed by the limit of a Riemann sum, and we have
K lim Atr0
n sl Ll i:l 2h
(At,'* Bt,+ c) L,x (Ax' * Bx + C) dx
 $Ax3+ LBx' +
 +A(ro* 2h)'
2h)2+ C(ro + 2h)  (*Axr' + +Bxo'* Cxs)
 +hlA(6xo'* L2hx0+ th2) + B(6xo+ 6h) + 6Cl
(3)
Substituting from (2) in (3), we get K *hlyot 4y, * yrl
I
Let the function f be continuous on the closed interval [4, b]. Consider a regular partition of the interval la, bl of 2n subintervals (2n is used instead of z becausewe want an even number of subintervals). The length of each subinterval is given by Ar : (b  a) l2n. Let the points on the curve y: f(x) having these partitioning points as abscissasbe denoted by Po(xo,yr), Pt(xt, yr), , Pro(xzn,!zn)) see Fig. 77.9.2,where /(r) = O for all x in fa, bl. We approximate the segmentof the ctfivey : f (x) from Psto P2by the segment of the parabola with its vertical axis through Pr, Pr, and Pr. Then by Theorem 11.9.1the measureof the area of the region bounded by this parabola,the r axis, and the lines r:ro and x: xz,withh: Lx, is given by $ Ar(yo * 4y, * yr) or
* Ar[/(ro) + af k)
+ f (xr)]
In a similar manner, we approximate the segment of the curve
528
T E C H N I Q U EO S F INTEGRATION
N
I N
n
A:Xo.Xr
X2
i
Xs
X+
X5
X6
N F 1
I
\ N
HH
F i g ur e 1 1. 9 . 2
,
+
f1*2n L N
y : f (x) from P" to Paby the segment of the parabola with its vertical axis through P2,Ps,and Pn.The measure of the area of the region bounded by this parabola, the x axis, and the lines r: xs and x: xt is given by
i Lx(yr* 4ys* y,) or t Axff(xr)+ 4f(xr)+ f(x)l This processis continued until we have n such regions, and the measureof the area of the last region is given by
* Axlyror*4ar,'* ar) or t Lx lf(xr^) + +1@r*_r) + f(xr*)l The sum of the measuresof the areas of these regions approximates the measure of the area of the region bounded by thelurv" *io"u uq,rution is y : f(x), the r axis, and the lines r : a andx: b. The measureof the area of this region is given by the definite integral ft fG) dx. so we have as an approximation to the definite integral
t Ax[f(x)+af@)+f(x,)l+ t Axlf(x,) +4f(x")+f(x)]+ . . . + * AxLf@zn) t 4f(x,,r)+ f(xr^r)l+g axlf(xr^_r) + 4f(x2,) + f(xr^)l dx : * Arlf (xo) + a f Q , ) + z f ( x r ) + a f f t s ) + Z f (+ x .)
+ 2f(xzn_z) + 4f(xrn_J+ f Gill where Lx  (b  a)l\n. Formula (4) is known as Simpson,srule.
ExAMPLEL: Use Simpson,s rule to approximate the value of
soLUrIoN' *, and
Applying Simpson's rule with 2n: 4, we have Ar : +(1 
1, 12
lf (xi + af@r)+ 2f(xr)+ af@s)+ f (x)l
. . G)
RULE 1 1 . 9S I M P S O N ' S
with 2n: 4. Give the result to four decimal places.
The computation of the expression in brackets on the right side of (5) where f(x): U(x + l)' is shown in Table 1'!'.9.'l', Substituting the sum from Table 11'9'1in (5), we get
:
+
: Q.69325+ (8.31905)
T a b l e 1 7 . 9. 1
xi 0 t 2 3 41
kt
f@)
0 0.25 0.5 0.75
L.00000 0.80000 0.66567 0.57143 0.50000 4
kr'f@i)
 L.00000 3.20000 4 L33334 2 2.28572 4 10.50000
k,f(x)  8.31e06
i :0
Rounding off the result to four decimal places gives us F dy: J, x+t
Q v '.vG' vg g g
The exactvalue of y; dxl(x + 1) is found as follows:
I:h:'n
l r + t l ] , : r n 2  l n 1: r n 2
From a table of natural logarithms, the value of ln 2 to four decimal places is 0.6931,which agreeswith our aPProximllion in the first three 0.0002' plu."t. And the elror in our aPProximationis In applying simpson,s rule, the larger we take the value of.Zn, the smaller will be the vaiue of. Lx,and so geometrically it seemsevident that the greater will be the accuracyof the approximation, becausea parabola p"rJttg through three points of a curve that are close to each other will be close to the curve throughout the subinterval of width 2 Ar' The following theorem, which is proved in advanced calculus, gives a method for determining the error in applying Simpson's rule' The error is denoted by es. 11.9.2Theorem
' Let the function f be continuous on the closed interval la, bl, and f , f " , ftt' and fcv)all exist on la, bl lf It fb
€s: I f@)dxS Ja
530
TECHNIQUEO S F INTEGRATION
lvhere s is the approximatevalue of [! f (x) dr found by simpson'srule, then there is somenumber 11in fa, bl such that €,s: rto (b  a)f Gv\ (il (Ax)n EXAMPLN2:
Find the bounds for
(6)
SOLUTION:
the effor in Example1.
f(x): (x* 1)' f'(x) f " ( x ) 2 ( x* 1 )  ' f "' (r) : 5(x * t;+ f ( i v )( r ) : 2 4 ( x * t ;  s (x) :  120(x * 1) u 1 o rt . f al \ ljm q i: A t "/r*rr'r)+c.a>o dx I 18. al \ l@:=srnn 19. Find the length of the catenary y:
cosh x from (ln 2, *) to (In 3, t).
20. The graph of the equation
t7r:cginhl 'l\l'VAry' is called a tractrix. Provethat the slope of the curve at any point (x, y) is Vl{FT.
Polarcoordinates
13.1 THE POLARCOORDINATE SYSTEM
13.1.THE POLAR COORDINATE SYSTEM P(r, 0) Figure
F i g u r e1 3 . 1 . 2

7 61r
F i g u r e1 3 . 1 . 3
F i g u r e1 3 . 1 . 4
F i g u r e1 3 . 1 . 5
P(n,t,)
a.A.
1.1 70 F i g u r e1 3 . 1 . 6
Until now we have located a.point in a plane by its rectangular cartesian coordinates. There are other coordinate systems that can be used. Probably the next in importance to the cartesian coordinate system is the polar coordinatesystem.In the cartesian coordinate system, the coordinates are numbers called the abscissa and the ordinate, and these numbers are directed distances from two fixed lines. In the polar coordinate system, the coordinates consist of a distance and the measure of an angle relative to a fixed point and a fixed ray (or half line). The fixed point is called the pole (or origin), and it is designated by the letter "O." The fixed ray is called the polar axis (or polar line), which we label OA. The ny OA is usually drawn horizontally and to the right, and it extendsindefinitely (seeFig. 13.1.1). Let P be any point in the plane distinct from O. Let 0 be the radian measure of the directed angle AOP, positive when measured counters its initial clockwise and negative when measured clockwise, having aas side the ny OA and as its terminal side the ny OP. Then ift r is the undirected distance from O to P (i.e., ,: lOPl\, one set of polar coordinates of P is given by r and 0, and we write these coordinates as (r' 0)' o rLLUsrRArroN1: The point P(4, ta) is determined by first drawing the angle having radian measure $2, having its vertex at the pole and its initial side along the polar axis. Then the point on the terminal side, which is four uriits from the pole, is the point P (see Fig. 13.1.2).Another set of polar coordinatesfor this samepoint is (4,&rr); see Fig. 13.1.3.Furthermore, the polar coordinates (4, #rr) would also yield the same point, as ' shown in Fig. 13.L.4. Actually the coordinates (4, $n *2nn), where n is any integer, give the same point as (4,trr). So a given point has an unlimited number of sets of polar coordinates. This is unlike the rectangular cartesian coordinate system becausethere is a onetoone conespondence between the rectangular cartesiancoordinates and the position of points in the plane, whereas there is no such onetoone correspondence between the polar coordinates and the position of points in the plane. A further example is obtained by considering sets of polar coordinates for the pole. If r: 0 and 0 is any real number, we have the pole, which is designated by (0, 0). We consider polar coordinates for which r is negative. In this case, instead of the point being on the terminal side of the angle, it is on the extension of the terminal side, which is the ray from the pole extending in the direction opposite to the terminal side. So if P is on the extension of the terminal side of the angle of radian measure 0, a set of polar coordinates of P is (r, 0), where r: lOPl. is the same . rLLUSrRArroN2: The point (4, ttr) shown in Fig. 1"3.1'.5 &r), another set 1. Still and (4, fzr) in Illustration point as (4, *n), (4, of polar coordinatesfor this samepoint is (4, l*zr); seeFig. 13.1'6. . The angle is usually meastlred in radians; thus, a set of polar coordi
556
POLARCOORDINATES
nates of a point is an ordered pair of real numbers. For eachordered pair of real numbers there is a unique point having this set of polar coordinates. However, we have seen that a particular point can be given by an unlimited number of ordered pairs of real numbers. If the point p is not the pole, and r and0 arerestricted so thatr > 0 and0 < 0 < 2rr,then there is a unique set of polar coordinates for p. EXAMPLE1: (a) plot the point having polar coordinates (3, 3r). Find another set of polar coordinates of this point for which (b) r is negative and 0 (c) r rs positive and 0 (d) r is negative and 2n < 0 < 0.
solurroN: (a) The point is plotted by drawing the angle of radian measure 3a in a clockwisedirection from the polar axis. Becauser)0,p is on the terminal side of the angle, three units from the pole; see Fig. 1,3.1,.7a. The answersto (b), (c), and (d) are, respectively,(3,$r), (3,trl), and (3, En).They are illustrated in Fig. 1,3.1.2b, c, and d. 4_
5"
o/ / x
+" PQ,_3I
7r
* { p( 3,
\
+c
4 o(_r,
(b)
(a)
+c
(d)
F i g u r e1 3 . 1. 7
(,, (*,
often we wish to refer to both the rectangular cartesian coordinates and the polar coordinates of a point. To do this, we take the origin of the first system and the pole of the second system coincident, the polar axis as the positive side of the r axis, and the ray for which 0 : in as the positive side of the y axis. suppose that P is a point whose representation in the rectangular cartesian coordinate system is (x, y) and (r, g) is a polarcoordinate representation of P. we distinguish two cases:r > 0 and r < 0.In the first case, if r ) 0, then the point P is on the terminal side of the angle of 0 radians,and r: loPl. Such a caseis shown in Fig. 13.1.g.Then cos d: xl lOPl : xlr and sin 0 : VlloPl : Alr; and so xrcose
F i g u r e1 3 . 1 . 8
and yrsrn0
(1)
In the second case, tf r I 0, then the point P is on the extension of the terminal side and rlOTl (see Fig. 13.1.9).Then if a is the point ( x , y), we have x X X x cosg=
=l@T
So x rcos0
;Fl
:,:
r
(2)
13.1 THE POLARCOORDINATE SYSTEM
557
Also,
y) Q( x,
aava
sino:16T:l6FT:=:; Hence,
P
F i g u r e1 3 . 1. 9
y:rsin9 (3) Formulas (2) and (3) are the same as the formulas in (L); thus, the formulas (1) hold in all cases. From formulas (1) we can obtain the rectangular cartesiancoordinates of a point when its polar coordinates are known. Also, from the formulas we can obtain a polar equation of a curve if a rectangular cartesian equation is known. To obtain formulas which give a set of polar coordinates of a point when its rectangular cartesian coordinates are known, we square on both sides of each equation in (1) and obtain 'f: rz cosz0 and A2: 12 sinz0 Equating the sum of the left members of the above to the sum of the right members, we have **y':12
cos2 o * r z s i n zo
or, equivalently, * * Y': rz(sin2o * coszo) which gives us *+Yz:rz and so (4)
Fromthe equationsin (1) and dividinS,we have r s i ne  _ y ,'cosg i or, equivalently, :':: if lt*i+fi i$il1ii.::':::i:i'iiiiiiirii:i;i::i:li':'ii::i
(5)
ri;1i1r1i fffitl++i+++
. rr,r.usrRArrorv3: The point whose polar coordinates are (6, Znt) is plotted in Fig. 13.1.10.We find its rectangular cartesian coordinates. From (L) we have
.10 F i g u r e1 3 . 1
x  r c o s0  5 cosZn
yrsin0  6 stn tn
POLAR COORDINATES
3\n Sothe point is (3 \n,3\n). The graph of an equation in polar coordinates r and 0 consists of all those points and only those points P having at least one pair of coordinates which satisfy the equation. If an equation of a graph is given in polar coordinates, it is called a polar equation to distinguish if from a cartesianequation,which is the term used when an equation is given in rectangular cartesian coordinates. EXAMPLr'2: Given a polar equation of a graph is 12:4stn20 find a cartesian equation.
SoLUTIoN:
Because y2  x, + y, and
sin 20:2 sin 0 cose  zfulr)(xlr),
from (1) we have, upon substituting in the given polar equation,
x 2+ Y ' : 4 ( D + ' ; x2* y',:w r2 xz*Y': !!V = x'+y, (x'*y')'8xy
EXAMPTn3: Find (r, 0) tf r > 0 and0 limaEon has a shape similar to the one in Example 3.
EXAMPIn 3:
Draw
the graph of
3+2stn0 1_
7T
a sketch of
solurroN: The graph is symmetric with respect to the in axis because if (r,0) is replacedby (r, r  0), an equivalent equation is obtained. Table ti.2.2 gives the coordinates of some of the points on the graph. is drawn by plotting the A sketch of the graph is shown in Fig. 13.2.3.It '1,3.2.2 and using the symin Table points whose coordinates are given metry property. T a b l eL 3. 2 . 2
F i g u r e1 3 . 2 . 3
0
0tntnLniT
r
3
4
3+\E
5
3
&n
*n
En
2
3\tr
L
564
POLARCOORDINATES
The graph of an equation of the form 'racosn0
or rastnn?
is a rose,having n leavestf n is odd and 2nLeavesrf n is even.
EXAMPur4: Draw a sketch of the fourleafed rose r4cos20 1
7r
solurroN: In Illustration L we proved that the graph is symmetric with respect to the polar axis, the trr axis, and the pole. Substituting 0 for r in the given equation, we get c o s2 0 : 0 from which we obtain, for 0 < 0 12rr,0:*rr,Zrr,Zr, and #2. Table 13.2.3gives values of r for some values of d from 0 to jz. From these values and the symmetry properties, we draw a sketch of the graph as shown in Fig. 13.2.4. TableL3.2.3 0
0
#n
r
4
2\tr
tn
L+n
tn
#n
tn
2
0
2
z\tr
4
Figure 13.2.4
The graph of the equation 0:C where c is any constant, is a straight line through the pole and makes an angle of C radians with the polar axis. The same line is given by the equation 0:Crnn where n is any integer. general, the polar form of an equation of a line is not so simple In as the cartesian form. However, if. the line is parallel to either the poiar axis or the *zr axis, the equation is fairly simple. If a line is parallel to the polar axis and contains the point B whose cartesian coordinates are (0, b) and polar coordinates ^t" 1b,fz'), then a cartesian equation is y : b. If we replace y by r sin d, we have rsin0:b which is a polar equation of any line paraller to the polar axis. If b is posi
13.2 GRAPHSOF EQUATIONSIN POLARCOORDINATES 565
tive, the line is above the polar
It b is negative, it is below the polar
we have a sketchof the graph of the equa. rLLusrRArroN2: In Fig. 1,3.2.5 ']..3.2.6 g: we have a sketch of the graph of the 3, and in Fig. tion r sin 3. . equation r sin 0: If a line is parallel to the *zr axis or, equivalently, Perpendicular to the polar axis, and goes through the point A whose cartesian coordinates are (a, 0) and polar coordinates are (a, 0), a cartesian equation is r : a. Replacing xby r cos 0, we obtain
F i g u r e1 3 . 2 . 5
rcos0:a
rsin0: Figure cO
tl +tfc)f, L,0 Let llAllbe the norm of the partition A; that is, IlAllis the measureof the largest Lr?.Then if we let A square units be the area of the region R, we define
#*ffi .......#=+U a,o I+r, ft$, Figure13.5.2
The limit in (t) is a definite inte gral, and we have
#$
EXAMPLE1: Find the area of the region bounded by the graph of
r:2+2cos0.
(2)
The region together with an element of. areais shown in Fig. ::l]t"_"' 13.5.3.Becausethe curve is symmetric with respectto the polar axis, we take the 0 limits from 0 to z which determint the area or tn" region bounded by the curve above the polar axis. Then the area of the entire region is determined by multiplying that area by 2. Thus, if A square units is the measure of the required area,
A :2 ,!ip i+ fr *2 l l a l l  o : n 1
4
costi) , Aio
r
+(2 + 2 cos 0)2 d0
I F i g u r e1 3 . 5 . 3
(1)
( 1 + 2 c o s 0 + c o s z 0 )d 0
sin 0 + te + * sin *]:
13.5 AREA OF A REGIONIN POLARCOORDINATES
 4(n* 0 + in * 0  0) :6n Therefore, the area ts 6n square units.
ExAMPLE2: Find the area of the region inside the circle r:3
sin0
3 sin02
sin0
sin0:Lz
and outside the limagon r:2
To find the points of intersection,we set
solurroN:
So 0:tr
sin0 r :, f (0)
Li1 ,0:Ei
' t,) $(Eil (g(€;),tr)
and 0:8n
The curves are sketched and the region is shown together with an element of area in Fig. 13.5.4.   sin 0, then the equation of the If we let f @):3 sin 0 and g(0) I circle is r: f (0), and the equation of the limagon is r: 8(0). The measure of the area of the element of area is the difference of the measuresof the areas of two circular sectors.
tlf G)l' toe ils?)l' 7 :, g(0)
: +([/(fn)]'  [g(f') ]') aoo Ln?
The sum of the measuresof the areas of r such elements is given by n
A'o > +(t/(f')l'[g(f,)]')
C=1
Hence, if A square units is the area of the region desired, we have F i g u r e1 3 . 5 . 4
A: tisr i +ttrtE lt' [g(f,)]')Aoo llalloEi This limit is a definite integral. Instead of taking the limits tr to *rr, we or" thu property of symmetry with respect to the ttr axis and take the limits from Szrto ir and multiply by 2.We have, then, fn12
A:2. 1
( t / ( 0 ) l '  l s @ ) l ' zd)e
J t16
:
f::re
sin
0)'l d0
8
e d0+4 s i n0 d e  4 f: f:sinz 8  c o s 2 0 )d o + f:(1 :402sin204
cos o  Ae'n'' lnrc
576
POLAR COORDINATES
sin 20 4 cos _ (2 sin 7T 4 cos
(2 sin tn  4 cos En)
2.+\E+4.+\n  3{g Therefore,the area is 3 \n square
13.5 Exercises In Exercises L throu gh 6,find
the area of the region enclo sed by the graph of the given equation.
1. r:3cos0
2.rZsin0
3. r4
c o s3 0
4.r:4sin2t0
5. r2:4sin20
6. r:4
s i n 20 c o s 0
In Exercises 7 through 10, find the area of the region encl sed by one loop of the graph of the given equation. 7. r:
3 cos20
9. r"1,+3sin0
8. r:a 10. r:
sin30 a ( l  2 c o s0 )
In Exercises LL through "1,4,find the area of the intersection of the regionsenclosedby the graphsof the two given equations. 1 1 [r2 ^^' lr  3  2 cos 0 sin20 4 rr'^ l r 3 3 cos 2o lr
l r' : 4 : s i n 0 12..t lr4cos0 (2  2 cos20 1,4.{ ' I r 1
In ExercisesL5 through 18,find the areaof the region which is inside the graph of the first equation and outside the graph of the secondequation. 15. [ r o l r a ( l  c o s0 )

1 7 . ft 2 sin o Lrsin0*cosg
 2a srn o L6. {r lra  4 stn 20 j.g. r \ " r [r' L t \n
19. The face of a bow tie is the region enclosed by the graph of the equation 12: 4 cos 20. How much material is necessary to cover the f.aceof the tie? 20. Find the area of the region swept out by the radius vector of the spiral r: not swept out during its first revolution.
a0 duing its second revolution which was
21 Find the area of the region swept out by the radius vector of the curve of Exercise20 during its third revolution which was not swept out during its second revolution.
ReaiewExercises (Chapter13) In Exercises7 and'2, find a polar equation of the graph having the given cartesian equation. l. x2*y'9x+8y:0
Z.yn:*(aryr)
REVIEWEXERCISES
In Exercises3 and4, find a cartesian equation of the graph having the given polar equation. 4. r:
3. r9sinzi0
a tanz 0
3/d (reciprocal spiral) and (b) r: 013(spiral of Archimedes). 5. Show that the equations r: L I sin 0 and /: sin 0  t have the same graph. y'lios 0f. 7. Dtaw a sketch of the gaph o1 7 : 5. Draw a sketch of the graph of (a) r:
8. Draw a sketctr of the graph of r:
{@{E
9. Find an equation of each of the tangent lines at the pole to the cardioid' r: 3 3 sin 0' 10. Find an equation of each of the tangent lines at the pole to the fourleafed rose r:4 cos20. 11. Find the area of the region inside the graph of r:2a
sin 0 and outside the graph of r:
a'
a cos 0 and 12. Find the area of the intersection of the regions enclosed by the graphs of the two equations /: > 0. a 0), cos r: a(l r: stco(k > 0) as 0 varies from 13. Find the area of the region swept out by the radius vector of the logarithmic spiral 0 to 2tr. :2 sin 20 and outside the graph of the circle r: 1' 14. Find the area of the region inside the graph of the lemniscate 12 : (o , n) . 15. Find a polar equation of the tangent line to the curve r 0 at the point the two given equations' In Exercises16 and 17, find all the points of intersection of the graphs of
L6{;:l i;lo,e
aF,  ,2(l+ Lt' lrr2cos0
cos 0)
of points of intersection. Also prove that the 1g. prove that the graphs of r: a0 and.r0: a have an unlimited number and find these points' intersection, tangent lines are perpendicular at only two of these points of each point of intersection of the graphs of the equa19. Find the radian measure of the angle between the tangent lines at 6 cos 0)' cos 0 and,t:2(1 tions r: : point (1, *n)' 20. Find the slope of the tangent line to the curve r 6 cos0 2 atthe * 2 cos 0) and also the area of the region en21. Find the area of the region enclosed by the loop of the limagon r:4(7 closed by the outer part of the limagon' n is a positive integer' 22. Find.the area enclosed by one loop of the curve r: a sin n0, wherc is 23. Prove that the distance between the two points PJh, 0) and'PzQz' 0)
@
24. Find the points of intersection of the graphs of the equations /: tan 0 and r: col 0; (2, tr,) and the line 0 : 0 25. (a) Use the forrrula of Exercise23 to find a polar equation of the parabola having its focus at of the paraboh hJving its focus at (0,2) and the r axis as its directrix' as its directrix. (b) Write a cartesian "qtr"tiotr Compare the results of parts (a) and (b). law of cosinesto 25. Find a polar equation of the circle having its center at .(rs,!6) and a radius of a units. (rur'rr: Apply the (r, (ro, and 0)') 0), pole, at the the triangle hiving vertices
14.1 THE PARABOLA
14.I THE PARABOLA
A conic section is a curve of intersection of a plane with a right cirorlar cone of two nappes. There are three types of curr/esthat occur in this way: the parabola, the ellipse (including the circle as a special case), and the h;rperbola. The resulting cuwe depends on the inclination of the axis of the cone to the cutting plane. In this section we study the parabola. Following is the analytic definition of a parabola.
1,4.1.1Definition
A parabolais the set of all points in a plane equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix. We now derive an equation of a parabola from the definition. In order for this equation to be as simple as possible, we choose the r axis as perpendicular to the directrix and containing the focus. The origin is taken as the point on the r axis midway between the focus and the directrix. It should be stressedthat we are choosing the axes (nof the parabola) in a specialway. SeeFig.1'4.1"1" x:
l
a( p,v)
.1 14.1 Figure
then P is on the parabola if and only if
lFTl:lOPl Because and
lilPl: t/(.T\YTTW P is on the parabola if and onlY if
\/G=pYTT: {GTff
580
THE CONIC SECTIONS
By squaring on both sides of the equation, we obtain xz  2px * p' * y' : )c2* 2px * p' Y' : 4Px This result is stated as a theorem.
14.1.2Theorem
An equation of the panbola having its focus at (p,0) and as its directrix the line y:p is (1)
F(p,o)
F i gur e 14 . 1. 2
14.1,.3Theorem
'!.4.'1..'!,, In Fig. p is positive; p may be negative,however, becauseit is the directed distance Of. Figure 14.1.2shows a parabola for p < 0. From Figs. 14.1.1and 1,4.'1,.2, we seethat for the equation yz :4px the parabola opens to the right if p > 0 and to the leftif p < 0. The point midway between the focus and the directrix on the parabola is called the ztertex.The vertex of the parabolasin Figs. 14.1.1and 14.1.2is the origin. The line through the vertex and the focus is called the axisof the parabola. The axis of the parabolasin Figs. 14.1.1and 1.4.1.2isthe r axis. In the above derivation, if the r axis and the y axis are interchanged, then the focus is at the point F(0, p), and the directrix is the line having the equation y:p. An equation of this parabola is *:4py, and,thi result is stated as a theorem. An equation of the parabola having its focus at (0, p) and as its directrix the line y : p is (2)
If p ; 0, the parabola opens upward as shown in Fig. 74.1.2; and ifp as shown in Fig. 14."1,.4. In each case the vertex is at the origin, and the y axis is the axis of the parabola.
F(0,p)
F i g u r e1 4 . 1. 3
Figure 14.1.4
14.1THE PARABOLA
581
When we draw a sketch of the graph of a parabola, it is helpful to draw the chord through the focus, perpendicular to the axis of the parabola. This chord is called the latus rectum of the parabola. The length of the latus rectum is lapl. (SeeExercise1.7.) EXAMPLE1: Find an equation of the parabola having its focus at (0, 3) and as its directrix the line y  3. Draw a sketch of the graPh.
sor,urroN: Since the focus is on the y axis and is also below the directrix, the parabola opens downward, and p: 3. Hence, an equation of the parabola is *:12y The length of the latus rectum is l4(3)  : 12
Q,
Qt
Qt
u
c
o
A sketch of the graph is shown in Fig. 1'4.1.5. Ary point on the parabola is equidistant from the focus and the directrix. In Fig. 74.L5, three such points (Pr, Pr, an'd Pg) are shown, and we have
 lre, I : lPre, tFP,llka,l lFtr,l IFP, F i g u r e1 4 . 1 . 5
EXAMPLE 2. Given the Parabola having the equation
solurroN:
The given equation is of the form of Eq. (1); so
4p7
u ': 7 x find the coordinates of the focus, an equation of the directrix, and the length of the latus rectum. Draw a sketch of the graPh.
opens to the right. The focus is at the point Becausep  2. The length of the latus F(+, 0). An equation of the directrix is x rectum rs 7. A sketch of the graph is shown in Fig. 1'4.1.6.
Directrix
F i g u r e1 4 . 1 , 6
THE CONIC SECTIONS
L4.1 Exercises For each of the parabolas in ExercisesI thnrugh 8, find the coordinates of the focus, an equation of the directrix, and the length of the latus rectum. Draw a sketch of the curve. 1. x2:4A 4. xz I5y
2' a2:6x
3. y' : 8x
5.f +A:0
6.y2 5r:0
7.2y'9x:0
8.3x2*4y0
In Exercises 9 through 16, find an equation of the parabola having the given properties. 9. Focus, (5, 0); directrix, y: 5. 10. Focus, (0, 4); directrix, A: 4. 11. Focus, (0, 2); directrix, y  2:0. 12. Focus, (8, O); directrix, 5  3r: 0. 13. Focus,(*, 0); directrrx,2xl l:0. 14. Focus, (0, i); directrix,Sy t2:0. 15. Vertex, (0, 0); opens to the leff length of latus rectum: 6. 15. Vertex, (0, 0); opens upward; length of latus rectum: 3. 17. Prove that the length of the latus rectum of a parabola is l4pl. 18' Find an equation of the parabola having its vertex at the origin, the r axis as its axis, and passing through the point (2, 4\. 19' Find an equation of the parabola having its vertex at the origin , the y axis as its axis, and passing through the point (2, 4). 20' A parabolic arch has a height of 20 ft and a width of 36 ft at the base. If the vertex of the parabola is at the top of the arch, at what height above the base is it 1g ft wide? 21' The cable of a suspension bridge hangs in th9 form of a parabola when the load is uniformly distributed horizontally. The distance between two towers is rsoo ft, the points ;i;;;6; of the cable on the to*"r, arc 220fr above the road
il1til*:T#ffiit:HfiJ?ff31""T"70rt
aloveth",oa'd*"v. Findthevertical distance tothecabre rromapoint
22' Assume that water issuing from the end of a horizontal pipe,25 t lb9",u the ground, describesa parabolic curve, the vertex of the parabola being at the end of the pipe.r, ui u'poit t 8 ft below trr"etine of the pipe, the flow of water has curved outward 10 ft beyond a vertical line throuih th; ;"e iiir" pip", how far beyond this vertical line will the water strike the ground? 23' A reflecting telescopehas a parabolic mirror for which the distance from the vertex to the focus is 30 ft. If the distance acrossthe top of the mirror is il in , how deep is the mirroiai ttre centerz 24' Using Definition 14'1.1,find an equation of the parabola having as its directrix the line a :4 and.asits focus the point (3, 8). 25' using Definition t4'l'1, hnd' an equation of the parabola having as its directrix the line r:3 point (2,5).
and as its focus the
26' Find all points on the parabola y2: 8r such that the foot of the perpendicular drawn from the point to the directrix, the focus, and the point itself are verrices of an equirateiJil;gil,' 27' Find' an equation of the circle passing through the vertex and the endpoints of the latus rectum of the parabola ,,  gy.
14,2 TRANSLATIONOF AXES
2g. prove analytically that the circle having as its diameter the latus rectum of a parabola is tangent to the directrix of the parabola. 29. A focal chord of a parabola is a line segment through the focus and with its endpoints on the parabola. If A and B are the endpoints of ifocal chord of a parabola, and ii C is the point of intersection of the directrix with a line thrcugh the vertex and point A, prove that the line through C and B ii parallel to the axis of the parabola. is half the fl). prove that the distance from the midpoint of a focal chord (see Exercise 29\ of. a parabola to the directrix length of the focal chord.
14.2 TRANSLATION OF AXES
The shape of a curve is not affected by the position of the coordinate axes; however, an equation of the curve is affected' . rllusrRArrorv 1: If a circle with a radius of 3 has its center at the point (4, 1), then an equation of this circle is
(x4)'+@+1)':e **Y'8r*2Y*8:0 Howevef, if the origin is at the center, the same circle has a simpler equation, namelY, ' f+Yz:g Ifwemaytakethecoordinateaxesasweplease,theyaregenerally chosen in such a $/ay that the equations will be as simple as possible. Iftheaxesaregiven,however,weoftenwishtofindasimplerequation of a given curve referred to another set of axes' Ingeneral,ifintheplanewithgivenrandyaxes,Itewcoordinate a chosen parallel to the given ones, we say that there has been "re" """s ttanslation of axesin the Plane' y' In particular' let the gi*tett x and'y axes be translated to the x' and that assume Also, axes. axes,hiving origin (h, k)withrespect to the givgn x' and y' the positive no*b"rs are on the iame side of the origin on the axes as they are on the x and'y axes (see Fig' 1'a'2'1)'
(h,k)
A, l
N,
I I I I
Figure14.2.1
A point P in the plane, having coordinates (x, y) wilh resPectto the the given coordinate axes; wiil have coordinates (x' , y') with resPectto
584
THE CONIC SECTIONS
new axes.To obtain relationships between these two sets of coordinates, we draw a line through P parallel to the y axis and the y' axis, and also a line through P parallel to the r axis and the I' axis. Let the first line intersect the r axis at the point A and the r' axis at the point A', and the second line intersect the y axis at the point B and the y' axisat the point B'. With respect to the r and y axes,the coordinates of P ate (x, y)' the coordinates of A are (r, 0), and the coordinates of A' are (r, k). Because A,p:nM,wehave y':Yk or, equivalently, y:y'*k With respect to the x and y axes, the coordinates of B are (0, y), and the coordinates of B' are (h, y). BecauseB'P : BP  BB' , we have X,:Xh
or, equivalently, x:x'*h These results are stated as a theorem. 14.2.1Theorem lf (x, y) representsa point P with respectto a given set of axes,and (x' ,y') is a representation of P after the axesare translated to a new origin having coordinates (ft, k) with respect to the given axes, then
yv' +k
(1)
y'y k
(2)
ot,
Equations (1) or (2) are called the equationsof translating the axes. If an equation of a curve is given in x andy, then an equation in r' and y' is obtained by replacing x by (x' t h) and y by (y' f k).The graph of the equation in r and y, with respect to the x and y axes,is exactly the same set of points as the graph of the corresponding equation in r' and y' with respect to the x' and y' axes. ExAMPLE1: Given the equation x2 + Llx * 6y + 19: 0, find an equation of the graph with respect to the x' and y' axes after a translation of axes to the new origin (5, 1).
soLUrIoN: A point P, represented by (x, y) with respect to the old axes, has the representation(r', y') with respectto the new axes.Then by Eqs. (1), with h: 5 and k : 1, we have x'5
and yy'
+1
Substituting these values of r and y into the given equation, we obtain
( x '  5 ) , + 1 0 ( r ' , 5 ) + 6 ( y ' , +1 ) + 1 g : 0
14.2 TRANSLATIONOF AXES
x'2  "1.0x' + 25 * l}x'
v+ x'
Figure 14.2.2
50 * 6y' + 6 + 19  0 )c'2: 6Y'
The graph of this equation with respect to the r' and y' axesis a parabola with its vertex at the origin, opening downward, andwith 4p :6. The graph with respect to the x and y axes is, then, a parabola having its vertex at (5, 1), its focus at (5, *), and as its directrix the line y : $ (seeFig. 1,4.2.2).
The above example illustrates how an equation can be reduced to a simpler form by a suitable translation of axes. In general, equations of the second degree which contain no term involving xy can be simplified by a translation of axes. This is illustrated in the following examPle.
EXAMPLE2: Given the equation * 32y + 37: 0, 9xz * 4y'  "1,8x translate the axes so the equation of the graph with respect to the )c' and y' axes contains no firstdegree terms.
vy'
solurroN: We rewrite the given equation: 9(*  2x) + 4(yz * 8y) :37 ' Completing the squares of the terms in parenthesesby adding 9 I' and 4 ' 16 on both sides of the equation, we have
9(*  2x* 1) + +(y' * 8yr 1'6):37 + 9 + 64 9 ( x  L ) ' + 4 ( y + 4 ) 2: 3 6 Then, if we let r'
1 and y' : y + 4, we obtain
9x'2* 4y'' :35 : yI 4 From Eq. (2), wesee that the substitutions of r' : x  L andy''l'4.2.3' 4). we In Fig. result in a translation of axes to a new origin of.(1, have a sketch of the graph of the equation in r' and y' with resPect to the r' and v' axes. Figure14.2.3
:# Figure 14.2.4
We shall now apply the translation of axes to finding the general equation of a parabola having its directrix parallel to a coordinate axis and its vertex at the point (h, k). In particular, let the directrix be parallel to the y axis. If the vertex is at point V(h, k), then the directrix and the focus is at the point F(h *p, k). x, has the equation x:hp, Let the .r' and y' axes be such that the origin O' is at V(h, k) (see Fig.1,4.2.4). An equation of the parabolain Fig. 1'4.2.4with respectto the r' and *u' axesis y'':
4px'
THE CONIC SECTIONS
To obtain an equation of this parabola with respect to the x and y axes,we replacex' by (x  h) andy' by (y  k) from Eq,.(2),whichgives us (Yk)':4P(xh) The axis of this parabola is parallel to the I axis. Similarly, if the directrix of a parabola is parallel to the r axis and the vertex is at V(h, k), then its focus is at F(h, k + p) and the directrix has the equation A : k  p, and an equation of the parabola with respect to the x and y axes is ( x  7 1 2 : 4 P ( Y k ) The axis of this parabola is parallel to the y axis. We have proved, then, the following theorem. 1.4.2.2Theorem
If p is the directed distance from the vertex to the focus, an equation of the parabola with its vertex at (h, k) and with its axis parallel to the x axis is
'i{.p'.H. ; Aik..ru}*.
fr)
(3)
A parabola with the same vertex and with its axis parallel to the y axis has for an equation (r  h)' : 4P{V k)
EXAMPLE 3: Find an equation of the parcbola having as its directrix the line U : 1,and as its focus the point F( 3, 7), F( 3,7)
v( t, +)l
(4)
soLUrIoN: Since the directrix is parallel to the x axis, the axis will be parallel to the y axis, and the equation will have the form (4). since the vertex v is halfway between the directrix and the focus, v has coordinates (3, 4).The directed distance from the vertex to the focus is p, and so P:74:3 Therefore, an equation is (r * 3)'z:12(y  4) Squaring and simplifying, we have * * 6 x  ' l  2 y* 5 7 : 0
Figure14.2.5
ExAMPrn 4: Given the parabola having the equation
y'+6x*8y*1:0
A sketch of the graph of this parabolais shown in Fig. 1.4.2.5.
sol.urroN:
Rewrite the given equation as
y ' + 8 y  6 x  1 Completingthe squareof the terms involving y on the left side of this
OF AXES 14.2TRANSLATION find the vertex, the focus, an equation of the directrix, an equation of the axis, and the length of the latus rectum, and draw a sketch of the graph.
v
equation by adding L5 on both sides/ we obtain
y'+8y*'1.6:6x+L5 (y + 4)' 6(x B) Comparing this equation with (3), we let
Directrix
h:E
k4 and 4P6
or
P:g
Therefore,the vertex is at ($, 4); an equation of the axis is y:4;the focus is at (7,4); an equation of the directrix is x:4; and the length of the latus rectum is 6. A sketch of the graph is shown in Fig. 14'2.6.
Figure 14.2.6
14.2 Exercises In Exercises1 through 8, find a new equation of the graph of the given equation after a translation of axesto the new origin as indicated. Draw the original and the new axes and a sketch of the graph. 2. x2+ y2 L}x * 4y + 13 0; (5,2) L . x 2+ y ' + 6 x * 4 Y : 0 ; (  3 ,  2 )
3.y26x*9:0; (8,01 5. xz+ 4y' * 4x * 8y + 4 : 0; (2, l) 7 . y  4  2 ( x 1 ) t ;( 1 ,4 )
4. y' + 3x  2y * 7  0; (2, t) 6 . 2 5 x z* y '  5 0 r * 2 0 y  5 0 0 : 0 ; ( 1 ,  1 0 ) 8. (Y+ l)' : 4(x  2)'; (2' I)
contain no In Exercises9 through 12, translate the axes so that an equation of the graph with respect to the new axeswill graph. the of and a sketch axes new the firstdegree terms. Draw the original and 10. l5x2 * 25y2 32x  l00y  284 0 9. x2 + 4y'  l5x * 24y + 84 0 L2. x:2 y' + L4x 8y 35  0 L1..3x2 2y' * 6x 8Y LL : 0 In Exercises13 through 18, find the vertex, the focus, an equation of the axis, and an equation of the directrix of the given parabola. Draw a sketch of the graph. 1 5 .y ' + 6 x * 1 0 y + 1 9  0 1 , 4 .A x z 8 r * 3 Y 2 : 0 L 3 .x 2 * 5 x * 4 y * 8 : 0 1 8 .Y : 3 x 2  3 x * 3 17. 2y' 4y 3x 16.3y'8xl2Y40 In Exercises19 through 28, find,an equation of the parabola having the given properties. Draw a sketch of the graph. 19. Vertexat (2,4); focusat (3,4). 20. Vertex at (1, 3); directrix, Y: 1. 21. Focusat (7,7); directtu, Y:3. 22. Focusat (*,4);
directrix, x:*.
588
THE CONIC SECTIONS
23. Vertex at (3 , 2); axis, r : 3; length of the latus rectum is 5. 24. Axis parallel to the r axis; through the points (I,Z), (S,Z), and (11,4). 25. Vertex at (4,2); acis,y :2; through the point (0, 6). 25. Directrix, x: 2; a0.It can be shown bycompleting the squares in r and y that an equation of the form (5) can be put in the form
( x h ) ' 11
(v k)'
A
C
(6)
CONIC SECTIONS
If. AC > 0, then A and C have the same sign. If G has the same sign as A and C, then Eq. (5) can be written in the form of (2) or (3). Thus, the graph of (5) is an ellipse. 2: Suppose we have the equation o ILLUSTRATToT.T 6**9y224x54y*51:0 which can be written as 6(f4x)+9(y26y):5L Completing the squares in r and y, we get 6(f 4x* 4) +9(y'6y t 9):51 +24+81 or, equivalently, 6(x2)2+9(V3)2:54 which can be put in the form (x2)'
_r
(y 3\' , ]:
J?
This is lr, un,r",roi of the form of Eq. (6).By dividing on both sidesby 54 we have ( x  2 ) ' _ ( yz: 3)'_.,
9 
'
which has the form of Eq. (2).
o
If in Eq. (5) G has a sign opposite to that of A and C, then (5) is not satisfied by any real values of r and y. Hence, the graph of (5) is the empty set. . rLLUsrRArroN3: Supposethat Eq. (5) is 6**9y224x54y*1L5:0 Then, upon completing the squares in r and y, we get 6(x2)z a9(! 3)':115 + 24+8L which can be written as (x2\' i
(y 3)' T, =___i_:
_10
(7)
T h i s i s l r , n " f o r  t E q , . ( 6 ) , w h e rGe :  1 0 , A : 6 , a n d C : 9 . F o r a l l values of r and y the left side of Eq. (7) is nonnegative; hence, the graph of (7) is the empty set. . If G:0
in (6), then the equation is satisfiedby only the point (h, k).
14.6 THE ELLIPSE
609
Therefore, the graph of (5) is a single point, which we call a pointellipse. o rLLUsrRATroN4: Becausethe equation 6f*9y224x54Y*105:0 can be written as
(*
,2)'*(y *+
_ . 9 )_' o '
its graph is the point (2, 3).
If the graph of Eq. (5) is a pointellipse or the emPty set, the graph is said to be degenerate. lf. A: C in (5), we have either a circle or a degeneratecircle, as mentioned above. A circle is a limiting form of an ellipse. This can be shown by considering the formula relating a, b, and.e fot an ellipse: bz: az('[,_d) From this formula, it is seent$at as e approacheszero, b2approaches 42. lf b2: az,Eqs. (2) and (3) become (x h)'+ (y  k)': az which is an equation of a circle having its center at (h, k) and radius a. We see that the results of Sec.1.6 for a circle are the sameas those obtained for Eq. (5) applied to an elliPse. The results of the above discussion are summarized in the following theorem. 14.5.1Theorem
If in the generalseconddegreeEq. (4) n:0 and AC > 0, then the graph is either an ellipse, a pointellipse, or the empty set. In addition, if A: C, then the graph is either a circle, a pointcircle, or the empty set'
Determine the graph of the equati on 25x2* I6Y2 * 150r  L28y  LIL9: Q.
soLUrIoN: From Theorem'1,4.5.7, becauseB:0 and ng:(25)(t0): 4O0> 0, the graph is either an ellipse or is degenerate. Completing the squares in r and y, we have 25(x2* 6x + 9) + l6(y',  8y * 16)  1119+ 225+ 256
EXAMPLE1:
2 5 ( x+ 3 ) ' + l 6 ( y  4 ) ' : (x*3)2 (y+)':1 100 54
L500 *
(8)
Equation (8) is of the form of Eq. (3), and so the graPh is an ellipse having its principal axis parallel to the y axis and its center at (3' 4).
610
T H E C O N I CS E C T I O N S
ExAMPLE2: For the ellipse of Example 1.,find the vertices, foci, directrices, eccentricity, and extremities of the minor axis. Draw a sketch of the ellipse and show the foci and the directrices.
solurroN: From Eq. (8) it follows that a:'1.0 and b : 8. Becausethe center of the ellipse is at (3, 4) and the principal axis is vertical, the vertices are at the points V'(3, 5) and V(3,l4). The extremitiesof the minor axis are at the points B'(_1t'1.,4)and B(5,4). Becauseb: a{1V, we have
8:101T=7 and, solving for e,weget e: $. Consequently,ae:6 and ale:ry. Therefore, the foci are at the points F'(3, 2) and F(3, L0). The corresponding directrices have, respectively, the equations y : + and y: 92. A sketch of the ellipse, the foci, and the directricesare in Fig. 14.6.1.
v V:
directrix
3
( z,t+) v
\
(3, L0) F
(  1 1 , 4)
B',
?3,4)
\t   r I I r
B (s,4) tl
r, r,J
F,
/ .J
(3, 6) V',
y:
 T38
directrix
F i g u r e1 4 . 6 . 1
EXAMPLE3: Find an equation of the ellipse for which the foci are at ( 8, 2) and (4,2) and the eccentricity is 3. Draw a sketch of the ellipse.
soLUrIoN: The center of the ellipse which is halfway between the foci is the point (2, 2). The distance between the foci of any ellipse is 2ae, and the distance between (8,2) and (4, 2) is L2. Therefore, we have the equation Zae: 12. Replacing e by *, we get 2a() : 12, and so a:9. Because b : a{177, we have
b:9lT=@:9lT=4:3rE The principal axis is parallel to the x axis; hence, an equation of the ellipse is of the form of Eq. (2).Because(h, k): (2,2), a:9, and b : 3V5, the

14,6THE ELLIPSE
611
required equation is
( x! 2 ) ' 8L
*
(y _z)'_ 45
1
A sketchof this ellipse is shown in Fig. "1.4.6.2.
( 2,2+ 3\/s)
\ ,2)
(  1 1 , 2)
lrrrrltl \
 ___L!.?)____ 7, 2)
lttlll/
o ./
(2,2  3\/s) F i g ur e 1 4
We conclude this section with a theorem that gives an altemative definition of an ellipse. The theorem is based on a characteristic property of the ellipse. 14.6.2 Theorem
d'
/(+,vp
r
/h o .,,/
,p,
An ellipse can be defined as the set of points such that the sum of the distances from any point of the set to two given points (the foci) is a constant. pRooF: The proof consists of two parts. In the first part we show that the set of points defined in the theorem is an ellipse. In the second part we prove that any ellipse is such a set of points. Refer to Fig 14.6.3 in both parts of the proof. Let the point Pt(rt, Ar) be any point in the given set. Let the foci be the points F'(ae,0) and F(ae,0), and let the constant sum of the distances be 24. Then
lrEl + 1fi1:2o Using the distance formula, we get
Y   A ' L e
F i g u r e1 4 . 6 . 3
{G=7*Tt7
* {Q,*i{tT y}:26 \/@=1d"TT : 2a \/Cn ad"W
Squaring on both sides of the above equation, we obtain x(  2Ae4 * azE * yl : 4a2 Aa\/(it + aAY + y7 * xr' * 2aex1+ a2e2* yr'
7 612
THE CONIC SECTIONS
or, equivalently,
\/GfidrTt!:
a+ exl
Squaring on both sides again, we get xr2+ 2aeh I azez* yr' : a2+ 2nex,* dxr2 ('l' E)x12* Yt': a2(1' e2) x,2
:
71,2
J !::1
a2(l  8) Becauseb2: az(l  e), and replacing 11and yrby x and y, respectively, we obtain a2
f
1t2
7+ fr:1 which is the required form of an equation of an ellipse. Now consider the ellipse in Fig. 1.4.6.3;Pr(xr,yr) is any point on the ellipse. Through P, draw a line that is parallel to the r axis and that intersects the directrices d and d' at the points Q and R, respectively. From Definition 14.3.1it follows that
lFql: elR4l and lFEl: 'ltr'al Therefore,
lrEl + lFEl:'(lR4l+ lFrel)
(e)
RPt and PrQ are both positive because an ellipse lies between its directrices. Hence,
:2 (3):'+ Ro
(10)
Substituting from (10)into (9),we get
l F E l+ l E l : 2 a which proves that an ellipse is a set of points as described in the theorem. I
o rLLUsrRArrow 5: The ellipse of Example L has foci at (3, 2) and (3, 10), and 2a: 20. Therefore, by Theorem '1.4.6.2 this ellipse can be defined as the set of points such that the sum of the distancesfrom any point of the set to the points (3, 2) and (3, 10) is equal to 20. Similarly, the ellipse of Example3 can be defined as the set of points such that the sum of the distancesfrom any point of the set to the points (8,2) and (4,2) is equal to 18.
14.7THE HYPERBOLA 013
Exercises L4.6 In Exercises1 through 5, find the eccentricity, center, foci, and directrices of each of the given ellipses and draw a sketch of the graph. 3.5x2* 3y'3yL2:0 2. 9x2* 4y'  l8x * l6y  1L : 0 1. 5x2+ 9y' 24x 54y+ 51  0 5. 4x2* 4y' * 20x 32y + 89 0 6. 3x2* 4y' 30r * L6y + 100 0 4. Zxz* 2y'  2x * L8y + 33 0 In Exercises7 through 12, find. an equation of the ellipse satisfying the given conditions and draw a sketch of the graph. 7. Foci at (5,0) and (5,0);
one directrix is the line x:20.
8. Vertices at (0, 5) and (0, 5)
and passing through the point (2,tr6l
9. Center at (4, 2) , a vertex at (9 , 2), 10. Foci at (2,3) and (2,7) LL. Foci at (L,l)
.
and one focus at (0 , 2).
and eccentricity of &.
and (1 ,7) and the semimajor axis of length 8 units.
L2. Directrices the lin es y  3 + +&s and a focus at (0 , 2) .
13. The following graphical method for sketching the graph of an ellipse is based on Theorem 14.6.2.Todraw a sketch of the graph of the ellipse 4* * y2: 15, first locate the points of intersection with the axes and then locate the foci on the y axis by use of compassesset with center at one point of intersection with the r axis and with radius of 4. Then fasten thumbtacks at each focus and tie one end of a string at one thumbtack and the other end of the string at the second thumbtack in such a way that the length of the string between the tacks is2a:8. Place a pencil against the string, drawing it taut, and describe a curve with the point of the pencil by moving it against the taut string. When the curve is completed, it will necessarilybe an ellipse, becausethe pencil point describes a locus of points whose sum of distances from. the two tacks is a constant. 14. Use Theorem 14.6.2to find an equation of the ellipse for which the sum of the distancesfrom any point on the ellipse to (4, 1) and (4, 7) is equal to 12. 15. Solve Exercise 14 if the sum of the distances from (4, 5) and (5, 5) is equal to 16. 15. A plate is in the shape of the region bounded by the ellipse having a semimajor axis of length 3 ft and a semiminor axis of length 2 ft. If the plate is lowered vertically in a tank of water until the minor axis lies in the surface of the water, find the force due to liquid pressure on one side of the submerged portion of the plate. 77. lI fhe plate of Exercise 15 is lowered until the center is 3 ft below the surface of the water, find the force due to liquid pressure on one side of the plate. The minor axis is still horizontal.
74.7 TII'E HYPERBOLA
In Sec.14.5 we leamed that an equation of a hyperbola having its center at the origin and its principal axis on the r axis is of the fotm *laz  y'lb': 1..Interchanging x and y in this equation, we obtain
(1) which is an equation of a hyperbola having its center at the origin and its principal axis on the y axis.
614
THE CONIC SECTIONS
o ILLUsrnnrroN L: The hyperbola with an equation
v' 9
x2  1 4 1,6
has its foci on the y axis because the equation is of the form (1).
o
Note that there is no general inequality involving a andb corresponding to the inequality a > b f.or art ellipse. That is, for a hyperbola, it is possible to have a 1b, as in Illustration 1.,where a:3 and b:4; or it is possible to have a ) b, as for the hyperbola having the equation y'125*lt6:1, where a:5 and b:4. If.,for a hyperbola,a:b, then the hyperbola is said to be equilateral We stated in Sec.1.4.4that a hyperbola has asymptotes, and we now show how to obtain equations of these asymptotes. In Sec.A.2hofizontal and vertical asymptotes of the graph of a function were defined. What follows is a more general definition, of which the definitions in Sec.4.2 are special cases. 14.7.1Definition Thegraphof theequation y:f(x) hastheliney:mx*basan tote if either of the following statements is true:
asymp_
(i) lim
lf(x)(mx+b)l:0 (ii) lim l f ( x ) ( m x + b ) l : 0 t*@
t@
Statement (i) indicates that for any € > 0 there exists a number N > 0 such that
lf(x)
( m x+ b ) l < e
whenever r > N
that is, we can make the function value /(r) as close to the value of mx * b as we please by taking r large enough. This is consistent with our intuitive notion of an asymptote of a graph. A similar statement may be made for part (ii) of Definitron 1,4.7.1.. For the hyperbola f la'  y'lV : 1.,upon solving for y, we get
\m So if
f(x)
b a
\tr4
we have lim ff*@
r i mP \ f f i  l r l
r*6
LA
A
 ffii(ffi*r) :_ b , l:r_ m O \m* 6r r+a
J
x
14.7THEHYPERBOLA615 b ,.
oP
;"11ffi+x :Q
Therefore, by Definition'1.4.7.1, we conclude that the line A: bxla is an asymptote of the graph of. y:bt@:Fla. Similarly, it can U:_glovvn that the line y:bxla is an asymptote of the graph of y:bV*azla. Consequently, the line y:bx/a is an asymptote of the hyperbola f laz y'lV:1.. [n an analogousmanner, we can demonstrate that the line y:bxla is an asymptote of this same hyperbola. We have, then, the following theorem. 14.7.2Theorem The lines b Y: o*
and
b y:x,
are asymptotes of the hyperbola
f _u't a2W
Figure 14.7.1shows a sketch of the hyperbola of Theorem'1.4.7.2together with its asymptotes. In the figure note that the diagonals of the rectangle having vertices at (a, b), (a, b) , ( a, b), and ( a, b) are on the asymptotes of the hyperbola. This rectangle is called the auxiliary rectangle of the hyperbola. The vertices of the hyperbola are the points of intersection of the principal axis and the auxiliary rectangle.A fairly good sketch of a hyperbola can be made by first drawing the auxiliary rectangle and then drawing the branch of the hyperbola through each vertex tangent to the side of the auxiliary rectangle there and approaching asymptotically the lines on which the diagonals of the rectangle lie. There is a mnemonic device for obtaining equations of the asymptotes of a hyperbola. For example, for the hyperbola having the equation 1, if the right side is replaced by zeto, we obtain flaz *la'y'lbz y2lb':0. Upon factoring,this equationbecomes(xla Vlb) (xla * ylb) : 0, which is equivalentto the two equationsx/a ylb:0 and xla * ylb:0, which, by Theorem 74.7.2,are equations of the asymptotes of the given hyperbola. Using this device for the hyperbola having Eq. (1), we see and that the asymptotes are the lines having equations yla xlb:0 yla* xlb:0, which are the same lines as the asymptotesof the hyperbola with the equation *lbz  f lo': 1. These fwo hyperbolas are called conjugatehyperbolas. The asymptotes of an equilateral hyperbola (a: b) are perpendicular to each other. The auxiliary rectangle for such a hyperbola is a square, and the transverse and conjugate axes have equal lengths.
616
THE CONIC SECTIONS
If the center of a hyperbola is at (h, k) and its principal axis is parallel to the x axis, then if the axes are translated so that the point (h, k) is the new origin, an equation of the hyperbola relative to this new coordinate system is *u2
7F:
t
If we replace 7 by x  h and ! by V  k, this equation becomes (2) Similarly, an equation of a hyperbola having its center at (h, k) and its principal axis parallel to the y axis is (3)
EXAMPLE1: The vertices of a hyperbola are at (5 , 3) and (5 , I), and the eccentricity is \8. Find an equation of the hyperbola and equations of the asymptotes. Draw a sketch of the hyperbola and the asymptotes.
solurroN: The distance between the vertices is 2a, and so ,, : L. For a hyperbola, b: alV  7, and therefore b: t{5  7: 2. Becausethe principal axis is parallel to the y axis, an equation of the hyperbola is of the form (3). The center (h, k) is halfway between the vertices and is therefore at the point (5,2). We have, then, as an equation of the hyperbola (y+2)'_(r*5)2_,, 1, 4:r Using the mnemonic device to obtain equations of the asymptotes, we have lu*2
x*5\ly*2*r*5\:n
\=u/t
1
2 t
which gives y*2:*(r+5) Figure 14,7.2
and y *2:*(x+5)
A sketch of the hyperbola and the asymptotes are in Fig. 1,4.7.2. If in Eqs. (2) and (3) we eliminate fractions and combine terms, the resulting equations are of the form A* + Cyz* Dx * Ey l F:0 (4) where A and C have different signs; that is, AC < 0. We now wish to show that the graph of an equation of the form (4), where AC < 0, is either a hyperbola or it degenerates.Completing the squares in x and y in Eq. (4), where AC < 0, the resulting equation has the form a2(xh)2
B'(yk)':H
(s)
14.7 THE HYPERBOLA 617
lf H ) 0, Eq. (5) can be written as (xh)'
(Vk)'_.
T_T:I q:2
B'
which has the form of Eq. (2). o rr,r,usTnarron2: The equation 4xz l2yz I 24x * 95y  18L 0 can be written as 4(* * 6x)  lz(y'  8y) : L8L and upon completing the squaresin r and Ar wE have , 4(x' * 5x * 9)  tZ(y'  8y + 16) : 181+ 36  192 which is equivalent to 4(x+3)212(y4)2:25 This has the form of Eq. (5), where H: (x*3)'
25 > 0. It may be written as
(y4)'_.,
TT:I which has the form of Eq. (2). If in Eq. (5), H < 0, then (5) may be written as (xh\' Uk)' lE[lE[r q2 which
.1
B',
has the form of Eq. (3).
o rLLUsrRArron3: Suppose.thatEq. (4) is 4*12f*24xI96y131:0 Upon completing the squares in r and y,we get 4(x*3)2tZ(Y4)2:25 This has the form of Eq. (5), where H:25 (y4)'_(x*3)2 2n at
26 T
( 0; and it may be written as
_,
which has the form of Eq. (3). It H = Q in Eq. (5), then (5) is equivalent to the two equations a(x  h)  BQ  k) : 0 and u(x h) + P(y  k) : 0
618
THE CONIC SECTIONS
which are equations of two shaight lines through the point (h, k). This is the degeneratecaseof the hyperbola. The following theorem summarizes the results of the above discussion. 14.7.3 Theorem
If in the general seconddegreeequation Af + Bxy t Cyz* Dxt Ey * F:0 B : 0 and AC < 0, then the graph is either a hyperbola or two intersecting straight lines.
ExAMPLE2: Determine the graph of the equation 9x2 4y' 18r  l6y + 29 0
soLUTroN: From Theorem 14.7.3, because B:0 and AC: (9) (_4) _ 36 ( 0, the graph is either a hyperbola or two intersecting straight lines. Completing the squares in r and A, we get
9 ( x '  2 x + 1 )  + ( y ' * 4 y+ 4 )   2 9 + 9  1 , 5
(6)
Equation (6) has the form of Eq. (3), and so the graph is a hyperbola whose principal axis is parallel to the y axis and whose center is at (1, 2). ExAMPLE3: For the hyperbola of Example 2, find the eccentricity, the vertices, the foci, and the directrices. Draw a sketch of the hlryerbola and show the foci and the directrices.
v(L,1)
//
/ /a v :
,2 *, o
nt/t3
x,
t..! F' d
I I \ I V'(L,  5 ) I I I I
2 
O

nVts
Figure14.7.3
solurrorT: From Eq. (5) we see that a : 3 and b:2. b: a{e2  1; thus,
For a hyperbola,
2:31F1 and solving for e, we obtain , : 4l/I3. Becausethe center is at (1, 2), the principal axis is vertical, and a:3, it follows that the vertices are at the points V'(1,p andV(1,1). Becauseae: {8, the foci are at the points F' (1, 2  Vi3) and F(1, Z + tfr). Furthermore, a/e: #ffi; hun.", the directrix corresponding to the focus at F' has as an equationy:2  rtv13, and the directrix corresponding to the focus at F has as an equation y : 2 * r+Vi3. Figure 14.7.3 shows a sketch of the hyperbola and the foci and directrices.
14.7 THE HYPERBOLA 6 1 9
Just as Theorem 1.4.6.2 gives an altemate definition of an ellipse, the following theorem gives an alternate definition of a hyperbola. 1'4.7.4Theorem
A hyperbola can be defined as the set of points such that the absolute value of the difference of the distances from any point of the set to two given points (ihe foci) is a constant. The proof of this theorem is similar to the proof of Theorem 14.5.2 and is left as an exercise (see Exercises13 and 14). In the proof, take the foci at the points F'(ae,0) and F(ae,O),and let the constant absolute value of the difference of the distances be 2a. See Fig. 14,7.4.
Pt(x1, y)
o rlr.usrnarron 4: Because the hyperbola of Example 2 has foci at (1,2 Vtgl and (1, 2+ \ffi) and2a:6,itfollowsfromTheoremll7.4 that this hyperbola can be defined as the set of points such that the absolute value of the difference of the distancesfrom any point of the set to the points (1,2  vEZTand ('1,,2 + rffil is 6. The graph of the general seconddegreeequation
F'( ae, o)
F(ae, O)
Af*Bxy*Cy'*Dx*Ey+F0
(7)
'1.4.6. In this section we where B : 0 and AC > 0, was discussed in Sec. have so far considered the equation when B : 0 and AC < 0. Now if for Eq. (7), B : 0 and AC:0, then either A: 0 or C: 0 (we do not consider A: B: C:0 becausethen Eq. (7) would not be a quadratic equation). Supposethat in Eq. (7), B:0, A:0, and C # 0. The equation becomes Figure 14.7.4
Cy'*Dx*Ey+F0
(8)
If D + 0, (8) is an equation of a parabola. lf D : 0, then the graph of (8) may be two parallel lines, one line, or the empty set. These are the degenerate cases of the parabola.
o rllusrRArrow 5: The graph of the equhtion 4y'  9: 0 is two parallel lines; 9y2* 6y * 1 : 0 is an equation of one line; and 2y' * y * 1 : 0 is . satisfied bv no real values of u. R ,irrritu, discussionf,ofi if B :0,C summarized in the following theorem. 14.7.5Theorem
: 0, and A # 0. The resultsare
If in the generalseconddegreeEq. (T),B:0 and either A:0 and C # 0 or C: 0 and A + 0, then the graph is one of the following: a parabola, two parallel lines, one line, or the empty set. From Theorems 1.4.6.'1., 14.7.3, and 1.4.7.5,it may be concluded that the graph of the general quadratic equation in two unknowns when B:0 is either a conic or a degenerate conic. The type of conic can be deterrrined from the product of A and C. We have the following theorem.
620
THECONICSECTIONS 14.7.6 Theorem
The graph of the equation Af * Cy2* Dx t Ey * F: 0, where A and C are not both zero, is either a conic or a degenerateconic; if it is a conic, then the graph is (i) a parabolaif either A: 0 or C: 0, that is, iL AC: 0; (ii) an ellipseif A and C have the same sign, that is, if. AC > 0; (iii) a hyperbolaif A and C have opposite signs, that is, if AC < 0. A discussion of the graph of the general qu4dratic equation, where B + 0, is given in Sec.14.8.
Exercises 14.7 In Exercises1 through 6, tind the eccentricity, center, foci, directrices, and equations of the asyrnptotesof the given hyperbolas and draw a sketch of the graph. 1. 9xz l8y' * 54x 36y + 79 0 2. xz  y' + 6x * l}y  4 0 3. 3y' 4x2 8x  24y  40  0 4 . 4 x 2 y ' + 5 6 x * 2 y + 1 9 5 0
5 . 4 y '  9 x 2 * I 5 y * 1 8 r: 2 9
6. y 2  x 2 + 2 y  2 x  L : 0
In Exercises7 through 12, find an equation of the hyperbola satisfying the given conditions and draw a sketch of the gruph. 7. One focus af (26, 0) and asymptotes the lines l2y : +5r. 8 . Center at (3, 5) , a vertex at (7, 5) , and a focus at (8, 5). 9. Center at (2, l), a focus at (2, 14), and a directrix the line 5y : SS. 10. Foci at (3,5) and (3,0) and passingthrough the point (5,3 + 6l\/5). 1 L . One focus at (3  Zt/B,1), asymptotesintersectingat (3, 1), and one asymptotepassingthrough the point (1,7). 12. Foci at (1,1)
and (7,4) and eccentricityof 3.
13. Prove that the set of points, such that the absolute value of the difference of the distancesfrom any point of the set to two given points (the foci) is a constant, is a hyperbola. 14. Prove that any hyperbola is a set of points such that the absolute value of the difference of the distancesfrom any point of the set to two given points (the foci) is a constant. 15. Use Theorem 14.7.4Io find an equation of the hyperbola for which the difference of the distances from any point on the hyperbolato (2,1) and (2,9) is equal to 4. 16. Solve Exercise15 if the difference of the distances from (8,4)
and.(2,4)
is equal to 5.
17, Three listening posts are locatedat the points A(0, 0), B(0, +) , and C(T, 0), the unit being 1 mile. Microphones located at these points show that a gun is * mi closer to A than to C, and I mi closer to B than to A. Locatethe position of the gun by use of Theorem 14.7.4. 18. Prove that the eccentricity of an equilateral hyperbola is equal to \n.
14.8 ROTATION OF AXES
We have previously shown how a translation of coordinate axescan simplify the form of certain equations. A translation of axes gives a new coordinate system whose axes are parallel to the original axes. We now
14.8 ROTATIONOF AXES
F i g u r e1 4 . 8 . 1
621
consider a rotation of coordinate axes which enables us to transform a seconddegreeequation having an ry term into one having no such term. Suppose that we have two rectangular cartesian coordinate systems with the same origin. Let one system be the ry system and the other the ry system. Suppose further that the i axis makes an angle of radian measure c with the r axis. Then of course the y axis makes an angle of radian measure a with the 3i axis. In such a case,we state that the xy system of coordinates is rotateil through an angle of radian measure a to form the iy system of coordinates. A point P having coordinates (r, y) with respect to the original coordinate system will have coordinates (x' y) wlth respect to the new one. We now obtain relationships between these two sets of coordinates. To do this, we introduce two polar coordinate systems, each system having the pole at the origin. In the first polar coordinate system the positive side of the r axis is taken as the polar axis; in the second polar coordinate system the positive side of the f axis is taken as the polar axis (see Fig. 1a.8.1).Point P has two sets of polar coordinates, (r, 0) and (7, 0), wherc 7:r
(1)
and e0a
The following equations hold: x7cos0
(2)
and ylsin0
Substituting from (1) into (2), we get x  r cos(e q and y  r sin(O q.)
(3)
Using the trigonometric identities for the sine and cosine of the difference of two numbers, Eqs. (3) become 2: r cosOcos a * r sin 0 sin a and ! : r sin0 cos a / cos 0 sin a Becauser cos0 : x and r sin 0 : y, we obtain from the above two equations xrcosa+ysLna
(4)
yxsina+ycosa
(s)
and Solving Eqs. (a) and (5) simultaneously for x and y in terms of i and y, we obtain (5) '
(7)
It is left as an exercise to fill in the steps in going from (4) and (5) to (6) and (7) (see Exercise L).
T H E C O N I CS E C T I O N S
ExAMPLE7: Given the equation xy  L, find an equation of the graph with respect to the x and y axes after a rotation of axes through an angle of radian measure *r.
sol,urroN: Taking a: *r in Eqs. (6) and (7), we obtain
* : $ x1_f i Y
1_
a n a A : n q1 x * f i . 1Y_
Substituting these expressionsfor r and y in the equation rU : l, we get
(#+,)(#c+#,) 1
or/ equivalently,
This is an equation of an equilateral hyperbola whose asymptotes are the bisectors of the quadrants in the ry system. Hence, we conclude that the graph of the equation ry:1is an equilateral hyperbola lying in the first and third quadrants whose asymptotes are the x anil y axes (see Fig. 1a.8.2).
Figure
'1,4.7 In Sec. we showed that when B: 0 and A and C are not both zero, the graph of the general seconddegreeequation in two unknowns,
Af*Bxy*Cy'*Dx*Ey+F0
(8)
is either a conic or a degenerate conic. We now show that if B * 0, then any equation of the form (8) can be transformed by a suitable rotation of axes into an equation of the form
At, + ey,+ Dr+ Ev+ F o
(e)
where A and C are not both zero.
If the ry system is rotated through an angle of radian measule a, then to obtain an equation of the graph of (8) with respect to the xy system, we replace x by x cos c  ! sin a *d y by r sin c * ! cos a. We get
Af+Bxy+Cy'+Dx+Ey+F0
(10)
where A
A cosza + B sin a cos a + C stnz a
BZA
sinacos a+ B(cosa z
CAstnzaB
sinza)+2Csinacosa
(11)
s i n C I c o sa + C c o s z a
We wish to find an a so that the rotation transforms Eq. (8) into an
14.8ROTATION OF AXES equation of the form (9). SettingB from (11) equal to zero, we have B(cosz a  sinz a) + (C  A)(2 sin d cos a) : 0 or, equivalently, with trigonometric identities, B cos2a * (C  A) sin 2a:0 Because B * 0, this gives
#"Eil$#$F...'$js+++++li $
(r2)
We have shown, then, that an equation of the form (8), where B + 0, can be transformed to an equation of the form (9) by a rotation of axes through an angle of radian measure a satisfying (12). We wish to show that A and C in (9) are not both zero. To Prove this, notice that Eq. (1'0) is obtained from (8) by rotating the axes through the angle of radian measure c. AIso, Eq. (8) can be obtained from (10) by rotating the axes back through the angle of radian measure (a). If A and C in (10) are both zero, then the substitutions r:r
c o sa * y s i n a a n d ! :  x s i n
o * y c o sa
in (10) would result in the equation D1r cosa* y sin Q + E(xsin a * y cose) * F: 0 which is an equation of the first degree and hence different from (8) because we have assumed that at least B # O. The following theorem has, therefore, beBn proved. 14.8.1 Theorem
If B + 0, the equation Af * Bxy_* Cyz_* Dx_* Ey_* F: 0 can b_etrans; formed into the equation A* + Cyz + Di + E, + F :0, where A and C are not both zero, by a rotation of axes through an angle of radian measure d for which cot2a: (e_ C)IB. By Theorems L4.8.1and'14.7.6,it follows that the graph of an equation of the form (8) is either a conic or a degenerateconic. To determine which type of conic is the graph of a particular equation, we use the fact that A, B, and C of Eq. (8) and A, B, and C of Eq. (10) satisfy the relation Bz  4AC:
82  4AC
(13)
which can be proved by substituting the expressionsfor A, B, and C given in Eqs. (11) in the right side of (13). This is left as an exercise(see Exercise 15).
The expression 82  4AC is called the discriminantof Eq..(8). Equation (L3) states that the discriminant of the general quadratic equation in two
F
I
THECONICSECTIONS
variables is inaariant under a rotation of axes. If the angle of rotation is chosen so that B:0, then (13)becomes Bz MC:4AC (14) From Theorem 14.7.6it follows that if the graph of (9) is not degenerate, then it is a parabolaif. Ae : 0, an ellipse it Ae > 0, and a hypeibola if AC < 0. So we conclude that the grapLof (9) is a parabola, an ellipse, or a hyperbola depending on whether4AC is zero, negative, orpositive. Becausethe graph of (8) is the same as the graph of. (9), it follows from (14) that if the graph of (8) is not degenerate, then it is a parabola, an ellipse, or a hyperbola depending on whether the discriminant 82  4AC is zero, negative, or positive. We have proved the following theorem. 14.8.2 Theorem
The graph of the equation A * * B x y* C y ' * D x * E y * F : 0 is either a conic or a degenerateconic. If it is a conic, then it is (i) a parabolaif.82 MC:0; (ii) an ellipseif 82  MC < 0; (iii) a hyperbolaif 82  4AC > 0.
EXAMPLE2: Given the equation lTxz  l2xy * 8y' 80 : 0, simplify the equation by a rotation of axes. Draw a sketch of the graph of the equation showing both sets of axes.
solurroN: 82 MC: (12),  4(17) (8) : 400 < 0. Therefore,by Theorem 14.8.2,the graph is an ellipse or else it is degenerate.To eliminate the xy term by a rotation of axes,we must choose an a such that AC .F' cotzd.: B
: L78 _12:4
3
There is a 2o. in the interval (0, zr) for which cot2q.: $. Therefore, o is in the interval (0, trr).To apply (6) and (7) it is not necessaryto find a so long as we find cos a and sin a. These functions can be found from the value of cot 2a by the trigonometric identities cos Because cot
1+ cos2a
2 and 0
and
sin ot:
So
L
\tr sln q. 
2
\E
Substitutingx : il {5  2V \tr andy : 2il \fS + yI \B in the given equa
REVIEWEXERCISES 625
tion, we obtain
/#AIy+4Y,'\_rz 17[. s /
)cy 2y' 5
+8(ry80o
Upon simplification, this equation becomes x2+4Y':t6 or, equivalently,
r' *!'  1 ' 1,5 E So the graph is an ellipse whose maior axis is 8 units long and whose minor axis is 4 units long. A sketch of the ellipse with both sets of axes is shown in Fig. 14.8.3.
Figure 14.8.3
74.8 Exercises y terms of I and'y' 1. Derive Eqs. (6) and (7) of this section by solving Eqs. ( ) and (5) for x and in axes. Draw a sketch of the graph In Exercises 2 through 8, remove the xy term from the given equation by a rotation of and show both sets of axes. 2. x2*xY+A2:3
3.24xy7y'+360
4. 4xy * 3x2: 4
5. xy 8
6. 5x2* 6xy * 5y' 9
7. 3Lx2+ 106*Y * 2tY2:'!'M
8. 5x2* Z}l.6xy * 25y' :324 Draw a sketch of the graph In Exercises9 throu gh t4, simplify the given equation by a rotation and translation of axes. and show the three sets of axes. 10. x 2  L } x y * y ' * x * y + 1  0 9. x2+xy+y23y6:0 L2. 3 x 2  4 x y * 8 r  1  0 11. 17x2 L2xy * 8y'  68x* 24y  12 0 14. 19xz* 6xy * tyz  25x + 38Y+ 31  0 13. Llx2 24xy * 4y'* 30r + 40y 45 0  4AC is invariant under a 15. Show that for the general seconddegreeequation in two variables, the discrimin ant 82 rotation of axes.
Reaiew Exercises(ChaPter14) In Exercises L through of the graph.
6, find a cartesian equation of the conic satisfying the given conditions
1. Vertices at (1, 8) and (1,4);
e:
?.
2. Foci.at (5 , L) and (1, !); one vertex at (4, t). 3. Center at the origin; foci on the x axis; e 2; containing the point (2,3).
and draw a sketch
THE CONIC SECTIONS
4. Vertexat (4,2); focusat (4,3); e:1. 5. A focus at (5, 3); directrix: x: 3; e: *. 5. A focus ar (4, 2); directrix: ! :2; e: *. In Exercises7 through 10, find a polar equation of the conic satisfying the given conditions and draw a sketch of the graph. 7. A focus at the pole; vertices at (2, rr) and (4, r). 8. A focus at the pole; a vertex at (6, ln); e : I. 9. A focus at the pole; a vertex at (3, En); e: l. 10. The line r sin d: 5 is the directrix corresponding to the focus at the pole and e: *. In Exercises11 through 14, the equation is that of a conic having a focus at the pole. In each Exercise,(a) find the eccentricity; (b) identify the conic,' (c) write an equation of the directrix which coresfonds to the focus at tfre pole; (d) draw a sketch of the curve.
In Exercises15 through 18, the equation is that of either an ellipse or a hyperbola. Find the eccentricit5z,center, foci, and directrices, and draw a sketch of the graph. If it is a hyperbola, also find Lquations of the asymptotes. 1.5.4f + y' + 24x l6y + 84  0 17. 25x2 y2 * 50r + 5y  9 0
16. 3x2 2y, + 6x  gy + 1L : 0 18. 4x2* gy' + 32x lgy + 37  0
In Exercises19 and 20, simplify the given equation by a rotation and translation of axes.Draw a sketch of the graph and show the three sets of axes. 19.3f 3xyy25y
0
20. 4x2* 3xy * y,  Gx* 12y 0
21. Find a polar equation of the parabola containing the point (2,*rr), whose focus is at the pole and whose vertex is on the extension of the polar axis.
22. lf lhe distance between the two directrices of an ellipse is three times the distance between the foci, find the eccentricity. 23. Find the volume of the solid of revolution generated if the region bounded by the hyperbola f la"  yrlbr:1. and the line r: 2a is revolved about the y axis. 24. Show that the hyperbola f  y': 4 has the same foci as the ellipse x2 + 9yz :9. 25. Find an equation of the parabola having vertex at (5, 1), axis parallel to the y axis, and through the point (9,3). 25. Show that any equationof theform xy I ax*by f c:0canalwaysbewrittenintheform x,y,:kbyatranslationof the axes, and determine the value of k. 27. Any section of a parabolic mirror made by passing a plane through the axis of the mirror is a segment of a parabola. The altitude of the segment is L2 in. and the length of the base iJ ts in. A section of the mirror made by pi"^" p"rpendicular to its axis is a circle. Find the circumference of the circular plane section if the plane perpendicular " to the axis is 3 in. from the vertex. 28. The directrix of the parabola y2 :4pr is tangent to a circle having the focus of the parabola as its center. Find an equation of the circle and the points of intersection of the two curves. 29. A satellite is traveling around the earth in an elliptical orbit having the earth at one focus and an eccentricity of The *.
REVIEWEXERCISES
closest distance that the satellite gets to the earth is 300 mi. Find the farthest distance that the satellite gets from the earth. 30. The orbit of the planet Mercury around the sun is elliptical in shape with the sun at one focus, a semimajor axis of length 35 million miles, and an eccentricity of 0.206.Find (a) how close Mercury gets to the sun and (b) the greatest possible distance between Mercury and the sun. 3 1 . A comet is moving in a parabolic orbit around the sun at the focus F of the parabola. An observation of the comet is made when it is at point Pr, 15 million miles from the sun, and a second observation is made when it is at point Pr, 5 million miles from the sun. The line segments FP, and FP2are perpendicular. With.this information there are two possible orbits for the comet. Find how close the comet comes to the sun for each orbit.
32. The ardr of a bridge is in the shape of a semiellipse having a horizontal span of 40 ft and a height of 15 ft at its center. How high is the arch 9 ft to the right or left of the center? 33. Points A and B are 1000ft apart and it is determined from the sound of an explosion heard at these points at different times that the location of the explosion is 500 ft closer to A than to B. Show that the location of the explosion is restricted to a particular curve and find an equation of it. 3 4 . F i n d t h e a r e a o f t h e r e g i o n b o u n d e d b y t h e t w o p a r a b o l a s : r : 2 1( l  c o s 0 )
andr:21(J+
cosd).
35. A focal chord of a conic is a line segment passing through a focus and having its endpoints on the conic. Prove that if two focal chords of a parabola ar perpJndicul,ar, the ium of the reciprocals of the measures of their lengths is a constant. (HrNr: Use polar coordinates.) 35. A focal chord of a conic is divided into two segments by the focus. Prove that the sum of the reciprocals of the measures of the lengths of the two segments is the same, regardlessof what chord is taken. (nrNr: Use polar coordinates.) 37. Prove that the midpoints of all chords parallel to a fixed chord of a parabola lie on a line which is parallel to the axis of the parabola.
Indeterminateforms, improperintegrals, and Taylorbformula
FORMO/O 629 15..1 THEINDETERMINATE 15.1 THE INDETERMINATE Limit theoremg (2.2.9)statesthat if lim /(r) and lim g(r) both exist, then FORM O/O
'
f(x) I'qrt'l
i'gm:ffi'(') provided.nu,rri. r,t ) + 0. There are various situations for which this theorem cannot be used. In particular, if lirn g(r) : 0 and lim /(r) : k, where k is a constant not equal to 0, then Limit theorem 12 (2.4.4) can be applied. Consider the and lim g(r) :0. Some limits of this type case when both lim f(x):0
have previously been considered. 1 TLLUSTRATTONL: We wish
to find
fxt2 llm t=rtX1JX+ Here, lim (x'  x  t2) :0 r4
and lim (f  3x  4) : 0. However, the nur4
merator and denominator
can be factored, which gives
,. fxL2 ,um . k4)(x*3) n\ ir lx _ 4: G_ 4Xx,+ii Because r + 4, the numerator and denominator can be divided b y ( x  4 ) , and we obtain 1.
x2x12
x*3
,.
fim:lTr+1:B
7
 0 and lim x  0; and by Theorem "1,0.2."1,, o ILLUSrnerroN 2: lim sin x Jfr0
l0
lim
o
frO
L5.L.L Definition
(x)  0 and lim S@): 0, then II f and g are two functions such that ftg f the function f lg has the indeterminate form 0/0 a t a . IA
15.1.1, (x'  x  l2)l(x2  3x  4) ; ILLUSTRATToN3: From Definition has the indeterminate form 0/0 at 4; however, we saw in Illustration 1 that ,. x2x!2 ;;xz3x4 rrFr:
7 5
Also, (sin xl x) has the indeterminate form 0/0 at 0, while lim (sin x/ x) : 1', ro as shown in Illustration 2. o We now consider a general method for finding the limit, if it exists,
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
of a function at a number where it has the indeterminate form 0/0. This method is attributed to the French mathematician Guillaume Frangois de L H6pital (7561L707),,,vho wrote the first calculus textbook, published in 7695.It is known as L'H6pital'srule. 15.1.2 Theorem L'Hopital's Rule
Let / and g be functions which are differentiable on an open interval I, except possibly at the numb er a in I. Supposethat for all x # a in I, g, (x) + 0. Then if lim /(r) :0 and lim g(r) :0, and if
,. f' ( x)
lllll*:
L
"" 8'\x) it follows that llm Ya
f(x)
;i:
L
8\X)
The theorem is valid if all the limits are righthand limits or all the limits are lefthand limits. Before we prove Theorem 15.1.2,we show the use of the theorem bv the following illustration and examples. . rLLUsrRArror 4: We use L'H6pital's rule to evaluate the limits in  x12):0 Illustrations 1 and 2. In Illustration 1, because and l2@, lim (l 3x 4): 0, we can apply L'H6pital's rule and obtain r.
x2x12
2xl
7
;:;2x3
5
r.
riffr:rllft:
;:;xz3x4
We can apply L'Hdpital's rule in Illustration 2 because lim sin x  0 IO
and lim x: 0. We have, then, JfO
r.
llm t0
EXAMPLE
Find
sin r X
soLUrIoN:
r.   I llm .Z*0
cos .r I
Because lim )c: 0 and lim (l  e") ffO
IO
be applied, and we have
l i m ; x ;  l i m + : a   1 L ;;te*
;;er
SOLUTION:
l i m ( 1  x + l n r )  1 , L + 0  0 fil
if it exists.
L'Hopital's rule can
FORMO/O 631 15,1THEINDETERMINATE and lim (r33x*2):l3*2:0 al
Therefore, applylng L'H6pital's rule, we have r  x +x + l nnxr 1 r. r'il .liii l r  : r l m  
1 +1 x ' ili3* 3
,r ..
x" 3 x*z Now,becaur" (t + llx) :0 andlim (3f  3) : 0,weapplyL'H6piItT tal's rule again giving
ftr 3f

3
Therefore,we conclude r. Lx+lnr um;; x'd3x*2
L 5
To prove Theorem 15.1.2 we need to use the theorem known as Cauchy's meanualuetheorem,which extends to two functions the meanvalue theorem (4.7.2)for a single function. This theorem is attributed to the French mathematician Augustin L. Cauchy (17891857),who was a pioneer in putting the calculus on a sound logical basis. L5.L.3 Theorem Cauchu's MeanV alue Theorem
lf. f and I are two functions such that (i) f andgare continuous on the closedinterval la, bl; ( i i ) / a n d g a r e differentiable on the open interval (a, b); (iii) for all r in the open interval (a, b), g'@) * 0 then there exists a number z in the open interval (a, b) such that .
'(z) f(u) f(a):f s@)g(o)g'(z) pRooF: We first show that g(b) + g(a). Assume g(b) : g(a). Becauseg satisfies the two conditions in the hypothesis of the meanvalue theorem (4,7.2), there is some number c in (a, b) such that g'(c) : IS(b) S@)ll(b a). But 1f g(b) : g(a), then there is some number c in (a, b) such that g' (c) : 0. But condition (iii) of the hypothesis of this theorem statesthat for all r in (a,b), g'@) * 0. Therefore, we have a contradiction. Hence, our assumption that BQI: g(a) is false. So, g(b) + g(a), and consequent$ g(b) g(a) * 0. Now let us consider the function h defined by
: f(x) f(a)ffi1 h(x)
frt,l s@)l
632
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
Then,
h'(x)f'(x)lffils,(,1
(1)
Therefore, ft is differentiable on (a, b) because/ and g are differentiable there, and ft is continuous onfn, b] becausef and g are continuous there.
 f(a) l#=ffi1 rrr,l 8(a)l : f(a) :o h(a)  f(a) U%=#l urrr g(a)l : f(b) :o h(b) Hence, the three conditions of the hypothesis of Rolle's theorem are satisfied by the function ft. So there exists a number z in the open interval (a, b) such that h'(z) : 0. Thus, from (1) we have
f '(,)
'\?! f (!), : o s @ ) s \ at s'Q)
(2)
Becauseg' (z) + 0 on (a, b), we have from (2)
f(b)

f(q) _ f' (r)
M:76
where z is somenumber in (a, b).This provesthe theorem.
r
We should note that if g is the function such that g(r) : r, then the conclusion of Cauchy's meanvalue theorem becomes the conclusion of the former meanvalue theorem because then g'(z) : t. So the former meanvalue theorem is a special caseof Cauchy's meanvalue theorem. A geometric interpretation of Cauchy's meanvalue theorem is given in Sec. 77.4. It is postponed until then becauseparametric equations are needed. o rLLUsrRArroN 5: Supposef(x):3f *3x  L andg(r): f  4x * 2. We find a number z in (0, L), predicted by Theorem 15.1.3.
f
' ( x )  6 x+ 3
g'(x)3x24
Thus, / and g are differentiable and continuous eve4rwhere, and for all r in (0, 1), g'@) # 0. Hence,by Theorem 15.1.3,there existsa z in (0,1) such that /(1) /(0) _ 6 z * 3  4 s(1) g(0) 3* Substitutingf(1):5, g(1) :1", /(0):1, and g(0):2, and solving for z, we have 5 (1) _62*3 l2 3224
FORM O/O 15.1 THE INDETERMINATE
633
5 2 2* 5 z  5 = 0
t2 _5+2\B T2  6 3 + \ B Only one of these numbers is in
(0, 1), namely,z  +(3 + t@l.
o
We distinguish We are now in a position to Prove Theorem 1'5.1'.2. three cases:(i) r + a+; (i1) x+ a; (iii) r + a. P R o o Fo F T H E o R E M1 5 . 1 . 2 ( 1 ) : Because in the hypothesis it is not assumed that / and I are define d at a, we consider two new functions F and G for which F(r):f(x)
1 fx #
G(r):g(r)
tfx#a
a
a n d F ( a ): 0 and G(a) : 0
(3)
Let b be the right endpoint of the open interval I given in the hypothesis. Becausef and g are both differentiable on 1, except possibly at d, we conclude that F and G are both differentiable on the interval (4, rl, where a < x < b. Therefore, F and G are both continuous on (a, rl. The functions F and G are also both continuous from the right at a because G(a). F(a), and lim G(r) : lim 8(r) :0: lim F(r) : lim f(x):0: I
t A+
A+
Therefore, F and G are continuous on the closed interual la, xf. So Fand G satisfy the three conditions of the hypothesis of Cauchy's meanvalue theorem (Theorem L5.L.3) on the interval la, xf. Hence, F(x)  F(a) _ F' (z) G(r)  G(a) G'(z)
(4)
where z is some number such that a < z < r. From (3) and (4) we have
f@f'(z)
(s)
ilDm
Because a < z I x, it follows that as x + a+, z + a+; therefore,
lim f9=hm f'Q)
t  e , +8 \ x )
'
fg
ra+ 86:'I.M
(5)
But by hypothesis,the limit on the right side of (6)is L. Therefore, lim f]
r_a+ g\X )
L
I
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
The proof of case (ii) is similar to the proof of case (i) and is left as an exercise (see Exercise 27). The proof of case(iii) is based on the results of cases(i) and (ii) and is also left as an exercise (see Exercise28). L'H6pital's rule also holds if either r increases without bound or r decreaseswithout bound, as given in the next theorem. 15.1.4 Theorem Let f and g be functions which are differentiable for all x > N, where N L'H6pital'sRule is a positive constant, and suppose that for all x > N, g'(r) + 0. Then if
 o and  o andif Jl* f@) Jn s(x) f' (x\
r.
J1treL it follows that
,. f@) lim L r*@ 8\X)
The theorem is also valid if "x >*q"
is replaced by "x 's'6."
pRooF: We prove the theorem for x + +6. The proof for r + oo is left as an exercise(seeExercise29). For all r > N, let r: Uf; then t:Llx. Let F and G be the functions defined by F(f) :f(Ut) and G(f) : g(Llt),if.t + 0. Then/(r): F(t) and g(r):G(f), where r )N and 0 < t < UN. From Definitions 4.1.1and 2.3.1,it may be shown that the statements
/tt) : rra and lim F(f) :14
lT
have the same meaning. It is left as an exercise to prove this (see Exercise 30). Because,by hypothesis, lim f (x):0 and lim g(r) : 0, we can rio t+@ conclude tJrat lim F(f) : $ and lim G(1) : g
(7)
Considering the quotient F'(t)lG' (t), we have, using the chain rule,
1r,/1\
F'Gl
Ft \tl
\ t, ' E \t)
P6 \tl
\t/
eit:m:m:N o
f'(x)
Becauseby hypothesis lim f'(x)lg'(x):L, r*a
it follows from the above
that
lim* to+
tr'(t) _ t P
\,
(f)
Becausefor all x ) N, g'Q) # 0,
(8)
FORMO/O 635 15.1THE INDETERMINATE
f o r a l l0 < t . L
G'(t)#0
N
that From (n, (8), and (9), it follows from Theorem 1.5.1..2 r:
F(f) :
I
I1T@IL But becauseF(t)lc(t) : f(x)lS@) for all x > N and t + 0, we have
.. f(x) llm 'i:
a+ 8\r)
L
and so the theorem is proved.
I
Theorems 15.1.2and 15.1.4also hold if L is replacedby *o or @. The proofs for these casesare omitted. EXAMPIn 3:
Find
solurroN:
lim sin (Ilx)  0 and lim tanl (Ux) : Q.So applying L'H6I*a
, ff*o
.L
sln
pital's ryle.we get 1.
lim t*@
cos' sin 1 x 4,zi: lim tanl
(l)
$*@
\r/
Because lim cos 1= t .?*@
ryz
and
r.
1
lltfu:,ritfil,  7
.r
it follows that 1, coslim =* 1 t*@
rya *
f+r Therefore, the given limit is L.
t *E
L5,1 Exercises In Exercises1 through 4, find all values of.z in the given interval (a, b) satrsSng the conclusion of Theorem 15.1.3for the given pair of functions. 1. f (x) : v3S , ( r ) : x 2 ;( a , b ) : ( 0 , 2 )
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
636
s i n x , g ( r ) : c o st r ; ( a , b ) : ( 0 , n )
3. f(x):
4 . f ( x ) : c o s2 x , g ( r )  s i n x ; ( a , b )  ( 0 , t n )
In Exercises5 throu gh 24, evaluatethe limit, if it exists. v5".; l; it m a n,x*
,
.v ( ' l P
O.
,ir, 2
,.
tanrx
;;
x sinr
riffr
7. lim I*@
I
x )r
8. lim=
r2 z
9. lim
x
fi
r)11. lim ln(sin @  2x)' ""i, e2" 
1
14. lim*sinz r ;;
20. lim ro
llII'
uFr
;:; ,.
ufi
;;
10. lim ro
€t
ercosr
"1,0t
^{ ,.
cosh r
,\, r. L=..
X2
t\ f7. . lli _m;t2
tan3 0
19. lim IO
L+cos2x
sinh x
X
+n_2n
0sin0
zL. Ilmr_7t rslnI
I
sin1x
. 2 . r _ 3 r 13. lim X ro
r sin r
tan\ 2x 1g. lim ;:; tanh x
)C
ro
(9.
l .
Lz..
rJ.
cos .r 
.r.r r. 1wr.
d n
iF
o.hm+ sin f2 ;; vet 
q
( t + x ) r r s ( 1  x ) r r s (1+x)rrs(1 x)trs
1, 1 +2tanr*x2  22. lim = 1 x I*a
sinr
llrrl
n o
sin3 x
A. electrical circuit has a resistance of R ohms, an inductance of L henries, and an electromotive force of E volts, where R, L, and E are positive. If I amperes is the current flowing in the circuit f sec after a switch is tumed on, then
I:fr(1 eH,) For specific values of t, E, and L, find lim I E0+
25. In a geometric progression, if a is the first term, r is the common ratio of two successiveterms, and S is the sum of the first n terms, then tf r # l,  1) o _ a(r" rL
p ) 
Find lim S. Is the result consistent with the sum of the first n terms if r : L? Tl
27. Prove Theorem 15.1.2(ii). 28. Prove Theorem 15.1.2(iii). 29. Prove Theorem 15."1,.4 for x + oo. 30. Suppose that f tt a function defined for all r ) N, where N is a positive constant. If t: Llx and F(f)  f(Llt), where t + 0, prove that the statements lirh f (x)  M and lim F(f )  M have the same meaning. .T
I5.2 OTHER INDETERMINATE FORMS
*@
f0+
Suppose that we wish to determine if cannot appty the theorem involvi.g
(sec2xlsecz3x) exists. We ):_T,, the limit of a quotient because
FORMS 15.2 OTHERINDETERMINATE
lim seC r:
637
*o and lim seC 3x: *oo. In this casewe say that,the funcrnl2
cnl2
tion defined by seC xlsef 3x has the indeterminate form (*oo/*o) at x:+,T. t'H6pital's rule also applies to an indeterminate form of this and (*o/o). This is given by the type as well as to (olo;, (l**) following theorems, for which the proofs are omitted because they are beyond the scopeof this book. 15.2.1 Theorem Let / and g be functions which are differentiable on an open interval I, L'H6pital'sRuleexcept posiibty at the number a inl, and supposethat forall x * a inl, , and lim g(r): *o or oo, and if 8'@) * 0' Then if lim f (x): *o or
f'(x)
12 s'TJ:' it follows that (t \
lim r*:
L
ea $\X)
The theorem is valid if all the limits are righthand limits or if all the limits are lefthand limits. EXAMPLE
SOLUTION: Because lim ln r "'o+ tal's rule and get
Find
@ and lim (Ux)  +@,w€ apply L'H6piJr.O+
L
limry hm+7 0+I7
15.2.2Theorem L'H6pital'sRule
_7,
hm(r) 0
0+I3.0+
,
Let f and I be functions which are differentiable for all x ) N, where N is a positive constant, and suppose that for all r > N, g' @) + 0. Then if lim f (x)  +co or @, and lim S@) +m or @, and if t*a
The theorem is also valid 7f "x + +m" is replaced by '1.5.2.2 Theorems "1.5.2.'/. also hold if L is replaced by f m or and and the proofs for these cases are also omitted.
638
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
ExAMPrn 2:
Find
soLUrIoN: Becauselim ln(2 * e") : *oo and lim 3r :]6, f*o
by applying
.T+@
L'H6pital's rule we obtain 7
..
rlm
ln(2 * e'\ ;::
..
o*o
5X
r*a
rlm
2 * e" 
.o*
J
:"t11Tf* ec: la *d,1*
Now becaur" Jit
(6 * 3e") : *@, we applyL'H6pital's
rule again and get
gr":
: :5 "11 &" "1T"3
,.e',.er,.11,
"11 o+ Therefore,
ln(2 t e')
..
1
;:
ulrl

3x
;;;
solurroN:
3
lim seC r:
*o and lim seC3x: *o. So by applying L'H6
Itlz
ITl2
pital's rule we get seczx a;:
r. Ilm
r. Irm
,ili2secz 3x
;:;;r6
lim 2 sec2r tan x:
2 secz x tan x seC 3r tan 3r and
*a
cnlz
lim
6 sec23r tan 3r:
*o
rnl2
It may be seen that further applications of L'H6pital's rule will not help us. However, the original quotient may be rewritten and we have seC x
,.
,.
llm : antzS€.C3X
cos23x
llm av12COS' X
Now, because lim cos23r:0
and lim cos2x:0,
rnl2
we may apply
Iil2
L'H6pital's rule giving r. Ilm lIIl
.
""ir
cosz 31 :
1. llm
e,
cosz)c
irir
6 cos 3r sin 3r 2 cos x sin x
r:_ 3(2 cos 3r sin 3x) :um r7ft2 (z cos )c st'a x) r.  3 sin 6x :rtm trrl2
Sln
ZX
Becauselim 3 sin 6x  0 and lim sin 2x :0, we apply L'H6pital's I rrl2
I 7rl2
FORMS 15,2 OTHERINDETERMINATE
639
rule again and we have L8 cos5x  18(1)  o r: 3: sin 6r  r:UIrl T:7 cos2x 2(Ll sin 2x rnt2"i"irZ Therefore, rlm
1:__ t""" : llm ,,A 5)C rzrl2Se9
g r
The limit in Example 3 can be evaluated without L'H6pital's rule by using Eqs. (11) in Sec.10.1and Theorem 10.2.1.This is left as an exercise (seeExercise28). In addition to 0/0 and +aF@, other indeterminate forms are 0'(*o), {o  (*o), 0, 1t;0, and 11. These indeterminate forms are defined analogously to the other two. For instancei if lim /(r) : *o and lim g(r) : 0, then the function defined by f (xftot has the indeterminate form (*o)o at a. To find the limit of a function having one of these indeterminate forms, it must be changed to either the form 0/0 or 1ol+o before L'H6pital's rule can be applied. The following examples illustrate the method.
EXAMPLE
Find
solurroN:
Becauselim sinr r : 0 and lim csc x: *a, the function de.f0+
lim sinl r csc tr t
O+
if it exists.
fOr
fined by sinl l csc r has the indeterminate form 0 . (+"o; at 0. Before we can apply L'H6pital's rule we rewrite sinl r cscr as sinl r/sin r, and consider lim (sin1 r/sin r). Now lim sinl r : 0 and lim sin r: 0, and so I O+
.?0+
I O+
we have the indeterminate form 0/0. Therefore, we apply L'H6pital's rule and obtain l 1,
EXAMPLE
Find
SOLUTION: 1.
BCCAU SC L
lim +Jf;O
'lr3 *
and
fm
I
,.
rlli:l
r0 f
"
:TJC
SeC X
we have the indeterminate form {owe have .. ll 1 \ secrl rrm co
l' \I
r
SeC Xl
lim ( s e cx  1 ) : 0
rrifi. aq
l:
and
t0
Ltg
f
(fo).
Rewriting the expression,
S€C I
(x' secx)
so we apply L'H6pital's rule
and obtain ,.
s e cx  l
llm:llm gO f SeC )C
r. 7O
sec x tan x 2x secx + rP sec x tan x
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
640
tan r
!.
:rliFr
;;2x*x2tanx
Ltg
tanr: 0
and
lim (2x * x2 tan x) I0
and so we apply the rule again and obtain seczx
:hm limtalr ;; 2x * xz tan x ro 2 + 2 x t a n x * x 2 s e c 2 x Therefore,
l i m e  , 1xz secxf\ : 12 "o
\r'
If we have one of the indeterminate forms, 00, (t*)0, or L!*, the procedure for evaluating the limit is illustrated in Example 6. EXAMPLE 6:
llt
Find
(x * 1)cotr
soLUrIoNl Becauselim (r * 1) : L and lim cot x  *@, we have the to+ determinate form r*J.ilut y:
if it exists.
(1)
(r * t)cotr
Then l n Y : c o tx l n ( r + 1 ) 'lnY ln(r * 1) tan r So lim ln Y: lim ot
t o+
t
ln(r * 1) tan
(2)
X
Because lim ln(r + L) : 0 and lim tan x : 0, we may apply L'H6pi
tal'sruleto tfJ?gnt sideot (2)ur,a"Jiltuir, ln(r+L)
r.
riifi
:
.?o+
tan X
,r l i.f i
1 x + 1 o:
no+ SeC' X
1
Therefore,substituting L on the right side of (2), we have lim lny  1
(3)
IV
Becausethe natural logarithmic function is continuous on its entire domain, which'is the set of all positive numbers, we may apply Theorem 2.6.5and we have from (3) ln lim U:1 0*
(4)
t
Therefore (because the equation ln a:
b is equivalent
to the equation
LIMITSOF INTEGRATION641 INTEGRALS WITHINFINITE 15.3IMPROPER n:
eb),it follows from Eq. (4) that lim U: er
I O*'
But from (1), y  (x * 1)cot', and so we have lim (r*1)cot''s Jf0*'
L5.2 Exercises In Exercises L throu gh 27, evaluate the limit, if it exists. ,v2
1. lim :, t+ 0*
r) z. lim 1"(."t
l'1
t_nt2ln ( tan x)
4.,tlt,ti'
5. lim x csc x f *0+
7 . r i m , J _ _xL L) l
8.
;:;\lnr
lim
10) t;n (r + eza)ttn
L1.. l i m ( x '  f f i ) tt@
w J1t *'tt 16. lim
9. lim tsin r
(sinh x)tanr
f 0+
L4. l i m ( 1 + x ) t n t
/ s 12. l i m i * , , ;;\r'*x6 @,
J/0*
*rtrn r
17. lim (1 + ax)rt";a + 0
f 0+
&
.T0
/
;r*@ \
hm Q2) taninx
21,.
t2
23. lim [ (ru * 3x5+ 4)rt6 xf
22. lim (cos x)rrt'
k" * x)zrt 1 \r2
18. l i m ( 1 + ; l
t0
19. lim (1 + sinh x)zrt
J1*
L\ x2t
ft*
2x/
[(cos x)et2tzfuue
24. l i m x " "
t *a
W
lim t *o
ln(r * e") 3;x
u"t1i#
Evaluatethe limit in Example3 without using L'H6pital's rule, but by using Eqs. (11)in Sec.10.1and Theorem10.2.1. (a) Prove that lim (ertc2lxn): 0 for any positive integer n. (b)lI f (x): erto',use the result of (a) to prove that the limit of f and all of its derivatives, as r approaches0, is 0.   lim P+\',:9,frnd,n. 30. v v ' If  rl ;;;
\nx
31.Suppose f(x):
f
:r,hndn. e$t/iF+tittand,g(r): xnes,.rf,t* tffi]
15.3 IMPROPER INTEGRALS WITH INFINITE LIMITS OF INTEGRATION
In defining the definite integral Il f @) dx,the function/was assumedto be defined on the dosed interval [a,bl.We now extend the definition of the definite integral to consider an infinite interval of integration and
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
call such an integral an improper integral.In sec. 15.4 we discuss another kind of improper integral. o rLLUsrRArroN1: Consider the problem of finding the area of the region bounded by the curve y: sr, the r axis, the y axis, and the line x:b, where b > 0. This region is shown in Fig. 15.3.1.Letting A square units be the area of this region, we have A_
lD
: e, I
lo :7  eb If we let b increase without bound, then f
[o e' ilx: lim (1
D+6 Jo
so1
b*o
rb
lim I e" dx:l
L
(1)
b+q Jo
We conclude from Eq. (1) that no matter how large a value we take for b, the area of the region shown in Fig. 15.3.1will always be less than L square unit. . Equation (L) states that if b > 0, for any e ) 0 there exists an N > 0 such that ltb I lf ecilx11 <e
tJo
wheneverblN
I
In place of (1) we write f+*
I
e, dt:L
JO
In general, we have the following definition. L5.3.L Definition
It f is continuous for all x f+*fb
I
Ja
I f(x) dx f ( x ) d x  bl i m +* Ja
if this limit exists. If the lower limit definition.
of integration is infinite,
we have the following
15.3 IMPROPERINTEGRALSWITH INFINITELIMITS OF INTEGRATION 643
15.3.2Definition
If / is continuous for all r < b, then rb
I
J*'
fb
f(x) dx: rim I f(x) dx a@ Ja'
if this limit exists. Finally, we have the casewhen both limits of integration are infinite. 15.3.3Definition
If / is continuous for all values of r, then f+a
fo
fb
I f 1' L9. Determine if it is possible to assign a finite number to represent the measure of the area of the region bounded by the curve whose equation is y : 11 1r" * e") and the x axis. If a finite number can be assigned, find it. 20. Determine if it is possible to assign a finite number to represent the measure of the area of the region bounded by the x axis, the line x:2, and.the curve whose equation is y:11 1rz 1). If a finite number can be assigned,find it.
a
Determine if it is possible to assign a finite number to represent the measure of the volume of the solid formed by revolving about the r axis the region to the right of the line r: 1 and bounded by the curve whose equation is ! : Tlxt/i and the r axis. If a finite number can be assigned, find it.
22. Determine f+*/
a value of n for which nxz
J, \ffi

the improper
integral
1. \ j sx+r)o*
is convergent and evaluate the integral for this value of n. 23. Determine a value of. i for which the improper integraL
l.*(T==?)4* 2r+n1 \x*1
J'
is convergent and evaluate the integral for this value of n.
15.4 OTHER IMPROPERINTEGRALS
647
Exercises24,25, and.25 show an interesting application of improper integrals in the field of economics. Suppose there is a continuous flow of income for which interest is compounded continuously at the annual rate of 100l percent and /(f) dollars is the income per year at any time f years. If the income continues indefinitely, the Present value, V dollars, of all future income is defined by l+@
V:
I
Jo
f ( t ) e  i ld t
2/. A eontinuous flow of income is decreasing with time and at f years the number of dollars in the annual income is 1000 . 2t. Find the present value of this income, if it continues indefinitely using an interest rate of 8 percent compounded continuously 6
fA. continuous flow of profit for a company is increasing with time, and at f years the number of dollars in the profit per year is proportional to t. Show that the present value of the company is inversely proportional to i2, where 1.001 percent is the interest rate compounded continuously.
8. fn British Consol is a bond with no maturity (i.e., it never comes due), and it affords the holder an annual lumpsum payment. By finding the present value of a flow of payments of R dollars annually and using the current interest rate of 1001percent, compounded continuously, show that the fair selling price of a British Consol is R/i dollars.
15.4 OTHER IMPROPER INTEGRALS
Figure 15.4.1shows the region bounded by the curve whose equation is y : U!x, the r axis, the y axis, and the line x: 4.lf it is possibleto assign a finite number to represent the measure of the area of this region, it would be given by lim llallo If this limit exists, it is the definite integral denoted by
t
dx
(r)
\fr
However, the integrand is discontinuous at the lower limit zero. Furthermore/ lim UVx : *@, and so we state that the integrand has an infinite discontinuity at the lower limit. Such an integral is improper, and its existencecan be determined from the following definition.
F i g u r e1 5 . 4 . 1
L5.4.1 Definition
If / is continuous at all r in ttie interval half open on the left (a, b], and if lim /(r) : t@/ then
L
f(x) dx:
f(x) dx
if this limit exists. . rLLUSTRArroul.: We determine if it is possible to assign a finite number to represent the measure of the area of the region shown in Fig. 15.4.1. From the discussion preceding Definition 15.4.1 the measure of the area of the given region will be the improper integral (1) if it exists. By Defini
648
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
tion 1,5.4.L, we have fn dx
f4 dx
1.
Jo&::'_TJ. &
: ttT'*t''1 4 f,  lim 14 2*l :40 t Therefore, we assign 4 to the measure of the area of the given region.
o
If the integrand has an infinite discontinuity at the upper limit of integration, we use the following definition to determine the existence of the improper integral.
15.4.2Definition
If f rs continuous at all x in the interval half open on the right la, b) and if lim f (x)  f m, then I  b
f be
fb
f(x) dx I ff*l dx lim eo*Ja Ja' if this limit exists.
If there is an infinite discontinuity at an interior point of the interval of integration, the existence of the improper integral is determined from the following definition. 15.4.3 Definition
If / is continuous at all r in the interval fa,bl exceptc when a I c 1 b and if lim l/(r)  : *o, then fb
fce
fb
I f(x) dx:rim I f@) dx*rim I f(x) dx Do*Jr+6' eo+Ja Ja'' if these limits exist. If Ilf@) dx is an.improper integral, it is convergent if the corresponding limit exists; otherwise, it is divergent. EXAMPLE1.: Evaluate fz
t 
dx
J o@  r ) ' if it is convergent.
SOLUTION: ThC iNtcg 15.4.3, we
Definition f2
dx
 I lr.m
Jo1, lim €0+
15,4 OTHER IMPROPERINTEGRALS 649
:tl1
}
t] * t'_T lt.*]
Because neither of these limits exist, the improper integral is divergent.
. rLLUsrRArrou 2: Suppose that in evaluating the integral in Example 1 we had failed to note the infinite discontinuity of the integrand at 1. We would have obtained L12 r1lo
1 , 1 :  '1 L
^
This is obviously an incorrect result. Becausethe integrand 1.1(x7)2 is never negative, the integral from 0 to 2 could not possibly be a negative number.
EXAMP
fr
rn 2: Evaluate
I rln JO
SOLUTION:
irg
xdx
if it is convergent.
Definiti
Th ne integrand has a discontinuity at the lower limit. Applyion n 15 . 4 . 1 w , e h ave fr
fr
:  l i m I x l n r dx I x l n ,x ddx e0+ J e Jo t
 lim ltr' €0+
L
€  tr'l
: tt5 [] ln Hence, fr
I xlnr dx:0+tlime2
Jo
ln€0
€*o+
To evaluate lim ezln €  lim I!s €0+
€0+ L e2
rule because lim ln
we apply L'H6pital's
€0+
have
T lne r. llm T: €0+ r e2
1. llm €0+
€
lim 2
€ 0
€3
Therefore, from (2) we have f1
I x lnx dx*
Jo
0
and l i m l l e ' : €0+
+@. We
AND TAYLOR'SFORMULA FORMS,IMPROPERINTEGRALS, INDETERMINATE
650
EXAMPIn 3:
Evaluate
dx
,f** 
J' )*FT.
SOLUTII Of N : Fr f,or: this integral, there is both an infinite upper limit and an :onltinuity of the integrand at the lower limit. We proceed infinit,te: d iisco as foll< OIws. +€ r+ @
I Jr
if it is convergent.
dx :rilrl x!x2 
r.
f2 t a:Ja
1
::
dx
, r.
fb 
l l ]l r
dx

u**J, xlFT.
xlF'
t. lD  lim l sect r 12 l + lim l secl r r ar* L
)"
D*m L
)z
(secl 2  secl a) * lim
:
lt_T.  lim secl a * lim secl b a!+
(secl b  secl 2)
b*a
0 **n  l * 2 u
Exercises L5.4 In ExercisesI through 20,determine whether the improper integral is convergent or divergent. If it is convergent,evaluateit. 1. f
d
2'
J o! t  x
4r+
Jo \/L5
x2
7 ' Jlntz, l  sdti n r
f2
dx
Jrffi
5I:# 8 ' Jfz , f f dx i f+
3' 6. (1
fnl2
J,,n frl2
J,
secx dx tan0 d0
f4dx
t
Jo x22x3
dX 10.f+J, ffi
L1' rnxdx J,
12.f' 4x ' Jr
l 3 ' rJo r f fdit
H.fL
4r
Jo\m
16. J, fta*
17. J,,,ffi
L L/"s f L x\FT.
z oJo f LxlR
7+* otlt
J,
f'
dt
rc'
f2
dx
J, *'
18I*
2L. Evaluate, if they exist:
4+ (a) l'T [/, l,+) I:,*,(b)
by the 22. Show that it is possible to assign_afinite number to represent the measure of the area of the region bounded a finite curve whose equation is y : tit/i, the line x : 1, and ihe r and y axes,but that it is not possible to assign about number to represent the ileasure of the volume of the solid of revolution generated if this region is revolved the r axis.
15.5 TAYLOR'SFORMULA
651
In Exercises23 through 25, find. the values of.n lor which the improper integral converges and evaluate the integral for these values of n. 23.
J:
24.
xn dx
15.5 TAYLOR'S FORMULA
L5.5.1 Theorem
['
xn lnz x dx
*" Ln x dx
Values of polynomial functions can be found by performing a finite number of additions and multiplications. Howevet, there are other functions, such as the logarithmic, exponential, and trigonometric functions, that cannot be evaluated as easily. We show in this section that many functions can be approximated by polynomials and that the polynomial, instead of the original function, can be used for computations when the difference between the actual function value and the polynomial approximation is sufficiently small. There are various methods of approximating a given function by polynomials. One of the most widely used is that involving Taylor's formula named in honor of the English mathematician Brook Taylor (16851731).The following theorem, which can be considered as a generalization of the meanvalue theorem (4.7.2), gives us Taylor's formula. Let / be a function such that / and its hrst n derivatives are continuous on the closed interval fa, bl. Furtherrnore, let f{n+r)(x) exist for all r in the open interval (a, b). Then there is a number f in the open interval (a, b) such that
fO)  f(a)
(b a)
(ba)'+ i(n+l\( t'\
. f@(a) ,, (b a)"+ffi1i *';
(b 61"+r (1)
Equation (1) also holds if b < a; in such a case/la,blis replacedby fb, al' and (a, b) is replacedby (b, a). Beforeproving this theorem, note that when n:0,
(1) becomes
f(b):f(a)+f'G)@a) where f is between a and b. This is the meanvaluetheorem (4.7.2). There are several known proofs of Theorem 15.5.1,although none is very well motivated. The one following makes use of Cauchy's meanvalue theorem (15.1.3). pRooFoF THEoREM15.5.1: Let F and G be two functions defined by
F(r): f@l f@) f' (x)(bx)'+
ffi
(b r)'
ry (b x)*(2) (b v1nr
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
and
G(r):ffi
(3)
It follows that F(D) :0
and G(b):0.
Differentiating in e), we get
F ' ( r ) :  f ' ( x ) * f ' ( x )  f " ( x ) ( b  x ) * 2 f " ( x )  ( b x ) 2! _f"' (x)(b x)2, 3f"' (x)(b x)2 (n 
(b  x)"z
7)f@\(x) rE____6=l[l_ rnf@(x)(b n!
frtv)(x)(b x)g
f(n)(x)(b v1"r
 x)n1 _f("+r)(x)(b x)n n!
Combining terms, we see that the sum of every oddnumbered term with the following evennumbered term is zero, and so only the last term remains. Therefore, F,(x):
 f("*D.(x)  )c)n @ n!
Differentiating
(4)
in (3), w€ obtain
G'(x):+,@x),
(s)
Checking the hypothesis of Theorem 15.1.3, we see that (i) F and G are continuous on Ia,bl; (ii) F and G are differentiable on (a,b); (iii) for all r in (a,b),G'(x) + 0. So by the conclusion of Theorem 1.5.1..3it follows that F(b)  F(a) _ F'({)
G(b)G(a)c'(f)
where f is tn (a, b). But F(b)  0 and G(b)  0. So
F(il GryG(a) \J* '(f)
(5)
for somef in (a,b). Letting tc: a irt (3), x : ( in (4), and r : f in (5) and substituting into (6), we obtain
F(a):
f(n+l)(t\
or, equivalently,
(b  €)"

nt
l €€)"j
]
(b_
a)"*,
(n + 1)!
15.5 TAYLOR'SFORMULA
'  a)n+r F(il   f,'*:'(i)= (n+ 1)! @ Letting x  a in (2), w€ obtain
(7)
F ( i l f ( u )  f ( a )  f ' ( o ) ( b  a )  ' + @  a ) ' 
" '
_,ffi@_a)n=_,#@_a)n (8) Substituting from (7) into (8), we get
f ( u ) f ( o )+ f ' ( a ) ( b a )* t : #
'
@ a)'+. . . f(',+l)(f)
,7 , i  @  a ) "+ f f i
*f{")(a)
(, 7b  a ) ' + r
which is the desired result. The theorem holds if b < a becausethe conI clusion of Theorem 15.1.3 is unaffected if. a and b are interchanged. If in (1) b is replaced by x, Taylor's formula is obtained. It is
(e) where f is between a and x. The condition under which (9) holds is that / and its first n derivatives must be continuous on a closed interval containing a and x, and the (tc* 1)st derivative of / must exist at all points of the corresponding open interval. Formula (9) may be written as f (x) : Pn(x) * R,(r)
(10)
where f" (a)
( x  a ) * ' t @  n ) r, + . . .
P"(x) : f (a)
+  f(")(a)Q  a ) " n!
and
R,(r) :m
Q
a)n+t
where f is between aand r
( 1 1)
(12)
P,(r) is called the zthdegree Taylor polynomialof the function / at the number a, and Rr(r) is called the remainder.The term R,(r) as given in (12) is called the Lagrangeform of the remainder, named in honor of the French mathematician Joseph L. Lagrange (17361813).
654
INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA
ExAMPLE 1:
Find the thirddegree Taylor polynomial of the cosine function at L+rratrd the Lagrange form of the remainder.
soLUrIoN: Letting f (x) : cos r, we have from (11.)
p,(r) i).LP (.o)'*r+ (,o)" : re). r'G)Q Because/(x) : cos x, f(*rr) : +{Z; f' (x) :sin r, f' (*zr): +t/2; rlf} :cos : : (x) (x) (*d:'4r2.Therefore, x, " x, sin f "' f f " Gd f "' : P,(r) +\/2 *{2(x Ln) i{2(x *o), + jrt/2(x  *o)' Because /(to)(r) : cosr, we obtain from (12) wheref is between*r andx &(r) : 2l(cos€) (x  In)n Because lcosf  = 1, we may concludethat for all x, lRr(r)  < *(x  *rr)n. Taylor's formula may be used to approximate a function by means of a Taylor polynomial. From (10) we obtain
lR"(r)  lf(x)  P,(r)I
(13)
If. P"(x) is used to approximate f(x), we can obtain an upper bound for the error of this approximation if we can find a number E ) 0 such that lR"(r)   E or, because of (13), such that lf (x)  P"(x)l < E or, equivalently, P"(x)ffQ)=P"(r)+E ExAMPLE2: Use the result of Example L to compute an approximate value of cos 47" and determine the accuracy of the result.
sor.urroN: 4T  f{or radians. Thus, in the solution of Example L, we takex:f&r and r in:#n, and we have
cos4T : *12L'  #rr  L(#d' + 6(#zr)'l + &(rttzr) where Rr(ttrr) ;r
cosf (#n)n
within N
l2n+U
 1 l: l z n  Z n  L l :   r I t + l2n rl lW  14"'r11 4 n * 2
has the limit #.
I n
Hence, w€ must find a number N > 0 such that for every integer n > N But
i s e q u i v a l et o nt 2n*l r+
#.€
which is equivalent to nrL2e 4e So it follows that
1l I n + 1 r l l2n
for every integer n u'l'
 2e 4e
Therefore, if N: (1  2e)14e,Definition 16.'!,.2holds. In particular, if e: t, N  (1  +)l+: g. So
.*
for every integer n , ,
1 6 . 1S E Q U E N C E S
For instance,
o rLLUsrRArron3: Consider the sequence{(1)*'lnl,. Note that the nth element of this sequence is (l)+rln and 11;"+t is equal to *L when n is odd and to 1 when z is even. Hence, the elements of the sequencecan be written L 1 1L "t ,  2 ' i '  Z ' E ' '
' 11' 1 n + t
Itr Fig. 16.1.5 are plotted points corresponding to successive elements of this sequence.In the figure, ar:1, az:t, as:t, a+:*, aa:*, ae:t, az:*, ae:t, as:*,4r0:*. The limit of the sequenceis 0 and the elements oscillate about 0. 1. 7 rf, 3  5
t
1 dS:
As
:
L 9
t2 5
Figure 16.1.5 Compare Definition 16.1.2with Definition 4.7.'1,of the limit of.f(x) as x increases without bound. The two definitions are almost identical; however, when we state that /tr) : L, the function / is defined for "l1a all real numbers greater than some real number R, while when we consider an, n is restricted to positive integers. We have, however, the nlim following theorem which follows immediately from Definition 4.1.1. 15.1.3 Theorem . If lim /(x) : L, and / is defined for every positive integer, then also L when n is any positive integer. ygiAl: The proof is left as an exercise (see Exercise20). o rLLUsrRATroI.t4: We verify Theorem 16.1.3for t'he sequence of Example L. For that sequencef (n) : nl(2n * 1). Hence,f (x) : xl(2x * 1) and
I N F I N I T ES E R I E S
It follows then from Theorem 16.1.3 that
when z is any ]!!_f(n):* positive integer. This agreeswith the solution of Example 1. o 16.1.4 Definition
ExAMPLE2: sequence
Determine if the
[4n'I l2n'+ L) is convergent or divergent.
If a sequence {a,} has a limit, the sequenceis said tobe conaergent,and, we say that an conoergesto that limit. If the sequenceis not convergent, it is said to be diaergent. solurroN:
We wish to determin" if
4*lQf
* 1) and investigate "tl1/(r). ,.4*..4 Irm;];: lim :2 I I I)+@e ,*2**
,It
4n2l(2nz* 1) exists.Let /(r) :
Therefore, by Theorem 16.1,3, Ol: 2. We conclude that the given ]1g_f sequenceis convergent and that4n2l(2n2 * 1) converges to 2. ExAMPu 3: Prove that if lrl the sequence{r"} is convergent and that rn convergesto zeto.
O: 0. Hence, Jl* the sequenceis convergent and the nth element converges to zero. If 0 < Irl < t, we consider the function / defined by f (x): f, where r is any positive number, and show that lim fc :0. Then from Theorem solurroN: First of all, if.r:
0, the sequenceis {0} and
1.6.L.3itwill follow that lim r^: O wneii.tl
ury positive integer.
n+@
To prove that
(0 < lrl < L), we shall show that for any r,:0 "ljt e > 0 there exists a numberN > 0 such that lfOl
<e
wheneverr)N
(4)
Statement (4) is equivalent to wheneverr>N
lrl' 0 there exists a numberN > 0 such that ln" Ll ( e for every integer n ) N. In particular, when 6: f, there exists a number N > 0 such that for every Lntegern > N
lo" Ll < + or, equivalently, ,+< nnL nN
(6)
Becausean: 0 if n is odd and an: 2 if. n is even, it follows from (5) that
+ anq11
ExAMPLEL: For each of the following sequences determine if it is increasing, decreasing, or not monotonic:
(a) {nlQn + 1)} (b) {Un} (c) t ( L)"*'ln\
solurroN:
(a) The elements of the sequence can be written
12 3 4
n
3'E'r'9''"761'
n+!
zn+g''''
Note that no*, is obtained from an by replacing n by n * l. Therefore, becausa eo:nl(2n*L), n+l an+r:ffiffi:m
n+l
Looking at the first four elements of the sequence,we see that the elements increase as n incteases. Thus, we suspect in general that
n n+l zn+r= h+g
(1)
Inequality (1) can be verified if we find an equivalent inequality which we know is valid. Multiplying each member of inequality (1) by (2n + 1) (2n + 3), w€ obtain n ( Z n+ 3 )
(2)
668
I N F I N I T ES E R I E S
Inequality (2) is equivalent to inequality (1) because (1) may be obtained from (2) by dividing each member by (2n + t)(Zn * 3). Inlquatity (2) is equivalent to 2n2*3n=2n2*3n+L
(3)
Inequality (3) obviously holds becausethe right member is L greater than the left member. Therefore, inequality (1) holds and so the given sequence is increasing. (b) The elements of the sequencecan be written
for all n, the sequenceis decreasing. (c) The elements of the sequencecan be written
( ; 1, )R' ' +,r. . .(  1 ) t t + 2
, ,  21'1i ,  4 '1
,
ar:l and az:*, and so ar) az.But ar:t, and so az< as.In a more generalsense,consider three consecutiveelementsan: (l)n+rln, an+r: (1)"*'l@ * 1), and anrz: FL)"*"(n+2).If. n is odd, an) an*, and an+r1 aaa2,?trd, if. n is evett, an 1 an*, attd anal) ar+z,Hence, the sequence is neither increasing nor decreasing and so is not monotonic.
16.2.2 Definition
The number C is called a lower bound of the sequence {a"} if. C < an for all positive integers n, and.the number D is called arr upper boundof the sequence {a"} if. an = D for all positive integers n. o ILLusrRATrouL: The number zero is a lower bound of the sequence {nl(2n * 1)} whose elementsare 12
3 4
g's'7't"
"
r f f inr
Another lower bound of this sequence is *. Actually any number which is less than or equal to $ is a lower bound of this sequence. . O ILLUSTRATION
"' ,L ' t
1
L 'g ' 4 "
For the sequence {1,1n} whose elements are
'L " A"
1 is an upper bound; 26 is also an upper bound. Any number which is greater than or equal to f. is an upper bound of this sequence,and any nonpositive number will serve as a lower bound. o
16.2 MONOTONICAND BOUNDEDSEQUENCES 669
From Illustrations 1 and 2, we see that a sequence may have many upper and lower bounds. 16.2.3 Definition
If A is a lower bound of a sequence {an} and if A has the property that for every lower bound Cof {a"}, C  A, then A is called the greatestlower bound of the sequence. Similarly, if B is an upper bound of a sequence {an} and if B has the property that for every uPPer bound D of.{an}, B = D , then B is called the least upper boundof the sequence. . rLLUsrRArroN3: For the sequence {nl(2n + 1)} of Illustration 1, the greatest lower bound is * because every lower bound of the sequenceis less than or equal to *. Furthermore,
2n*L
z+!
.t
for all n, and * is the least upper bound of the sequence. In Illustration 2, we have the sequence{Uz} whose leastupper bound is 1 because every upper bound of the sequenceis greater than or equal o to 1. The greatest lower bound of this sequenceis 0. 1,6.2.4Definition
A sequence{a"} is said to beboundedif and only if it has an upper bound and a lower bound. Becausethe sequence {1,1n}has an upper bound and a lower bound, it is bounded. This sequence is also a decreasing sequenceand hence is a bounded monotonic sequence.There is a theorem (1,6.2.6)that guarantees that a bounded monotonic sequence is convergent. In particular, the sequence {Un } is convergent because lim (1/n) :0. The sequence {n} whose elements are L,
n,
monotonic (because it is increasing) but is not bounded (because there no upper bound). It is not convergent because lim n  f o.
For the proof of Theorem 16.2.6we need a very important property of the realnumber svstem which we now state. 16.2.5 The Axiom of Every nonempty set of real numbers which has a lower bound has a greatCompleteness est lower bound. Also, every set of real numbers which has an upper bound has a least upper bound. The axiom of completenesstogether with the field axioms (1.1.5 through 1.1.11)and the axiom of order (1.1.14)completely describe the realnumber system. Actually, the second sentence in our statement of the axiom of completenessis unnecessarybecauseit can be proved from
I N F I N I T ES E R I E S
Ar flz
As da
fl5&5Tz
Aa As
flto
D
F i g u r e1 6 . 2 . 1
16.2.5Theorem
the first sentence. It is included in the axiom here to expedite the discussion. Suppose that {ar} is an increasing sequence that is bounded. Let D be an upper bound of the sequence.Then if the points corresponding to successive elements of the sequence are plotted on a horizontal axis, these points will all lie to the left of the point corresponding to the number D. Furthermore, because the sequence is increasing, each point will be either to the right of or coincide with the preceding point. See Fig. 1,6.2.1,. Hence, as n increases,the elements increase toward D. Intuitively it appears that the sequence {a"} has a limit which is either D or some number less than D. This is indeed the case and is proved in the following theorem. A bounded monotonic sequenceis convergent. PRooF: We prove the theorem for the casewhen the monotonic sequence is increasing. Let the sequencebe {a"}. Because{a"} is bounded, there is an upper bound for the sequence. By the axiom of completeness(15.2.5), {a,} has a least upper bound, which we call B. Then if e is a positive number, B  e cannot be an upper bound of the sequencebecauseB  e ( B and B is the least upper bound of the sequence.So for some positive integer N B € < a1s
(4)
BecauseB is the least upper bound of {a"}, by Definition 1,6.2.2it follows that an < B
for every positive integer n
(s)
Because{ao} is an increasing sequence,we have from Definition 16.2.1(i) an s an11
for every positive integer n
and so flysan
whenevetn>N
(5)
From (4),(5), and (6), it follows that B  € < ay from which we get B € < an
whenever n > N
or, equivalently, € < fln B which can be written as
l a " B l
(7)
16,2 MONOTONICAND BOUNDEDSEQUENCES
But by Definition 1.6.1.2,(7) is the condition that lim an:B. Therefore, the sequence {a"} is convergent. To prove the theorem when {a"} is a decreasing sequence, we consider the sequence { a"}, which will be increasing, and apply the above results. We leave it as an exercise to fill in the steps (see Exercise 13). t '!,5.2.6 states that tf {a"} is a bounded monotonic sequence, Theorem there exists a number L such that lim an: L, but it does not state how fl*@
to find L. For this reason, Theorem 16.2.6 is called an existencetheorem. Many important concepts in mathematics are based on existence theorems. In particular, there are many sequences for which we cannot find the limit by direct use of the definition or by using limit theorems, but the knowledge that such a limit exists can be of great value to a mathematician. In the proof of Theorem 76.2.6we saw that the limit of the bounded increasing sequenceris the least upper bound B of the sequence.Hence, if D is an upper bound of the sequence, lim an: I < D. We have, then, the following theorem. 15.2.7 Theorem
Let {aJ be an increasing sequence,and suppose that D is an upper bound of this sequence.Then {a"} is convergent and lim an2
(e)
When n:1, inequality (9) becomes2:2, and (9) obviously holds when n ) 2. Because inequality (8) is equivalent to (9), it follows that the given sequenceis decreasing and hence monotonic. An upper bound for the given sequence is 2, artd.a lower bound is 0. Therefore, the sequence is bounded. The sequence {2ln!} is therefore a bounded monotonic sequence, and by Theorem 1,6.2.6it is convergent. Theorem 16.2.5states that a sufficient condition for a monotonic sequence to be convergent is that it be bounded. This is also a necessary condition and is given in the following theorem. 15.2.9 Theorem
A convergent monotonic sequenceis bounded. pRooF: We prove the theorem for the casewhen the monotonic sequence is increasing.Let the sequencebe {an}. To prove that {a"} is bounded we must show that it has a lower bound and an upper bound. Because{ar} is an increasing sequence,its first element serves as a lower bound. We must now find an upper bound. Because{an} is convergent, the sequencehas a limi! call this limit L. Therefore, lim an: L, and so by Definition 76.1.2,for any e > 0 there exists a number N > 0 such that
l a " Ll
N
or, equivalently, €
N
or, equivalently, L€
N
Because {an} is increasing, w€ conclude that an
16.3 INFINITESERIESOF CONSTANTTERMS
Therefore, L * e will serve as an upper bound of the sequence {ar}. To prove the theorem when {a,,} is a decreasing sequencewe do as '1.6.2.6:. Consider the sequence {a,'}, suggested in the proof of Theorem which will be increasing, and apply the above results. You are asked to do this proof in Exercise L4.
16.2 Exercises L through In Exercises
12, determine if the given sequence is increasing, decreasing, or not monotonic.
,4' [ 3 n  I l
3.
2. {sin nn}
lan * 5)
z^ [' 2 " I lt+zJ
6.
7 ' . ["tI L3'J
9. L2.
10. {n' * (l)"nI
nn
n!
r.3.5
13. Use the fact that Theorem 15.2.5holds for an increasing sequenceto prove that the theorem holds when {co} is a decreasing sequence.(nrrt: Consider the sequence {a"1.) 14. ProvglTheorem 15.2.9when {an} is a decreasing sequenceby a method similar to that used in Exercise 13. In Exercises15 through 21, prove that the given sequenceis convergent by using Theorem76.2.6, 15. The sequenceof Exercise 1.
16. The sequenceof Exercise4'
17. The sequenceof Exercise 5.
18. The sequenceof Exercise8.
19. The sequenceof Exercise 11.
20. The sequenceof Exercise 12'
,1 '  ' 1ttl 2)
16.3 INFINITE SERIES OF CONSTANT TERMS
L5.3.L Definition
The familiar operation of addition applies only to a finite set of numbers. We now wish to extend addition to infinitely many numbers and to define what we mean by such a sum. To carry this out we deal with a limiting process by considering sequences. If t un\ is a sequence and n
sn:
). ur:
L l r* u , * u g *
' ' '*lln
i:l
then the sequence{s"} is called an infinite series. Thenumbetsu1,u2rxs, . . . arecalled thetermsof theinfiniteseries. We use the following symbolism to denote an infinite series:
674
I N F I N I T ES E R I E S
)
un: ut* uz* us* . . ' + un* . . .
(1)
Given the infinite series denoted by (1), sr: Llr, sz: tt, * ttr, ss: ur * uz * ^nd in general :r,
s k:
Ltr* ur* us* . . . * urc
Z"r:
(2)
where sa, defined by (2),is called the kth partial sum of the given series, and the sequence {sr} is a sequence of partial sums. . . + iln_t Because snt: Llr* u, * . . . + lln_r and sn: Ltr* u r * . * ,r, we have the formula Sn:
EXAMPLE
Given the infinite
Snt *
sot,urroN: Applying formula (3), we get
series
f ":,z#n find the first four elements of the sequence of partial sums {sn]t, and find a formula for sn in terms of n.
(3)
Un
St: ut:
32:
Srt
L1 L 2:,
,L12
U2:;f
z
2'3
r
3'4
,213
SB:SztUs:;t
:
:
3 4
314 S + : S s* L l 4 : ; + f t : ; By partial fractions,we see that 1 1 wrk, _ k ( k + 1 )  k_ t _
1
k+L
Therefore, .1 Ltr: 1 
i.,
ur:
11 1, 4 1, ' Us: ; ;, 3
n*l
: Llr * u, *+ . . . + Llnr* un,we have Thus, because s?z
,.:('il*(,t*)*(}r,). .(,,t).eh) Upon removing parenthesesand combining terms, we obtain 1,n sn:ln11:rt7 Taking n:
1
l, 2, 3, and 4, we see that our previous results agree.
Note that the method of solution of the above example applies only
16.3 INFINITESERIESOF CONSTANTTERMS
to a special case. In general, it is not possible to obtain such an expression forsr. We now define what we mean by the "sum" of an infinite series. 16.3.2 Definition
f"t j
unbe agiven infinite series, and let {s,} be the sequenceof partial
n=l
t" exists and is equal to S, ,lit we say that the given series is conuergentand that S is the sum of. the sn does not exist, the series is said to be digiven infinite series. If ,lin oergentand the series does not have a sum. sums defining this infinite series. Then if
Essentially Definition 16.3.2statesthat an infinite seriesis convergent if and only if the corresponding sequenceof partial sums is convergent. If an infinite series has a sum S, we also say that the series converges to S. Observe that the sum of a convergent series is the limit of a sequence of partial sums and is not obtained by ordinary addition. For a convergent series we use the symbolism +@
2o'. n=t to denote both the series and the sum of the series. The use of the same symbol should not be confusing because the correct interpretation is apparent from the context in which it is used. EXAMpLE2: Determine if the infinite series of Example L has a sum.
SOLUTION: IN thc
solution of Example L, we showed that the sequence
of partial sums for the given seriesis {s"} : {nl (n * 1)}. Therefore, lim sn: n
*@
lim ll
*@
nr.L L
::uur:r
ni
h_.re 1+l
n
So we conclude that the infinite series has a sum equal to 1..We can write
. . . +_+:* ' s_L:1+!+L+l+ ' n(ntL) 2' 6 12' 20
f_,n(n*l)
. :1
As we mentioned above, in most cases it is not possible to obtain an expression for sn in terms of. n, and so we must have other methods for determining whether or not a given infinite serieshas a sum or, equivalently, whether a given infinite series is convergent or divergent. 16.3.3Theorem If the infinite series !
a, is convergent,then lim un:0.
676
I N F I N I T ES E R I E S
PRooF: Letting {s,} be the sequenceof partial sums for the given series, and denoting the sum of the series by S, we have, from Definition 16.3.2, Therefore,for any € >0 there exists a number N>0 such "lit*:S. that lS snl ( *e for every integer n > N. Also, for these integers n > N we know that lS  sn+rl< te. We have, then, lun*rl: ls,*,
 s,l : lS sn*sn*l Sl s lS r,l * ls"*r Sl
So
l ' n * r l< t e * l e Therefore, lim ur:
e
for every integer n > N
0.
I
The following theorem is a corollary of the preceding one.
1.6.9.4Theorem
If
lln
unLsdivergent.
un * 0, then the series 5
n:l
PRooF: Assume that i
un is convergent. Then by Theorem 16.3.3
n:l
lim un:0.
But this contradicts the hypothesis. Therefore, the series is
lI*@
I
divergent.
EXAMPLE3: Prove that the followitrg two series are divergent.
(a)yry z+Z+# *#* +@
(b) ':t33+33+
soLUrIoN:
(a) lim un: tt t@
lim It *@
n2 +'l' nz
_ lim n*@
+0 Therefore,by Theorem15.3.4the seriesis divergent. (1)n*t3, which does not exist. Therefore,by (b) JiT" "": l*_ Theorem15.3.4the seriesis divergent. Note that the converseof TheoremL6.3.3is false.That is, if
un: 0, olim it does not follow that the series is necessarilyconvergent. In other words, it is possible to have a divergent series for which lim zn: 0. An example of such a series is the one known as the lror*onit'rln1es, which is (4)
16,3 INFINITESERIESOF CONSTANTTERMS
gI7
0. In Illustration 1 we prove that the harmonic series ):*tl": diverges and we use the following theorem, which states that the difference between two partial surls sa and s7of a convergent series can be made as small as we please by taking R and T sufficiently large. Clearly,
15.3.5 Theorem
Let {sn} be the sequence of partial sums for a given convergent series +@
2 u". Then for any € > 0 there exists a number N such that lsas7l< e
w h e n e v e r R> N a n d T > N
pRooF: Because the series i
unis convergent, call its sum S. Then for
any e ) 0 there exists an N = O such that lS  rrl < te whenevet n > N.
Therefore, if R > N and T > N, l s r  r . l : l s r  S + S  s r l < l s r  S l+ l S  s r l < i e * * e So lsas7'< l e
I
wheneverR>NandT>N
. rLLUsrRArroN L: We prove that the harmonic series (4) is divergent. For this series, sn:L++.. and szn:l +
+
.
So szn sr:
L,L,L
"*l
r+Z
(s)
n+3
lfn
11,1L1,111 n+lntrr n+3
' T
2n>mr
hTmr
' ' ' Tm
(oi
There are n terms on each side of the inequality sign in (6); so the right side Lsn(ULn) : *. Therefore,from (5) and (5) we have sznsnlt
whenevetn
(7)
But Theorem 1.6.3.5states that if the given series is convergent, then szn sn may be made as small as we please by taking n large enough; that is, if we take e  *, there exists an N such that sro sn 1* whenever 2n > N and n ) N. But this would contradict (7). Therefore, we conclude that the harmonic series is divergent even though lim Lln:0. ll*@
678
INFINITE SERIES
A geometricseriesrs a series of the form +co
'
.LJ n:L
ornr a* ar* arz*.,.+arnL*
(8)
The nth partial sum of this series is given by sn:a(l
*r*r2*
' ''+r"t)
(e)
From the identity 1  vn:
(1  r) (1 * r * r2 I
. . . + r"t)
we can write (9) as a(r/)_ sn: _1 _ r
16.3.6 Theorem
ifr#L
The geometric series converges to the geometric series diverges if lrl
(10)
sum al(1  r) if lrl
PRooF: In Example 3, Sec. 1,6.1,, w€ showed that lim r' : 0 if lrl Therefore, from (10) we can conclude that if lrl ,.4
,t]1t':1, So if lrl ( L, the geometricseriesconverges,and its sum is al(1. r). lf. r:1, sn: na.Then lim s, : fo if.a > 0, and lim s,r: @ if a < 0. n *@ lf. r:l,the geometricseriesbecomesa a * a  . .  * (Ll"to +' ' ' ; so sz:0if ziseven,andsr:aif.n isodd.Therefore, rll1t does not exist. Hence, the geometric series diverges when lrl : 1. If lrl > 1,,li1 ay'tt:, Clearly, rnt * 0 becausewe )!T_r"1. ,li1 can make lro11as large as we pleaseby taking nlarge enough. Therefore, I by Theorem 16.3.4 the series is divergent. This completes the proof. h*@
* +++1** which is the geometric series, with a : 1 and r: !. Therefore, by Theorem 15.3.5the series is convergent. Becauseal(L  r) : Ll(l  L) :2, . the sum of the series is 2. The following example illustrates how Theorem 15.3.6 can be used to expressa nonterminating repeating decimal as a common fraction.
16.3INFINITE S E R I E SO F C O N S T A N TT E R M S
EXAMPw 4: Express the decimal 0.3333. . as a common fraction.
SOLUTION:
3333 , 10 100 L,000 10,000 We have a geometricseriesin which a: fo and r : $. Becauselrl < 1,,It follows from Theorem 1.6.3.6that the series converges and its sum is al (L  r). Therefore,
The next theorem states that the convergence or divergence of an infinite series is not affected by changing a finite number of terms. 1,6.3.7Theorem
two infinite series, differing only in their ffust m
If
,.;,
(i.e. , oi! rrif k
series diverge. PRooF: Let {sr} and {t"} be the sequencesof partial sums of the series +@
+@
n:l
n:1
Then Sn:
At*
Az*
*
fl* *
a**,
*
*an
A*'r2 *
and ' ' ' + b**b**t+b*+z+
tn:bt*br*
' '
*bn
Becausl ak: brcrf k > m, then rf n sn tn: @r* az* . . . + a*)
(br*br*
' ' +b*)
So Sn
tn:
t*
S*

whenevet n
We wish to show that either both lim sz and jlt
(11)
f, existor do not
exist. Supposethat lim fn exists.Then from Eq. (11)we have ll*@
sn: tn* (s* t*)
whenever n > m
and so lim sn: n*@
lim tn * (s*  t*) n*@
Hence, when lim tn exists, lim sn also exists and both series converge. ll
*@
fl,*@
SERIES INFINITE
Now suppose that lim f, does not exist and lim s, exists. From Eq. (11.)we have tn: sn* G
s^)
whenevern > m
Because,lit * exists,it follows that lim fn: lim sn* (t^ s*) n+6
n+@
fo does to exist, which is a contradiction.Hence,if ,111 )!I_t"has I not exist, then lim s, does not exist, and both series diverge. and so
glven serles ls
EXAMPLE5: Determine whether the infinite series
s1 n ? : r f l+ 4 is convergent or divergent.
which can be written as
o+o+o+o+1*!*1 * 5'6'7'
+ !n+
(12)
Now the harmonic series which is known to be divergent is
1+l++I+f+l++
+ !n+
(13)
Series (L2) differs from series (13) only in the first four terms. Hence,by '1,6.3.7, Theorem series (12)is also divergent.
EXAMPLE6: Determine whether the following infinite series is convergent or divergent:
solurroN:
The given series can be written as
ry*%*W*%
+@ n:l
_  31 L' 2 r gr'
L r t + 1 L2 L2 2 _' ' ' ' ' 3t gn' 3t 3u gr' 38
Consider the geometric series with a  3 and 2,2,2,2,2,2,2,2,
5 y' F y' FF*
y F ' '
(r4) l.  _ r3.
(15)
This series is convergent by Theorem 16.3.6.Becauseseries (L4) differs from series (15) only in the first five terms, it follows from Theorem16.3.7 that series (14) is also convergent.
1 6 . 3 I N F I N I T ES E R I E SO F C O N S T A N TT E R M S
681
As a consequenceof Theorem 1,6.3.7,for a given infinite series, we can add or subtract a finite number of terms without affecting its convergence or divergence. For instance, in Example 5 the given series may be thought of as being obtained from the harmonic series by subtracting the first four terms. And because the harmonic series is divergent, the given series is divergent. In Example 6, we could consider the convergent geometric series )))
(15)
L t L ) L I
36'37'38r
and obtain the given series (1a) by adding five terms. Becauseseries (L6) is convergent, it follows that series (14) is convergent. The following theorem states that if an infinite series is multiplied term by term by a nonzero constant, its convergence or divergence is not affected. 16.3.8 Theorem
Let c be any nonzero constant. +@
(i) If the series n  l
+m
(ii) li'.nu seriesX Ltnisdivergent,then the seriesi cu,is also rt:r n:r divergent.
F@
be sn. Therefore, 7rr" ' * un. The nth partial sum of the series 5 cun is
PRooF: Let the ntt: partial sum of the series ' sn:ur*uz* ' ' ' * u)  csn. c(urlur* +co
nenr (i): If the series Therefore,
h'*a
ft:l
lim csn c l i m s n : c ' S n *@
lLl@
+@
Hence, the series 2 ,un is convergent and its sum is c ' S. n=l
ranr (ii):
If the series )
z, is divergent, then lim s, does not exist' Now n+@
n:l +@
suppose that the series n:I sn: csrlc, and so lim sn:
lim
1 L ( c s " )  ; c ):\""
ll'*@
I N F I N I T ES E R I E S
Thus, lim sn must
which is a contradiction. Therefore, the series
flt'a
+@
I n:1
ExAMPLE7: Determine whether the infinite series
$1
solurroN'h:l+$+ ,!r+ft+ ,i Because 2 tl" is the harmonic series which is divergent, then by The
4n
"4r is convergent or divergent.
orem 1,5.3.8(ii)with c  *, the given series is divergent.
Note that Theorem 15.3.8(i) is an extension to convergent infinite series of the following propefiy of finite sums: nn
2 , o * : , 2k : l o o
k:l
Another property of finite sums is nnn k:l
k:l
k:7
and its extension to convergent infinite series is given by the following theorem. +@
L5.3.9 Theorem
If )
n:l
+co
a" and )
bn are convergent infinite series whose sums are S and R,
respectively,then {m
(i)
is a convergentseriesand its sum is S * R; \::
(ii) @,bn)is a convergentseriesand its sum is S u
R.
The proof of this theorem is left as an exercise(seeExercise11). The next theorem is a corollary of the above theorem and is sometimes used to prove that a series is divergent.
L6.3.L0 Theorem
If the series 5 an is convergentand the series n:l
the series X @, * b,) is divergent. n:l
+@ n:l
1 6 . 3 I N F I N I T ES E R I E SO F C O N S T A N TT E R M S
pRooF: Assume that i
683
@** b,) is convergentand its sum is S. Let the
n=l
sum of the series )
a"be R. Then because
n=l
+@
+6 A " : 2 ) l( a"* b") a"l n:l n=7
+@
bo is convergent and its sum is
it follows from Theorem 16.3.9(ii)that \ n=l
+m
S  R. But this is a contradiction to the hypothesis that )
b, is divergent.
nl
+@
Hence
I
+m
8: Determine whether EXAMPTE the infinite series
s(t
1\
A\n*n)
solurroN:
Ll4n is divergent.
In Example 7 we Proved that the series n:l
Because the series i
U4' is a geometric series with
lrl 
+ < 1.,it is con
vergent. Hence, by Theorem 16.3.10the given series is divergent.
is convergent or divergent.
Note that if both series5 an and i i
b, are divergent,the series
!e, tf an Un and (ant bn)mayor may not ol'.or,""rr;r1 For examp
T,t: Lln,then a, * bn: 2lnand b n :  U n , t h e n a nl b n : 0
divergent.But tf a, : Lln and
2rtris
+m
and
L6.3 Exercises In Exercises1 through 6, find the first four elements of the sequenceof partial sums {s,} and find a formula for snin terms of n; also,determine if the infinite series is convergent or divergent, and if it is convergent, find its sum' r'
o 5t"h
a S l
?:, (2n 1) (2n + 1)
6'>,7 +oo
In Exercises 7 through 10, find the infinite series which is the given r::;:"ce nite series is convergent or divergent, and if it is convergent, find its
8.{s,}:{h}
,n_l
of partial sums; also determine if the infi
s. {s,} : {1}
lz)
10. {s,}  {3n}
684
I N F I N T ES E R I E S
11. Prove Theorem 15.3.9. Iir Exercises12 through 25, write the first four terms of the given infinite series and determine if the series is convergent or divergent. If the series is convergent, find its sum.
12. L L ' $ l . i !rn+
,?'
t?
17. (1)"*'* ,;, "z .f !
22.$
i!t
sinhn n
+@r+@
2tz+ 1 1? Lv' $ A3n+2
L4.s+ A3n
18.$ r, n1
+00
l r t5 . ! 1
L6.
?:,
n=l +@
fm+co
20. .L/ '
L9. Z.J S en
zs.i (2n+3,)
sln rrn
n:l
n:l
"?t
2 3nr
+@
21,.
cos nTn n:l
+@
24. n:l
In:l
In Exercises25 through 29, express the given nonterminating repeating decimal as a common fraction. 26. 0.2727 27 . . .
27. 1.234234234 . . .
28. 2.04545 45 . . .
29. 0.4653 46534553 . . .
E\ A ball is dropped from a height of.12 ft. Each time it strikes the ground, it bounces back to a height of threefourths the distance from which it fell. Find the total distance traveled by the ball before it comes to rest.
16.4 INFINITE SERIES If all the terms of an infinite series are positive, the sequence of partial OF POSITM TERMS sums is increasing. Thus, the following theorem follows immediately from Theorems15.2.5and'1,6.2.9. 16.4.1 Theorem
An infinite series of positive terms is convergent if and only if its sequence of partial sums has an upper bound. pRooF: For an infinite series of positive terms, the sequenceof partial sums has a lower bound of 0. If the sequenceof partial sums also has an upper bound, it is bounded. Furthermore, the sequenceof partial sums of an infinite series of positive terms is increasing. It follows then from '1,6.2.6 Theorem that the sequence of partial sums is convergent, and therefore the infinite series is convergent. Supposenow that an infinite series of positive terms is convergent. Then the sequence of partial sums is also convergent. It follows from 'l'6.2.9 Theorem that the sequenceof partial sums is bounded and so it has an upper bound. r
EXAMPLE 1: Prove that the series \{1 n! ,.4:, is convergent by using Theorem 15.4.'1,,
SOLUTION:
WC
MUSI
find an upper bound for the sequence of partial
f@
sums of the series LIS U n ! .
s r : ! , s z t + # ,
16.4 INFINITESERIESOF POSITIVE TERMS
sn:r+h+#+ffi+ Now consider the first n and rt:
*' 1 L'2 '3'
. . . 'n
geometric series with A : " 1 ,
1
$ l
(1)
1 + ; * 8.l  * 1 * 2r* #_r2*4
(2)
By Theorem 16.3.6 the geometric series with a : 1 and 1: $ has the sum a/(1  r) : U G  t) :2. Hence, sumrnation (2) is less than 2. Observe that each term of summation (1) is less than or equal to the corresponding term of summation (2); that is, 11 kI zkr This is true becausek!   ' 2 ' tor L contains k 1factors each
' k, which in addition to the facthan or equ al to 2. Hence,
n Zr
From the above we see that sn has an upper bound of 2. Therefore, by '1.6.4."1. Theorem the given series is convergent.
In the above example the tenrrs of the given series were compared with those of a known convergent series. This is a particular caseof the following theorem known as the comparisontest. 16.4.2 Theorem Comparison Test
Let the series5 un be a series of positive tenns. n:l
+o
(i) rf n:l
+m
vergent,andllnSanforallpositiveintegersn,then> n:r convergent. {o
(ii) If
is a series of positive terms which is known to be diA*" +@ vergent, and un 2 vonfor all positive integers n, then n:l divergent.
PRooF oF (i): +@
Let {sr} be the sequence of partial sums for the series
sums for the series 5 T)n.Beh:t
I N F I N I T ES E R I E S +@
cause
a, is a series of positive terms which is convergent, it follows ,) from Theorem 15.4.1that the sequence {t } has an upper bound; call it B. Becauseun < onfor all positive integers n, we can conclude that sn  tn < B for all positive integers n. Therefore, B is an upper bound of the sequence {sr}. And becausethe terms of the series !
un areall positive, it follows
n:l
*oo
from Theorem 16.4.7that
convergent. n:l
PRooFoF (ii): Assume that i + co
+@
n:l
n:1
r,tnis convergent. Then because both
n:l
of positive terms and znn
positive integers n, it follows from part (i) that i ,,is convergent. However, this contradicts the hypothesis, and assumption is false. "Jirr" Therefore,
I
As we stated in Sec. '1.6.3,as a result of Theorem 76.3.7, the convergence or divergence of an infinite series is not affected by discarding a finite number of terms. Therefore, when applying the comparison test, if ui s ut or tti 2 rDi when i > m, the test is valid regardless of how the first m tenrts of the two series compare.
ExAMPLE2: Determine whether the infinite series \'4 n?:r3" + 1 is convergent or divergent.
soturroN: The given serles rs a
a
4,4,4,4 n+ l,o 2g g2+ comparing the nth term of this series with the nth term of the convergent geometric series 4+4+4 +4 3'.9',27',9L'
L
r*.
1
we have 44 3n 3n+1 for every positive integer z. Therefore, by the comparison test, Theorem 1,6.4.2(i),the given series is convergent.
1 6 . 4 I N F I N I T ES E R I E SO F P O S I T I V E TERMS
ExAMPLE 3: Determine whether the infinite series
sol,urroN:
The given series is
L+ t:+
L
$L
rtr
AW,:4+V*.6*
+@1
s^: r.?:t \/n
is convergent or divergent.
687
1
Yn
Comparing the nt}:.term of this series with the nth term of the divergent harmonic series,we have
+YLnLn, +@
So by Theorem \6.4.2(ii) the given series l:I
The following theorem, known as the limit comparisontest,is a conand is often easier to aPply. sequenceof Theorem "1.6.4.2 +@
15.4.3Theorem Let )
Limit ComparisonTest
n=r
un and \
a"be two seriesof positive terms.
n=r
(i) If lim (unla*): c ) 0, then the two serieseitherboth converge fl*'@
or both diverge. (ii) If lim (unla)  0, and if
+@
+@
n:l
n:l
lltcp
un converges.
un converges, then )
+oo
+oo
(iii) If lim (unla)  f m, and if n:l
fl*@
an diverges, then )
un dlerges.
n:\
pRooFoF (i)t Because lim (u*lu) : c, it follows that there exists an N>0suchthat
ll a* n l z d . *
f o ra r r n >N
or, equivalently,  t N
or, equivalently,
c 'u"N
(3)
From the righthand inequality (3) we get u, l Ecan
(4)
+6
lf \
+6
a" is convergent, so is \
{ca,. It follows from inequalif
(a) and
the comparison test that 5 un LSconvergent. From the lefthana ifi:"tquatity(3) we get
(s) +@
tt E
un rs convergent,so is
son test it follows that )
)
) ,n.From inequality (5) and the comparii
u, is convergent.
n=l
+@
It \
n:t
+@
a" is divergent, we can show that )
un is divergent by assum
n=7
*
ing that \ u^ is convergentand getting a contradictionby applying inequality (5) and the comparisontest. In a similar manner,tf
u,is divergent,it follows that j
oois di
vergent becausea contradilltr, i, obtained from inequ"firy 17')and the +@
comparison test if )
u, is assumed to be convergent.
we have trru"JrL proved parr (i). The proofs of parts (ii) and (iii) are left as exercises(see Exercises19 and 20). I A word of caution is in order regarding part (ii) of rheorem 1,6.4.g. Note that when
nlim
(uJa,):0,
the divergenceof the series
f
,, do",
ll:l
not imply that the series 5 un diverges. ExAMPLE 4: Solve Example 2 by using the limit comparison test.
soLUrIoN: Let un be the zth tenn of the given series )
4/(3, + 1) and
n=l
c,nbe the nth term of the convergentgeometricseries +rc".Therefore, f n=t 4
" lim 3:
lim ii*a" ;;;
en+L e
&
7n : lim 3" * L ,+
TERMS 16.4 INFINITESERIESOF POSITIVE
Hence, by part (i) of the limit comparison test, it follows that the given series is convergent.
+m
EXAMPLE5: Solve ExamPle 3 bY using the limit comParison test.
Let un be the nth term of the given series
sorurroN:
Ll \fr and anbe
n:l
the nth term of the divergent harmonic series. Then 1 lim na*
un:.lim 7)v
n*@
V: I
hm \n:*m n*cn
n Therefore, by part (iii) of the limit comparison test, we conclude that the given series is divergent.
+6
6: Determine whether EXAMPLE the series
In Example 1, we proved that the series )
solurroN:
Using the limit comparison test with un: rf lnt
$ng
: Unt',we have
ns
3'n! is convergent or divergent.
",,a5l
llnl is convergent.
tr
nl
1im :2 : lim T: na
An
n+o
I
lim ns : *a 't*@
nl part (iii) of the limit comparison test is not applicable becaur" 5 o,,.or,verges. However, there is a way that we can use the limit "3:rrlpu"iror, test. The given series can be written as
l" *2"*3t*4 t * I'+ . ' ' 4 ! 5! Tt3!
. *4*.' nt
. .
Because Theorem 15.3.7 allows us to subtract a finite number of temrs without affecting the behavior (convergence or divergence) of a series, we discard the first three terms and obtain
4'*5'*{* 4!F6!'
+' ( (n r* t3?) !) l + .
Now lettin E un: @ + 3)sl@ + 3) ! and, as before, letting an: llnl' we have u, r. Irm J
n1 * An
INFINITESERIES
(n+3)ln! : lim i+ n!(n+ 1)(n + 2) (z + 3) : rim  @*3)2 . ili:@+L)(n+2)
: , ',1. T n" 2F* 6+ng* 9n + 2 :rrl
r+9+Z n n'
n  +L + t + 4
:l It follows from part (i) of the Limit comparison Test that the given series is convergent.
o rLLUsrRATron 1.: Consider the geometric series
1+tr+Ln** ft*$*. . .+f,*. . .
(5)
which convergesto 2 as shown in Illustration 2 of Sec.16.3.Regrouping the terms of this series, we have
+(#*+;+ (,.;)*(l*rJ.(#*#)* which is the series 3 9*9*3*. .r 2E92r'Tr4*rr.
 L . .. .
(7)
S"1": fi) is the geometricserieswith o: g andr: *. Thus,by Theorem 1,6.3.6 it is convergent,and its sum is a : I ,:" '1,r t + We see,then, that series (7), which is obtained from the convergent series (6) by regrouping the terms, is also convergent. Its sum is the sameas that of series (5). . Illustration 1 gives a particular caseof the following theorem. +@
16.4.4 Theorem
If
is a given convergent series of positive terms, its terms can be )u" grouped in any manner, and the resulting series also will be convergent and will have the same sum as the given series.
16.4 INFINITESERIESOF POSITIVETERMS
pRooF: Let {s"} be the sequenceof partial sums for the given convergent series of positive terms. Then lim sn exists; let this limit be S. Consider +@
are obtained by grouping the terms of > ttn
a series
n:t
n:l
+oo
may be the series
in some manner. For examPle,
u r * ( u ,* u r ) + ( u n* u s* " " r +
(ur*us*ug*uro) + ' ' '
or it may be the series (ur*ur)*(ue*u)+(ur*u)+(uz*u")+'''
+@ and so forth. Let {t*} be the sequenceof partial sums for the series 2 o". Each partial sum of the sequence {f,,} is also a partial sum of a["t r"qrtettc" {sn}. Therefote, as m increaseswithout bound, so does n' Because I t,  S, *e conclude that lim t*: S' This proves the theorem' ,11 Theorem 1.6.4.4and,the next theorem state properties of the sum of a convergent series of positive terms that are similar to properties that hold for the sum of a finite number of terms' +@
15.4.5 Theorem
If
given convergent series of positive terms, the order of the
n:l
' terms can be rearranged, and the resulting series also will be convergent and will have the same sum as the given series.
pRooF: Let {sr} be the sequenceof partial sumsfor the given convergent seriesof positive tems, and let lim s,: S. Let i
o, U" a seriesforrred
n=1
+oo
by rearranging the order of the terms of.)
n:l
+@
un' For example, 2 ao may n:l
be the series
u+*us*uz*ut*ug*us*
' +@
Let {t"I be the sequenceof Partial sums for the series n:L sum of the sequence{t"}
T]l terms of the infinite series
O" less than S becauseit is the sum of.n Therefore, S is an uPPer bound of the
sequence {t"}. Furthermore, because all the terms of the series
n:l
positiV€, {t"} is a monotonic increasing sequence. Hence, by Theorem < S. Now because 16.2.7 the sequence {t"} is convergent, and lim tn: T
SERIES
j
the given ,"ri",
Ltn can be obtained from the series
)
ing the order of?u terms, we can use the same urg.r;:"lt that S and S that S  T. This proves the theorem.
anby rearrangand conclude I
A series which is often used in the comparison test is the one known as the p series,or the hyperharmonicseries.Itis
$+$+$+
where p is a constant
(8)
In the following illustration we prove that the p series diverges if p 1.. . rLLUsrRATrow2: rf p : 1., the p series is the harmonic series, which diverges. If p < 1.,then no  n, and so 11 > n" i
tor every positive integer z
Hence, by Theorem 1,6.4.2(ii)the p series is divergent if p < 1.. lI p > 1.,we group the terms as follows:
*#* . ++,. (e (h.+*h*+).(# +.(+*#)* Consider the series
1*Z+4+8+
(10)
Lp'2p,4o'gor
This is a geometric series whose ratio is 2l2o  tl2?1, which is a positive number less than 1,.Hence, series (10) is convergent. Rewriting the terms of series (10),we get
1 , lr , 1\ , /1 *L+l*1\*/1*a*...
1'\F2,)
\FF@F/*\a"
8p
_, s 1o /\ ,
(11)
comparing series (9) and series (1r.), we see that the group of terms in each set of parentheses_ after the first group is less in Jum ior (9) than it is for (11). Therefore, by the comparison test, series (9) is convergent. Because(9) is merely a regrouping of the terms of the p rlri", whenp> 1, we conclude from Theorem '1,6.4.4that the p series is convergent ifp>l,. . .1.; Note that the series in Example 3 is the p series where p: t < therefore, it is divergent.
TERMS 1 6 . 4 I N F I N I T ES E R I E SO F P O S I T I V E
7: Determine whether EXAMPLE the infinite series {o1 sr
?:, (n' + 2)rrs is convergent or divergent.
solurroN: Becausefor large values of n the number nz + 2 is closeto the numb et n2, so is the number U (n' + 2) 1/3closeto the numb er tlnzl3.The +@ because it is the p series with p  3 < L. series Using ;;" hmit comparison test with un: Ll(n' + 2)1/3 and an: Llnzrs we have
T Therefore, the given seriesis divergent.
16.+ Exercises In ExercisesL through 18, determine if the given series is convergent or divergent. +oo 1 1 . Lst n )^n n:l
!S
n?:rt/2n + L
'"
+@
1
rs
nl +@
nz
,?:r4nt + L
,?:r!nz + 4n
+oo
+@1
nl
7 " .tfr(n+Z)l
6. +@
f, ln3 + 1.
n
n:l
j3 $1ryl ln
1 r^'1 $ 4 (zn)I f, r\f
L
il4t . \7' 3_r+ \fr
13# r ( n + 2 ) ( n + 4 )
L7sg_ 2n'+2
19. Prove Theorem "l'5.4.3(ii).
20. Prove Theorem t6.4.3(iii).
n:l
1 nn
lsinn I nz
n:l
9. n:l +@
+@
1_ 10. ).Lt sin 
n?:,
+@
3.
12. n:l +co
L5. n:2 .
t)
n 5n2*3 2n nl.
1 n\/nz  1.
+co
n:l
3n 
cos t4
694
INFINITESERIES +@
+6
21. If > an and )
b, are two convergent series of positive terms, use the Limit Comparison Test to prove that the series
n=l *.n=t
is alsoconvergent' Zo"o" 22. Suppose/ is a function such that f(n) > 0 for n any positive integer.Furthermore,supposethatif p is any positive numbernlimnpf(n)existsandispositive.Provethattheseries) if p>landdivergentif.o
"1.;hence, the integral tes st can can be applied. f *+mm
is convergent or divergent.
dx
I u l
:
lim D *m
:
lim D *m
:*rc
We conclude that the given series
16.6INFINITE SEBIES OF POSITIVEAND NEGATIVETERMS
L6.5 Exercises In Exercises1 through 12, determine if the given series is convergent or divergent.
2.S I
1 '' . S b a n
3.
f,:rnlnn
3,
+6
+@
6.
5. \nze"
4. \ne"'
n=l
n=l
+@ ^il'r
+@
8.\F+1
7.)cschn
n
e.
n=L
n:l
*. /n+j\ 11. ) ln \T) n=l
+€ 10. ) sech2n n:L
13. Prove that the ,"ri"
t
1
*
>ffiW
L4. Prove that the series )
12.
is convergent if andonlyif p > 1' is convergent if and only if p > 1.
t6ilh.,,
15. If s;,is the kth partial sum of the harmonic series, prove that ln(k+1) <s* 0 for all positive integers n, then the series +@
n:l
and the series +@
) .
t r \
ur" 5"a
(l)nan:
 ar*
fl2
as* fla
' ' ' + (l)"an*
ahernatingseies.
The following theorem gives a test for the convergence of an alternating series.
INFINITESERIES
16.5.2Theorem If thenumb€tst\tu2tus, . . . ,un, . ..arealternatelypositiveandnegaTest 'ive, Alternatingseries lun*rl < lrrl for all positive integers n, and lim un: 0, then the altemating r"ri", j
uo is convergent.
PRooF: *" that the oddnumbered terms of the given series are positive "rr;;" and the evennumbered terms are negative. This assumption is not a loss of generality because if this is not the case, then we consider the series whose first term is z, because discarding a finite number of terms does not affect the convergenceof the series.So urn, ) 0 antdu2n( 0 for every positive integer r. Consider the partial sum szn: (u, * u) + (u, * u) +
+ (urn, * urr)
(1)
The first term of each quantity in parentheses in (1) is positive and the second term is negative. Because by hypothesis lun*rl < lu"l, w€ conclude that each quantity in parentheses is positive. Therefore,
0 We also can write s2nds szn: ut *
(ur*ur) + (ua*u) +
+ (urn, * urrr) * urn
(3)
Becauselrn*rl so also is u2n.Therefore, Szn 1
l,lt
for every positive integer n
(4)
From (2) and (a) we have 0 < s2o< u,
for every positive integer n
!o the sequence {srn} is bounded. Furthermore, from (2) the sequence {sz,} is monotonic. Therefore, by Theorem 1,6.2.6the sequence{s2n}is convergent. Let szo: S, and from Theorem '1,6.2.7 we know that S = ur. ,lim Becauseszn+r: s2n]u2n411 we have olim
s2a1:"li1
srn*
IT_urn*,
but,by hypothesis, lim u2o*r:0, and ro
szn+r:,lt
rrr.Therefore, "li1 the sequence of partial sums of the evennumbered terms and the sequence of partial sums of the oddnumbered terms have the same limit s. We now show that lim sn: S. Because lim szn: S, then for any
€ > 0 t h e r e e x i s tan s integer Nl > 0 such that whenever 2n > N1 ltrn sl < € And becaur" szn+r: S, there exists an integer N2 > 0 such that Jlt whenever 2n * 1, > N, ltrn*r sl < e
16.6 INFINITESERIESOF POSITIVEAND NEGATIVETERMS
699
If N is the larger of the two integers Nt and N2, it follows that if n is any integer, either odd or even, then lsrSl <e
whenevern>N +@
Therefore, lim sn: S, and so the series )
EXAMPLEL: Prove that the alternating series + o (_ 1 ) n + 1 srr Ltn n:l
r
The given series is
solurroN:
1+i !+ BecauseU(n + 1.) < Un for all positive integers n, at:rdlim (Un): 0, it follows fromTheorem1,6.6.2that the given altemating seriesis convergent.
is convergent.
EXAMPLE
z, is convergent.
n:l
n+@
Determine if the
series
)"ffi :l(1 is convergent or divergent.
sor,urroN: The given series is an alternating series, with
un: (1)"
n+3 a n d u n + r : (  1 ) n + l( n * I ) ( n + 2 )
ffi
L*2
n+) r. rrm
#:limryo L) n**
n*@n\nf
t*;
Before we can apply the altematingseries test we must also show that lun*rl < lunl or, equivalently, lun*tlllunl < 1'.
l r n * r_l
l'"1
n+3 (n*1)(n+2)'
. n ( nn + * LL) )  n ( n+ 3 ) n+2
(n*2)'
Then it follows from Theorem1,6.6.2that the given series is convergent.
15.G.3Definition
If an infinite series j
,, is convergent and its sum is S, then the re
mainderobtainedUy l]tpro*imating the sum of the seriesby the kth partial sum s7,is denoted bY Ra and R*:Ss* 15.5.4Theorem Suppose 2u^ i" an altemating series, lrn*rl < lunl, and ]l}rn:O. the sum of the approximating by obtained Then, U r{ttr the remainder < first k terms, seriesby the sum of the lRol luo*tl' pRooF: Assume that the oddnumbered terms of the given series are positive and the evennumbered terms are negative. Then, from (2) in
700
I N F I N I T ES E R I E S
'J.6.6.2, we see that the sequence {s2,} is increasing. the proof of Theorem given of the series, we have if S is the sum So
(s)
Szrc1 Szrc+z To show that the sequence {srnt} is decreasingr w€ write sznr:ur*
(u2*ur) + (un*ur) + ' ' ' + (urnr*urnr)
(6)
The first term of each quantity in parentheses in (5) is negative and the second term is positive, and becaus" lun*rl < lu,l, it follows that each quantity in parentheses is negative. Therefore, because u, ) 0, we conclude that s1 ) ss ) s5 ) {srnt} is decreasing.Thus, (7)
S From (7) we have S  srrc S  sn .Therefore,
and from (5) we have 0
0
(8)
From (5) we have S 1szrc; hence, szkt S cause from (7) it follows that 0 1 sn r  S, we have

Sac
Llzk. Be
(e)
0
Becausefrom Definition 16.6.3,Rrc: S  to, (8) can be written as 0 we lRr,,l< luro*rl and (9) can be written as 0 have lRol I
EXAMPLE 3:
A series for com
puting ln ( I  x) rf x is in the open interval (1 ,I) is
l n (I  x ) : S  4 f,:r n Find an estimate of the error when the first three terms of this series are used to approximate the value of ln L.L.
SoLUTION: Using the given series with x
0.1ry*ql)lln1.1
0.L, we get
(oil)*+
This series satisfies the conditions of Theorem'15.6.4; so if R, is the difference between the actual value of ln 1.1 and the sum of the first three terms, then lR'l < lunl:O.OOOOZS Thus, we know that the sum of the first three terms will yield a value of ln L.l. accurateat least to four decimal places. Using the first three terms, we get ln l.L : 0.0953
Associated with each infinite series is the series whose terms are the absolute values of the terms of that series.
16.6 INFINITESERIESOF POSITIVEAND NEGATIVETERMS
16.5.5 Definition
Tlie infinite ,"rie, !
z, is said to be absolutelyconoergent if. the series
n:l
t@
)
701
lut is convergent.
.: ,r"arSTRATroN1: Consider the series +@
(10)
3n'
reries will be absolutely convergent if the series
T#
$3
2'2'Z+Z+...+Z*... T3.
Ar:5*ytFFis convergent. Becausethis is the geometric serieswith r: $ < 1, it is convergent. Therefore, series (10) is absolutely convergent.
o
. rLLUsrRArroN2: A convergent serieswhich is not absolutely convergent is the series €
(l;',*'
A" In Example 1 we proved that this series is convergent. The series is not absolutely convergent because the series of absolute values is the har. monic series which is divergent. The series of Illustration 2 is an example of a "conditionally convergent" series. 16.6.6 Definition
A series which is convergent, but not absolutely convergent, is said to ergent. be conditionallyconz: It is possible, then, for a series to be convergent but not absolutely convergent. If a series is absolutely convergent, it must be convergent, however, and this is given by the next theorem.
16.6.7 Theorem
If the infinite series j
u, is absolutely convergent, it is convergent and
'f v*1"=' 1,5,',1 pRooF: Considerthethreeinfinite s"ri", j
,^,f,lu^l,"na
j
@,* lunl),
of partialsumsbe iti, tT"i, unaU,Iill.spectively. and let their sequences For everypositiveintegern, un* lurl is either0 or 2lunl;so we havethe inequality 0 =un+lr"l
1 or lim lun*rlunl:*@, then in either casethere is an integer N ) 0 such that lun*rluol > t for all n > N. Lettingz take on the successivevaluesN,N+t,N+ 2, . . .,andsoon, we obtain fl*@
ll*6
lz"*tl> larl lr**rl> lz"*rl> lurl lr"*rl > lu**rl>lu*l So we may conclude that lu"l
for all n > N. Hence, lim un * 0, ?Lt@
and so the given series is divergent. PRooF oF (iii):
If we apply the ratio test to the p series, we have
:rT.l(#.J'l :' Becausethe p series diverges if p  1 and converges if p > 1.,we have shown that it is possible to have both convergent and divergent series for which we have lim lun*rlunl : 1. This proves part (iii). I
16.6 INFINITESERIESOF POSITIVEAND NEGATIVETERMS
ExAMPLE 5: series
Determine if the
iln:
+ € (_ I ) n + t n sr\ ./J n:l
(  1 ) " * ' n l 2 ' a n d u n + r : (  l ) n + z ( n * 1 )l z * t
nIL
2n
)n
2n
is convergent or divergent.
Therefore, by the ratio test, the given series is absolutely convergent '1.6.6.7, it is convergent. and hence, by Theorem
EXAMPLE
In Example 2, the
series
;(1)"ffi
solurroN: To test for absolute convergencewe apply the ratio test. In the solution of Example 2 we showed that lu"*rlllu"l: (nz * 3n)I @2t 4n * 4) . Hence,
1 + fl3 t., , z :l na_unIn_+_t+j+ft 1
was shown to be convergent. Is this series absolutely convergent or conditionutly convergent?
lun+rl: fim
So the ratio test fails. Because n*2 , , lu"l: n@:41.
n*2
1.1 n n
we can apply the comparison test. And because the series i
tt" is the
harmonic series, which diverges, we conclude that ,h" rJ;,
5 l,'l t,
+d
divergent and hence )
z" is not absolutd
convergent. Therefore, the
n:l
series is conditionally convergent.
It should be noted that the ratio test does not include all possibilities lun*rlu,l becauseit is possiblethat the limit does not exist and is ,li1 not *oo. The discussion of such casesis beyond the scope of this book. To conclude the discussion of infinite series of constant tetms, we suggest a possible procedure to follow for determining the convergence un * 0, we may conor divergence of a given series. First of all, if "1111 clude that the series is divergent. If lim un:0 and the series is an alterfot
INFINITESERIES
706
nating series, then try the alternatingseries test. If this test is not applicable, try the ratio test. If in applying the ratio test, L:1, then another test must be used. The integral test may work when the ratio test does not; this was shown for the p series.Also, the comparison test can be tried.
Exercises 16,6 In ExercisesL through 8, determine if the given alternating serles is convergent or divergent.
I irD"#
+@
4.
nz
3.
n
n:2
+@
+oo
7T
2.
n:l +@
ln n sn+r
on
5. .LJ ) (t)" n n:l
n:l +a
7.t (1)"# n:l
In Exercises9 throu gh L2, find the error if the sum of the first four terms is used as an approximation to the sum of the given infinite series.
s.i (1 )*'I
10 (1)"*. >
n:l
+@1
1 1 . 5 (  1) n + , # D ,
12. .LJ ).
(11n+r \ /
vllr
n:l
In Exercises L3 through 15, find the sum of the given
infinit" ,Jl"r, accurut! tothreedecimalplaces. L (2n)I
13. (1)nr1 > # "::
1,6 5(1)"ffi1,
ls.,e (1)n+1 #,
In Exercises17 through 28, determine if the given series is absolutely convergent, conditionally convergent, or divergent. Prove vour answer. +@
18. n:l
1. ( 2 n r ) I
+@
L9. n:l
n2 nl,
+co
22. n:l +@
23.
+a
3n
n!
25.
26. 2 S ' n 2e i l n c o s = n
28.
n:2
+@
+@
n:l
n:l
29. Prove by mathematicalinduction that Un! = 1l2nt +@
30. Prove that if i n:l
un is absolutely convergent and,un # 0 for all n, then
Lllu"l is divergent. n:\
16.7 POWERSERIES
3 1 . Prove that if > un is absolutely convergent,then )
unzis convergent.
32. Show by means of an example that the converseof Exercise3L is not true.
16.7 POWER SERIES We now study an important type of series of variable terms called "power series." L5.7.1Definition
A power seies in (x  a) is a seriesof the form co* cr(x a) * cz(x a)'+''
I cn(x a)"+'''
(1)
We use the notation 5 ,"(t  a)n to representseries (1). (Note that x : a, for conveniencein writing the genwe take (x  a)o: 1, "lr"rr"#hen eral term.) If r is a particular number, the power series (1) becomes an infinite series of constant terms, as was discussed in previous sections' A special case of (1) is obtained when a: 0 and the series becomes a power series in x, which is +@
(2)
n:o
In addition to power series in (r = a) and t, there are Power series the form of +@
c"l6@)1"+''' ) t " t 0 ( t ) l ' : c o *c r 6 @+) c r l f @ ) 1 z + ' ' ' + n:0 where d is a function of r. such a series is called a power series in {(r). In this book, we are concerned exclusively with Power series of the forms (1) and (2), and when we use the term "power series," we mean either of these forms. In discussing the theory of power series, we confine ourselves to series (2). The more general power series (1) can be obtained from (2) by the translation x: i  a; therefore, our results can be applied to series (1) as well. In dealing with an infinite series of constant terms, we were concemed with the question of convergence or divergence of the series. In considering a power series, we ask, For what values of x, if any, does the power series converge? For each value of r for which the power series conve"ges, the series represents the number which is the sum of the series. Therefore, we can think of a Power series as defining a function. The function f, with function values
f(x):
(3) n:o
INFINITESERIES
has as its domain all values of.x for which the power series in (3) converges. It is apparent that every power series (2) is convergent for r: 0. There are some series (seeExample 3) which are convergent for no other value of r, and there are also series which converge for every value of r (seeExample 2). The following three examples illustrate how the ratio test can be used to determine the values of.x f.orwhich a power series is convergent. Note that when n! is used in representing the zth term of a power series (as in Example 2), we take 0!: L so that the expression for the nth term will hold when z:0.
EXAMPLE1: Find the values of x for which the power series +@
' S el) ' n+l !,, zr n3n n:l )flyn
is convergent.
solurroN: un :
For the given
, ( 1 )n+r
rio, lful: ttn
Znxn
W
rnd. (  1t )\n+2 n + 2 2n+rx6n+t and u Ltn+.t n +:r : (
TffiTI
 2n+rxn+t. n3 l  ,, 2 ,., n
hm np*l I ,_*l lffiF'Nl:,li1ilxlfr_l:i
2
lxl
Therefore, the power series is absolutely convergent when $lrl < 1 or, eguivalently, when lrl < *, The series is divergent when flrl )'1 or, equivalently,when ]rl = *. When Slrl:1 (i.e., when x:!8), the ratio test fails. When x: $, the given power series becomes
+I+* I+. . . + (1)n+, 1+ which is convergent, as was shown in Example 1 of Sec. 16.6. When x:t, we have
which by Theorem 15.3.8is divergent. we conclude, then, that the given power series is convergent when * < x < 9. The series is absolutely convergent when * < .r < * and is conditionally convergent when r: g. If r < * or x > 8, the series is divergent.
EXAMPLE 2: Find the values of x for which the power series +m
yftt
\* ,?on! rs convergent.
solurroN: For the given series, ttn: tnln! and un*r: xn+tl(n * 1) !. So by applpng the ratio test, we have
 ,"*' .ntl,, 'Fl: l'l rio,lgul: un 1im
n+*  I z;; l1;Tln
1
Jii;ft:
o< r
we conclude that the given power series is absolutely convergent for all values of r.
16.7 POWERSERIES
ExAMPLE3: Find the values of x for which the power series +@
S nlxn
LJ n:0
(n * l)lx"+r. Apply
soLUrIoN: For the given seriesun: nlN,nandur*r: ing the ratio test, we have t" I fim l'n+11: lim l ( n * ' l ' ) l x " + 1 1 n+lunl n+*lMl
:
is convergent.
:.1
709
+ 1)rl ,111l(n f 0 if r:0
t**
if.x*0
It follows that the seriesis divergentfor all valuesof r except0. 16.7.2 Theorem
If the power series f
,r*" is convergent for r : xr (xr + 0), then it is ab
n=O
solutely cohvergent for all values of r for which lrl < pRooF: ff. f
cnxr" is convergent, then
l*r,rr":0.
l r 'l . Therefore, if we
take e : L in Definition 4.1.1,there exists an integer N > 0 such that Itnxr"l N
Now if x is any number such that lrl < lrtl, we have 
nl
lrln
ltln
. lc,x"l: lc"x,^ ful: lt"tr'l lil El
n> N whenever
(4) (s)
is convergent becauseit is a geometric series with r : lxlxl < 1 (because lrl < lrtl).Comparing the series )
lc^x",where ltl < lr'1, with series
n:N
+@
lcnx"l is convergent ") for lrl < ltrl. So the given power series is absolutely convergent for all r values of.xfor which ltl < lrtl.
(5), we see from (4) and the comparison test *",
o rLLUsrRArrou1: An illustration of Theorem'l'6.7.2is given in Example 1. The power seriesis convergentfor r:8 and is absolutely convergentfor all values of.x for which lrl < g. The following theorem is a corollary of Theorem 16.7.2. L5.7.3 Theorem
'
If the power series )
cnxnis divergent for r:
values of x for*r,t"fr=itt> lxrl'
xr, it is divergent for all
710
I N F I N I T ES E R I E S
pRooF: Suppose that the given power series is convergent for some number r for which lrl > lrrl. Then by Theorem 16.7.2the series must converge when r: rz. However, this contradicts the hypothesis. Therefore, the given power series is divergent for all values of x for which
.
lrl > ltrl.
I
o rLLUsrRArror 2: To illustrate Theorem 16.7.3 we consider again the power seriesof Example 1. It is divergent for x: B and is also divergent . for all values of x for which tl > l91. '1.6.7.2 From Theorems and t6.7.3, we can prove the following important theorem.
,
15.7.4 Theorem
fet i
coxnbe a given power series. Then exactly one of the following
"o.iioror,,
holds:
(i) the seriesconvergesonly when x: 0; (ii) the series is absolutely convergent for all values of x; (iii) there exists a number R > 0 such that the series is absolutely convergent for all values of x for which lrl for all values of.x for which lrl
pRooF: If we replacex by zero in the given power series,we havecs* 0 + 0+ ' , which is obviously convergent.Therefore,every power +@
series of the form )
c;cn is convergent when x:
0. If this is the only
value of r for whicil1l" ,"ri", converges, then condition (i) holds. Suppose that the given series is convergent for r: rr where x1 * 0. Then it follows from Theorem1,6.7.2 that the series is absolutely convergent for all values of r for which lrl < lrrl. Now if in addition there is no value of x f.ot which the given series is divergent, we can conclude that the series is absolutely convergent for all values of r. This is condition (ii). If the given series is convergent fot x: xr, where xr* 0, and is divergent for x:.rr, where lx"l> l*rl, it follows from Theorem 15.7.3that the series is divergent for allvalues of xfor which lrl > lrrl.Hence, lrrl is an upper bound of the set of values of lrl for which the series is absolutely convergent. Therefore, by the axiom of completeness (1.6.2.5),this set of numbers has a least upper bound, which is the number R of condition (iii). This proves that exactly one of the three conditions holds. I
serresconvergent
for lxl < R
Theorem 1,6.7.4(iii) can be illustrated on the number line. See Fig. 1,6.7.1. seriesdivergent for lrl )
Figure 16.7.1
R
+@
If instead of the power series )
+@
cnxnwe have the series )
cn(x  n)n,
16.7 POWERSERIES
711
in conditions (i) and (iii) of Theorem'1.5.7.4,r is replacedby x  a. The conditions become aR
n
a+R
seriesdivergentfor lx  ol > R Figure16.7.2
(i) the series converges only when x: ai (iii) there exists a number R > 0 such that the series is absolutely convergent for all values of. x for which l*  ol < R and is divergent for all values of.x for which lr ol > R. (SeeFig. 16.7.2 for an illustration of this on the number line.) The set of all values of.x for which a given Power series is convergent is called the interztalof conoergenceof the Power series. The number R of 'J,6.7.4 is called the radius of conaergenceof the condition (iii) of Theorem power series. If condition (i) holds, we take R: 0; if condition (ii) holds, we write R: *o. 3: For the power seriesof Example 1, R: $ and the interval o TLLUSTRATTOTI is of convergence (8, g7.In Example2, R: ao, and we write the interval o of convergenceas (o, *oo). +@ If R is the radius of convergenceof the power series ) cnro, the irrn=0 terval of convergenceis one of the following intervals: (Oi}), [R, R], (R, R], or [R, R). For the more general Ponrer series
Ar"@a)n, (aR, a*R), following: of the one the interval of convergence is R ) . R , a * la R, a * R], (a R, a* R], or laA given power series defines a function having the intenral of con.r"rg"rrJ" as iis domain. The most useful method at our disposal for deteriining the interval of convergence of a Power series is the ratio test' However, the ratio test will not reveal anything about the convergence
which it converges (see Exercise 22). There are casesfor which the convergence or divirgence of a power series at the endpoints cannot be detirmined by the methods of elementary calculus. ExAMPLE 4: Determine the interval of convergence of the Power series +6 n:l
n(x 2)"
solurloN: The given seriesis ( x  2 ) + 2 ( x  2 ) , + . . . + n ( x  2 ) " + ( n * 1 ) ( x 2 ) n +*r ' ' ' Applying the ratio test, we have
+ 1 ) ( r = ? ) " . ':l limlwl hml(n n(xz)" W  z l t i m# :nl x  2 1 
;;;lu,
l
"+;l
";;
712
I N F I N I T ES E R I E S
The given series then will be absolutely convergent if lx  2l equivalently,1 1x (3. equivalently,L < x2 When x : l, the series is
jsu"
+oo
# 0. When x:3, the seriesis 2 n, which is alsodivergentben=l
cause lim un * 0. Therefore, the interval of convergence is (1, 3). So the ll*'@
given power series defines a function having the interval (L, 3) as its domain.
EXAMPLE5: Determine the interval of convergence of the power series +6 Sl
yn J+
t,ffi
n:l
SoLUTIoN: The given series is x,x2,x3l
Applying

'
2t, 1zT 2+22
2+y
ry,?t
'
l * ,  ,
2*nz'
Xn+L
2+ (n*'1,)2
I
the ratio test, we have
2*nz ,., r. Tl: ltl ri l1*l : fim l, T'*l ^. .2*n2l n+tun  ,*lz+ @+ty ,li* 2ffi:lxl So the given series will be absolutely convergent if lrl < L or, equivalently, l < x < 1. When x:1, we have the series L,L,L ' 2+12' 2+22' 2+92 __I
BecauseU (2 + n2)< 7ln2for all positive integersn, and,because ,fu, f is a convergent p series, it follows from the comparison test that the given power series is convergent when r : L. When x: 1, we have the series +@
>
(t)"1(2*
n), which is convergent becausewe have just seen that
liti, convergent. Hence, the interval of convergence of the given"Urot.rtely power seriesis [1, 1].
Exercises L6,7 In ExercisesL through20, find the interval of convergence of the given power series. +@
1.P(1)n+'fu1, *2n_l
4.X(1)"# n:l
r.)(lvffi*_, +6
1 6 . 8D I F F E R E N T I A T I O ON F P O W E RS E R I E S 7 1 3 +@
9.
t*11n (1)nrtT
+oo
10.
n:l
n:l $co
*co
11.
12. LlI
( sinh 2n) x" n:0
n:l
+@
( 1\^ ' z+1
L3.
\
n:2 +@
15. n:l +co
17. n:l
+@
xn
1,4.
n(ln n)z
n:l
xn
ln(n + 1) (x * 5)zt n2
n:l +@
nlxn nn
1,6. n:l
ln n(x  5)"
xn nn
+@
18.L/s n"(x  3)
n+l
lI:1 +@
+@
19.
(x* 2)" (n * 1)2
2'4'6'.
. .'2n
20.
xzn+l
n:1
(  1 ) n + r 1' 3 ' 5 ' 2'4'5'.
. . . ' (znl) . '2n
xn
21. Prove that if a power series convergesabsolutely at one endpoint of its interval of convergence,then the power series is absolutely convergent at each endpoint. 22. Prcve that if a power series converges at one endpoint of its interval of convergenceand diverges at the other endpoint, then the power series is conditionally convergent at the endpoint at which it converges. 23. prove that if the radius of convergence of the powercerie, j
is {r. 2ru,*2n 24.Provethat if lim +M  L (L + 0), thenthe radiusof
unxnisr, then the radius of convergenceof the series
+!p
convergence of the power series
unx'is LlL. n:l
A power series 2 cnx" defines a function whose domain is the interval 16.8 DIFFERENTIATION N:O OF POWER SERIES . of convergence of the series. o rLLUsrRArrow1: Consider the geometric series with a:1 +o
which is )
and r:x,
r". By Theorem 16.3.6 this series converges to the sum
 )c)if ltl 1.1(1,
)cndefines the func
tion f forwhich f (x) 1l(1  x)and ltl
we can write
1+ x+x2+),3+ ' ' '+ xn+ ' ' ':
1 1 x
if lrl
The series in (1) can be used to form other power series whose sums can be determined. o rLLUSrRArroN2: If in (1) we rePlace x by *,
1x+x2x3+
+ (1)"x"* ' ' '
we have 1
L*x
if lrl
714
I N F I N I T ES E R I E S
Letting )c: x2 in (1), w€ get 1 + )cz+ xa * xG+ . . . + x2' *
. . .
if lrl
(3)
lf x is replaced by  x2 in (1) we obtain 1 x2+x4x6+
' ' ' + (1)'x'"+
if lrl
In this section and the next we learn that other interesting series can be obtained from those like the above by differentiation and integration. We prove that if R # 0 is the radius of convergence of a power series which defines a function /, then / is differentiable on the open interval (R, R) and the derivative of / can be obtained by differentiating the power series term by term. Furthermore, we show that / is integrable on every closed subinterval of (R, R), and the integral of f is evaluated by integrating the power series term by term. We first need some preliminary theorems. +@
L5.8.1 Theorem
If ) cnx, i s a power series having a radius of convergence of R ) 0, then Eo +co the series LI ncnxnl also has R as its radius of convergence. n:l
This theorem states that the series, obtained by differentiating term by term each term of a given power series, will have the same radius of convergenceas the given series. pRooF: Let r be any number in the open intewal (R, R). Then lrl < R. Choose a number r, so that lrl < lrrl < R. Becauselxrl < R, j
corrois
convergent.Hence,
rn:0. So if we take e : L in Oefiniiln A.t.t, ]!I_rn there exists a number N > 0 such that lcn*rnl< 1
whenevern >N
Let M be the largestof the numbers Icfirl , lcrxrrl, lrr*rtl, Then lrnrrnl = M
for all positive int egerc n
, lcNxrNl,1. (5)
Now
: lnc,x'' : lrr,.# . xrnl " W l;l*' From (5) and the above equatiorr, we get
="#l;l' lncnx,,
(5)
16.8 DIFFERENTIATION OF POWERSERIES 715
Applytng the ratio test to the series M t{ lr lo1 ri )l n l:l lxrl #r lrrl
(7)
we have
,,_ l z,*rl:_ llm ,:_ l( n+ 1) lxl'. lr ,l.'  tl llm ll
,il'i!* u"  ;].i; I :
lr'1"
nlxl"r1
n * L
l r l llrl ,. lt
lrrl n_+
n
: l _l =r ll ( 1 lrrl Therefore, series (7) is absolutely convergenf so from (6) and the comparison test, the series )
ncnxnris also absolutely convergent. Because
r is any number i" tnlntl, it follows that if the radius of convergenceof +@
> R. \ ncnx"ris R', then R' n=t To complete the proof we must show that R' cannot be greater than R. Assume lhat R' > R and let 12 be a number such that R < lr2l < R'. Becauselrrl > R, it follows that +@
(8) n'o
Becausel*rl ( R', it follows that i
ltcnx2nLis absolutely convergent.
Furtherrnore, +@
*rc
l*rl and so from Theorem 16.3.8 we may conclude that +@
(e)
positiveinteger "", (10) Itn*rnl= nlcnxr"l lncnxr"l From statement (9), inequality (L0),and the comparison test it follows that
If n,
+@
)
+@
lc,xr"l is convergent.Therefore, the series 2 ,n r" is convergent, which
that R' > R is false. Iot,o"Ur",, statement (8). Hence, ttre assffition Therefore, R' cannot be greater than R; and because we showed that I R' ) R, it follows that R' : R, which proves the theorem.
716
I N F I N I T ES E R I E S
. rLLUsrRArroN 3: We verify Theorem 16.8.1.for the power series S
tn*t
_,f
?^@*:'*T*e
+d+
*
.
xn+r ,n*, r_. +@T+G+'/r+'
_L
. .
To find the radius of convergence,we apply the ratio test.
, ln'* 2n* 1,1  r"r l : Hml@+ t!?*+:.1 lxl ::T17;+ 4n+ 4l : lrl ,*l l@lWl
rio,l
n+* iln 
Hence, the power series is convergent when lxl < 1., and so its radius of convergenceR: L. The power series obtained from the given series by differentiating term by term is $(n+1)r':t{_j'
.
x.f
*4Try:,17"+t:r+r+T*
.fr...r
++''
n*t + nx+nt +, x; + 2+'''
Applyirg the ratio test for this power series,we have
*,L) !n+tt
:  ln+ r l hm l( F l'l nJ; ;i;l @+z)*l: I* lffil: l'l l 
l i m lwl: un
This power series is convergent when lrl vergenceR' 1. BecauseR  R', Theorem 16.g.lis verified. 1.6.8.2Theorem
If the radius of convergence of the Power series
of cono
+@ n:O
is also the radius of convergence of the series
X n(n 1)cnx,r.
n:2
+co
pRooF: If we apply Theorem 16.8.1to the series ncnxnl, w€ have the desired result. n:l I We are now in a Position to prove the theorem regarding termbyterm differentiation of a power series.
1'6.8.3Theorem
Let ) coxnbe a power series whose rad.iusof convergenceis R > 0. Then tr 1irirr" function defined by +€
f(x)
( 1 1)
' f (x) exists";revery r in the open interval (R, R), and it is given by +@
f'(x)
SERIES 717 OFPOWER 16,8DIFFERENTIATION pRooF: Let r and a be two distinct numbers in the open interval (R, R). Taylor's formula (formula (9) in Sec.15.5),with n: L, is (" (l\
('(o\
f(x): f(a\+Lff (x a)+'f,a k a)' Using this formula with f (x) : xn, we have for every positive integer z (12) { : nn* nonr(x a) + in(n  l) ((")"2(x a)z where f, is between a and r for every positive integer z. From (11)we have
f(x)  f(a) :2
+@
+00
2 cna" n=o "n*" n0
: co*5 ,* ce5 ,,o" n:l nl +@
:)cn(x"a") n=l
(becausex*
Dividing by xa
+@
a) and using (12)' we have from the
n  z ( x a ) ' f c n l n a n  t ( r a ) + + n ( n 1 ) ( f , )
n:l
+@
+@
ncnanr++(xa) n:l
n(n  1)c"(tn)"'
(13)
n:2
Because a is in (R, R), it follows from Theorem 16.8.1that )
ncnanr
is absolutely convergent. K> 0 Becauseboth a and x are in (R, R), there is some number "l'6.8'2 that Theorem suchthat lal < K < R and lrl < K < R. It follows from +@
is absolutely convergent. Then because ln(n
L)c,(€*)"'l < ln(n L)cnK'l
for each €n, we can conclude from the comparison test that +@
,,
";;utely
conversent.
(14)
SERIES
It follows from (13) that lf(x)f(a\ t# xa I
ro
.l I
n:r
l. I
+@ . n:2'\'?
However, from Theorem 1,6.6.T we know that if vergent, then Itg
.
L,vna
I I
*o
un is absolutely conn:l
+

l',1 l},""l=p,
Applying this to the right side of (15),w€ obtain
f@) ll M ' * J c  a
I
{ { n c r a ".l r=l + l *   ' t + @ > 'trv ol
#,
I
(16)
"1:,
From (14) and (16) we get
(x) f(n)lf t A , n(n 1)lc,lKn, +lx  watj ; ncnan,1= I xa  A "n
(r7)
where 0 < K < R. Becausethe series on the right side of (L7)is absorutely convergent, the limit of the right side, as r approaches a, is zero. Therefore, from (17) and Theorem 4.3.3 it follows thit
y ^ f @ )  f ( a )  *' \' n c n a n  r c'a
xA
or, equivalently, +@
f'(a):
\
ncna"r
and becausea may be any number in the open interval (R, R), the theorem is proved. I
ExAMrLE 1: Let b" the function f defined by the power series of Illustration 3. (a) Find the domain of f; (b) write the power series which defines the functi on , and f find the domain of f ,.
The domain of f is the intervar of convergenceof the power series. In Illustration 3 we showed that the radius ofof *re pou/er series is 1; that is, the series converges when "or,rr"rg"n"e lrl < rl we now consider the power serieswhen lrl : 1. Whenr: 1, the seriesis
1+ l+!r*
719
OF POWERSERIES 16.8 DIFFERENTIATION
which is convergentbecauseit is the p serieswithp:2. +@
we have the series > (1)'*Y(n*2)2,
When x:L,
which is convergentbecauseit
is absolutety converg:.1',t.Hence, the domain of f is the interual [1, 1]. ' (b) From Theorem15.8.3it follows that f is defined by
f'(x)$+ +1 *?:^n
(1s)
and that f '(x) exists for every x in the open interval (1, L). In Illustration 3 we showed that the radius of convergenceof the Power series in (18) is L. We now consider the power series in (18) when x +1. When x: !, the series is
' . '+' n * \1=*"' 1+****1* z 3 4 which is the harmonic series and hence is divergent' When t: series is 1,
r , L 1 , i;rs
1, the
+ (1vftr+
which is a convergent alternating series. Therefore, the domain of f the interval [  1 , 1 ) .
is
Example 1 illustrates the fact that if a function / is defined_by a power series and this Power series is differentiated term by term, the resulting ',has the same radius of convergence but power series, which defines f not necessarily the same interval of convergence. ExAMPrn2: obtain a Powerseries representation of
solurroN:
if ltl
+'/,,1*x*x2*13*...+)cn+ x
L
@
From (1) we have
using Theorern 1.5.8.3,differentiating on both sides of the above, we get L
ffi1*2x*3x2* EXAMPT,n 3:
Show that
solurroN: +6
+@
vsr
q2fl
S + n!
#,
:L* for all real values of x.
I
* nxnr +
if ltl
In Example 2 of Sec. \6.7 we showed that the Power series
x"lnl is absolutely convergent for all real values of r. Thereforc, 7f.f is
lil rrrr,.tior,definedby (1e)
I N F I N I T ES E R I E S
the domain of / is the set of all real numbers; that is, the interval of convergence is (rc , +m). It follows from Theorem 16.8.3 that for all real values of x
f'(x):S+n! "1
Becausenln!:
(20)
 l)1, (20) can be written as U (n
,*''. f'(*):S #, (n L)I or, equivalently,
f'(x):fSo n+!
(21)
Comparing(19)and (27),we seethat f ' (x) : f (x) tot all real valuesof x. Therefore,the function / satisfiesthe differentialequation du
fr:y for which the general solution is y: Cec. Hence, for some constant C, f(x) : Ce'. From (19)we seethat/(0) : 1. (Rememberthat we take r0: L even when r: 0 for convenience in writing the general term.) Therefore, C : 7, and so f (x) : ,','and we have the desired result. ExAMPtn 4: Use the result of Example 3 to find a powerseries representation of st.
SOLUTION: e &:  4
If we replac e x by x in the series for er, it follows that n2
ng
1 )c+nfr+
vTI
+ (r)"\n ! +
for all real values of )c. EXAMPLE5: Use the series of Example 4 to find the value of et correct to five decimal places.
solurroN:
Taking x  1 in the series for er, w€ have
,1 3.r'f8ff    1 + 0.5 0.1.66667  0.008333 + 0.041667 + 0.001389  0.000198  0.000003 + 0.000025 + 0.0000003 We have a convergentaltemating seriesfor which lur*rl < lnrl. So if we usethe first ten termsto approximatethe sum,by Theorem16.5.4the
OF POWERSERIES 16.8 DIFFERENTIATION
errof is less than the absolute value of the eleventh term. Adding the first ten terms, we obtain 0.357880.Rounding off to five decimal places gives er  0.36788 In computation with infinite series two kinds of errors occur. One is the error given by the remainder after the first n terms. The other is the roundoff error which occurs when each term of the series is approximated by a decimal with a finite number of places.In particular, in Example 5 wewanted the result accurateto five decimal places;so we rounded off each term to six decimal places.After computing the sum, we rounded off this result to five decimal places. Of course, the error given by the remainder can be reduced by consideriirg additional terms of the series, whereas the roundoff error can be reduced by using more decimal places.
L6.8 Exercises exerciie do the following: (a) Find the radius of In Exercises1 through 8, a function / is defined by a power series.In each series which defines the function f' and the (b) write of domain the and Power power series f; convergence of the given Theorem 15.8.1);(c) find the domain of /'' find its radius of .orr',rerg"nceby using methods of Sec. 16.7 (thus verifying +@
+co
2' f(x)
L.f(x) n:l +co
s. f (x)
4. f(x)
yn
3. f (x) 
n
,: +@
2#"
XZnt
l\
n:2
+co (x _
7.f(x):ZT
+@
8. f(x)
1)"
n:2
 x)3' 9. Use the result of Example 2 to find a powerseries representation of U(1 of e€. 1.0.Use the result of Example 3 to find a powerseries representation series (2) term by term' 11. Obtain a powerseries representation of U(l l x)2 if lrl < r by differentiating series (4) term by term' 12. Obtain a powerseries representation of fl\ + *)2 it lxl < 1 by differentiating decimal places' 13. Use the result of Example 4 to find the value of u {i correct to five M. rt f (x):
j n=u
{r)'
$,
ma/'(*)
correctto four decimal places'
of (a) sinh x and (b) cosh r' 15. Use the results of Examples 3 and 4 to find a powerseries representation 15 can be obtained from the other by termbyterm 15. show that each of the power series in parts (a) and (b) of Exercise differentiation. 17. Use the result of ExampIe Zto find the sum of the series ;
n:l
#. 
 1)lx' 18. (a) Find a powerseries rePresentation for (e" (b) By differentiating
term by term the power series in part (a), show that
I N F I N I T ES E R I E S
19. (a) Find a powerseries representation for fec. (b) By differentiating term by term the power series in part (a), show t h a t S ( _ 2 ) r + 1" \ 2 : n! f=,
n. +@
20. Assume that the constant 0 has a powerseries representation
cnxn,where the radius of convergence R > 0. Prove n:o
that cn: 0 for all n.
21..Supposeafunctionfhasthepowerseriesrepresentation!
cnro,wheretheradiusof convergenceR) O.lf f,(x):f(x)
and/(O) : 1, find the power seriesby using only propertrlitf
po*"r reries and nothing about the exponential function. 22. (a) Using only properties of power series, find a powerseries representation of the function / for which /(r) > 0 and '(x) :2xf (x) for all r, and f /(0) : 1. (b) Verify your result in part (a) by solving the differential equation D,y :2xy having the boundary condition y: 1 when r: 0.
15.9 INTEGRATION The theorem regarding the termbyterm integration of a power series OF POWER SERIES is a consequenceof Theorem 76.8.9. 16.9.1 Theorem
fet i
coxnbea power series whose radius of convergenceis R > 0. Then
if 1 ir=tU""function defined by +@
f(x):2
cnx"
n=0
/ is integrable on every closed subinterval of (R, R), and we evaluate the integral of f by integrating the given power series term by term; that is, if r is in (R, R), then +@ fc dt: ' S cn ,nir l 'ro \ ' ' Jo fton+l*
Furthermore, R is the radius of convergenceof the resulting series. pRooF: Let g be the function defined by tr,rn*, .s(r):$ flon + l Becausethe terms of the powerseries representation of f(x) are the derivatives of the terms of the powerseries representation of g(r), the two series have, by Theorem 16.8.']., the same radius of convergence.By Theorem 16.8.3it follows that for every r in (R, R) f (x) '1,6.8.2, By Theorem it follows that /'(x): g,, (r) for every r in (R, R). Because/ is differentiable on (R, R) , / is continuous there; consequently, / is continuous on every closed subinterval of (R, R). From Theorem g'(x):
SERIES 723 OF POWER 16.9INTEGRATION 7.6.2ilt follows that if r is in (:R, R), then fr IJ o f ft) itt: s@) s(o): s(x)
or, equivalently, ' [' nl $ h ,n*r t t , t *dr: *Lon*L
r
Jo
Theorem 15.9.1 often is used to comPute a definite integral which cannot be evaluated directly by finding an antiderivative of the integrand. Examples 1 and 2 illustrate the technique. The definite integral j{ to dt appeaing in these two examples rePresentsthe measure of the area of a tegiot under the "notmal probability curve'" EXAMPLE1: Find a Powerseries representation of fr
I et' dt
Jo
SoLUTIoN: In Example 4of Sec.'1.6.8we showed that H
€0:
)'
(l)nxn n!
"A
for all values of )c.Replacing x by t2, we get
+# fi*''' et'1 1z
+ (1Y#*
for all values of. t
we integrate term by term and obtain Applying Theorem 1,6.9.1, fx
J,
et'dt:
f
lzn
ft
x3,
x5
'2! '5 3
LJ
x7
3!'7
nZn*L
'
+(l)"ffi+
The power series represents the integral for all values of x. ExAMPLE2: Using the result of Example L, comPute accurate to three decimal Places the value of frl2
I et' dt Jo
SOLUTION:
Replacing x by t in the power series obtained in Example L,
we have f rt2 ' ' ' IJ O t" d t = +  # + #  ; = b a +
: 0 . 5  0 . 0 4 1 . 7 + 0 . 0 0 310 . 0 0 0 2 + ' ' ' We have a convergent alternating series with lao*tl < lu"l' Thus' if we use the first three terms to approximate the sum, by Theorem 15.6.4 the error is less than the absolute value of the fourth term. From the first three terms we get [t'' ,o dt  0.461 Jo
I N F I N I T ES E R I E S
ExAMPLE 3: Obtain a powerseries representation of ln(1 * r).
solurroN: Consider the function / defined by f (t) : U (l + t). A powerseries representation of this function is given by series (2) in sec. 16.g, which is
+ (1)"t"+
1zt3+
#1t+
if lr l
Applying Theorem 16.9.1,we integrate term by term and obtain
f' .d, _i{
r
Jo'].+t f,u.,
( 1)"t" dt
and so
rn(r +r): x$*tf;*
iflrl 0 , x < 0 , a n d .r : lfx)0,then0
0.
t,r'=r, o v a\ , *n*r \ v (n* 1)!*
(18)
(n+l)I
From (16) it follows that lim )cn+rl(n* 1)l:0,
and so
rn+l
lim et zj;::0 r*e ln+ I)l Therefore, from (18) and the squeeze theorem (4.3.3) we can conclude that Rz(r) : 0. nlim If r < 0, then x 1 tn< p4,
o.6i1;1
x n +.
l(s,2)l: s i ne   +
\E
and cos
So from (1) we have
(5,2)  \E Figure 1 7 . 2 . 3
/5.2.\
\aEr@t)
\E
AND SCALARMULTIPLICATION755 ADDITION 17,2PROPERTIES OF VECTOR
17.2.3 Theorem
If the nonzero vector A: ari * ari, then the unit vector U having the same direction as A is given by (2)
PRooF:
(2),
From 11
U: ri
lAl
@j* a2j):;T
(A)
lAl
Therefore,U is a positive scalartimes the vector A, and so the direction of U is the stune as the direction of A. Furthermore,
:,t(ft)'*(ft)' tur _\ffi lAl _ lAl lAl 1 Therefore,U is the unit vector having the same direction asA, and the theorem is proved. ExAMPrn3: Given A  (3, 1) and B  , find the unit vector having the same direction as AB.
solurroN: A  B  (3, 1)  (2, 4) :
So we may write
AB5i3i Then
lABl :ffi\E By Theorem 17.2.3,the desired unit vector is given by
u#ifti Exercises17.2 In Exercises1 through 8, let A 2i + 3i and B 4i i.
L.FindA+B
2. FindAB
4 .F i n dl A l l B l 7. Find l3A  2Bl
5. Find lA + Bl 8. Find l3Al l2Bl
9. Given A:
8i * 5j and B :3i

3. Find 5A  68 6. Find lAl + lBl
i; find a unit vector having the same direction as A * B.
I
VECTORSIN THE PLANEAND PARAMETRICEQUATIONS
10. Given A :8i + 7i; B : 5i 9i; C: i i;find l2A  38  Cl. 11. Given A : 2i+ j; B : 3i  2i; C : 5i  4j; hnd scalarsh and,k such that C : hA * kB. In Exercises12 through 15, (a) write the given vector in the form r(cos 0i * sin 0i), where r is the magnitude of the vector and 0 is the radian measure of the angle giving the direction of the vector; and (b) find a unit vector having the same direction. 12. 3i3i 13. 4i+ 4\/3i 14. r6i rs. 2i 16. Prove Theorem 17.2.1(ii).
17. Prove Theorem 17.2.1(v).
18. Prove Theorem 17.2.1(vii).
19. Prove Theorem 17.2.7(iii)and (viii).
20. Two vectors are said to be independentif. and only if their position representations are not collinear. Furthermore, two vectors A and B are said to form a basisfor the vector spaceV, if and only if any vector in V, can be written as a linear combination of A and B. A theorem can be proved which statesthat two vectors form a basis for the vector spaceV, if they are independent. Show that this theorem holds for the two vectors (2, 5) and (3, 1) by doing the following: (a) Verify that the vectors are independent by showing that their position representations are not collinear; (b) verify that the vectors form a basis by showing that any vector ari l a2ican be written as c(2i + 5i) + d(31 il, where c and . d are scalars.(nrwr: Find c and il in terms of a, and ar.\ 21' Refer to the first two sentenies of Exercise 20.A theorem can be proved which statesthat two vectors form a basis for the vector spaceV, only if they are independent. Show that this theorem holds for the two vectors (9,2) and (6,4) by doing the following: (a) Verify that the vectors are dependent (not independent) by showing that their position representations are collinear; (b) verify that the vectors do not form a basis by taking a particular vector and ihowing that it cannot be written as c(3i + 2j) + d(6i + 4i), where c and d are scalars.
17'3 DOT PRODUCT
17.3.1 Definition
In Sec.17.1 we defined addition and subtraction of vectors and multiplication of a vector by a scalar.However, we did not consider the multiplication of two vectors. we now define a multiplication operation on two vectors which gives what is called the ,,dot product.,, If A : (at, azl and B : (bt , brl are two vectors in Vr, then the dot product of. A and B, denoted by A . B, is given by A . B:
( a r , a r l . ( b r , b r l : a r b r* a " b ,
The dot product of two vectors is a real number (or scalar) and not a vector. It is sometimes called tlrrescalarproduct or inner product. o rLLUsrRArroN 1: If A:
(2,9) and B : (+,4), then
A . B : (2,3) . (i, +>: (2)(+) + (_3)(4): _13
.
The following dot products are useful and are easily verified (see Exercise 5). i.i:L
(1)
i .i  L i'i:o
(2) (3)
17.3 DOT PRODUCT
757
The following theorem states that dot multiplication is commutative and distributive with respect to vector addition. 17.3.2 Theorem
If A, B, and C are any vectors in V2, then (i) A. B: B . A (ii) A' (B+C) :A.
(commutativelaw) (distributivelaw) B+A.C
The proofs of (i) and (ii) are left as exercises(see Exercises5 and 7). Note that becauseA. B is a scalar, the expression (A . B) ' C is meaningless. Hence, we do not consider associativity of dot multiplication. Some other laws of dot multiplication are given in the following theorem. 17.3.3 Theorem
If A and B are any vectors in V2 and c is any scalar, then (i) c(A. B) : (cA)'B; (ii) 0'A:0i (iii) A' A: lAl'. The proofs are left as exercises(seeExercises8 through 10). We now consider what is meant by the angle between two vectors, and this leads to another expression for the dot product of two vectors.
17.3.4 Definition
Let A and B be two nonzero vectors such that A is not a scalarmultiple of B. If OFis the position representation of A and O? is the position rePresentation of B, then the angle betweenthe aectorsA and B is defined to be the angle of positive measure between O? and OQ imerior to the triangle POQ.If A: cB, where c is a scalar,then if c > 0 the angle between the vectors has radian measure 0; if. c < 0, the angle between the vectors has radian measure ?r. It follows from DefinitionL7.3.4 that if a is the radian measure of the angle between two vectors, then 0 < d. < z. Figure 17.3'1 shows the angle between two vectors if A is not a scalar multiple of B. The following theorem is perhaps the most important fact about the dot product of two vectors.
F i g u r e1 7 . 3 . 1
17.3.5Theorem
If a is the radian measure of the angle between the two nonzero vectors A and B, then A . B:
l A l l n l c o sa
(4)
thepositionrePreari* ariandB :bi*b2i'LetOFUe sentation of A and oQ be the position representation of B. Then the angle
pRooF: LetA
THE PLANEAND PARAMETRIC EQUATIONS
Q(br, bz) P(or, a z)
between the vectors A and B is the angle at the origin in triangle POQ (see Fi9.77.3.2); P is the point (ar, ar) and Q is the point (01,b2).In triangle OPQ, lAl is the length of the side OP and Inl is the length of the side Oe. So from the law of cosines we obtain lvog=_
l A l ' + l B l , l P Q l , 2lAllBl
_ (arz+ ar2)I (brz* br2).. .l(a,  br)z* (a,  br)zf
2lAllBl
Figure17.3.2
_ 2arb,* 2arbz
2lAllBl
_ lrbr * arb,
lAllBl
Hence, cosa:
,A;, B,
lAllBl
from which we obtain
A . B : l A l l n lc o sa
r
Theorem L7.3.5statesthatthe dot product of two vectors is the product of the magnitudes of the vectors and the cosine of the radian measure of the angle between them. o rLLUsrRArroN2: If A : 3i  2i, B :2i * i, and a is the radian measure of the angle between A and B, then from iheorem 17.3.5,we have
cosc: I 'l B, lAllBl GX2) + (2)(1)
\M\M
62 ____,V13 V5
:+y65
0
we leamed in sec. 17.2 that if two nonzero vectors are scararmultiples of each other, then they have either the same or opposite directions. We have then the following definition. 17.3.6 Definition
Two vectors are said tobe parallel if. and only if one of the vectors is a scalar multiple of the other. . rLLUsrRArrow3: The vectors ; B : (  4 , 3 > 3 .A  2 i  i ; B : i * 3 i 5 . S h o wt h a ti . i : 1 . ; i , i : l ; i 8. Prove Theorem 17.3.3(i).
2 .A  ( + ,  + ) ; B(:* , * > 4.42i;Bi+i .
i:0.
6. ProveTheorem17.3.2(i).
7. Prove Theorem 17.3.2(ii).
9. ProveTheorem17.3.3(ii).
10. Prove Theorem 17.3.3(iii).
17.4 VECTORVALUED FUNCTIONSAND PARAMETRICEQUATIONS
In Exercises11 through 14, if. a is the radian measure of the angle between A and B, find cos a. 1 1 .A :
(4,31;8: (1,1)
1 2 .A :
1 3 .A : 5 i  1 2 j ; B : 4 i * 3 i
(2,31;8:
(3,2>
1 4 .A : 2 i + 4 j ; B :  5 i
15. Find k so that the radian measure of the angle between the vectors in Example 1 of this section is *a. 16. Given A:
ki2i
and B:
17. Given A:5iki;B:ki+5j, are parallel.
ki* 6i, where k is a scalar.Find k so that A and B are orthogonal. where k is a scalar.Find (a) k so that A andB are orthogonal;(b) k so thatAandB
18. Find k so that the vectors given in Exercise15 have opposite directions. 19.GivenA:5i+12j;B:i*kj,wherekisascalar.Findksothattheradianmeasureof
theanglebetweenAandBislzr.
20. Find two unit vectors each having a representation whose initial point is (2,4) and which is tangent to the parabola Y : 12 there' 21,.Find two unit vectors each having a representation whose initial point is (2, 4) and which is normal to the parabola Y  12 there' 22. If A is the vector ari. arj, find the unit vectors that are orthogonal to A. 6i, frnd the vector projection of A onto B. 23. If A: 8i * 4i and B:7i24. Find the vector projection of B onto A for the vectors of Exercise23. 25. Find the component of the vector A : 5i  6i in the direction of the vector B:7i
+ i. direction of vector A. 25. For the vectors A and B of Exercise 25, find the component of the vector B in the
27. A vector F represents a force which has a magnitude of 8 lb and *z as the radian measure of the angle giving its direction. Find the work done by the force in moving an object (a) along the r axis from the origin to the point (5, 0) and (b) along the y axis from the origin to the point (0, 6). Distance is measured in feet. 28. Two forces represented by the vectors F1 and F2 act on a particle and cause it to move along a straight line from the point (2, 5) to the point (7, 3). If Fl : 3i  i and F2: 4i + 5j, the magnitudes of the forces are measured in pounds, and distance is measured in feet, find the work done by the two forces acting together. 29. If A and B are vectors, prove that
( A + B ) ' ( A + B ): l ' '
A+2A' B+B'B
30. Prove by vector analysis that the medians of a triangle meet in a point. 31..Prove by vector analysis that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is onehalf the length of the third side. 32. Prove by vector analysis that the line segment joining the midpoints of the nonparallel sides of a trapezoid is parallel to the parallel sides and its length is onehalf the sum of the lengths of the parallel sides. 33. Prove that two nonzero vectors are parallel if and only if the radian measure of the angle between them is 0 or z.
17.4 VECTORVALUED FUNCTIONS AND PARAMETRIC EQUATIONS
We now consider a function whose domain is a set of real numbers and whose range is a set of vectors. Such a function is called a "vectorvalued fimction." Following is the precise definition.
17.4.1 Definition
Letl and I be two realvalued functions of a real variable f. Then for every
764
VECTORS IN THE PLANEAND PARAMETRIC EQUATIONS number f in the domain common to / and g, there is a vector R defined by
R(f):f(f)i+s(f)i
(1)
and R is called a aectoraalued function. o ILLUSTRATToNL: Suppose
R(f): \Fi+
(t 3)'i
Let a n dg ( f ) : ( t  3 )  '
f (t): \F
The domain of R is the set of values of f for which both /(t) and g(f) are defined. The function value /(t) is defined for f * 2 and g(f) is defined for all real numbers except 3. Therefore, the domain of R is
{tltz,t+S}.
.
If R is the vectorvaluedfunction defined by (1), as f assumesall values in the domain of R, the endpoint of the position representation of the vector R(f) haces a curye C. For each such value of f, we obtain a point (x, y) on C for which
xf(t)
and ysG)
(2)
The curve c may be defined by Eq. (1) orEqs. (2). Equation (1) is calleda aectorequationof c, and Eqs. (2) are called parametricequationsof C. The variable t is aparameter.The curve C is also called a graph; that is, the set of all points (x, y) satisfying Eqs. (2) is the graph of the vectorvarued function R. A vector equation of a curve, as well as parametric equations of a curve, gives the curve a direction at each point. That is, if we think of the curye as being traced by a particle, we can consider the positive direction along a curve as the direction in which the particle moves as the parameter f increases.In such a caseas this, f may be taken to be the measure of the time, and the vector R(t) is called the positionaector.sometimes R(f) is referred to as the radius aector. If the parameter f is eliminated from the pair of Eqs. (2), we obtain one equation in r and y, which is called a cartesianequationof c. It may happen that elimination of the parameter leads to a cartesian equation whose graph contains more points than the graph defined by either the vector equation or the parametric equations. This situation occurs in Example 4. EXAMPLE
Given the vector
equation R(f)
cos ti + 2 srn ti
SOLUTION: The domain of R is the set of all real numbers. We could tabulate values of x and y for particular values of t. However, if we find the magnitude of the position vector, we have for ev ery t
l R ( f )l t / 4 c o s z i + Z ; i p i _ 2 m _ z r
(a) draw a sketchof the graph of
17.4 VECTORVALUED FUNCTIONSAND PARAMETRIC EQUATIONS
this equation,and (b) find a cartesian equation of the graph. v P(2 cost, 2 sin t)
Therefore, the endpoint of the position representation of each vector R(f) is two units from the origin. By letting f take on all numbers in the closed interval 10,2nl, we obtain a circle having its center at the origin and radius 2. This is the entire graph becauseany value of f will give a point on this parametric equations of circle. A sketch of the circle is shown in Fig. 17.4.1,. the graph are x,:2cosf
and y:2sint
A cartesian equation of the graph can be found by eliminating f from the two parametric equations, which when squaring on both sides of each equation and adding gives 17.4.1
* + Yz:4 As previously stated, upon eliminating f from parametric equations (2) we obtain a cartesian equation. The cartesian equation either implicitly or explicitly defines y as one or more functions of r. That is, if x: f (t) andy : g(f), then y : h(x).lf.h is a differentiable function of r and / is a differentiable function of.t, it follows from the chain rule that DtA: (D"y)(Dtx) or
g,G): (h,(x))(f,(t)) or, by using differential notation, dy :dy dx dt dx dt If. dxldt # 0,we can divide on both sides of the above equation by dxlilt and obtain
(3)
Equation (3) enables us to find the derivative of y with respect t o x directly from the parametric equations.
ExAMPun2: Given x:3t2 and y  4tB, find dy ldx and dzyldxz without eliminating t.
solurroN:
APplying (3), we have
EQUATIONS VECTORSIN THE PLANEAND PARAMETRIC
d(y') dt dx dt Becausey' :2t,
d(y')ldt:2;
thus, we have from the above equation
dry dx2
ExAMPrn 3: (a) Draw a sketch of the graph of the curye defined by the parametric equations of Example 2, and (b) find a cartesian equation of the graph in (a).
solurroN: Becauser:3P, we conclude that r is never negative. Table gives values of r and y for particular values of f. BecauseDry:2t, 17.4.1, we see that when t:0, Dry: 0; hence, the tangent line is horizontal at the point (0,0). A sketch of the graph is shown in Fig. 17.4.2.From the two parametricequationsr:3f2 and y:4t3, we getf :27F andy2:'),6f . Therefore,
*"a'
6 5 4
27
or, equivalently, 15xs 27y'
3 2
16
which is the cartesianequation desired.
T
o
T a b l e7 7 . 4 . 1
1_
2 3
000
4
24'
5
134 21232
6 Figure17.4.2
131
+*+
L 2
4 32
3 12
o rr,lusrru,rrox 2: If in Eq. (4) we differentiate implicrtly, we have 4gx2 
dlt
s4y d*
and solving for dyldx, we get dV _ 8*' dx 9y
17.4VECTORVALUED FUNCTIONS ANDPARAMETRIC EQUATIONS767 Substituting for r and y in terms of t fuom the given parametric equations, we obtain dy :: dx
8(3t2)2 9(4ts)
'''
which agreeswith the value of dyldx found in Example 2. ExAMPrn 4: Draw a sketch of the graph of the curue defined by the parametric equations xcoshf
and y:sinhf
(5)
Also find a cartesian equation of the graph.
.
solurroN: Squaring on both sides of the given equations and subtracting, we have *  y': cosh2f  sinlf f From the identity cosh2t  sinh2 f : 1, this equation becomes (5)
fy':l
Equation (5) is an equation of an equilateral hyperbola. Note that for f any real number, cosh f is never less than 1. Thus, the curve defined by parametric equations (5) consists of only the points on the right branch of the h;nperbola.A sketdr of this curve is shown in Fig. 17.4.3.A cartesian equationis *  A2:1, where x >'1..
Figure17.4.3
The results of Example 4 can be used to show how the function values of the hyperbolic sine and hyperbolic cosine functions have the samerelationship to the equilateral hyperbola as the trigonometric sine and cosine functions have to the circle. The equations xcosf
and ysinf
(7)
are a set of parametric equations of the unit circle becauseif f is eliminated from them by squaring on both sides of each and adding, we obtain
Figure17.4.4
f * Y': cos2f * sin2f: 1 The parameter f in Eqs. (7) can be interpreted as the number of radians in the measure of the angle between the r axis and a line from the origin to P(cos f, sin f) on the unit circle. See Fig. 17.4.4 Becausethe area of a circular sector of radius r units and a central angle of radian measure f is
768
VECTORSIN THE PLANE AND PARAMETRICEQUATIONS
cosh f, sinh f
given by *r2f square units, the area of the circular sector in Fig. 17.4.4 is +t square units becauser: 1. In Example 4 we showed that parametric equations (5) are a set of parametric equations of the right branch of the equilateral hyperbola f U2:1. This hyperbola is calledtheunit hyperbola.LetP(cosh f, sinh l) be a point on this curve, and let us calculatethe areaof the sectorAOP shown in Fig. 17.4.5.The sector AOP is the region bounded by the x a>b.lfth" origin is at the center of tne fixea circle, f is the number of parameter and the circles, two the of of tangency point moving is the tact with the fiied circle, B are hypocycloid of the equations parametric that ptove AOB, angle radians in the
x:(ab)cosf*bcos+, and
y:(ab)sinf bsn+t of this cuwe are x: 24. If a : 4b in Exercise 23, we have a hypocycloidof f our cusps.Show that parametric equations : t. y a sirts 4 cossf and of four cusps, and draw a 25. Use the parametric equations of Exercise24 to find a cartesian equation of the hypocycloid equation. resulting of the graph sketch of the
26. Parametric equations for the ttactfix arc
x : t  a t a n hi
,:o"""hl
Draw a sketchof the curve for a:4' intercept of the tangent line. 27. provethat the parameter t in the parametric equations of a tractrix (seeExercise26) is the r line from the point 2g. Show that the tractrix of Exercise26 is a curve such that the length of the segment of every tangent to a' and equal constant is r axis of tangenry to the point of intersection with the (16). 29. Find the area of the region bounded by the r axis and one arch of the cycloid, having Eqs. 30. Find the centroid of the region of Exercise29.
772
VECTORS IN THEPLANEANDPARAMETRIC EQUATIONS
17.5 CALCUTUS OF VECTORVALUED FUNCTIONS !7.5.1 Definition
We now discuss limits, continuity, functions.
and derivatives of vectorvalued
Let R be a vectorvalued function whose function values are given by
R(f)  f(f)i + s?)i Then the limit o/ R(t) as t approaches tr is defined by
rimR(f): tt
+ [ln fuli [n,
(f)]t
and g@ bothexist. 1tTf@ 1T
o rr.r.usfRArroN L: If R(f) : cos fi + 2t j, then
lim R(r;: (ti3 cosf)i + (liT 2e,)i: i+ zi 17'5'2 Definition
The vectorvalued function R is continuousat trifand only if the following three conditions are satisfied: (i) R(fr) exists; (ii) lim R(f) exists; ttr
(iii) lim R(f) : R(r,). 2tr
From Definitions 17.5.'Land'],7.s.2,it fonows that the vectorvalued functionR, defined uy n(t) : f (t)i + g(t)i, is continuous at f, if and onry if / and g are continuous there. In the following definition the expression R(f+Af)R(r) At is used' This is the division of a vector by a scalarwhich has not yet been defined. By this expression we mean 1 6 tntt+ ar) R(r)l
L7'5'3 Definition
If R is a vectorvalued function, then the deivatiae of R is another vectorvalued function, denoted by R, and defined by R ' (t ) : l i m
R(t + Af)  R(f)
^t
ir thistimit I:,,. The notation DrR(t) is sometimes used in place of R'(f).
FUNCTIONS773 17.5CALCULUS OFVECTORVALUED The following theorem follows from Definition 17.5.3and the definition of the derivative of a realvalued function. 17.5.4 Theorem
If R is a vectorvalued function defined by (1)
R(f):f(f)i+s(f)i then
R ' ( f ) f ' ( t ) i +
g'G)i
if f ' (t) andg' G) exist. PRooF: From Definition
17.5.3
R(f+af)R(f) R ' ( t ) : lim Lt A/0
lim
lf ( t + af) i+ s( t + At) il [/( t) i + s( t) i] t
A/0
:lim
i+lim Af O
Af  0
: f,(t)i+g,(f)i
I
The directionof R'(f) is given by d (0 < e < 2n), where dy dt _ dY e'(t) tan d: dx dx f'(t)
E
From the above equation, we see that the direction of R'(f) is along point the tangent line to the curve having vector equation (1) at the
tation of R(f) as f assumesall values in the domain of R. Let OP be the position representation of R(f) and OQ be the position replesentation of
F i g u r e1 7 . 5 . 1
sentations tangent to the curve C at the point P.
VECTORSIN THE PLANEAND PARAMETRIC EQUATIONS
o rt,t,usrnerron 2: If R(f)  (2 * sin f)i + cos fj, then R'( t) : cos tr  sin fj Higherorder derivatives of vectorvalued functions are defined as for higherorder derivatives of realvalued functions. so if R is a vectorvalued tunction defined by R(f) : f (t)i + g(t)j, the secondderivative of R, denoted by R"(t), is given by R"(f) : Dr[R'(f)] We also have the notation DrrR(t) in place of R,,(t).By applying Theorem 17.5.4to R'(t), we obtain
R " (f) : f" ( t) i* g" G) i if f" (t) andg"(f) exist. o rLLUsrRArroN3: rf R(r): (ln f)i + (]) i, tr,""
R',(t):ltii and
 +i+ ii R"(t) 17.5.5 Definition
A vectorvalued function R is said to be differentiableon an interval if R'(f ) exists for all values of f in the interval.
The following theorems give differentiation formulas for vectorvalued functions. The proofs are based on Theorem 17.5.4 and theorems on differentiation of realvalued functions. 17'5'6 Theorem
If R and Q are differentiable vectorvalued functions on an interval, then R + Q is differentiable on the interval, and
D,[R(r)+ Q(r)] : D,R(r)+ D,Q(f) The proof of this theorem is left as an exercise (see Exercise 16). EXAMPLE 1:
If
R(f) tzi+ (tL)i and Q(f) : sin fi + cos fi verify Theorem 77.5.6.
SOLUTION:
D'[R(t) + e(r)] :Dt(Jtzi+ (t 1)i] + [sin ti+cos ti]) : DtlG, * sin f)i + (t  1 + cost)il  (2t+ cosf)i + (1  sin f)j D'R(r) + DIQG): Dtltzi+ (r  1)il * D,(sinfi + cosri)
FUNCTIONS 17.5 CALCULUSOF VECTORVALUED
: (zti + i) + (costi  sin fi)  ( 2 t + c o s f ) i + ( 1 s i n f ) i D'[R(f)+ Q(f)l DrR(f)* D'Q(t). Hence, 17.5.7 Theorem
If R and Q are differentiable vectorvalued functions on an intervd, then R' Q is differentiable on the interval, and
D , [ R ( r ) .Q ( r ) ] : [ o , n ( t ) ] ' Q 0 ) + R ( t ) ' [ D ' Q ( t ) ] and Q(f) :f2!)i+8r(t)i' Then by PRooF:Let R(f) :flt)i+8,(t)i Theorem17.5.4 DE(f) : fl (t)i+8,'(t)i and D'Q(f): fz'G)i+ 82'G)i
R(f). Q(f): 11,(t)l lf,(t)l + [s'(t)][s'(t)] So
.Q(r)] D,[R(r) + [s'(f)][g,'(t)] : lfr'(r)lt/r(f)l + t/,(r)llf,'(t)l* [8,'(t)]lsr(t)]  {lf,'(r)lt/2(r)l + {[/,(t)]lf,'(t)l+ [s'(t)][g"(t)]] * [8,'(r)]ts,(r)l] I : [P'n(t)]' Q(t)+ R(t)' tD'Q(f)l EXAMP:n2: Verify Theorem 17.5.7 for the vectors given in Example 1.
SOLUTION:
R ( r ) ' Q ( t ) : f 2s i n t + ( t  1 ) c o sf '
Therefore,
Q ( f ) l  2 f s i n t + f 2 c o st + c o s t + ( f t ) (  s i n f ) D,[R(r) Q ( f ) l  ( f + 1 ) s i n t + ( f 2 + 1 ) c o sf
D,[R(f)
Because
D,R(r) D,ltzi+ (t  1)il  zti+ i [D,R(f)] . Q(t) : (2ti+ i)' (sin fi + cosfi)  2 t s i n t + c o sf Because D,Q(t) : D,lsin fi + cos til:
cos ti sin fi
R ( f ) . [ D , Q ( f ) ] : [ t z i + ( t  1 ) i ] ' ( c o st i  s i n f i ) : t2 cos t  (f  1) sin f Therefore, : (2t sin t + cosf) tD,R(f)l . Q(f) + R(f) ' [D'Q(f)] + [f2 cost
(f  1) sin f]
VECTORSIN THE PLANEAND PARAMETRIC EQUATIONS
Thus, [D,R(f)]. Q(f) + R(r) . [ D , Q ( f ) ] : ( f + 1 ) s i n f + ( t 2 + L ) c o s f (3) ComparingEqs. (Z) and (3), we see that Theorem 17.5.T holds. 1'7.5.8Theorem
If R is a differentiable vectorvalued function on an interval and f is a differentiable realvalued function on the intenral, then
Dilf@ltR(f)ll:
[D,/(f)]R(t)+ f(t) D,R(f)
The proof is left as an exercise(seeExercise1Z). The following theorem is the chain rule for vectorvalued functions. The proof which is left as an exercise (see Exercise 1g) is based on Theorems 2.6.6 and 3.6.1, which involve the analogous conclusions for realvalued functions. 17'5'9 Theorem
Suppose that F is a vectorvalued function, h is a realvalued function such that 6: h(t), and G(f) : F(/r(O).If ft is continuous at f and F is continuous at h(t), then G is continuous at f. Furthermore, D1s if and DoG(f) exist, then D$(t) exists and is given by D$(t):
lDoG(t))Dt6
we now define an indefinite integral (or antiderivative) of a vectorvalued function. 17.5,10 Definition
If e is the vectorvalued function given by
Q(f):f(t)i+s@i then the indefinite integral of eG) is defined by f.
lr
d t : i f G )d t + i s @a t J a(r) J J
(4)
This definition is consistent with the definition of an indefinite integral of a realvalued function because if we take the derivative on both sides of (4) with respect to f, we have rff
D , Jq ( t )d t : i D t f @ d t + i D t s @ a t J J which gives us f
D, Q@ itt:if(t) + wG) J For eachof the indefinite integrals on the right side of (4) there occurs anarbitrary scalar constant. when each of theJe scalarsis muttiplied by either i or i, there occurs an arbitrary constant vector in the sum. so we
17,5 CALCULUSOF VECTORVALUED FUNCTIONS
have f
I Q(f) dt _ R(f)
+c
J
where DrR(t) : Q(f) and C is an arbitrary constant vector.
ExAMPrr3: Find the most general vectorvalued function whose derivative is
If D'R(t)  Q(t), then R(t; : ,l"Q(f)dt, or
SOLUTION:
ff
R ( r ; : i I sin t dt  3i' J I cosf dt J
Q(f) : sin ti 3 cos fj
i(cos t+ C') 3i(sin t+cr) cos ti  3 sin ti + (C,i  sCri) cos ti 3 sin ti+C
EXAMPTu 4: Find the vector R(0 for which DB(t):
solurroN:
t dt+i a a, I
R(f)
So
eti* e'i
and for which R(0) : i + i.
R(f)  i (e' * C') + i k' + Cr) Because R(0) : i + i, we have i + i: i(1 + C') + i(1 + Cr) So Cl L:1
and
Cr*1.:1.
Therefore, Cr: 2
and
Cz: 0
So
R(t;: (e,+2)i+eti
The following
l7.5.ll
Theorem
theorem will be useful later.
If R is a differentiable vectorvalued function on an interval and lR(t)l is constant for all t in the interval, then the vectors R(t) and DrR(f) are orthogonal. pRooF: Let lR(01: k. Then by Theorem 17.3.3(iii),
R (t)' n1 4:12 Differentiating on both sides with respect to f and using Theorcm17.5.7
VECTORSIN THE PLANEAND
PARAMETRIC EQUATIONS we obtain
. R(r)+ R(r). [D,n1r;1 :s [D,R(r)] Hence, 2R(f)'D'R(t):0 Becausethe dot product of R(f) and DlR(f) is zero, it follows from Definition 77.3.7that R(f) and D1R(f)are orthogonal. I Df R(f)
The geometric interpretation of Theorem 17.5.1.7is evident. If the vector R(t) has constant magnitude, then the position representatiotr OF of R(t) has its terminal point P on the circle with its center at the origin and radius k. So the graph of R is this circle. BecauseDrR(t) and R(f) are +
orthogonal, OP is perpendicular to a representation of DE(t).
Figrre
17.5.2shows a sketch of a quarter circle, the position representation OP of R(t), and the representationPB of DE(f).
Figure 17.5.2
Exercises 77.5 In Exercises L through 4, find the indicated limit, if it exists.
1 . R ( f ) ( t  2 ) i +
2. R ( f )  e t + t i+ l f + 1 . l i ; t i m R ( f )
=i,rr,gR(r)
3. R(f)  2 sin fi * cos,t,
4. R(f):
) l:nR(f In Exercises 5 throughL0,find R'(f ) and R" (f ). tL.
t2.
s. R(f) : (t2 3)i + (2t+ r)i
6 .R ( f ) : f i i + i i,
8. R(f) : coszti + tan fi
9. R(f) : tanr ti + zti
In Exercises11 and 12, find.D,lR(f) l. (2t)i 1 1 .R ( t ) : ( f  l ) i +
122t3.,t25t+6.1
fi:i
+
timR(f) f _ 3i i;
7. R(f) : szti+ ln fi
1 0 .R ( t ; : \ t r t + r i + ( t  l ) ' i
1 2 .R ( f ) : ( e t * l ) i + ( e '  l ) j
In Exercises13 through 15, find R'(t) . R"(t). 1 3 .R ( t ) :
(2P l)i+ (f +3)i
15. Prove Theorem 17.5.5.
1 4 .R ( t ) :  c o s
2ti*sin2ti
17. Prove Theorem 17.5.8.
ls. R(t):ezti+ezti 18. Prove Theorem 17.5.9.
In Exercises19 through 22, find, the most general vector whose derivative has the given function value.
t s .t a n r i+ i
20.hi+t,i
sint fi + 2 coszti, andR("t) :0, find R(f). 24. If R'(f) : et sin ti+ et cos ti, and R(0)  i  i, find R(f). 23. If R'(t):
2t. ln fi + Pi
22. }ti  zti
17.6 LENGTHOF ARC
25. Given the vector equation R(t) : cos fi * sin fj. Find a cartesian equation of the curve which is traced by the endpoint of the position representationof R'(f). Find R(t) . R'(t). Interpret the result geometrically. 26. Given R(t):zti+ find Dp(t).
(tz 1)j
and Q(t):3ti.
lf. a(t) is the radian measureof the angle between R(t) and Q(f),
27. Suppose R and R' are vectorvalued functions defined on an interval and R' is differentiable on the interval. Prove D ' [ R ' ( t ) . R ( t ) ] : l R ' ( t )  ' z+ R ( t ) ' R " ( t ) 28. If lR(t)  : h(t) , prove that R(t) . R'(t) : lh(t)llh' (t)1. 29. If the vectorvalued function R and the realvalued function / are both differentiable on an interval and /(t) * 0 on the interval, prove that Rff is also differentiable on the interval and
(r) /'(f)R(f) _ /(r)R', l] fRC)..l 'LfG) lf@l' J 30. Prove that if A and B are constant vectors and / and g are integrable functions, then fff
+ Bs(t)litt: A f@ at+n s@ d.t J J tA/(t) J (nrNr: Express A and B in terms of i and i.)
17.5 LENGTH OF ARC
In Sec.8.10 we obtained a formula for finding the length of arc of a cunre having an equation of the form y : f (x).This is a special kind of cunre becausethe graph of a function /cannot be intersected by a vertical line in more than one point. We now develop a method for finding the length of arc of some other kinds of curves. Let C be the curve having parametric equations xf(t)
and y:g(f)
(1)
and suppose that / and g are continuous on the closed interval [a, b]. We wish to assign a number L to represent the number of units in the length of arc of C from t: a to t: b. We proceed as in Sec.8.10. Let A be a partition of the closedintenral [a, b] formed by dividing the interval into n subintewals by choosing n  L numbers between a and,b. L'et fo: a and tn: b, and let tr, tz, . . . , tnr be intermediate numbers: to1t, The dth subinterval is [/i1, t1] and the number of units in its length, denoted by Llt,is \ f6r,where i: L,2, . . . , n. Let llAllbe the norm of the partition; so each A/ = llAll. Associatedwith each number f1is a point P{f (t),9(4)) on C. From each point P1, draw a line segment to the next point P1 (see Fig. 77.6.7). The number of units in the length of the line segment from P1, to P; is denoted by lPprPTl. From the distance formula we have
IFFIP{l :
e)
780
VECTORSIN THE PLANEAND PARAMETRICEQUATIONS
v Pi(f(t),s(f ;))
P"(f(b),s(b))
Pit(f (t it), s$ rr))
F i g u r e1 7 . 6 . 1
The sum of the numbers of units of lengths of the n hne segments is n
(3) tl
Our intuitive notion of the length of the arc from f : a to f : b leads us to define the number of units of the length of arc as the limit of the sum in (3) as llAllapproacheszero. 17.5.1 Definition
Let the curve C have parametric equations r : /(t) and y : g!).Then if there exists a number L having the property that for any € > 0 there is a E>0suchthat ln
I
l) lP,_rPiILl l?_r' for everypartition A of the intenral la, bl for which llAll
L lim i lml s f:1
(4)
llall
and L units is called the length of arc of the curue C from the point (f (a), g@)) to the point (f (b), g(b) ).
The arc of the curve is rectifiable if the limit in (4) exists. If f ' and g' are continuous on la, bl, we can find a formula for evaluating the limit in (a). We proceed as follows. Because/' andg' are continuous on [a, bl,lhey are continuouson each subinterval of the partition A. So the hypothesis of the meanvalue theorem (Theorem 4.7.2) is satisfied by f andg on each lti,_r,tl; therefore, there are numbers zi and ari in the open interval (tr_r, f1) such that
' f (t,)  f (tr)  f (2,) Aot
(s)
gG)  s1r)  g'(w) L,t
(6)
and
17.6LENGTHOF ARC
781
Substituting from (5) and (5) into (2) , we obtain
tmt: or, equivalently, (7)
Lit
lffil:
inthe openintervil (tor,tu).Thenfrom (4) and (7),if where z.land,wieltE the limit exists, n
L
(8)
lim ' l l a l l  of i
The sum in (8) is not a Riemann sum becausezi and wl N€ not necessarily the same numbers. So we cannot apply the definition of a definite integral to evaluatethe limit in (8). Howevet, there is a theorem which we can apply to evaluate this limit. We state the theorem, but a proof is not given becauseit is beyond the scope of this book. You can find a proof in an advanced calculus text. 17.6.2 Theorem
If the functions F and G are continuous on the closed interval la, bl, then the function \/FtTtr, is also continuous on la,bf , and'if A is a partition I tr, < ti < o f t h e i n t e r v a fl a , b l ( A , a: : t o I t , I ' ' ' (f11, f1), then in and z1 artd wl are any numbers Lit:
['
dt
(e)
' Applytng (9) to (8), where F is f and G is g', we have
t:
fb
J"
dt !lf,(t)1,+ 18,G)12
We state this result as a theorem. 17.5.3 Theorem
Let the curve C have parametric equations r: l(t) and y : 8G), and suppose that f' andg' are continuous on the closed interval [4, D]. Then the length of arc L units of the curve C from the point (/(a), g(a) ) to the point (f (b), S(b) ) is determinedby (10)
EXAMPLE1: Find the length of the arc of the curye havingParametric equations x: f 3 and y : 2t2 in each of the following cases:(a) from t  0 to t  1; (b) from t2tot0.
sor,urroN: A sketch of the curve is shown in Fig. L7.6'2' (a) Letting a n d l e t t i n gy : 8 G ) , 8 ' ( t ) : D , Y : 4 f ' s o f r o m x: f(t), f'(t):Dp:3P; Theorem L7.6.3,if L units is the length of the arc of the curve from f : 0 to t:1,
VECTORSIN THE PLANEAND PARAMETRIC EQUATIONS fr
:l
t\ffidt
JO l1
: + . ? ( g t r + 1 5 ) 3 / 21 lo
;r[(25)Btz  (16)3/2] L0864Z
F i g ur e 1 7. 6 . 2
+(12s64) n61 
(b) If L units is the length of the arc of the curye from we have from Theorem 17.6.3
Ll
fo
f0
I \F\ffi Jz
ffidt:
Jz Because 2 < t S 0, \F ro
L  t\ffi Jz
2tot:0
dt
 t. So we have
dt
 ;?
lo (9t' + I6)st2  Jz
 ;r
[ (16)3t2
(50) 3/2]
;r(2s0 \D  64) : 10.7
The curve C has parametric equations (1). Lets units be the length of arc of C from the point (f(t),g(tr)) to the point (/(r), gG)), and let s increase as f increases.Then s is a function of f and is given by ft
s l Jto
du
(11)
From Theorem 7.6.1, we have
A vector equation of C is
R(t):f(f)i+ s?)i
(13)
Because
R'(t):f'(t)i+ g'U)i we have (14)
17.6 LENGTHOF ARC
783
Substituting from (14) into (12),we obtain
(1s)
lR'(f)l
From (15) we conclude that if s units is the length of arc of curve C having vector equation (13) measured from some fixed point to the point (f (t), SG)) where s increasesas f increases,then the derivative of s with respect to f is the magnitude of the derivative of the position vector at the point (/(f),9(f)). If we substitutefrom (14) into (10), we obtain f : "ljlR'(t)ldf. So Theorem 17.5.3 can be stated in terms of vectors in the following way. 17.6.4 Theorem
Let the curve C have the vector equation R(t) : f (t)i + SG)i, and suppose thatf'andg' are continuous on the closed interval [a, b]. Then the length of arc of C, traced by the terminal point of the position representation of R(f) as f increases fuom a to b, is determined by (15)
EXAMPrn 2: Find the length of the arc traced by the terminal point of the position rePresentation of R(f) as f increases from L toAtf R(f) : et sin fi + et cos ti
sorurroN: R'(t) : (e'sin t + et cost)i + (et cos t et sin t)i. Therefore, 2
l R ' (t ) l : e '
sin f cos t*
stnz t
 et{z From (16) we have
f,et dt: nr)^r \D,(en e)
An altemate form of formula (10) for the length of an arc of a curve C, is obtained by reand y:8(t), having parametric equations x:f(t) placing f'(t) by dxldt and g'U) by dyldt.Wehave dt
(r7)
Now supposethat we wish to find the length of arc of a curve C whose (x, y) is the cartesianrepresentationof a polar equation is r:F(0).If. point P on C and (r, 0) is a polar rePresentation of P, then xrcose
and yrsrn0
(18)
Replacirg r by F(0) in Eqs. (18), we have x  F(0) cos 0
and
F(0) sin 0
(le)
VECTORSIN THE PLANEAND PARAMETRIC EQUATIONS
Equations (19) can be considered as parametric equations of c where 0 is the parameter instead of f. Therefore, if.E' is continuous on the closed interval fo,Ff , the formula for the length of arc of the curve c whose polar equation is r:F(d) is obtained from (12) by taking f :9. So we harre d0
(20)
From (18) we have dx dr UEr d0_cos
sin0 and
du dr d0:sin0 dn*rcos0
Therefore,
ffi: * 2r sin 0 cosd
EXAMPLE3: Find the length of the cardioid r  2(1 + cos A).
fi*
r2 cosz 0
solurroN: A sketch of the curve is shown in Fig. 17.6.9.To obtain the length of the entire curve, we can let d take on values from 0 to 2zr or we can make use of the symmetry of the curve and find half the length by letting 0 take on values from 0 to z. Becauser:2(l * cos d), drld: 2 sin 0. Substitutinginto (21), integrating from 0 to z, and multiplying by 2, we have frr
L'J,
dE de
: 4 ! 7 f n  t / t + c o s e a e JO
Figure 17.6.3
To evaluatethis integral, we use the identity cos2*g:+(1+ cos 0), which gives vT +cos 0 : V2 lcosggl.Becauseb = 0 = 7T,0 0. To each value of s there corresponds a unique point P on the curve C. Consequently, the coordinates of P are functions of s, and s is a function of f. In Sec.17.6we showed that lD'R(f)  : Dts
(7)
Substituting from (7) into (1), we get
T(f):ry D,R(t) : DrsT(f) If the parameter is s instead of f, we have from the above equation, by taking f : s and noting that Dp : L, DR(s) : T(s) This result is stated as a theorem. 17.8.3Theorem
If the vector equation of a curve c is R(s) : /(s)i + g(s)i, where s units is the length of arc measured from a particular point Poon C to the point P, then the unit tangent vector of C at P is given by
Now suppose that the parametric equations of a curve C involve a parameter f, and we wish to find parametric equations of C, with s, the number of units of arc length measured from some fixed point, as the
AND ARC LENGTHAS A PABAMETER AND UNITNORMALVECTORS 17.8THEUNITTANGENT
parameter.Often the operations involved are quite complicated. However, the method used is illustrated in the following example. ExAMPLE2: Suppose that parametric equations of the cunre C a r ex  f 3 a n d y : t 2 , w h e r e f > 0 . Find parametric equations of C having s as aparameter, where s is the number of units of arc length measured from the point where t:0.
solurroN: If Peis the point where t:O,Po is the origin. The vector equation of C is R(f) : tsi+ t2i BecauseD6s: lDlR(f)1, we differentiate the above vector and obtain DrR(t):3t2i+2ti So
lD,R(r)l: y9FT@ : {F 6FTT Becauset > 0, lF
: f. Thus, we have
l D ' R (lr:) t @ + + Therefore, Dos,:115P14 and so ft
s: I u\/9uz*4 du JO
 * J, fot t8u\m
du
: iv (9u,+ 4rt,r]. We obtain (S)
t:419t2*47ttz* Solving Eq. (8) for f in terms of s, we have (9f+.41"r':27s*8 9p + 4: (Z7s* 8;zra Becauset > 0, we get
t:t{@iT?ryEZ Substituting this value of f into the given parametric equations for C, we obtain ,:fi1(27s*8)za4ltrz
and y:*l(27s*8)ztrn,
(9)
Now becauseD"R(s) : T(s), it follows that if R(s) : x(s)i + y(s)i, then T(s) : (Drr)i + (Dry)i. And becauseT(s) is a unit vector, we have (10) ( D , r )z 1 ( D ,y ) '  1
796
VECTORSIN THE PLANEAND PARAMETRICEQUATIONS
Equation (10) can be used to check Eqs. (9). This check is left as an exercise (see Exercise 11).
Exercises 17.8 In Exercises1 through 8, for the given curve, find T(f) and N(f ), and at f : fr draw a sketch of a portion of the curve and draw the representationsof T(tr) and N(fr) having initial point at t: h. 1. x:*t3t,y:F;tr:2 2. x:itz,y:+f;tt:l 3. R(f) : etiI e ti;
\:0 4 . R ( t ) : 3 c o sf i * 3 s i n 4 ; h : t r 5. r : cos kt, y : sin kt, k > 0; tr: nlk
6 . x : t  s i n f , y :   c o sf i \ : n 7. R(t) : ln cos fi * ln sin ti,0 < t < tn; tt: ir 8. R(t) : f cos fi * f sin ti; \:0 9 . I f t h e v e c t o r e q u a t i o n o fc u r v e C i s R ( f ) : 3 t 2 i * ( t t  3 t ) i , vectorsR(2) and T(2).
f i n d t h e c o s i n e o f t h e m e a s u r e o ft h e a n g l e b e t w e e n t h e
10. If the vector equation of curve C is R(f ) : (4  3P)i + (ts  3t)i, find the radian measure of the angle between the vectorsN(1) and D,'?R(l). 11. Check Eqs. (9) of the solution of Example2 by using Eq. (10). In Exercises12 through 15, find parametric equations of the curve having arc length s as a parameter, where s is measured from the point where f : 0. Check your result by using Eq. (10). 72. x: e cost, !:
a sin t
1 3 .x : 2 * c o st , y : S * s i n f 1 4 . x : 2 ( c o s f * f s i n t ) , V : 2 ( s i n t  f c o sf ) l.5. One cusp of the hypocycloid of four cusps:R(t) : a cossfi * a sin3 ti, 0  t =tr. 16. Given the cycloid x:2(tsin from the point where f : 0.
f), l:2(1cos
t), expressthe arc length s as a functionof t, where sismeasured
L7. Prove that parametric equations of the catenary y : a cosh (xl a) where the parameter s is the number of units in the lengthofthearcfromthepoint(0,a)tothepoint(x,y)and.s=0whenx=0ands(0whenx(0,are )casinhr!
and y\m
17.9 CURVATURE
Let @ be the radian measure of the angle giving the direction of the unit tangent vector associated with a curve C. Therefore, S is the radian measure of the angle from the direction of the positive r axis counterclockwise to the direction of the unit tangent vector T(f ). See Fig. 17.9.1,.
17.9CURVATURE BecauselT(t)l:
L, it follows from Eq. (1) in Sec.l7.2 that
T(f) : cos 0i + sin rfi
(1)
Differentiating with respect to Q, we obtain DoT(f)
(2)
0i + cos,fi
 1 , D a T ( t ) i s a unit vector. B e c a u s leD r T ( t ) l : that BecauseT ( f ) has constantmagnitude it follows from Theorem 17.5.11 DoT(t) is orthogonalto T(f). Replacingsin f by cos(*n*d) and cos f by sin(*zr*f), we write (2) as F i g u r e1 7 . 9 . 1
,
(3)
DoT(f) cos(ln + 0)i + sin(tn + 0)i
From (3) and the previous discussion, the vector DoT(f) is a unit vector orthogonal to T(t) in the direction*zr counterclockwise from the direction of T(t). The unit normal vector N(t) is also orthogonal to T(f). By the chain rule (Theorem 17.5.9),we have D;I(t):
(4)
[Dor(f)]DtQ
Becausethe direction of N (f ) is the same as the direction of.D it (t), we see from (4) that the direction of N(t) is the sameas the direction of DoT(f) if D# > 0 (i.e., if T(f) turns counterclockwiseas f increases),and the direction of N(t) is opposite that of D6T(t) if D# < 0 (i.e., if T(f) tums clockwise as f increases). Becauseboth D6T(f) and N(t) are unit vectors, we conclude that
Dar(r): t_N[t] D,6>o
ifD#>0 ifD$ 0, and in c and d,DtO < 0. The positive direction along the curve C is indicated by the tip of the arrow on C. In each figure are shown the angle of radian measure { and representations of the vectors T(f),
D,6( o F i g u r e1 7 . 9 . 2
VECTORSIN THE PLANE AND PARAMETRICEQUATIONS
DoT(t), and N(f). The representation of the unit normalaectorN(f) is always on the concaaesideof the curT)e. o Consider now DrT (t) , where s units is the arc length measuredfrom an arbitrarily chosen point on C to point P and s increasesas f increases. By the chain rule D,T(r)  DaT(t)D,o Hence,
: lDrT(f)lD,dl lD,T(r)  : lDrT(f)D,dl But becauseD6T(f) is a unit vector, lDrT(f)  :
thus, we have
lD,T(r)l:lD,4l
(5)
The number lDdl is the absolute value of the rate of change of the measureof the angle giving the direction of the unit tangent vector T(f) at a point on a curve with respect to the measure of arc length along the curve. This number is called the curaatureof the curve at thJpoint. Befo.e giving the formal definition of curvature, let us see that taking it as this number is consistent with what we intuitively think of as theturvature. For example, at point P on C, { is the radian measure of the angle giving the direction of the vector T(f ), and s units is the arc length from a poi"t po on c to P. Let Q be a point on C for which the radian measureof the angle giving the direction of T(f * Af) at e is d * A@and s * As units is the arc length from Poto Q. Then the arc length from p to e is As units, and the ratio AdlAs seemslike a good measure of what we would intuitively think of as the aaeragecurvature along arc PQ. o ILLUsrRArroN2: See Fig. 17.9.3a,b, c, and d: In a, A0 > 0 and As > 0; in b, Af ) 0 and As ( 0; in c, A{ ( 0 and As > 0; and in d, 46 ( 0 and As0
find the curvature vector and the curuature at any t.
soLUrIoN: The vector equation of the circle is R(f)  a cos ti* a sin fi So DrR(t) : a sin ti + a costi
l D , R ( r ) l : (
a sin t)' + (a cost)2  a
, r(f):#!tl lD,R(f)l D;T(t) : cos ti  sin fi QTJil lpRbl
cosf , sinf , ra ol
So the curyature vector
1 ' K ( f ) :  l c o s t i t  A s m t. t. a and the curyature
K(r;:lK(ill:+
:
VECTORS IN THEPLANE ANDPARAMETRIC EQUATIONS The result of Example L statesthat the cunrature of a circle is constant and is the reciprocal of the radius. Suppose that we are given a curve C and at a particular point p the curyature exists and is K(f), where K(t) + 0. consider the circle which is tangent to curve C at P and has curvature K(f) at P. From Example 1.,we know that the radius of this circle is UK(f) and that its center is on a line perpendicular to the tangent line in the direction of N(f). This circle is called the circle of curaature,and its radius is the radiusof curoatureof. C at P. The circle of curvature is sometimes refered to as the osculating circle. 17.9.2 Definition
If I((t) is the curvature of a curve C at point P and K(t) + 0, then the radiusof curlature of C at P, denoted by p(t), is defined by
p(r): fri, 1
ExAMPrn 2: Given that a vector equation of a curye C is R(f) :2ti * (t'
SOLUTION:
R ( f ): 2 t i * ( t '  L ) i D , R ( f ) 2 i + z t i
l)i
find the cunrature and the radius of curvature at t  1. Draw a sketch of a portion of the curye, the unit tangent vector, and the circle of cunrature at t : t.
lD,n(f)l21ffi2
1=i+ t ( f ') : , ? ' l l t ) rt \ m lD,R(fl irffii
'
D;r(r)G*fui+&i DTG)lD,n(f)l
2(L+t), r'2(1 +t)rt  I sz , l.o\ K (t): r ) l g  _ " ) : tlVg, r 4(r*tzy+
hffitfiTl
K ( t') r) 
x(t):
1
+b
L
20
+ t2)3t2
pill) :4t/I
and 11
r(1):1ri+Vri Figure17.9.4
Figure
17.9.4 shows the sketch that is required.
The accompanying
17.9CURVATUREE01 T able 17.9.'1.gives the corresponding values of r and ! for t': 2, 'L , 0, 1, and2. Table L7.9.7
2 1.
43 20
0 L 2
0 20 43
L
We now find a formula for computing the cunrature directly from parametric equations of the cun/e, x f(t) and y: g(f). From Eq. (5)
IDJ(f)  : lD,,$l,and so
K(r; lD,ol Assuming that s and f increase together, we have
So
do Dr6:
dt
ffi
(10)
To find dfildt, we observe that because@is the radian measureof the angle giving the direction of the unit tangent vector,
(11)
Differentiating implicitly with respect to f the left and right members of (11), we obtain
802
VECTORSIN THE PLANEAND PARAMETRICEQUATIONS
So d0_ dt
(#)(#)(HGry)
But
Substituting this expression for sec2
(#)(#)(#)(#)
(14)
ExAMPr,n 3: Find the cunratureof the curuein Example2by using formula (14).
soLUTroN: Parametric equations of C are x  2t and y : t2 'j... Hence,
dx  " ,
d2x
dtL
dt2
0
du +:2t dt
d2u a/
_n
dtzL
Suppose that we are given a cartesian equation of a curye, either in the form A : F(x) or x: G(y). Specialcasesof formula (L4) can be used to find the cunrature of a curve in such situations. If y : p 1r1 is an equation of a curve C, a set of parametric equations of C is r: f and y:F(t). Then dxldt:1, dzxldt2:0, dylilt: dyldx, and dzyldt2: dzyld*. Substituting into (14), we obtain
(ls)
17,9CURVATURE 803
C is r
Similarly,
G(y), we obtain
(15)
EXAMPTS4: If the curye C has an equati on )cy  1, find the radius of cunrature of C at the point (1, 1) and draw a sketch of the curye and the circle of cunrature at (1, 1).
solurroN: Solving for !, we obtain A:Ux. dzyldf :21x3. Applying formula (15), we have
K
l*l
2xu:ry:ry
So dylik11f
and
Zxa
v Becausep:
LIK we have
lrl(# + 1)3/2 p:ESo at (1, l), p:tE.fne
required sketch is shown in Fig.77.9.5.
ure17.9.5
Exercises 17.9 In Exercises L through 4, find the curvature K and the radius of curvature p at the point where f  fr. Use formula (9) to find K. Draw a sketch showing a portion of the curve, the unit tangent vector, and the circle of curvature at l: fi. 2. R(f)  (sz zt)i+ (f3 t)i; tr 1 1. R(f) : 2ti* (t'  r)i; h: L 3. R(f )  Zeti+ Zeti)h:
4. R(f ) : sin fi + sin 2fi; h:
0
Ln
In Exercises5 and 6, find the curvature K by using formula (14).Then find K and p at the point where f : fr and draw a sketch showing a portion of the curve, the unit tangent vector, and the circle of curvature at t: ty 5. )c+
{y:
t_ t;
6. X: et* er, y  et_ et. tr 0
fr: o
In Exercises7 through t2, find the curvature K and the radius of curvature p at the given point. Draw a sketch showing a portion of the curye, a piece of the tangent line, and the circle of curvature at the given point. (+,t) 9. a : e'; (0, 1) 8. 7. y : 2{x; (0, o)
a':f;
10. 4x2* 9y' : 36; (0, 2)
11. x : sin y; (*, tn)
12. x:
tan y; (l, Ln)
VECTORSIN THE PLANEAND PARAMETRICEQUATIONS
In ExercisesL3 through L8, find the radius of cunrature at any point on the given curue. 1 3 .y  s i n  r x 15. xuz+ yuz auz
ln secx L5. R(f)  et sin fi * et cos fj
17. The cycloid x  a(t  sin f), A : a(l  cos f)
L8. The tractrixx t  a tanh
14. y:
1,,
 a r".h 1
19. Show that the cuwature of the catenary y: a cosh (xla) at any point (x, y) on the curve is aly2. Draw the circle of curvature aI {0 , a) . Show that the curvature K is an absolute maximum at the point (0, a ) without referring to K' (r) . In Exercises20 through 23, find a point on the given curve at which the curvature is an absolute maximum. 2 0 .y : s t
2 1 .y : 6 x  f
2. y:
2 3 . R ( r ) : ( 2 t 3 ) i + ( p  l ) i
sin x
24' lf a polar equation of a curve is r:
F(0), prove that the curvature K is given by the formula
K _ lr' + ?@tae), _r\g!do,)l (drlde)zf'n lr' *
In Exercises25 through 28, find the curvature K and the radius of curyature p at the indicated point. Use the formula of Exercise 24 to find K. 25. r: 4 cos 20; e  #rr 2 6 . r : ' / . . s i n 0 ; 0  0 2 7 .r : A s e c 2 t 0 ; 0 : 3 n
2 8 .r  a 0 ; 0 : l
29. The center of the circle of curvature of a curve C at a point P is called the centerof curcature at P. Prove that the coordinates of the center of curvature of a curve at P(x, y) are given by
xc: x  @vld*)lf,t=7vldx)'l d2yI dxz
, (dyldx)z+ L A":Y+@
In Exercises30 through 32, find the curvature K, the radius of curvature p, and the center of curvature at the given point. Draw a sketch of the,curve and the cirde of curvature.
30. y : cosr; (tn, t)
3 1 .y  f
x'; (0,0)
3 2 .y : l n
x; (1,0)
In Exercises33 through 35, find the coordinates of the center of curvature at any point. 33. y':
4px 35. R(f)  a cos fi +b sin fi
I7.IO TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION
34. At : a'x 36. R(f)  a coss fi + a sins ti
If a particle is moving along a curve C having the vector equation
R ( f ): f ( t ) i + s @ i from Definition
(1)
17.7.1, the velocity vector at P is given by
v(f)  D/R(f)
(2)
From Sec.17.8,if T(t) is the unit tangent vector atP, s is the length of arc of C from a fixed point P0 to P, and s increasesas f increases,we have D,R(t)  DtstT(f)l
(3)
OF ACCELERATION 805 AND NORMALCOMPONENTS 17.10TANGENTIAL
Substituting from (3) into (2) , we have
v(f)  Dtstr(f)l
(4)
Equation (4) expressesthe velocity vector at a point as a scalartimes the unit tangent vector at the point. We now proceed to expressthe acceleration vector at a point in terms of the unit tangent and unit normal vectors at the point. From Definition17.7.2 the accelerationvector at P is given by A(f) : Dr'zR(f)
(s)
From Eq. (6) in Sec.17.8we have
+ lD' R( t) llD' r ( t) lN( f) D ,'zR (t):[D' lD' R( t) l]r ( f)
(6)
Because D t s : l D r R ( f )
(7)
if we differentiate with resPect to t, we obtain
Dr,s: DD,n(t)
(8)
Furtherrnore,
lD,R(r)P tD,R(r)llD,r(r)l: l#fr&l
t'l
Applying (7) above and Eq. (9) of Sec. 17.9 to the right side of (9), we have lD,n(f)  lD,T(f) : (Dts)'K(t)
(10)
Substitutingfrom (5), (8), and (10) into (6), we obtain A(r;  (D,rs)T(f)+ (D,s),K(f)N(f)
(11)
Equation (1L) expressesthe accelerationvector as the sum of a scalar times the unit tangent vector and a scalartimes the unit normal vector. The coefficient of T(f) is called the tangentialcomponenfof the acceleration vector and is denoted by Ar(t), whereas the coefficient of N(t) is called the normal componentof the accelerationvector and is denoted by A1s(t). So
(r2) and (13)
(14)
806
VECTORS IN THEPLANE ANDPARAMETBIC EQUATIONS Becausethe number of units in the speedof the particle at time f units is lv(f )  : DF, ArG) is the derivative of the measureof the speed of the particle and,4"(f) is the square of the measureof the speed divided by the radius of curvature. From Newton's second law of motion F:mA where F is the force vector applied to the moving object, m is the measure of the mass of the object, and A is the accelerationvector of the object. In curvilinear motion, the normal component of F is the force t or11121 to the curve necessaryto keep the object on the curve. For example, if an automobile is going around a curye at a high speed, then the normal force must have a large magnitude to keep the car on the road. Also, if the curve is sharp, the radius of curvature is a small number, and so the magnitude of the normal force must be a large number. Equation (4) indicates that the tangential component of the velocity vector is Dp and that the normal component of the velocity vector is zero. Substituting from (12) and (13) into (11), we have A(f) : Ar(f)T(f) * Ar(f)N(f)
(1s)
from which it follows that
l A ( r ) l : lAr(f)lz a lA;wP Solvin g for tAr(f ), noting from (13) that A*(t) is nonnegative, w€ have iiiii
ExAMPrn L: A particle is moving along the curye having the vector equation
R ( r ; ( t '  l ) i + ( * r '  t ) i Find each of the following vectors: V(f ), A( t) ,T (t), andN(f ). Also, find the following scalars: lV(f) l, Ar(t) , A*(t) , and K(f ). Find the particular values when t: 2. Draw a sketch showing a portion of the curye at the point where t  2, and representations of v (2), A (2), Ar(2)T (2), and Ar(2)N(2), having their initial point at t: 2.
(15)
SOLUTION:
v(r):DrR(t)  zti+ (t'L)i A(f) DN(r)  2i+zti
lv(r)l: lA(t)l:\M
m:t2+1 zt/L+t2
Dts:lv(t)l:t2+L Ar(t):D,'s2t From (16) A*(t)=
@2
r(r):ffi:hi+#i To find N(f), we use the following formula which comes from (11):
(r7)
EXERCISES REVIEW
r (t) = 2i+ zti r, (#i A(r) (D,'s) ht(1 A(f) (D,'s)r(t)
+#
1\i+ ztil
t) 1ra)
From (17) we see that N(t) is a scalar times the vector in (18). BecauseN(f) is a unit vector, N(f) can be obtained by dividing the vector in (18) by its magnitude. Thus, we have N(f): =
6 5
A(2)
4 3 2 1
v(2)
rvN(z),"
s,3)
1 3
4
5
6
7I
2t ! ': t  S z r' +'  l+t'l r+tz
K(f ) is found from (13), and we have
r((f)=ffi:& When t: 2, we obtain V(2) : 4i + 3i, A(2) : 2i + 4i, lVlZ;; : 5, Ar(2\: 4, AN(2):2, T(2) : *i + gi, N(2) : ti + €i, K(2): 4z The required sketchis shown in Fig. 77.10'1.
F i g u r e1 7 . 1 0 . 1
L7.10 Exercises In Exercises1 through 5, a particle is moving along the curve having the given vector equation. In each problem, find the vectorsV(t), A(f), T(t), and N(f), and the following scalarsfor an arbitrary value of f: lV(f) l, Ar, A*, K(f). Also find the particular values when f : fr. At f : fr, draw a sketch of a portion of the curve and representations of the vectors V(ft), A(tr), A"T(f'), and ArN(h). 3 . R ( f )  5 c o s3 f i + 5 s i n 3 f i ; h : t n 2 . R ( t ; : ( t  l ) i + t z i ;h : f I . R ( f ) : ( 2 t+ 3 ) i + ( t '  L ) i ; t r : 2 5. R(f ) : 3tzi+ 2Fi; tt: I 5 . R ( f ) : e t i * e  t i ;f r : 0 4. R(f) : cos tzi+ sintzi;tr: +\G 7. A particle is moving along the parabola A":8x and its speed is constant. Find each of the following when the particle is at (2,4): the poJitiotr vector, the velocity vector, the acceleration vector, the unit tangent vector, the unit normal vector, A7, and A11.  * : 9, such that D6, is a positive constant. Find each 8. A particle is moving along the top branch of the hyperb ola y' (4, position vector, the velocity vector, the acceleration vector, the unit 5): the of ihe fo[owing when the partid is at tangent vector, the unit normal vector, A7, arrd Ap
(ChapterL7) ReaiewExercises In Exercises1 through 12,let A:4i1. Find 38  7A 4. Find lsBl  l3cl
and C:9i  5i. 2. Find 4A + B  6C 5. Find (A  B) ' c
5i,B:i*7i,
7. Find a unit vector having the same direction as 2A + B. 8. Find the unit vectors that afe orthogonal to B.
3. Find 5B  3C 5. Find (A . B)C
808
VECTORSIN THE PLANEAND PARAMETRICEQUATIONS
9. Find scalarsh artd k such that A:
hB + kC.
10. Find the vector projection of C onto A. 11. Find the component of A in the direction of B. 12. Find cos a if a is the radian measure of the angle between A and C. In Exercises13 and 14, f.or the vectorvalued function, find (a) the domain of R; (b)
i'l 1 4 .R ( r ) l t  1 l i + l n ti
R(t);(c)DyR(t).
In Exercises15 and 16, find, equations of the horizontal and vertical tangent lines, and then draw a sketch of the graph of the given pair of parametric equations. 1 5 .r : 1 2  t 2 , y : l z t  t 3 t6.
x : T + t z , Y : T 2aF R,a 2at2
> 0
(the cissoid of Diocles)
17. Find the length of the arc of the curve R(f) : (2  t)i* pi from t: 0 to t: 3. 18. Find the length of the arc of the curve r: 3 sec d from 0 : 0 to 0: *r. t9. (a) Show that the curvedefined by the parametric equations, x: a sin t and.y: measure of the length of arc of the ellipse of part (a), ihow that
t=4
Frl2
Jo
at/t=FRl
b cos f, is an ellipse. (b) If s is the
At
where k2: (a2 b') I a' < 1. This integral is called an elliptic integral and cannot be evaluated exactly in terms of elementary functions. There are tables available that give the value of the integral in terms of k. 20. Draw a sketdr of the graph of the vector equation R(t) : eti* eti and find a cartesian equation of the graph. 2 1 . S h o w t h a t t h e c u r v a t u r e otfh e c u r v e ! : l n x a t a n y p o i n t (x,y)isxl (f*l;arz.Alsoshowthattheabsolutemaximum curvature ils2lry'i which occursat the point (+{r,^+ln2)."' 22. Find the curvature at any point of the branch of the hyperbola defined by x:a the curvature is an absolute maximum at the vertex. 23. Find the radius of curvature at any point on the curve r:
cosh t,y:b
sinh f. Also show that
a(sin f  f cos f). 24. Find the curvature, the radius of curyature, and the center of curvature of the curve : e" at the point (0, 1). A 25. Forthehypocycloidof fourcusps,r:4cosstandy:asinat,hnd,dyldxand,&yldfwithouteliminatingtheparameter. a(cos t * f sin t),y:
.26. A particle is moving alonga : 3ti + (4t  p)i. Find a cartesian equation of the curvehaving the vector equation R(t) path of the particle. Also find the velocity vector and the accelerationvector at t: l'. Find the tangential and nonnal components of the accelerationvector for the particle of Exercise25. If a particle is moving along a curve, under what conditions will the accelerationvector and the unit tangent vector have the same or opposite directions? In Exercises 29 and. 30, for the given curve find T(f) and N (f), and at t : h draw a sketch of a portion of the curve and draw the representations of T(fr) and N(fr) having initial point at t: tr.
29. R(t)  ( et+ et)i+ Zti; h:2 3 0 . R ( f )  3 ( c o st + t s i n f ) i + 3 ( s i n t  t c o s f ) i , t >
tr: tn
REVIEWEXERCISES f = 0, find parametric equations having the 31. Given the curve having parametric equations x:4t, y:*(2t*l1srz, measure of arc length sal a parameter, where arc length is measured from the point where f : 0. Check your result by using Eq. (10)of Sec.17.8 32. Find the radian measure of the angle of elevation at which a gun should be fired in order to obtain the maximum range for a given muzzle speed. 33. Find a formula for obtaining the maximum height reached by a projectile fired from a gun having a given mazAe speed of vn ftlsec and an angle of elevation of radian measure tr. 34. Find the position vector R(f) if the accelerationvector 1
A(r): fi ui andV(1) : i, and R(1) : *i + +i. 35. Proveby vectoranalysisthat the diagonalsof a parallelogrambisecteachother. : *V(AT); V(Ei) : 36. Given trianglez{BC,points D, E, and.F are on sidesAB, BC, andAC, respectively,and V(.D)
v(Al) + v(ti) + v(cE): o. *v(rZ);v(ai) :*vtc?l. Prove In Exercises37 and 38, find the velocity and acceleration vectorq, the speed, and the tangential and normal components of acceleration. 37. R(f ) : cosh zti + sinh 2fi
38. R(t; : (2 tanr s  f)i + ln(1 + t')i
39. An epicyctoidis the curve traced by a point P on the circumference of a cirde of radius b which is rolling extemally on a fixed circle of radius a. If the origin is at the center of the fixed circle,A(a,O) is one of the points at which the given point p comes in contact with the iixed circle, B is the moving point of tangency of the two circles, and the Parameter 1 ir the radian measure of the angle AOB, prove that parametric equations of the epicycloid are
x(a*b)cosfbcos+, and
y : ( a t b )s i nt  b s i n f f t
Vectors in threedimensionalspace and solid analytic geometr
NUMBER SPACE 811 18.1tr, THETHREEDIMENSIONAL 19.1 Rt, THE In Chapter 1 we discussed the number line Rl (the onedimensional THREEDIMENSIONAL number space) and the number plane R2 (the twodimensional number NUMBER SPACE space).we identified the real numbers in Rl with points on a horizontal axis and the real number pairs in R2with points in a geometric plane. In an analogous fashion, we now introduce the set of all ordered triples of real numbers. L8.1.L Definition
The set of all ordered triples of real numbers is called the threedimensional numberspace and is denoted by Rt. Each ordered triple (x, y, z) is called a point in the threedimensionalnumber sPace. To represent R3 in a geometric threedimensional spacewe consider the directed distances of a point from three mutually perpendicular planes.The planes are formed by first considering three mutually perPe,ndicular lineJwhich intersect at a point that we call the origin and denote by the letter O. These lines, called the coordinate axes,are designated as the r axis, the y axis,and the z axis.Usually the r axis and the y axis arc taken in a horizontal plane, and the z axis is vertical. A positive direction is selected on eachaxis. If the positive directions are chosen as in Fig. 18.1.1,the coordinate system is called a righthandedsystem.This terminology follows from the fact that if the right hand is placed so the thumb is pointed in the positive direction of the r axis and the index finger is pointed in the posiiive direction of the y axis, then the middle finger is pointed in the positive direction of the z axis.If the middle finger is pointed in the negative direction ofthe z axis, then the coordinate system is called lefthanded. A lefthanded system is shown in Fig. 78.1.2. In general, we use a righthanded system. The three axes determine three coordinate planes: tl]rexy plane containing the x and y axes,the xz plane containing the r and z ixes, and the yz plane containing the y and z axes.
F i g u r e1 8 . 1 . 1
Figure 18J.2
An ordered triple of real numbers (x, y, z) is associated with each point P in a geometricthreedimensional space.The directed distance of P
812
VECTORSIN THREEDIMENSIONAL SPACE AND SOLID ANALYTICGEOMETRY
v (o,2, o)
(
G , o ,o )
from the yz plane is called the x coordinate,the directed distance of p from the xz plane is called the y coordinate,and the z coordinateis the directed distance of P from the xy plane. Thesethree coordinates are called the rectangularcartesiancoordinates of the point, and there is a onetoonecorrespondence (called a rectangularcartesiancoordinatesystem)between all such ordered triples of real numbers and the points in a geometric threedimensional space. Hence, we identify R3 with the geometric threedimensional space, and we call an ordered triple (r, A, z) a point. The point (3, 2,4) is shown in Fig. 18.1.3,and the point (4, 2,S) is shown in Fig. 18.1,.4.The three coordinate planes divide the space into eight parts, called octants.The first octant is the one in which all three coordinates are positive. z A
F i g u r 1e 8 . 1 . 3 (4, 0, 0)
5) ;*tl(o,0,
(4,
F i g u r 1e 8 . 1 . 4 A line is parallel to a plane if and only if the distance from any point on the line to the plane i s the same. O ILLUSTRATION
1.: A line parallel to the yz plane, one parallel to the xz plane, and one parallel to the xy plane are shown in Fig. 18.1.5a,b, and c, respectively.
at'
aa'
Figure 18.'1 .5
AU L M B E RS P A C E 1 8 . 1 R 3 ,T H E T H R E E  D I M E N S I O NN
813
We consider all lines lying in a given plane as being parallel to the plane, in which casethe distance from any point on the line to the plane is zero. The following theorem follows immediately. 18.1.2 Theorem
,
18.1.3 Theorem
( i ) A line is parallel to the yz plane if and only if all points on the line have equal x coordinates. ( i i ) A line is parallel to the xz plane if and only if all points on the line have equal y coordinates. (iii) A line is parallel to the xy plane if and only if all points on the line have equal z coordinates.
In threedimensional space,if a line is parallel to each of two intersecting planes, it is parallel to the line of intersection of the two planes. Also, if a given line is parallel to a secondline, then the given line is parallel to any plane containing the second line. Theorem 18.1.3follows from these two facts from solid geometry and from Theorem 18.1.2. (i) A line is parallel to the r axis if and only if all points on the line have equal y coordinates and equal z coordinates. (ii) A line is parallel to they axis if and only if all points on the line have equal r coordinates and equal z coordinates. (iii) A line is parallel to the z axis if and only if all points on the line have equal r coordinates and equal y coordinates. . rLLUsrRArroN2: A line parallel to the x axis, a line parallel to the y axis, 'l'8.1.6a, b, and c, reand a line parallel to the z axis are shown in Fig. ' sPectivelY.
18.1.6 Figure The formulas for finding the directed distance from one point to another on a line parallel to a coordinate axis follow from the definition of directed distance given in Sec.1.4and are stated in the following theorem.
lII
VECTORS IN THREEDIMENSIONAL SPACEAND SOLIDANALYTIC GEOMETRY (i) If A(xr,y, z) and B(xr,y,z) are two points on a line parallelto
18.1.4 Theorem
the r axis, then the directed distancefrom Ato B, denoted by B, is given by AB:xzx1 (ii) If C(x, At, z) and D(x, Az,z) are two points on a line parallelto W V axis, then the directed distance from C to D, denoted by CD, is given by CD: Uz Ar (iii) If E(x, y, z) and F (x, y, zr) aretwo points on a line parallel to the
z axis, then the directed distance from E to F, denoted by EF, is given by EF: zz zr o rLLUsrRAtroN3: The directed distancePQ from the point P(2,5,4) to the point Q(2, 3, 4) is given by Theorem 1S.1.4(ii).We have
The following theorem gives a formula for finding the undirected distance between any two points in threedimensional space. 18.1.5 Theorem
B(xr,Ar,zrl Pt (xt , At, Zr) i..
A(x2, Uz, zt)
The undirected distance between the two points Pr(xr, yr, zr) and Pr(xr, Az,z) is given by
pRooF: Construct a rectangular parallelepiped having Pr and P, as opposite vertices and faces parallel to the coordinate planes ( see Fig. '/,8.'J,.7). By the Pythagorean theorem we have
INF;P:ITEI'+ I7F,I'
(1)
Because
lmP: lF,fl,+l$l, ,)/,/
F i g u r e1 8 . 1 . 7
Pz(x2,yr,zr)
) / ,/
(2)
we obtain, by substituting from (2) into (l) , Applying Theorem 18.1.4(i),(ii), and (iii) to the right side of (3), we obtain ffpr;z : (xr xr), I (Ar yr), * (zz zr)z So and the theorem is proved.
I
AU L M B E RS P A C E 1 8 . 1 F ' , T H E T H R E E  D I M E N S I O NN
l.: Find the undirected distance between the points P (3, 4,  1 ) a n d Q ( 2 , 5 ,  4 ) . EXAMPLE
815
soLUTroN: From Theorem 18.L.5, we have
(5  4)'
:m{gs
Note that the formula for the distance between two points in R3 is merely an extension of the corresponding formula for the distance between two points in R2 given in Theorem 14.I It is also noteworthy that the undirected distance between two points x2 and x1 in Rl is given by
lxrrrl:@ The formulas for the coordinates of the midpoint of a line segmentare derived by forming congruent triangles and proceeding in a manner analogous to the twodimensional case.These formulas are given in Theorem and the proof is left as an exercise(seeExercise15). 18.'/..6, 18.1.6 Theorem The coordinates of the midpoint of the line segment having endpoints Pr(xr,Ur,zr) and Pr(xr,Az,z2)aregiven bY
18.1.7 Definition
The graph of an equationin RBis the set of all points (x, y, z) whose coordinates are numbers satisfying the equation. The graph of an equation in RBis called a surface.One particular surface is the sphere, which is now defined.
18.L.8 Definition
A sphere is the set of all points in threedimensional space equidistant from a fixed point. The fixed point is called the centerof the sphere and the measure of the constant distance is called t}:retadius of the sphere.
18.1.9Theorem An equation of the sphere of radius r and centet at (h, k,l) is (4)
pRooF: Let the point (h, k,l) be denotedby C (seeFig. 18.1.8).The point P(x, y, z) is a point on the sphere if and only if
lcnl: t or, equivalently, @:7 Squaring on both sides of the above equation, we obtain the desired I result.
816
VECTORS IN THREEDIMENSIONAL SPACEAND SOLIDANALYTIC GEOMETRY
c(h,k,l)
If the centerof the sphereis at the origin, then L :k:l: equation of this sphere is * + y'l
0, and so an
22: 12
If we expand the terms of Eq. (a) and regroup the terms, we have * * y' * z2 Zhx  2ky  2lz * (h2+ P + 12 rz) : 0 This equation is of the form x2 + y',+ 22 + Gx * Hy * lz + I  0 P(*, v, z)
x F i g u r e1 8 . 1 . 8
(5)
where G,H,l, and/ are constants.Equation (5) is called the general formof an equation of a sphere, whereas Eq. (4) is called the centerradiusform. Becauseevery sphere has a center and a radius, its equation can be put in the centerradius form and hence the general form. It can be shown that any equation of the form (5) can be put in the form
( x  h ) , + ( y  k ) , + ( z t ) , : K
(6)
where
h*C
ktU
l:iI
K+(Gr+H2+tz41)
It is left as an exerciseto show this (see Exercise 16). If K > 0, then Eq. (5) is of the form of Eq. (4), and so the graph of the equation is a spherehaving its center at (h, k,l) and radius l'R.lt K: 0, the graph of the equation is the point (ft, k, l). This is calledapointsphere. If K < 0, the graph is the empty set becausethe sum of the squaresof three real numbers is nonnegative. We state this result as a theorem. 18.1.10 Theorem
The graph of any seconddegree equation in r, y, and z, of the form *+y,+22+Gx*Hy*tz*J:g is either a sphere,a pointsphere, or the empty set.
EXAMPLE2: Draw a sketch of the graph of the equation x2+y'+226x4y+222
solurroNt Regrouping terms and completing the squares,we have f  6x t 9 f y2  4y + 4 * z2* 2z * l: 2 * 9 + 4 + l ( r  3 ) ' + 0  2 ) 2* ( z * 1 ) 2: 1 5 So the graph is a spherehaving its centerat (3,2,1) and radius 4. A sketch of the graph is shown in Fig. 18.1.9.
F i g u r e1 8 . 1 . 9
SPACE 817 NUMtsER 18.1 R',THETHREEDIMENSIONAL
3: Find an equation of EXAMPT,n the sphere having the Points A (  5 , 5 ,  2 ) a n d B ( 9,  4 , 0 ) a s endpoints of a diameter.
solurroN: The center of the sphere will be the midpoint of the line segment AB. Let this point be C(7,y,2).By Theorem 78.'l'.6'we get 4+6a 95 ^ z=:0 : 2  L  . . L i:::z t:T: So C is the point (2,1,l).
The radius of the sphere is given by
rTherefore, from Theorem 18.L.9,an equation of the sphere is ( x  2 ) ' + Q  r ) ' ,+ ( z * r ) 2: 7 5 or, equivalently, x2 + y', + z2 4x  2y * 2z  69 : 0
18.1 Exercises parallelepiped, hawing its faces In Exercises 1 through 4, the given points A and B are opposite v,ertices of a rectangular the coordinates of the other sii find p"r*"r to the coordin"t" ptunlr. In each problem 1a)drari a sketch of the figure, 1b) vertices, (c) find the length of the diagonal AB. 2 . A ( I , 1, l ) ; B ( 3 ,4 , 2 ) 1 . A ( 0 ,0 , 0 ); B ( 7, 2 , 3 ) 4. A(2, 1, g); B(4, 0, l)
3 . A (  1 , l , 2 ) ; B ( 2 ,3 , 5 )
th9 first comer. (a) Draw a sketch 5. The vertex opposite one comer of a room is 18 ft east, 15 ft south, and 12 ft up from (c) find the coordinates of all eight vertices, joining opposite two diagonal of the length of the figure, (b) determine the vertices of the room. In Exercises 6 throu gh g,find segment joining A and B. 6. A(3, 4, 2); B(1,6,3) 5) 8. A(4, 3, 2); B(2, 3,
(a) the undirected distance between the points A and B, and (b) the midpoint
of the line
7 . A ( 2 ,  4 , 1 ) ;B ( t , 2 , 3 1 9. A(2, L,5); B(5,1, 4)
1,3) , (2, L,7) , and,(4,2,6) are the verticesof a right triangle, and find its area' 10. prove that the three points (7, of the points on this 11. A line is drawn through the point (6,4,2) perpendicular to the yz plane. Find the coordinates (0,4,0)' point the from 10 units of line at a distance 12. Solve Exercise11 if the line is drawn perpendicular to the ry plane. distancefonnula' 13. Prove that the three points (3,2,4), (6,1,2), and (12,9,5) are collinear by using the (0,3,4)' 14. Find the verticesof the triangle whose sides have midpoints at (3,2,3), (1,1,5), and 15. Prove Theorem 18.1.6' 1 5 . S h o w t h a t a n y e q u a t i o no f t h e f o r m * * y " * z 2 l G x ' l H y * l z + 1 : 0 * (z I)': X.
can be putin theform (xh\"+
In Exercises17 through 21, determine the graph of the given equation. L8. x2+ y', + 22 8y * 5z  25 : 0 17. x2+ y' + 228r * 4y * 2z 4 : 0
(yk)'
1 9 .x 2 + y ' + z , 2  6 2 * 9 : 0
818
VECTORSIN THREEDIMENSIONAL SPACE AND SOLID ANALYTICGEOMETRY
2L. x2+ y' + zz 6x * 2y  4z *L9 : 0
20. x 2 + y ' + 2 2  x  y  3 2 * Z   0
In Exercises22 through 24, find, an equation of the sphere satisfying the given conditions. 22. A diameter is the line segmenthaving endpoints at (6,2,5) and (4,0,2). 23. It is concentric with the sphere having equation f * y, * z2 2y I 8;z ) : g. 24. It containsthe points (0,0,4)', (2,1,2),and (0, 2,6) and,hasits centerin the yz plane. 25. Prove analytically that the four diagonals joining opposite vertices of a rectangular parallelepiped
bisect each other. 26. I!!: Q'4, S are four points in threedimensional space and A, B, C, and D are the midpoints of pe, T{ eR, RS,and SP, respectively, prove analytically that ABCD is a parailelogram.
18'2 VECTO_RSIN The presentation of topics in solid analytic geometry is simplified by the THREEDIMENSIoNAL SPACE use of vectors in threedimensional ,p".". irr" a"fir,itions and theolms given in Secs.17.1 and17.2 for vectors in the plane are easily extended. 18.2.1Definition
A aector in threedimensional spaceis an ord.eredtriple of real numbers (x, y, zl . The numbers tt, y, and z are called the componezlfs of the vector
(x,y, zl.
(r * ar,y * or,z *
as)
(x, v, z)
x F i g ur e 1 8 . 2 . 1
We let Vs be the set of all ordered triples (x, y, zl for which x, y, and.z are real numbers. In this chapter, a vector is always in v3 unless otfrerwise stated. Just as forvectors inv2, a vector in v, can be representedby a directed line segment. If A : (ar, a2, asl, then the directed line segmeni having its initial point at the origin and its terminal point at the polnt (ar, ar, af,)is the position representationof A. A directed line segment travinf its :{tdinitial point at (x, y, z) and its terminal point at (x * ar, y * a2,z+ ar) is also a representationof the vector A. SeeFig. 1g.2.1. The zeroaectoris the vector (0, 0,0) and is denotedby 0. Any point is a representation of the zero vector.  The magnitudeof a vector is the length of any of its representations.If the vector A: (ar, ez, fls), the magnitude of A is denotei by lAl, and it follows that
Wl: t/ar'177 t, og The directionof anonzero vector in v, is given by three angles, called the direction anglesof the vector. 18.2.2 Definition
The direction anglesof a nonzero vector are the three angles that have the smallestnonnegative radian measuresa, B, andT measuied from the positive x, A, and z axes, respectively, to the position representatio' o] the vector. The radian measure of each direction angle of a vector is greater than
SPACE 18.2VECTORSIN THREE.DIMENSIONAL
819
or equal to 0 and less than or equal to zr.The direction angleshaving radian meaiu"es ot,F, andT of the vector A: (ar, az, asl are shown in Fig' 18'2'2' In this figure the components of A are all positive numbers, and the direction angles of this .r"itot all have positive radian measure less than *2. From thl figure we see that triangle POR is a right triangle and
a1 lml loPl lAl
COSC:__:_
It can be shown that the same formula holds if. Ln = q. < 7r.Similar formulas can be found for cos B and cos 7, and we have (1) Figure18.2.2
The three numbers cos a, cos B, and cos 7 are called the directioncosinesof vector A. The zero vector has no direction angles and hence no direction cosines. . rLLUsrRArroN1: We find the magnitude and direction cosines of the vector A : (3, 2, 6>.
:!9T4+35:t/4s 7
lAl: m From Eqs. (1) we get
cosa+
cosy:+
cosF:?
If we are given the magnitude of a vector and its direction cosines,the vector is uniqud determined becausefrom (1) it follows that (2) n s : l A l c o s7 n z : l A l c o sB n r : l A l c o sa The three direction cosines of a vector are not independent of each other, as we see by the following theorem' 1'8.2.3Theorem
If cos a, cos B, and cos 7 are the direction cosines of a vector, then coszc*cos2F*cosz7:1 pRooF: If A : (a1,az,a.), then the direction cosinesof A are given by (1) and we have
y:ffi*#i.ffi d+cosz cos2 P+cos2
1
I
VECTORSIN THREEDIMENSIONAL SPACEAND SOLID ANALYTICGEOMETRY
o rLLUsrRArroN2: we verify Theorem1.8.2.3 f.orthe vector of Illustration 1. We have 'cos2 a+ coszB+ aG i:
The vector A: (ar, ar, a"l is a unit vector if lAl : t, and from Eqs. (1) we see that the comPonents of a unit vector are its direction cosines. The operations of addition, subtraction, and scalar multiplication of vectors in Vs are given definitions analogous to the corresponding definitions for vectors in V2. 18.2.4Definition IfA: bv
(ar,nz,ar)andB:(br,br,brl,thenthesumof thesevectorsisgiven
A+ EXAMPLE1: Given A  (5, 2, 6> and B  (8, 5, 4), find A + B.
(a, * br, flz * br, fls * br)
solurroN: A * B:
(5 + 8, (2) + (S),5 + (4)) : (13,7,2>.
The geometric interpretation of the sum of two vectors in v3 is similar to that for vectors in Vr. See Fig. 18.2.3.If p is the point (x, y, z), A: (a1, a2, ar) and tr a representation of A, then e is the point "? (x * ar, A * az,z * ar).Let B : (br, br,b3) and let e? be a representation of B. Then R is the point (r * (a1*br), Af @r*br), z* (ar*b")). Therefore, p? is a representation of the vector A + B, and the parallelogram law holds.
x
L8.2.5 Definition
Figure18.2.3
lf A: (ar, ar,ar), then the vector(ar,az,ar) tioe of.A, denoted by A.
is definedto be thenega
SPACE 18.2VECTORSIN THREEDIMENSIONAL
18.2.6 Definition
821
The difierenceof the two vectors A and B, denoted by A B, is defined by AB:A*
(B)
: From Definitions 18.2.5and 1'8.2.6it follows that if A (ar, ar, asl and B : ( b r , b r , b " l ,t h e n  B : (  b r ,  b 2 ,  b " l a n d A  B : \ a r  b 1 ,a 2  b 2 ,a s  b s l
ExAMPrn 2: For the vectors A and  B. B of Example 1, find A
SOLUTION:
AB(5,
4>
6 > ( 8 ,
5 , 4>
 ( 5 ,  2 , 6 >+ : (3 , 3, 10)
The difference of two vectors in Vais also interpreted geometrically as it is in vz. see Fig. 1,8.2.4.4representation of the vector A B is obtained point. by choosing representations of A and B having the same initial segment line Tilen a ,"pi"r"t t"tion of the vector A B is the directed point of from the terminal point of the representation of B to the terminal the representation of A. o rLLUsrRAuoN3: Givgl the points P(L,3,5) and Q(2,L, a)' Fig 19"2'5 shorys pi as well as OPand O0. We see from the figure that V(PQ): V(OQ)  v(OP). Hence,
x
5): (L,4, L> v(t[) : (2,1, 4> , (1, 1, 0), and (1, 1, 1) by doing the following: (a) Verify that the vectors are independent by showing that their position representations are not coplanar; (b) verify that the vectors form a basis by showing that any vector A can be written
1 8 . 3T H E D O T P R O D U C Tt N V 3
825
(6)
r ( 1, 0 , 0 ) * s ( 1, 1 , 0 ) + t ( ' l . , l , l > where r, s, and t are scalars.(c) If A : ,Fz: The scalarprojection of v(aF) onto V(Ai)
is then
v(,4Ft. v(aEr_ (4,2, 4) . (6,6, g') v36+ 36+e V(;;)  24+12+ 12
x
Figure 18.3.2
\ET
_+1 Thus,r^r): :ffi 6\ffi :3\ffi The definition of parallel vectors in V3is analogousto Definition17.3.6 for vectors inV2, and we state it formally. 18.3.5 Definition
Two vectors in V3 are said to be parallel if. and only if one of the vectors is a scalar multiple of the other.
828
SPACEAND SOLID ANALYTICGEOMETRY VECTORSIN THREEDIMENSIONAL
The following theorem follows from Definition 18.3.5and Theorem 18.3.4.Its proof is left as an exercise(seeExercise16). 18.3.6 Theorem
Two nonzero vectors in Vs are parallel if and only if the radian measure of the angle between them is 0 or zr. The following definition of orthogonal vectors in Vr corresponds to Definition 17.3.7f.orvectors in Vr.
18.3.7 Definition
If A and B are two vectors in Vs, A and B are said to be orthogonal if and onlvifA.B:0.
ExAMPln3: Prove by using vectorsthat the pointsA(4,9 , L) , B (  2 , 5 , 3 ) , a n d C ( 6, 3 ,  2 ) a r e the vertices of a right triangle.
solurroN: Triangle CAB is shown in Fig. 18.3.3.From the figure it looks as if the angle at A is the one that may be a right angle. We shall find
B( 2,6,9)
V(AB) and V(AC), and if the dot product of thesetwo vectorsis zero,the angle is a right angle. +
+
v(AB)v(oB)v(oA)  (  2 , 6 , 3 ) ( 4 , 9 , ' ] . , > : (_5,_3, 2>
v(A?)v(oa)v(o?) A(4, 9, L)
 ( 6 , 3 ,  2 ) < 4 , 9 , ' j . , > : (2,5, 3)
C(6, g,  z) F i g u r e1 8 . 3 . 3
v ( f r ) . v ( f r )  F 6 ,  g ,2 >. < 2 ,  6 ,  g)  1 2+ Therefore;V(AB) and V(;E) angle CAB is a right angle.
L8 6  0 are orthogonal,and so the angle at A in tri
18.3 Exercises 1. Prove Theorem 18.3.2(i).
2. Prove Theorem 18.3.2(ii).
3. Prove Theorem 18.3.2(iii).
4. Prove Theorem 18.3.2(iv)and (v).
InExercises5through14,letA:(4,2,4r,8:, from (2) and the above equation we obtain a ( x  x s )+ a Q  A ) * c ( z  z o ): 0 which is the desired equation.
F i g u r e1 8 . 4 . 1
ExAMPLEL: Find an equation of the plane containing the point (2, 1, 3)and having 3i  4i + k as a normal vector.
(3) I
sot,urroN: Using (1) with the point (xo, Uo,zo) (2,I, 3) and the vector (a, b, cl : (3,4,'].>,we have as an equation of the required plane
3(x2)4(v1)+
(z 3):0
or/ equivalently, 3x4y*z5:0 L8.4.3 Theorem
If. a, b, and c are not all zero, the graph of an equation of the form ax*by * cz* d:0
(4)
is a plane and (a, b, cl is a normal vector to the plane. PRooF: Supposethatb * 0. Then the point (0,dlb,0) is on the graph of the equation becauseits coordinates satisfy the equation. The given equation can be written as /s\ a(x o)+ b \a + t) + c(z o): 6 which from Theorem 18.4.2is an equation of a plane through the point (0, dlb,0) and for which (a, b, cl is a normal vector.This provesthe theorem if b # 0. A similar argument holds if b:0 and either a * 0 or c#0. r Equations ( 1) and (4) are calledcartesianequationsof a plane.Equa
18.4 PLANES
831
tion (1) is analogousto the pointslope form of an equation of a line in two dimensions. Equation (4) is the general firstdegree equation in three variables and it is called a linear equation. A plane is determined by three noncollinear points, by a line and a point not on the line, by two intersecting lines, or by two parallel lines. To draw a sketch of a plane from its equation, it is convenient to find the points at which the plane intersects each of the coordinate axes. The r coordinate of the point at which the plane intersectsthe r axis is called the x interceptof the plane; the y coordinate of the point at which the plane intersectsthe y axis is called the y interceptof the plane; and the z interceptof the plane is the z coordinate of the point at which the plane intersectsthe z axis. . rLLUsrRArroN 1: We wish to draw a sketch of the plane having the equation 2 x l 4 Y * 3 z : 8 Figure 18.4.2
By substituting zero for y and z, we obtain .x: 4; so the r intercept of the plane is 4. In a similar manner we obtain the y intercept and the z intercept, which are 2 and $, respectively. Plotting the points corresponding to theseinterceptsand connectingthem with lines, we have the sketch of the plane shown in Fig. 18.4.2.Note that only a portion of the . plane is shown in the figure. I rLLUSrRArrox2: To draw a sketch of the plane having the equation 3x*2Y62:0
F i g u r e1 8 . 4 . 3
Figure 18.4.4
we first notice that because the equation is satisfied when r, y, and z are all zero, the plane intersects each of the axes at the origin. If we set r: 0 in the given equation, we obtain y  32: 0, which is a line in the yz plane; this is the line of intersection of the yz plane with the given plane. Similarly, the line of intersection of the xz plane with the given Drawing a plane is obtained by setting A:0, and we get r22:0. sketch of each of these two lines and drawing a line segment from a point on one of the lines to a point on the other line, we obtain Fig. 18.4.3. . In Illustration 2 the line in the yz plane and the line in the xz plane used to draw the sketch of the plane are called the tracesof the given plane in the yz plane and the xz plane, respectively. The equation r : 0 is an equation of the yz plane becausethe point (x, y , z) is in the yz plane if and only if r : 0. Similarly, the equations y : 0 and z:0 areequationsof the xz plane and the xy plane, respectively. A plane parallel to the yz plane has an equation of the for:n x: k, where k is a constant. Figure 18.4.4shows a sketch of the plane having the equation x: 3. A plane parallel to the xzplarrehas an equation of the form A : k, and a plane parallel to the xy plane has an equation of the form
832
VECTORSIN THREEDIMENSIONAL SPACEAND SOLID ANALYTICGEOMETRY
z: k. Figures 1.8.4.5and 18.4.6show sketchesof the planes having the equationsy: 5 and z 6, respectively.
F i g u r e1 8 . 4 . 5
18'4'4 Definition
F i g u r e1 8 . 4 . 6
The anglebetweentwo planesis defined to be the angle between the nonhal
vectors of the two planes.
EXAMPTB 2: Find the radian measure of the angle between the two planes5x 2y * 5z l2:0 and 2x*y72* 11:0.
solurroN: Let N1 be a normal vector to the first plane and N1 : Si2i * 5k. Let N, be a normal vector to the second plane and N2 : 2i +  7k. i By Definition 78.4.4the angle between the two planes is the angle between N1 and N2, and so by Theorem 18.3.4if d is the radian measureof this angle, c o so 
N' 'N' :
lN,llN,l ffiffi
27
1
542
Therefore,
o &n 18'4'5 Definition
Two planes are parallel if and only if their normal vectors are parallel. From Definitions 18.4.5 and 18.3.5, it follows that if we have two planes with equations arx*bry+cfi+dt:0
nzx+ bzA* c2z* dr 0
18.4 PLANES
and normal vectorsNr : (4r, bt, ct) and Nz: (az, br, crl, respectively,then the two planes are parallel if and only if N1 : kN2
where k is a constant
Figure 18.4.7shows sketchesof two parallel planes and representations of some of their normal vectors.
L8.4.6 Definition
Two planes are perpendicularif and only if their normal vectors are orthogonal. From Definitions L8.4.6and 18.3.7it follows that two planes having normal vectors Nr and N, are perpendicular if and only if N1'N2:0
3: Find an equation of ExAMPLE the plane pe{pendicular to eachof the planesx y + z  0 and 2x* Y  4z 5  0 and containing the point (4, 0 , 2) .
Q)
soLUrIoN: Let Mbe the required plane and (a,b, cl be a normal vectorof M.Let Mr be the plane having equation te y + z: 0.By Theorem18.4.3a normal vector of.Mr is . Because (4, 0 , 2) is a point in M, it follows from Theorem "1,8.4.2 that an equation ofMis
a ( x 4 ) + z a ( y 0 ) * a ( z+ 2 ) or, equivalently, x*2Y*z2:0 Consider now the plane having the equation ax * by + d: 0 and the xy plane whose equation is z:0. Normal vectors to these planes are (a, b,0) and (0, O, 1.>,respectively. Because(a, b,0) ' (0, 0, 1):0, the two planes are perpendicular. This means that a plane having an equation with no z term is perpendicular to the xy plane. Figure 18.4.8 illustrates this. In a similar manner, we can conclude that a plane having an equation with no r term is perpendicular to the yz plane (see Fig. 1'8.4.9),and a plane having an equation with no y term is perpendicular to the xz plane (see Fig. 18.4.10). An important application of the use of vectors is in finding the undirected distance from a plane to a point. The following example illustrates this.
F i g u r e1 8 . 4 . 8
Figure 18.4.9 EXAMPI^.S4:
Find the distance
from the plane 2x  y + 2z * I0  0 to the point (l , 4, 5) .
F i g u r e1 8 . 4 . 1 0
soLUrIoN: Let P be the point ('1,,4,5) and choose any point Q in the plane. For simplicity, choosethe point Q as the point where the plane intersectsthe r axis, that is, the point (5, 0, 0). The vector having QP as a representation is given by
v(a?):6ir 4i+ 6k A normal vector to the given plane is
N:2ii+zk
18.4 PLANES P ( 1 ,4 , 6 )
The negative of N is also a normal vector to the given plane and N: 2i+ i 2k We are not certain which of the two vectors, N orN, makes the smaller angle with vector V(QIF). Let N' be the one of the two vectors N or N
5r0,0)
which makes an angle of radian measure 0 * 
(5)
k
+2
y?z+&
1 8 . 5 L I N E SI N R 3
from which we get 2x3Y*42*3:0
x2
1 For the line of intersectionof these two planes, find (a) a set of sym or, equivalently, metric equations and (b) a set of y+ x2 parametric equations. 3 2
z0
which is a set of symmetric equations of the line. A set of parametric equations can be obtained by setting eachof the above ratios equal to f, and we have
y:&+2t
x:2:3t ExAMPrr 3: Find the direction cosines of a vector whose rePresentations are parallel to the line of Example 2.
4; Find equations of the ExAMPLE line through the Point (L, t , !) , perpendicular to the line
solurroN: From the symmetric equations of the line in Example 2, we see that a set of direction numbers of the line is [3,2,3]. Therefore, the vector (g,2,3) is a vector whose representationsare parallel to the line. The direction cosines of this vector are as follows: cos a:3llD'
cosY : 3l\8.
cosF:21 \8,
solurroN: Let la, b, cl be a set of direction numbers of the required line. The equations 3r : 2y : z can be written as
x0
y0
x*Yz:0
z0
1 2
\x:Zyz and parallel to the Plane
z:3t
which ares;rmmetricequationsof a line. A setof directionnumbersof this line is lt,t, tl. Becaus6the required line is perpendicularto this line, it follows that the vectors(a, b, cl and (*, *, 1) are orthogonal'So
( a , b , c > ' ( + ,1+),: 0 or, equivalently,
ta**b+c:0
(5)
Anormal vector to the plane r * y z:0 is ' ( L , l ,  1 )  0 or, equivalently, a+bc0
(7)
Solving Eqs. (6) and (7) simultaneously for a and b in terms of c we get fl : gc and b  8c. The required line then has the set of direction
SPACEAND SOLID ANALYTICGEOMETRY VECTORSIN THREE.DIMENSIONAL
numbers l9c, 8c , cl and contains the point (1, metric equations of the line are x  " 1 ._ y + I %8cc
1). Therefore,sym
_rL
or, equivalently, xl 98
 . t 
ExAMPrn 5: If I, is the line through A(L, 2,7) and B(2,3, 4) and lrts the line through C (2, 'J,, 4) and D (5 , 7 , 3) , prove that I, and 12 are skew lines (i.e., they do not lie in one plane).
u*l
zL
sorurroN: To show that two lines do not lie in one plane we demonstrate that they do not intersect and are not parallel. Parametric equations of a line are x
xs* ta
yyo+tb
z zs* tc
where.[a,b, cf is a setof directionnumbersof theline and (16,Ao, zo)is any point on the line. BecauseVta7l: (3, l, !'!,>, a set of direction numbersof It is l3, 1, 111. TakingA as the point Po,we haveas para
metric equations of h xl3t
y2+t
z7ILt
(8)
Becausev(CD)  (3,8, 7>, and Iz containsthe point C, we have as parametric equations of lz x2*3s
y1
*8s
z47s
(e)
Becausethe setsof direction numbers are not proportional, l, andl2are not parallel. For the lines to intersect, there have to be a value of f and a value of s which give the samepoint (xr, Au zr) in both setsof Eqs. (8) and (9). Therefore, we equate the right sides of the respective equations and obtain l3t:2*3s 2* t: 1* 7 llt:47s
8s
Solving the first two equations simultaneously, we obtain s: f7 and t: tl.This set of values does not satisfy the third equation; hence, the two lines do not intersect. Thus, l, andl, are skew lines.
18.5 Exercises In Exercises 1 throu gh 5, find parametric and symmetric equations for the line satisfying the given conditions. L. Through the two points (I,2, t) and (5,  L , l ) .
1 8 . 5 L I N E SI N R 3
841
2. Through the point (5,3,2) with directionnumbers 14,t, 1]. 3 . Through the point (4,5,20) and perpendicularto the plane x*3y  5z8  0 . 4. Through the origin and pe{pendicular to the lines having direction numbers 1 4 , 2 , " 1a. n 1 d[  3 ,  2 , l f . 5 . Through the origin and perpendicular to the line i@  L0)  *y  tz at their intersection. 5 . Through the point (2,0,4)
and parallelto eachof the planes2x* y  z:0
and r * 3y * 5z:0.
7. Show that the lines
ry
 a * = 4: ' = 2 a n d ) c 2 35
y+1,4
253
z8 3
are coincident. 8. Prove that the line r * t:
t(y  6) : z lies in the plane 3x * y  z: 3.
The planes through a line which are perpendicular to the coordinate planes are called the projecting planesof the line. In Exercises9 through 72, find equations of the projecting planes of the given line and draw a sketch of the line. 1 0 .x * y  3 2 * ' 1 . : 0 2x y 32*'l'4:0
9.3x2yI5z30:0 2 x * 3 y  1 0 2 6 : 0
72.2xytz7:0 4x y *32 73:0
1 7 .x  2 y  3 2 * 6 : 0 x I y * zl:0
*4, and x:y*7, 13. Find the cosine of the measureof the smallestangle between the two lines r:2yl4,z:y 2y:y12. 14. Find an equation of the plane containing the point (6,2,4) and the line *(r  1) : t@ + Z) : i(z 3)In Exercises15 and 16,find an equation of the plane containing the given intersecting lines.
) c  Z   4y: + 3
1 L6 v'4L3
z+2
 (3x+2ylz+20 and )cy+22L:o t
x u2 r5.;:T:
x z"1, and L 1
y2 L
z1 1.
L7. Show that the lines
FxYz:0 _0 L8x2y32*t
and
[x3y+z+3:0 L3ryz+5:0
are parallel and find an equation of the plane determined by these lines. 1 8 . F i n d e q u a t i o n s o ft h e l i n e t h r o u g h t h e p o i n t ( 1 ,  1 , 1 ) , p e r p e n d i c u l a r t o t h e l i n e S x : 2 y : z , a n d . p a r a l l e l t o t h e p l a n e x+yz:0. 19. Find equationsof thelinethroughthepoint (3,5,4),intersectingthezaxis,andparalleltotheplanex 3y *526:020. Find equations of ttre line through the origin, perpendicular to the line x : A. 5 , z : 2y  S, u;iintersecting the line !:2xt1'z:x*2. 21,. Find the perpendiculardistancefrom the point (1,3, 1) to the line x22:7,Y:1. 22. Find the perpendicular distance from the origin to the line
x:2ltt
y:7it
z:4*trt
23. Prove that the lines
!=:L:Jl: 52orJ2
un4'2:y+:ry
are skew lines. 24. Find equations of the line through the point (3, 4, 5) which intersects each of the skew lines of Exercise23.
842
SPACEAND SOLID ANALYTICGEOMETRY VECTORSIN THREEDIMENSIONAL
Let A and B be two nonparallel vectors. Representations of these two vectors having the same initial point determine a plane as shown in Fig. 18.5.1.We show that a vector whose representations are perpendicular to this plane is given by the vector operation called the "cross product" of the two vectors A and B. The cross product is a vector operation for vectors in Vr that we did not have for vectors in Vr. We first define this operation and then consider its algebraic and geometric properties. F i g u r e1 8 . 6 . 1
L8.6.LDefinition
If A : (ar.,ar, a3) and B : (bt, br, b"l , then the crossproducfof A and B, denoted by A x B, is given by  atbs, fltbz 
arbtl
(1)
Becausethe cross product of two vectors is a vector, the cross product also is called the aectorproduct.The operation of obtaining the crossproduct is called aectormultiplication. . rLLUsrRArroul.: If A : (2, 1,,3, and B : (J,1,4), then from Definition 18.6.1we have A X B : (2, 1, 3) X 0. Therefore, from Eq. (5), we get
I n x n l : l A l l Bsl i n0
r
We considernow a geometricinterpretationof lA x Bl. Let p? be a representationof A and let p? f" a representationof B. Then the angle betweenthe vectorsA and B is the angle at P in triangleRPQ (seeFig. 18.6.3).Let the radianmeasureof this anglebe 0.Therefore,the numberof squareunits in the areaof the parallelogram having pT anaFO ur adjacent sidesis lAllnl sin 0 becausethe altitude of the parallelogramhas iength lBl sin 0 units and the lengthof the baseis lAl units. Sofrom Eq. (g) it follows that lA x Bl squareunits is the areaof this parallelogram.
F i g u r e1 8 . 6 . 3
EXAMrLE 1: Show that the quadrilateral having vertices at
P(1, 2, 3), Q(4,3, r) , R(2,2, 1,),andS(5, 7,3) is a
parallelogram and find its area.
solurroN:
Figure 18.6.4shows the quadrilateral peSR.
v ( P ? ): ( 4  ' t , s  (  2 ) , (  t )  3 ) : ( s , s ,  4 ; v ( F f t :) ( 2  ' ! . , 2  (  2 ), t  3 ): ( t , 4 ,  2 > v ( n  3 ) :( 5  2 , 7 2 ,  g  1 ) : ( 3 ,s ,  4 1 v(0) : (5 4,7 s,g  (1)) : (t, 4, 2l : V(n3) andV(P?) : V(Q*S),itfollows thatPQisparalBecauseV(FQ) lel to R3 and P? is parallel to Q3. Therefore,PQSRis a parallelogram. Let A: v(pT) and B: V(P!), then A x B : (i + 4i 2k) x (3i + si 4k)
/
b, R(2, z, !)
(4, 3,  L)
Figure18.6.4
(6)
S(s,7,  g)
:3(i x i) +5(i x 4(i x k) +72(ix i) +20(ix i) i)  t i ( i x k )  5 ( k x i )  1 0 ( kx + 8 ( k x k ) i) : 3(0) + s(k)  4(i) + 12(k) + 20(0) 16(i)  5(i)  10(i) + 8(0)
18.6 CROSSPRODUCT
847
2i7k Hence,
lexBl:m\@ The area of the parallelogram is therefore \@
square units.
The following theorem, which gives a method for determining if two vectors in V3 are parallel, follows from Theorem 18.6.5. 1g.6.7Theorem If A and B aretwovectorsinVg,AandB areparallelifand onlyif AxB:0' pRooF: If either A or B is the zero vector, then from Theorem 1'8.6.2, A x B:0. Becausethe zero vector is parallel to any vector, the theorem holds. If neither A nor B is the zero vector, lAl + O and lBl # 0. Therefore, from Eq. (3), lA x Bl :0 if and only if sin 0:0. BecauselA x Bl : 0 if and only if A X B : 0 and sin 0 : 0 (0' 0 = r) if and only if 0 : 0 ot r, we can conclude that AxB:0
if andonlyif 0:0otr
However, from Theorem 18.3.6, two nonzero vectors are parallel if and only if the radian measure of the angle between the two vectors is 0 or zr. I Thus, the theorem follows. The product A ' (B x C) is called the triple scalar product of the vectors A, B, and C. Actually, the Parenthesesare not needed because A ' B i s a s c a l a r / a n d therefore A ' B X C can be interpreted only in one way. 18.6.8 Theorem
If A, B, and C are vectors in Vs, then (7) Theorem 18.6.8 can be proved by letting A: (at, az, ag), : (br, br, b"l , and C : (cr, c2, ca) and then by showing that the number B on the left side of (7) is the same as the number on the right. The details are left as an exercise (see Exercise 17). 1, 21, B: 11rLLUSrRArroN 5: We verify Theorem 18.6'8 if A: (1,
( 3 , 4 , 2 > ,a n dC  (  5 , 1 ,  4 > . B x c (3i+ 4i 2k) x (5i +i4k)
 3k  12(i)  zo(k)  L6i+ loi  2(i) 1,4i+22i+23k
IN THREEDIMENSIONAL VECTORS SPACEAND SOLIDANALYTIC GEOMETRY
A . ( Bx C ): ( L ,  1 , 2 , . (  L 4 , 2 2 , 2 3 > :  L 4  2 2 + M :L0 A x B  ( i  i + 2 k ) x ( 3 i+ 4 i  2 k ) :4k2(i)3(k) +2i+6i +8 (i) :6i+8i +7k (A x B) . C: (5,8,7>. (5,1,4> :30+828 :L0 This verifies the theoremfor thesethreevectors. L8.5.9 Theorem
If A and B are two vectors in 7r, then the vector A x B is orthogonal to both A and B. PRooF: From Theorem 18.6.8we have
A.AXB_AXA.B From Theorem18.5.2(i), AxA:0. we have
Therefore, from the above equation
A'AxB:0.8:0 Becausethe dot product of A and A x B is zero, it follows from Definition 18.3.7that A and A x B are orthogonal. We also have from Theorem 18.6.8that
AxB.B:A.BxB Again applytt g Theorem 1,8.6.2(i),we get B x B : 0, and so from the above equation we have AxB.B:A.0:0 Therefore, since the dot product of A X B and B is zero, Ax B and B are orthogonal and the theorem is proved. I From Theorem 18.5.9we can conclude that if representations of the vectors A, B, and A X B have the same initial point, then the representation of A X B is perpendicular to the plane formed by the representations of A and B. EXAMPTu 2: Given the points P(1, 2, 3) , Q(2,1,0), and R(0, 5, L), find a unit vector
solurroN: Let A  V(p0) and g  v(p?). Then A  (2 (1),1 (2), 0  (s)) : (1 ,3,3>
18.6 CROSSPRODUCT
whose representations are Perpendicular to the plane through the points P, Q,,and R.
B (0 (1),5 (2),r (3)): (L,7,4> p0 and Pl, The plane through P, Q, and'R is the plane formed by any Therefore, and B. A vectors which are, respectively, representationsof plane. to this x perpendicular representation of the vector A B is
A x B : (i + 3i + 3k) x ( i+ 7i + 4k):  9i+7i 10k The desired vector is a unit vector parallel to A X B. To find this unit vector we apply Theorem 18.2.9and divide A x B by lA x Bl, and we obtain 7 9_rAXB _ ri__10
lffiI:n5r+
ExAMPln3: Find an equation of the plane through the Points P ( 1 ,3 , 2 ) , Q ( 3 ,  2 , 2 ) , a n d R(2,L,3).
1\/mt16x
solurroN: v(6) : i * 3i + k and v(PR) : i ?i + k. A normal vectorto the requiredplaneis the crossproductVtp?) x V(p?)' which is (i+ 3i + k) x (i 2i+ k) : si + 2i  k Soif Po: (1,3,2) and N: (5, 2,ll, equation of the required Plane
from Theorem18'4'2we have as an
5 (r 1 ) + z( V 3)  ( z 2) : 0 or, equivalently, 5x*2Yz9:0 A geometric interpretation of the triple scalar product is obtained by consideringa parallelepipedhaving edgesPn,Frt.,and fi, and letting A: V(P?), B:V(PR), and C:V(PS). SeeFig'18'6'5'The vectorA X B is pA. fne vector(A X B) is also a a normal vector to the plane of p0 and normal vector to this plane. we are not certain which of the two vectors,
PA F i g u r e1 8 . 6 . 5
(AxB)or_(AxB),makesthesmalleranglewithC.LetNbetheoneof radian ti\e two vectors (A x B) or (A x B) that makes an angle of their having C N and of representations the Then < with C. measure0 tn FR as shown F0 plane of of the initial points at P are on the same side "",{ in Fig.'18.5.5.The area of the base of the paralletepiped i9 lA X Bl square unitsl If h units is the length of the altitude of the parallelepiped, and if V cubic units is the volume of the parallelepiped,
vlAxBlh
(8)
"l'8.3.4,N . C _ Considernow the dot product N ' c. By Theorem B e c a u sNe is either l N l l g l c o s0 . B u th  l C lc o s0 , a n ds oN ' C : _ l N l h . (A Hence,we have x nl. x B), it followsthat lNl lR. tA r b) ot (e) N.e:lAxBlh
VECTORSIN THREE.DIMENSIONAL SPACEAND SOLID ANALYTICGEOMETRY
Comparing Eqs. (S) and (9) we have N.C:V It follows that the measure of the volume of the parallelepiped is either (A x B) ' c or  (A x B) ' c; that is, the measure of the volume of the parallelepiped is the absolute value of the triple scalarproduct A X B . c. ExAMPLE4: Find the volume of the parallelepiped having vertices P ( 5, 4 , 5 ) , Q ( 4 , 1 , 0 ,6 ) , R ( ' ! . 9 , ,7) , a n d S ( 2 , 6 , 9 ) a n d e d g e s& , , P ? ,
solurroN: Figure 18.6.6shows the parallelepiped.Let A: V(pQ) : (  1 , 6 , 1 ) , B : V t p ? ) : (  4 , 4 , 2 ) , a n dC, : V ( p 3 ): (  3 , 2 , 4 ) . T h e n A x B : (i + 6i + k) x (4i + 4i +2k) : 8i 2i+ 20k
andu;
Therefore, ( A x B ) . C : ( 8 , 2 , 2 0 > . (  9 , 2 , 4 l :  2 4 
4 * g0:52
Thus,the volume is 52 cubic units. s (2,6,9)
R (1,9,7)
Q @ , T O6, ) P (5, 4, 5)
F i g u r e1 8 . 6 . 6
EXAMPLE5: Find the distance between the two skew lines, /, and 12,of Example 5 in Sec. 18.5. N
solurroN: BecauseIt and /, are skew lines, there are parallel planes p, and Pt containing the lines l, and 12,respectively.see Fig.1,g.6.7. Let d
units be the distancebetweenplanesp, and pr. th" distaice between/, andl, is also d units. A normarvectorto the two planesis N: vterti x v(cD). Let u be a unit normarvector in the direction of N. Then
x v(ci; u: v(aB) x v(6)  lv(aB) Now we taketwo points,onein eachplane(e.g.,BandC). Thenthe scalar projectionof V(CT) on N is V(CTI . U, and
d: lv(c?) ul: Ittcll . v(AE)x v(cB) I
lv(AB)x v(cD)  I
Performing the computations required, we have
v(rt):3i+illk F i g u r e1 8 . 6 . 7
v(cB)3i+Birk
18.6 CROSS PRODUCT
851
kl rrl: zr(3i 2i k) 7 1
?izik
u
\n,
Finally,vtdBl : 4i + 4i 8k, and so
tI: # dlv(c?)'
lD 8+8: h:
+ ffi
1'8.6 Exercises
 1 ) , C : (  5 ,  3 , 5 ) , p  (  2 , L , 5 ) , E < 4 ' O '  7 > ' a n d F t, ihe cross as increase ellipse of the semiaxes 2: k is an ellipse and the lengths of the lkl increases. The cross sectionsof the surfacein the planes x: k arethe hyperbolas ,z1sz y2lb2: l. * k2 a2whose transverseaxesare parallel to thez axis' In a : similaifashion, the cross sections in the planes A k arc the hyperbolas axesare also parthe transverse given by z2lc2 f la2: 1 * Plbz forwhich allel to the z axis. If.a : b, the surfaceis a hyperboloid of revolution in which the axis is the line containing the transverse axis of the hyperbola' Each of the above three quadric surfacesis symmetric with respect to each of the coordinate planes and symmetric with respect to the origin' Their graphs are called ienttal quadricsand their center is at the origin. The graph of any equation of the form
Figure 18.8.3
+t*Y'*z' a2
b2
1 c2
where a, b, and c ate Positiv€, is a central quadric.
nls.rvrptE L: Dra* asketch of the graph of the equation
4x2y'+
soLUTIoN: The given equation can be written as
100
and name the surface.
which is of the form of Eq. (2) with y and z interchanged. Hence, the sur
SPACEAND SOLID ANALYTICGEOMETRY VECTORSIN THREEDIMENSIONAL
face is an elliptic hyperboloid of one sheet whose axis is the y axis. The cross sectionsin the planesy : k arethe ellipses*125 + z2l4:  + k21700. The cioss sectionsin the planes x: k are the hyperbolasz2l4 y'1100: lk2125, and the cross sections in the planes z:k anethp hyperbolas fl25  y21700: !  kzl4. A sketch of the surfaceis shown in Fig. 18.8.4.
F i g u r e1 8 . 8 . 4
ExAMPLE2: Draw a sketch of the graph of the equation 4x225y222:100 and name the surface.
solurroN:
The given equation can be written as
!t,'l_:1. 25 4 100 which is of the form of Eq. (3) with r and z interchanged; thus, the surface is an elliptic hyperboloid of two sheetswhose axis is the r axis. The cross sections in the planes x:k, where lkl >5, are the ellipses yzl4  1. The planes x = k, where * 221100:k2125 lkl < 5, do not intersectthe surface. The cross sections in the planes A:k are'the hyperbolas fl2  221100:1* k214,and the cross sections in the planes z:k are the hyperbolasf 125 U'14:1 + k,/100.The required sketchis shown in Fig. 18.8.5.
F i g u r e1 8 . 8 . 5
18.8QUADRIC SURFACES The following two quadrics are called noncentral quadrics. The elliptic p araboloid
fuzz
7* w:;
F i g u r e1 8 . 8 . 6
G)
where a and b are positive and c * 0. Figure 18.8.5 shows the surface if c>0. Substituting k for z in Eq. (4), we obtain *laz * y'lb' : k/c. When k: 0, this equation becomes *laz * y2lb2 0, which represents a single point, the origin. If. k + 0 and k and c have the same sign, the equation is that of an ellipse. So we conclude that cross sections of the surface in the planes z: k,where k and c have the samesign, are ellipses and the lengths of the semiaxesincreaseas lkl increases.If k and c have opposite signs, the planes z: k do not intersect the surface.The cross sections of the surface with the planes x: k and y : k arc parabolas' When c ) 0, the parabolas open upward, as shown in Fig. 18.8'6; when c < 0, the parabolas open downward. lI a: b, the surface is a paraboloid of revolution. The hyperbolicparaboloid u2   :  )c2 z
(s)
b2n2c
F i g u r e1 8 . 8 . 7
ExAMPLE3: Draw a sketch of the graph of the equation
where a and b are positive and c * 0. The surface is shown in Fig. 18.8.7 forc>0. The cross sections of the surface in the planes 2: ft, where k # 0, ate hyperbolas having their transverse €xes parallel to the y axis if k and c v t the samesign and parallel to the r axis if k and c have opposite signs. ""e The cross section of the surface in the plane z: 0 consists of two straight lines through the origin. The crosssectionsin the planesx: k ateparabolas opening upward if c > 0 and opening downward if c < 0. The cross sectionsin the pl".t"t y: k areparabolas opening downward if c ) 0 and opening upward if c < 0.
solurroN:
The given equation can be written as
3y'*1222:"1.6x and name the surface.
which is of the form of Eq. (4) with r and z interchanged. Hence, the graph of the equation is an elliptic paraboloid whose axis is the r axis. The f crois sections ir, the planes x: k ) 0 are the ellipses y2lL6 zz 4: kl3, and the planesx: k 10 do not intersectthe surface.The crosssectionsin the planes y : k are the parabolas L2z2:16x  3k', and the cross sections in
SPACEAND SOLID ANALYTICGEOMETRY VECTORSIN THREEDIMENSIONAL
the planesz : k arethe parabolas3yz: L5x Ue. A sketchof the elliptic paraboloidis shov,rnin Fig. 18.8.8.
F i g u r e1 8 . 8 . 8
EXAMPLE4: Draw a sketch of the graph of the equation 3y'
1222: l6x
and name the surface.
solurroN:
Writing the given equation as
tlz   :  22 x 1643 we see it is of the form of Eq. (5) with x artd z interchanged. The surfaceis therefore a hyperbolic paraboloid. The cross sections in the planes x: k * 0 are the hyperbolas y'lL5  z2l4: k/3. The cross section in the yz plane (r: 0) consistsof the two lines y : h ffid y  2z.In the planes z: k, the cross sections are the parabolas 3A, : L6x * l2t9; in the planes A : k, the clrosssections are the parabolas l2z2 : 3k2 t6x. Figure 1g.g.9 shows a sketch of the hyperbolic paraboloid.
z
Figure18.8.9
The elliptic cone
5.#$:o
F i g u r e1 8 . 8 . 1 0
(6)
where a, b, arrdc are positive (see Fig. 18.8.10). The intersection of the plane z: 0 with the surface is a single point, the origin. The cross sections of the surface in the planes z: k, *hete k * 0, are ellipses, and the lengths of the semiaxesincreaseas k increases. Cnrsssectionsin the planes x: 0 and,y: 0 are pairs of intersecting lines. In the planes x: k arrd U: k, where k * 0, the cross sections a"e hyperbolas.
18.8 QUADRICSURFACES
ExAMPLE5: Draw a sketch of the graph of the equation
soLUrIoN: x2
4xz  yz * 2522 0 and name the surface.
The given equation can be written as
25 which is of the form of Eq. (6) with y and z interchanged. Therefore, the surface is an elliptic cone having the y axis as its axis. The surface intersectsthe xz plane (y : 0) at the origin only. The intersection of the surface with the yz plane (r : 0) is the pair of intersecting lines U : +52, and the intersection with the xy plarre (z: 0) is the pair of intersecting lines y:+zx. The cross sections in the planes A:k+ 0 are the ellipses f 125+ z2l4:WlL00.In the planesx:k * 0 and z:k # 0, the crosssections are, respectively, the hyperbolas y2lt00z'14:l&125 and y21100  f IZS: k214.A sketch of the surfaceis shown in Fig. 18.8.11.
F i g u r e1 8 . 8 . 1 1
The general equation of the second degree in x, y, and z is of the form af t byz * czz* dxy * exz + fyz + gx * hy * iz * i : g where a,b, . . . , j areconstants. It can be shown that by translation and rotation of the threedimensional coordinate axes (the study of which is beyond the scope of this book) this equation can be reduced to one of the following two forms: Axz*By',*Czz+l0
(7)
A* + Byz* lz:0
(8)
or Graphs of the equations of the seconddegree will either be one of the above six types of quadrics or else will degenerateinto a cylinder, plane, line, point, or the empty set. The nondegenerate cur:vesassociatedwith equations of the form (7) are the central quadrics and the elliptic cone, whereas those associated with equations of the form (8) are the noncentral quadrics. Following are examples of some degeneratecases: f  y ' : 0 ; t w o p l a n e sx,  Y : 0 a n d r * Y = o
ANDSOLIDANALYTIC GEOMETRY IN THREEDIMENSIONAL SPACE VECTORS
864
z2:0;
one plane, the ry plane
* * Y': 0; one line' the z axis f * y'l z2:0; a single Point, the origin f t y' I z?* 1 : 0; the empty set
1.8.8 Exercises In ExercisesL through 12, draw a sketch of the graph of the given equation and name the surface. t.4*+9y2lz2:36
2.4*9yz"2*
3.4*+9y2*:36
4.4*9y2**:36
s.f:Vz2z
6.f:y,+2,
,.***:n,
a,.fi+{:+
,"2
9.
,2
1 0 .f : 2 y * 4 2
re*:gy
1 7 .f + 7 6 2 2 : 4 y 2  7 6
13. Find the values of k for which the intersection of the plane x + W: y' *  z2:1 is (a) an ellipse and (b) a hyperbola.
12.9y24z2t t8r:0
1 and the elliptic hyperboloid of two sheets
L4. Find the vertex and focus of the parabola which is the intersection of the plane y :2 y' _x" _1 1649
with the hyperbolic paraboloid
15. Find the area of the plane section formed by the intersection of the plane y : 3 with the solid bounded by the ellipsoid f7f22 *1 9254
:
*_:1
16. Show that the intersection of the hyperbolic phraboloid yzlbz * I a, : zl c and the plane z : bx I ay consists of two intersecting straight lines. 17' U1e the method of parallel plane sectionsto find the volume of the solid bounded by the ellipsoid rfl a2+ y17 * zzlcz: l. (The measure of the area of the region enclosed by the ellipse having semiaxesi and,b is nab.)
1'8.9 CURVES IN R3 We consider vectorvalued functions in threedimensional space. 18.9.1 Definition
Let f u fzi and1" be three realvalued functions of a real variable f. Then for every number f in the domain common to fr, fz, and fs there is a vector R defined by
R(r): fJt)i+ f,(t)i+ fr(t)u
(1)
and R is called a aectoroalued function. The graph of a vectorvalued function in threedimensional spaceis obtained analogously to the way we obtained the graph of a vectorvalued function in two dimensions in sec. 17.4. As f assumes all values in the domain of R, the terminal point of the position representation of the vector R(f) traces a curve C, and this curve is called the graph of (l). A point on the curve C has the cartesian representation (r, a, z) , where
18.9 CURVESIN R3
865
(2) y: fr(t) z: fr(t) Equations (2) are called parametricequationsof C, whereas Eq. (L) is called a aectorequationof.C. By eliminating f from Eqs. (2) we obtain two equations in x, y , and z. Theseequations are called cartesianequationsof.C. Each cartesian equation is an equation of a surface, and curve C is the intersection of the two surfaces. The equations of any two surfaces containing C may be taken as the cartesian equations defining C. x: fr(t)
o rLLUsrRArroN 1: We draw a sketch of the curve having the vector equation R(f) : a cos fi * b sin fi + tk Parametric equations of the given cuwe are ' x:acost z:t A:bsint To eliminate f from the first two equations, we write them as f : a2 coszt
and A2: V sinz t
from which we get x2 ;:
cosz t
and
Adding correspondingmembersof thesetwo equations,we obtain f a2
+a' :1 b2
Therefore, the cuwe lies entirely on the elliptical cylinder whose directrix is an ellipse in the xy plane and whose rulings are parallel to the z a> as a normal vector.
EI4
IN THREEDIMENSIONAL VECTORS SPACEAND SOLID'ANALYTIC GEOMETRY
EXAMPLE3: Find an equation in cylindrical coordinates for each of the following surfaces whose equations are given in cartesian coordinates and identify the surface: (a) xz + !' : z; (b) x2A':2.
F i g u r e1 8 . 1 0 . 6
ExAMPrn 4: Draw a sketch of the graph of each of the followirrg equations where c is a constant: (a) p:c, and c >0; (b) 0_ c; (c) 6c,and0
F i g u r e1 8 . 1 0 . 7
solurroN: (a) The equation is similar to Eq. ( ) of Sec. 18.8,and so the graph is an elliptic paraboloid.lf. * * y2 is replaced by r', the equation becomesf : z. (b) Ttre equation is similar to Eq. (5) of Sec.18.8with x and y interchanged. The graph is therefore a hyperbolig paraboloid having the z axis as its axis. When we replacex by r cos 0 and y by t sin 0, the equation becomesrz cosz0  rz sinz 0 : 4 andbecause cosz0  sinz 0: cos 20,we can write this as z:12 cos20.
In a sphericalcoordinatesystemthere is a polar plane and an axis perpendicular to the polar plane, with the origin of the z axis at the pole of the polar plane. A point is located by three numbers, and the sphericalcoordinate representationof a point P is (p, 0,S), where p : lOPl, I is the radian measure of the polar angle of the projection of P on the polar plane, and @is the nonnegative radian measure of the smallest angle measured from the positive side of the z axis to the line OP. SeeFig. 18.10.6. The origin has the sphericalcoordinaterepresentation(0,0,6), where 0 and $ may have any values. If the point P (p, 0, @) is not the origin, then p > 0 and 0 < 6 < zr, where Q:0 if P is on the positive side of the z axis and 6: r if P is on the negative side of the z axis.
soLUrIoN: (a) Every point P(p, 0, S) on the graph of p: c has the same value of p, 0 rnay be any number, and 0 = Q = zr.It follows that the graph is a sphere of radius c and has its center at the pole. Figure 18.1,0.2 shows a sketch of the sphere. (b) For any point P (p, 0, Q) on the graph of 0 : c, p may be any nonnegative number, d may be any number in the closed interval 10,nf , and 0 is constant. The graph is a half plane containing the z axis and is obtained by rotating about the z axis through an angle of c radians that half of.the xz plane for which r > 0. Figure 18.10.8shows sketchesof the half planes for 0:ir,0: & r r , 0 : t r r , a n dg :  * 2 .
1 8 . 1 0C Y L I N D R I C AALN D S P H E R I C A C L O O R D I N A T E S 875
(c) The graph of 6: c contains all the points P(p,0, d) for which p is any nonnegative number, 0 is any number, and S is the constant c. The graph is half of a cone having its vertex at the origin and the z axis as its axis. Figure 18.10.9aand b each show a sketch of the half cone for 0 < c < *n and izr < c < zr, respectively.
o: :27 0 , B : (  5 , 7 , 2 ) , a n d C  ( 4 , 6 ,  1 ). 1 9 .B . A  C 1,9.A+B.C 20.A.BC 21,.A8.C 24. If A is any vector, prove that A 
22.AXB.CXA (A. i)i+ (A . i)i + (A . k)k.
23.AxB.A+BC
25. Find an equation of the sphere concentric with the sphere * * y, + zz+ 4x + 2y  6z * 10 : 0 and containing the point (4, 2,5). 26. Find an equation of the surface of revolution generated by revolving the ellipse 9* * 422: 36 in the xz plane about the r axis. Draw a sketch of the surface. Determine the value of c so that the vectors 3i * ci  3k and 5i  4i * k are orthogonal. ShowthattherearerepresentationsofthethreevectorsA:5i+i3k,B:i*3i2k,andC:4i+2i*kwhich form a triangle. 29. Find the distance from the origin to the plane through the point (6,3,2) and having 5i  3j + 4k as a normal vector. 30. Find an equation of the plane containing the points (1,7 , 3) and (3, 1, 2) and,which does not intersect the r axis. 3r. Find an equation of the plane through the three points (1,2,l) , (1,4, 0), and (1,1,3) by two methods:(a) using the cross product; (b) without using the cross product.
REVIEW EXERCISES 879
32. Find two unit vectors orthogonal to i  3i * 4k and whose representations are parallel to the yz plane. 33. Find the distancefrom the point P(4, 6,4) to the line through the two points A(2,2,1) and B(4,3,1'). 34. Find the distancefrom the plane 9r  2y 'f5z * 44: 0 to the point (3, 2, 0). * 5kfindcos 0 intwo ways: i+ kandB:4i3i identity. and a trigonometric product cross producf (b) using the by dot the (a) by using
35. I f 0 i s t h e r a d i a n m e a s u roef t h e a n g l e b e t w e e nt h e v e c t o r sA : 2 i + 36. Prove that the lines +2 _x_ l_ : 2 : a2
z2
. x2 Y andn:3:
S
z5
are skew lines and find the distance between them' to each of the lines of 32. Find symmetric and parametric equations of the line through the origin and perpendicular Exercise36.  3k and 5i * 4k' 38. Find the area of the parallelogram two of whose sides are the position representationsof the vectors2i 1, 3) , (2,2' 1), and (1,l' 2) ' 39.Find the volume of the parallelepipedhaving verticesat (1, 3, O), (2,
40. Find the length of the arc of the curve y:tsint
l:fcost from f: 0 to t:
z:l
itr.
vector, and the speed at 4r. Ajparticle is moving along the curve of Exercise40: Find the velocity vector, the acceleration and acceleration t tn. Draw a sketch of iportion vectors there.
of the curve at t:
Ln and draw the representationsof the velocity
curve having the vector equation r12. Find the unit tangent vector and the curvature at any point on the R(t):
eti* etiI2tk
: equations: (a) (r * !)2 + t : z; (b')25* r +yz 1gg' 43. Find an equation in cylindrical coordinates of the graph of eachof the  4y2 of the equations: (a) r' i Y'+ 422:4; (b) 4f 44. Find an equation in spherical coordinates of the graph of each l9z2:36' with respect to f exist, prove that 415.If R, e, and W arethree vectorvalued functions whose derivatives D,[R(r).e(r)xw(f)]:D,R(f).Q(f)xw(t)+R(t)'DrQ(t)xw(t)+R(t)'Q(t)xD'w(t)
Differentialcalculus of functions of severalvariables
19.1 FUNCTIONSOF MORE THAN ONE VARIABLE
I9.I FUNCTIONS OF MORE THAN ONE VARIABLE
1r9.L.1Definition
19.1.2 Definition
In this section we extend the concept of a function to functions of n variables,and in succeedingsectionswe extend to functions of n variables the concepts of the timit of a function, continuity of a function, and derioative of a function. A thorough treatment of these topics belongs to a course in advanced calculus. In this book we confine most of our discussion of functions of more than one variable to those of two and three variables; however, we make the definitions for functions of n variables and then show the applications of these definitions to functions of two and three variables. W" utto show that when each of these definitions is applied to a function of one variable we have the definition previous$ given. To extend the concept of a function to functions of any number of vari' ables, we must first consider points in ndimensional number space.Just as we denoted a point in Rl by a real numb et x, apoint in R2by an ordered pair of real numterc (x, y), and a point in R3 by an ordered triple of real numbers (x, y, z), we rePresent a point in ndimensional number sPace' : R", by an'oriered ntaple of real numbers customarily denoted by P ( x ' y ) ; n:2'P: ( r r , * r , . . . , r n ) . I n p a r . t i c u l a irf,. n : 7 , w e l e t P : x i i f if.n: g, P : (x, y, z)t if.n: 6, P : (xr.,x2,xg,xa,x5,x6). The set of all ordered ztuples of real numbers is called the n'dimensional xr, . ' , xo) numberspace andis denoted by R'. Eachordered ntuple (1,r, space' number is called a point in the ndimensional Afunctionofnaariablesisasetoforderedpairsoftheform(P.al)inwhich P is a point in ,ro t"o distinct ordered pairs rpu." have the same fiist element. set of all The number' real and w is a ndimensional numbe, the set and function, the p"rriUi" values of P is called the domain of function' ff ail possible values of w is called the rnnge of the of.n variFrom this definition, we see that the domain of a function or, numbers feal of a set ables is a set of points in R" and that the range is of one function a have we of points in Rl' When rt: L' equivalently, " ""t of variable;thus,thedomainisasetofpointsinRlor'equivalently'asetof real numbers' Hence' we see that a set is range the and real numbers, .1..7.L d speciaicaseof Definition 19.1..2.lf.n:2, wehave a is Definition and the domain is a set of points in R3or, equivvariables, two of function is a set of a set of ordered pairs of real numbets (x' y)' The range ;iliit, real numbers. variables r and y be the set . rLLUsrRArroN1: Let the function f of two that such (P, z) of all ordered pairs of the form
z: {zs777l
y) for which 25  * The domain of I is the set of all ordered pairs (x' I * Az:25 y, > o.This is tire set of all points in_the xy plane on the circle and in the interior region bounded by the circle'
CALCULUSOF FUNCTIONSOF SEVERALVARIABLES DIFFERENTIAL
Becausez: ry'F=@Tfr, we see that 0 < z < 5; therefore, the range of / is the set of all real numbers in the closed interval [0, 5]. In Fig. 19.1.1we have a sketch showing as a shaded region in R2 the set of points in the domain of /. o . rLLUsrRArron2: The function g of two variables r and y is the set of all ordered pairs of the form (P, z) such that \/F t 7 ,5 F i g u r e19 . 1. 1
v The domain of g consists of all ordered pairs (r, y) for which * + yz  25 and y # 0. This is the set of points, not on the r axis, which are either on the circle * * y, :25 or in the exterior region bounded by the circle. Figure 19.7.2is a sketch showing as a shaded region in R2the iet of points in the domain of g. o
F i g u r e1 9 . 1. 2
F i g u r e1 9 . 1. 3
o rllusrRArror.r 3: The function F of two variables x au';rd, y is the set of all ordered pairs of the form (P, z) such that
z:yG+f,2s lf y:0, then z  0 regardlessof the value of r. Howeveg iI y # 0, then f * y' 25 must be nonnegative in order for z to be defined] Therefore, the domain of F consists of all ordered pairs (r, y) for which either y = g or rP * y'  E > 0. This is the set of all points on the circle rp * yr:25, all points in the exterior region bounded by the circle, and all poitrts on the r axis for which 5 . r < 5 In Fig. 19.1.3we have a sketclishowing as a shaded region in R2 the set of points in the domain of F. o o rr,r,usrRArror.r4: The function G of two variables r and y is the set of all
19.1FUNCTIONS OF MORETHANONEVARIABLE ordered pairs of the form (P, z) such that z
'
ffi
v
lf y:0, then z:0 provided that * * y' 25 + 0. lf. y # 0, then 12* y'  25 must be positive in order f.orz to be defined. Hence, the domain of G consists of all ordered pairs (x, y) for which x' * y'  25 > 0 and those for which 3r: 0 and x # +5. These are all the points in the exterior region bounded by the circle 12* y' :25 and the points on the r axis for which 5 ( r < 5. Figure 19.1.4is a sketch showing as a shaded region in R2 o the set of points in the domain of G. If f is a function of n variables, then according to Definition 19.'1"2' / i s a s e t o f o r d e r e d p a i r s otfh e f o r m ( P , w ) , w h e r e P : ( x r , x 2 , .  . , x n ) is a point in R" and ar is a real number. We denote the particular value of ar, which corresponds to a point P, by the symbol f (P) or f (xr, h, . . , xo). In particular, 7f.n:2 and we let P : (x, A) , we can denote the function value by either /(P) or f(x,y). Similarly, if.n:3 and P: (*, y, z), we denote lhe function value by either /(P) or f(x, y, z). Note that if n: l, P : xi hence, if f is afunction of one variable, f (P) : /(r). Therefore, this notation is consistent with our notation for function values of one variable. A function f of.n vaiables can be defined by the equation
F i g u r e1 9 . 1. 4
w:f(\,x2,.
.
,xn)
The variablES X1,Xz, . . ., xn ate called the independentaariables,arrd w is called tl:.edependentaariable. o ILLUsrRArron 5: Let / be the function of Illustration 1; that is,
f(x' Y): \/8=7f (9,
4) 
f (2, r) : f (u, 3a) : EXAMPLE1: The domain of a function I is the set of all ordered triples of real numbers (x, Y, z) such that
g(x, y, z)
Sxz* yz'
Find (a) S(1, 4, 2); (b) S(2a, b, 3c); (c) 8(x', y', z'); (d) Sg, z,x).
:ffig :ffi2\E _ffi
SOLUTION:
( a )8 ( 1, 4 ,  2 ) (b) B(2a,
: 1 2  5 ( 1 ) (  2 ) + 4 (  Z ) ' : 1 ' + 1 0+ 1 6 : 2 7 3c)  (2a)2 5(2a)(3c) + (b) (3c)' : 4a2 30ac 9bc2
rc4 5x222* y'zn (c) g@r, Ar, zr) : (x')2  5(x')(z') + (y')(z')2: : yz  Sy(r) * z(r)2 : y' + Sxy * xzz (d) SQ, z, )c)
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
19.1.3 Definition
If / is a function of a single variable and g is a function of two variables, then the compositefunction f ' I is the function of two variables defined by
(f ' s)(x,y) : fQ@'y)) and the domain of f " I is the set of all points (x, y) in the domain of g such that g@, y) is in the domain of /. EXAMPTr2: Given f (t)  ln f a n d g @ , y )  x 2* y , f i n d h ( x , y ) if h  f o S, and find the domain of. h.
SOLUTION]
h(x,y) : ( f " il@, y) : f k@' y) )
: 1 1 *+ y ) :ln(*
I y)
The domain of g is the set of all points in R2,and the domain of / is (0, +.;. Therefore, the domain of h is the set of all points (r, y) for which f*y>0. Definition 19.1.3 can be extended to a composite function of n vaiables as follows. 19.1.4Definition
If / is a function of a single variable andg is a functionof.n variables, then the compositefunction f " g is the function of.n vartables defined by U"il(xr,xr,.
. . ,x):f(g@r,x2,.
. , ,xn))
and the domain of f " g is the set of all points (x1, x2, . . .,r,) domain of g such thatg(rr, x2, . . . , x,\ is in the domain of f. EXAMPLE 3:
Given F(r)  sinl r
and G(x, A, z) find the function F o Q and its domain.
,
in the
(F.c)(x, y,z) :ii:##_r,
SOLUTION:
: sinl tp+f+Za The domain of G is the set of all points (x, y, z) in Rs such that I * y, * zz  4 > 0, and the domain of F is [1, 1]. So the domain of F. G is the set of all points (x, A, z) in R3 such that 0 <x2*yr*224 0 such that
lf(P) Ll
(e)
If f is a function of one variable and if in the above definition we take A: a in Rr and P: r, then (8) becomes
: f'T/t') r and (9) becomes 0suchthat ( x o ,y o, f ( x o , v o ) )
( x o ,U o ,o )
Figure 19.2.7
lf (x,Y)
 Ll < € whenever 0
(10)
In words, Definition 19.2.5 states that the function values f(x, v) (rq' yq) if the approach a limit L as the point (r, y) approachesthe point arbibemade I, can and a'bsolutevalue of the diffeience between f(x, V) (xo' yo) but to close (!,y) sufficiently tt*ify small by taking the point about said is nothing 19.2.5 Definition (io, yJ. frote that in "lf"J," ""i valul at the point (xo,yJ; that is, it is not necessarythat the the fulnction to exist' function be defined at (xo, /o) in order for ,,,rli1.,*, f(x,y) in Fig' A geometric interpretation of Definition 19.2.5 is illustrated having surface ( (ro, of the B Ao\;r) 1g.2.7.1heportion above the open disk lie will axis z the on that f(x, U). ug";,io" ,': f(*, y) is shown] W" t"" in plane is (r, ry the y) in point the 2 and L + e whenever between f tnatf(x'A) this.is stating of way tfr" op"r, disk 8((16, /o); S). Another on th" z axiscan be."ttti.t"d toliebetweenLe and L* e byrestrictiig ifr" point (r, y) in the xy planeto be in the open disk B((ro' /o);6)' that . rLLUsrRArroN1: We apply Definition l'9'2'5 to prove lim
(c,rr)(r,s)
(2x * 3Y) = ll
> 0 such that We wish to show that for any € > 0 there exists a 6
VARIABLES OF FUNCTIONS OF SEVERAL DIFFERENTIAL CALCULUS
l ( 2 x + 3 1 )  1 1< 1 e w h e n e v eOr < V G = I I + f y  g ) z < 6 Applying the triangle inequality, we get
l 2 x + 3 y 1 1 1 :l 2 x  2 t 3 y  e l = z l x 1 l+ s l V 3 1 Because
and lygl  I/G=TTQ=T
lx 1l  t/@ffiQ$ we cnn conclude that
0such that
l(3* +y) 51 < e wheneverO< V(7=lzTlyZ)t
0 such that < e whenever 0 ( ll(x,y)  (xo,/o)ll< S The above will be true if we further restrict (x, y) by the reguirement that (x, y) be in a set S, where S is any set of points having (io, yo) as an accumulation point. Therefore, by Definition 1.9.2.7, lf(x,y)Ll
lim
(t,u)(to,uo) (p in s)
f(x,A):L
and L does not depend on the set S through which (x, y) is approaching (xo, Ao).This proves the theorem. I The following is an immediate consequenceof Theorem 19.2.g. 19.2.9 Theorem
If the function / has different limits as (x,y) approaches (ls, yo) through two distinct sets of points having (xr, yJ as an accumulation point, thett does not exist' t"''li["'" f(x'Y) PRooF: Let 5r and Sz be two distinct sets of points in Rz havin g (xr, yo)
19.2 LIMITSOF FUNCTIONSO F M O R ET H A N O N E V A R I A B L E
895
as an accumulationpoint and let lim
(r,u)(ro,uo) (p in sr)
f (*, a) : Lt and
lim ( r ,d ( to,uo) (p in sz)
f (x, y) : Lz
f (x,y) exists.Then by Theorem 19.2'8Lt must equal L", but by hypothesis Lr * L2, and so we have a contradiction. (x'y) does not exist' I Therefore' ,r,rl1T,,rorf
Now assumethat
ExAMPtn 2:
Given
xY f ( x , a' ) : x 2 * y ' lim
find
f (x, y) if it exists.
(r,a)(o'o)
lim
(r,y)(ro,yo)
sol.urroN: Let S, be the set of all points on the r axis. Then lim
(r,u)(o,o) (p in sr)
m f (x,0) : f ( x , y )  lr io
0 t;9ffio r.
Let 52be the set of all points on the line y  r. Then lim (r,g)  (o'o) @ in sz)
f (x,y) :
Because
lim f(x,y) +
(r,s)*(o,o) (p in sr)
li*
(r'u)*(o'o)
f Q,y)
(p in sz)
it fotlows from Theorem1,9.2.9that ( r , ulim f (x , y) does not exist. )(o,o)In the solution of Example 2, instead of taking s1 to be the set of all the points on the x axis, we could just as well have restricted the points of S, to be on the positive side of the r axis becausethe origin is an accumulation point of this set. EXAMPLE 3:
Given
f(x,Y):ffi find
lim ( r ' g )  ( o , o)
f (x,y) if it exists
solurroN:
Let 51 be the set of all points on the x axis. Then
f (*'Y): t"T',h  o ,,,11%,or (p in sr) Let Sz be the set of all points on any line through the any point (x, y) in Sz, A : mx. :lim lim f (x, a) x' + m'x': (r,s)(o,o) 'o:{'}:^
r' tJT
that
for
3mx : o u Jt, ntz:
(p in sz)
Even though we obtain the same limit of 0,if. (x, y) approaches(0,0) through a set of points on any line through the origin, we..cannot conHowever' cludelhat f(*, y) exists and is zero (see Example4)' *,ll%., let us attempt to prove an",,,,llTo,orfG,V):0' FromDefinition 19'2'5'
CALCULUSOF FUNCTIONSOF SEVERALVARIABLES DIFFERENTIAL
if we can show that for any e ) 0 there exists a E > 0 such that
l%l
lx'*y'l
.\€\
' Yh r eneve w or< t W < D
then we have proved that
lim
f (x, A) : g.
Because w€have rp xz* yr"fi'ilj''! W, 3(x'+U\W:_g  3 * ' y^ 1: g l l v l W lry,l:m So if 6  *e, we can conclude that
l+l
lf+vl
. € w h e n e voe0. Cqse2: lf x
+ 0 , l f( * , y ) 0 l: l ( r + V) sin(Ur)l.
: V+vt lI ( x+ v ) sin*l 1""11 I
CALCULUSOF FUNCTIONSOF SEVERALVARIABLES DIFFERENTIAL
:zvf
+ y2
Then 
1l
+ v) sinjlx l < z .*e whenevero < lTf l@ l'
< te
So take 6: te. Therefore, in both caseswe have found a E > 0 for any e ) 0 such that lim f(x, y):0. that (16)holds, and this proves (c,g)(o,o)'
The limit theorems of Chapter 2 and their proofs, with minor modifications, apply to functions of several variables. We use these theorems without restating them and their proofs. o rLLUsrRArrow4: By applying the limit theorems on sums and products, we have lim G,ilGz,r)
(r'* 2*y  y' + 2):
(2)' + 2(2)'(1,)  (1)' * 2:7
o
Analogous to Theorem 2.6.5 fot functions of a single variable is the following theorem regarding the limit of a composite function of two variables. l9.2.l0 Theorem
If g is a function of two variables and
!i+
.g@, y):b,
and f is a
tunctionof a singlevariable continuous "gi?;tfi:"rl . l i + . ( f "d @ , y ) : f ( b ) (;rll)(co,aol
or, equivalently,
fQ@, D) :/(",,1iT".,", s@,v)) ,",,IiT".,u The proof of this theorem is similar to the proof of Theorem 2.6.5 and is left as an exercise (see Exercise20). ExAMPtn6: Use Theorem to find lim ln (xy  1). 19.2.10 (t,u)  ( 2,1)
soLUrIoN: Let g be the function such that g@, y)  ry  1, and / be the function such that f(t):ln t. lim (w L\ :'t G,i(z,i

and because/ is continuous at 1, we use Theorem 19.2.10and get
 t) : t"(,,.11T.,, (v  t)) rn(xv ",ll%,,, lnL 0
19,2 LIMITSOF FUNCTIONSOF MORE THAN ONE VARIABLE
Exercises19.2 In ExercisesL through 6,establish the limit by findi.g a 6 > 0 for any e ) 0 so that Definition 1,9.2.5holds. 1. lim (3x  4V)  1 2. lim (5x 3y)  z 3. lim (x' * A') :2 (c,u)(g,z)
4.
lim
(c,u)(z.g)
(x,u)(2,4\
(2x'  yz) : L
5. li
(r,g)(1,1)
(x'*2xy):4
5'
(t,g)(2,4)
(x' + yz 4x * 2y) : + ,r,jtT,r,
In Exercises7 through 12, prove that for the given function f ,,r,llTo,o,f (x,y) doesnot exist.
V1 7.f (x,A):.?x'+y'
8 .f ( x , a ) :xJ' + y '
xa*3*'y'*Zxyg
L 0 .f ( x ,A ) :
11.f (x,A): , =t'n * (x' + yn)t
(x' * y')'
L3through15,provethat,",llTo, @,U) exists. In Exercises orf * Yt :*t u. f(x,y) x'+ y'
13. f(x,a):ffi (,
1 5 .f ( x , a ) : l
L
L
( r +. y )\ s i n 2 s r n ' =r f x + 0 a n d y + 0 if eitherx:0 or y  0
L0
r c . f @ , y ) : l x[f
i" (xy) \ '.
lv
if x+0 rf.x0
In Exercises17 through 19, evaluatethe given limit by the use of limit theorems. 17.
lim (c,u)(2$)
(3x' * xy  2y')
18. lim v+m (c,a)(2,E)
lg.
e* * eu Iim (a,s)(o,o) COSX * Sin y
20. Prove Theorem 19.2.10 In Exercises2L through 23, show the application of Theorem 19.2.L0to find the indicated limit. 21.
lim
u
tanl 1
22.
X
(c,y)(2,2'1
lim (t,s)(2,g)
[5r + ty'\
23.
tl r lim \'l (x,u\(t,2, g*  +V
In Exercises24 through 29, determine if the indicated limit exists.
24. lim
(c,s)*(o,o>f
!'a' =
25. lim
!'a' , * Yn
.L sln x
i f x + 01h
ti" zs.f (x,a):{t i* '
.1 sln 
t f x + L a n d ' la * 0
L0
lim
sin(r + y)
(c,u)(2,2)
X+ y
22. tim
*: + Y=
(c,s)(o,o)X' + y2
l
f*v + 28f(x,a):wtv .
x
26.
lim
f(x,y) (e's)(o'o) if x  0J
if either x:0
I lim f(x,y) I (c'a)'(o'o) : or y o)
30. (a) Give a definition, similar to Definition 19.2.5, of. the limit of a function of three variables as a point (x, y, z) approaches a point (xo, yo, zo). (b) Give a definition, similar to Definition 19.2.7,of the limit of a function of three variables as a point (x, y, z) approachesa point (xo, yo, zs) in a specific set of points S in R3. 31. (a) State and prlove a theorem similar to Theorem 19.2.8f.or a function / of three variables. (b) State and prove a theorem similar to Theorem 19.2.9for a function f of three variables.
VARIABLES CALCULUS OFSEVERAL OF FUNCTIONS 9OO DIFFERENTIAL 30 and 31 to proveanu,,",r,ltTo,o @,A, z) does In Exercises32 through 35, use the definitions and theoremsof Exercises ,orf not exist. g2. f (x, y, z) '/ f
* Yx,'* z2x2 f + Yn* zn
*+y222
33.f(x,v,"):FTFr,
g5. f(x,y,z):Fffi
u . f ( x , A z,' ') : , O . *! " f + Y'+zn
In Exercises36 and 37, use the definition in Exercise30(a)to provethat
f @, y, z) exists. ,,.r.1l1o.o.o,
96.f (x,A,z) : *r!l;r( ,, ;): {e*y*z) L0
r i r ,1 r i r ,1 i f x + o a n d , y + o if eitherx :0 or y : 0
19.3 CONTINUITY OF FUNCTIONS OF MORE THAN ONE VARIABLE 19.3.1 Definition
We define continuity of a function of.n variables at a point in Rz
Suppose that / is a function of n variables and A is a point in R". Ttren / is said to be continuousat the point A if and only if the following three conditions are satisfied: (i) f(A) exists; (ii) lim /(P) exists; P.A (iii) lim f(P):f(A). If one or more of these three conditions fails to hold at the point A, then / is said to be discontinuousat A. Definition 2.5.1 of continuity of a function of one variable at a number a is a special caseof Definition 19.3.1. If / is a function of two variables, A is the point (re, ys), and.P is a point (r, y), then Definition 19.3.1becomes the following.
19.3.2 Definition
The function f of. two variables r and y is said to be continuousat the point (rs, ye) if and only if the following three conditions are satisfied: (1) f(xo, yo) exists; (ii) y) exists; lim f(x, (t,il(so,ud
(iii) _ !i+ . f(x, y) : f (xo,y). (a,i(h,sd' soLUrIoN: We check the three conditions of Definition 19.3.2 at the point (0, 0).
+ (0,0)  (0,0)
(i) f (0, 0) : Q.Therefore,condition (i) holds.
19.3 CONTINUITY OF FUNCTIONSOF MORE THAN ONE VARIABLE
determine If f is continuous at ( 0 ,0 ) .
(tt) f(r, v):,",llTo,o, ,,.11%,, ffi:0,
901
whichwasproved in Ex
ample 3, Sec.19.2.
(ttt),".ll1, u): /(0,o). rf(x' So we conclude that / is continuous at (0, 0). EXAMPLE 2:
LCt
the function f
be defined by
f*Y f(*,a):Tf Lo
solurroN: following.
Checking the conditions of Definition "1.9.3.2,we have the
(i) f (0, 0)  0 and so condition (i) holds. ( i i ) W h e n ( x , y ) + ( 0 , 0 ) , f ( x , U ) : x y l ( x ' + y ' ) . I n Example 2, Sec. '!.9.2, + y') does not exist and so we showed that ,r,llT, ,rr*Al@' lim f (x,y) doesnot exist.Therefore,condition (ii) fails to hold.
if (x, y) + (0, 0) i f ( x , y )  ( 0 ,0 )
Determine tf.f ts continuous at ( 0 ,0 ) .
(t,a)(o,o)
We concludethat / is discontinuousat (0, 0).
If a function f of.two variables is discontinuous at the point (ro, yo) tim f(x, y) exists, then / is said to have a remoaabledisconti(r,u)(co,ui' nuity at (xs, !) becauseif f is redefined at (xo, /o) so that f(xo, y): . li* f (x, y), then / becomescontinuous at (xs, y). It the discontibut
(t,il(co,uo)
nuity is not removable, it is called art essentialdiscontinuity. * yz), then g is discontinuous o rLLUsrRArron1: (a) If g(r, y):3x'yl(f at the origin becauseg(0, 0) is not defined. Howevet, in Example 3, Sec. * y'):0. Therefore, the discon19.2, we showed an"a or3*yl(xz ,,,11T0, tinuity is removable if g(0, 0) is defined to be 0. (Refer to Example 1.) (b) Let h(x, y):xyl(xz *y2). Then ft is discontinuous at the origin because h(0, 0) is not defined. In Example 2, Sec. 79.2, we showed that lim xyl(x2 * y) does not exist. Therefore, the discontinuity is essential. k,u)(0,0)
(Refer to Example 2.)
o
The theorems about continuity for functions of a single variable can be extended to functions of two variables. 19.3.3 Theorem
If f andg are two functions which are continuous at the point (16,yo), then (i) / + g is continuous at (xo, Y); (ii) /  g is continuous at (xo, Y); (iir) /S is continuous at (ro, yo); (iv) fl7 is continuous at (ro, /o), provided that g(xo, yr) * 0. The proof of this theorem is analogous to the proof of the corresponding theorem (2.6.1)for functions of one variable, and hence it is omitted.
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
19.3.4 Theorem
A polynomial function of two variables is continuous at every point in R2. PRooF: Every polynomial function is the sum of products of the functions defined by f (x, y) : x, g@, y) : y, and h(x, y): c, where c is a real number. Becausef , g, and ft are continuous at every point in R2,the theorem follows by repeated applications of Theorem 19.3.3, parts (i) and (iii). I
19.3.5 Theorem
A rational function of two variables is continuous at every point in its domain. pRooF: A rational function is the quotient of two polynomial functions f and g which are continuous at every point in R2 by Theorem 1,9.9.4.If. (xo, y) is any point in the domain of flg, then g(ro, y) # 0; so by Theorem 19.3.3(iv)f lS is continuous there. I
EXAMPTT3: Let the functi on f be defined by
f (x, y) : [ r ' * y ' L0
ifx2*y, rfxz*y,
0.Thenthefunctionhisthecompositefunctionf"gand'byTheorcmlg.3.T,iscontinuousatallpoints(x,y)inR2forwhichxyL)0or' region of equivalently , xy > 1.. The shaded region of Fig. 19.3.1 is the ' continuity of h' soLUrIoN: This is the same function as the function G of lllustration 4 in sec. 19.1. we saw there that the domain of this function is the set of all and the points in the exterior region bounded by the circle rc *y',:25 5 < < x 5' points on the r axis for which The function / is the quotient of the functions g and h for which The tunction g is a polynomial g@, y) : y and h(x, V): \/FW=E irnction and therefore is continuous everywhere. It follows from Theorem 19.3.7that h is continuous at all points in xP for which * * y2 > 25. Therefore, by Theorem 19.3.3(iv) we conclude that / is continuous at all :25' points in the exterior region bounded by the circle f * y' 5 < r < 5, that is, Now consider the points on the x axis for which 5 I a < 5. If Sr is the set of points on the line the points (a, 0) wherl x: a, (a, 0) is an accumulationpoint of Sr, but
f(x'Y):tu#x *.}T,r, (P tn Sr)
gO4
DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES
which does not exist becauseyl{ffi is not defined if lyl = \E. Therefore,/ is discontinuousit *re points on the r axis for which 5 ( r < 5. The shadedregion of Fig. 19.3.2is the region of continuity of /.
Figure19.3.2
Exercises19.3 In Exercises 1 through 6, discuss the continuity of /.
t+
1. f (r, V) : ]{Ff t0 (x'y
2 .f ( x , a ) :  f f i f.0
3 .f ( x , a ) : t # L0
ir @,y)+ (o,o) if (x,y) : (0,0)
i. r ( x ' Y *) ( o o' ) if (x, y) : (0, 0)
(nrNr: SeeExample 4, Sec.19.2.)
i f @ , v+) ( oo, )
(x, y) + (0,0) ( x , y ) : ( 0 ,o )
if (x, y) : (0, 0)
(ry
ir (x'Y)+ (o'o)
L0
i f ( x , y ) : ( 0 ,0 )
5.f(x,a)={lffiT
lnrNr: SeeExercise13,Exercises19.2)
if (x, y) + (0, 0) if (x, y) : (0, 0)
7. ProveTheorem19.3.7. In Exercises 8 through 17, determine the region of continuity of I and draw a sketch showing as a shaded regron in R2the region of continuity of /.
L1..f ( x , a ) :
W
L3. f (x, A) : r ln (xV')
19.4PARTIAL DERIVATIVES 905 L4. f (x, A) : sinl (xy)
15. f (x, A) : tanr L + secr(xV)
(*  y' li f".x' r* y 16.f(x,D:lxy ifx:y Lxy
rsin(x* u) ifx+y+O 1 t zj .. ft (_x ,"y.), :_t l, # x  t y ifx*y:g U
v
In Exercises18 through 27,provethat the function is discontinuousat the origin. Then determineif the discontinuityis removableor essential.If the discontinuityis removable,define/(0, 0) so that the discontinuityis removed. 1 9 f. ( x , A ) : ( x + y ) s i n l
v
2 1f.( x , A ) : Y
xu*yn
22. (a) Give a definition of continuity at a point for a function of three variables, similar to Definition 19.3.2.(b) State theorems for functions of three variables similar to Theorems 19.3.3and 79.3.7.(c) Define a polynomial function of three variables and a rational function of three variables. In Exercises23 through 25, use the definitions and theorems of Exercise22 to discuss the continuity of the given function.
23.f (x, A, z) :
24. f(x, y, z) : ln(36  4x2 y2  9z')
W Sxyz
(
2 5f.( x , y , z ) : ] W L0
if ( x , y , z ) # ( 0 , 0 , 0 ) if ( x , y , z ) : ( 0 , 0 , 0 )
I9.4 PARTIAL DERIVATIVES
19.4.1 Definition
' I xzv'
25.f (x, A,z)  fr' * y' * z2 L0
if
( x , y , z ) + ( 0 ,0 , 0 )
if
(x,y,z) (0,0,0)
We now discuss differentiation of realvalued functions of n variables. The discussion is reduced to the onedimensional caseby treating a function of n variables as a function of one variable at a time and holding the others fixed. This leads to the concept of a "partial derivative." We first define the partial derivative of a function of two variables. Let / be a function of two variables, r and y. The partial deriaatitseof f with respect to x is that function, denoted by Dtf , such that its function value at any point (x, y) in the domain of / is given by (1) if this limit exists.Similarly, the partial deriaatiaeof f u)ith respectt o y r s that function, denoted by Drf , such that its function value at any point (x, y) in the domain of / is given by (2)
The process of finding a partial derivative is called partial differentiation.
FUNCTIONS OFSEVERAL VARIABLES Drf is read as "D sub t of f ," and this denotes the partialderivative function. Drf (*, y) is read as '? sub L of.f of.x and yi' and this denotes the partialderivative function value at the point (x, y). Other notations for the partialderivative functionDrf arefr, f ,, and O/dr. Other notations for the partialderivative function value D/(r, A) arc fr(x, A), f ,(x, y), and f(x, y)lax. Similarly, other notations for Drf are fz, fa, and Aflay; other notations for Dd@, y) are fr(x, y), fu(x, y), and \f(x, V)laV.It z : f (x, A) , we can write 0zl0x for D rf (r, y) . A partial derivative cannot be thought of as a ratio of 0z md Er becauseneither of these symbols has a separate meaning. The notation dyldx can be regarded as the quotient of two differentials when y is a function of the single variable r, but there is not a similar interpretation for 0zl0x. SOLUTION:
find Drf (x,y) and Drf (x,y) by applying Definitio n'!.9.4.1.
Drf(r,/) r/\.'"' : lim f@
+ ax'Y.) f(x' Y)
4* * d^r1'2(x+ ax)y* y,  (g*  2ry!n
: ;;
Ar
Ano
: lim 3f*6x Ax+3(Ax)22ry:2y Lx+y23f+zryyz Ax
Aro
: :
 2u a'x 5xLx+ 3(4t!)2 ^tlll j lim (Gx+ Lx2y)
Ao0
:6x  2A Drf(x,/) : lim f @'v + AY) f(x' v) aao
Ly
: tim 3* Zx(y+ AV)+ (y * Ay), (gf  2ry * yr) auo
Ly _ r.^ 3* 2xy 2x A,! * y2'* 2y Ay + (Ay)z 3* I 2xy  yz Auo
: lim '
Ly
2x LY + 2YLY + (LY)z
aao
Ly
: lim (2r + 2y + Ay) AaO
:2x * 2A If (xr, y)
is a particular point in the domain of /, then
(3)
19.4 PARTIALDERIVATIVES
if this limit exists, and (4)
if this limit exists. o rLLUSrRArroNL: We apply formula (3) to find Drf(3,2) for the function / of Example 1.
D,f(3,2):ll* At + ^x'fI f@'2\ :,::, + L2+ 4 Lx * 4 43 : lim 27+ LBAr * 3(Ax)'? Ax
Ar*o
:lim
(18+3 Lxt4)
A.r0
:22 Altemate formulas to (3) and (4) for Drf (xo, ys) and Ozf(n,/o) given by
o are
(s) and (5)
if this limit exists. 2) for the function o rLLUSrRArrow2: We apply formula (5) to find Drf(3, f of Example 1.
D,f(3,2):n^t@AFA x 3 "'t
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
o rLLUsrRArrON3: In Example 1, we showed that Drf (x, V) :6x  2y Therefore, Dtf(9,2) :18 + 4 :22 This result agreeswith those of Illustrations 1 and 2.
o
To distinguish derivatives of functions of more than one variabre from derivatives of functions of one variable, we call the latter derivatives ordinary derioatiaes. comparing Definition 19.4.1with the definition of an ordinary derivative (3.3.1),we see that Drf (x, y) is the ordinary derivative of if is f f considered as a function of one variable x (i.e., y is held constant), and Drf (r, y) is the ordinary derivative of.f if/ is considered as a function of
ExAMPrn2: Given f(x, A) : 3x3  4x'y * 3xyz a Tx  By, find D r f( x , y ) a n d D r f( x , y ) .
soLUTroN: Considering f as a function of r and holding y constant, we have Dtf (x, A) :9*
 *ry + 3y2+ 7
Considering / as a function of y and holding r constant, we have Drf (x, A) : 4* * 5ry  8
solurroN:
+ (0,o) : ( 0 ,0 )
(a)f '(0,
(b)f ,(x, 0).
(a) If y # 0, from (5) we have
f@' Y)  f(o' Y) /,(0, y) lim r0
19.4 PARTIALDERIVATIVES 909
If y:
0, we have
f /'(0,0):liT
:tTT0
Because/, (0, y)  y i f y + 0 a n d / ' ( 0 , 0 ) : 0 ,  y for all y. /t (O, y) (b) lf x * 0, from (5) we have
we can conclude that
' fr ( x ,o ) : t J T  lim A'0
: lim U0
x3 x:2
If x:
0, we have
f ( 0 , y )  f ( 0 , 0 )_ lim fr(O,o)  lim y0 U0 Becausef ,(x, o ) : r , i f x # 0 a n d fr(O, 0) :0, f r ( x , 0 )  x f o r all x.
we can conclude that
Geometric interpretations of the partial derivatives of a function of two variables €uesimilar to those of a function of one variable. The graph of a function f of two variables is a surfacehaving equation z:f(x,U). If y is held constant (say,y : yo),then z : f (*,yo) is the equation of the trace of this surface in the plane y: yo. The curve can be represented by the two equations (a) x
a:ao
and zf(x,y)
(7)
becausethe curve is the intersection of these two surfaces. Then Dtf(rq, /o) is the slope of the tangent line to the curve given by Eqs. (7) at the point P6(16,Ao,f (xo,yo)) in the plane A : Ao.In an analogous fashion, Drf (xs, ye) represents the slope of the tangent line to the curve having equations x:xo (b) x Figure19,4.1
and z:f(x,y)
at the point Po in the plane r:xe. Figure L9.4.Laand b shows the portions of the curves and the tangent lines.
910
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
EXAMPIE4: Find the slope of the tangent line to the curye of intersection of the surface
SOLUTION:
(2,2, \E).
x
0z
z+@withthe planey:2at
The required slope is the value of 0zl0x at the point
thepoint
( 2 , 2 ,\ E ) .
A partial derivative can be interpreted as a rate of change. Actually, every derivative is a measure of a rate of change. lf. is a r,ri.tio'of the f two variables r and y, the partial derivative oi1*iit respect to r at the point P1(ro, /o) gives the instantaneous rate of change, at ps, of f(x, y) per unit change in r (r alone varies and y is held fixed at ye). similariy, the partial derivative of / with respect to'y atpo gives the instantaneous rate of change,at Po,of f(x,y) per unit changein y. EXAMPTn5: According to the ideat gaslaw for a confined gars, if P pounds per square unit is the pressure, V cubic units is the volume, and T degrees is the temperature, w€ have the formula
PV: kT
(8)
where k is a constant of proportionality. Suppose that the volume of gas in a certain container is 100 in.3 and the temperature is 90o,andk:8. (a) Find the instantaneous rate of change of P per unit change in T if V remains fixed at 100. (b) Use the result of part (a) to approximate the change in the pressure if the temperature is increased to 92". (c) Find the instantaneous rate of change of V per unit change in P if T remains fixed at 90. (d) Suppose that the temperature is held constant. Use the result of part (c) to find the approximate change in the vol
soLUrIoN: SubstitutingV:100, T:90, andk:8 in Eq. (g),we obtain P : 7.2. (a) Solving Eq. (8) for P when k : 8, we get
Therefore, a: P 8 aTv When T:90 and V: l0o,lPl0T:0.08, which is the answerrequired. (b) From the result of part (a) when T is increased by z 1ind,v rep is 2(0.09) :0.16. We conclude, mains an approximate increase in {ixed) then, that if the temperature is increased from 90" to 92", the increase in the pressure is approximately 0.1,6lblin.z (c) Solving Eq. (8) for V when k: 8, we obtain
Therefore,
_av : AP
8T P2
WhenT90andP
t25
19.4 PARTIALDERIVATIVES 9 1 1
ume necessary to produce the same change in the pressure as obtained in part (b).
which is the instantaneous rate of change of V per unit change in P when T: 90 and P :7.2 if T remains fixed at 90. (d) If P is to be increased by 0.L6 and T is held fixed, then from the result of part (c) the change in V should be approximately (0.16) (435) : #. Hence, the volume should be decreasedby approximately # in.3 if the pressureis to be increasedfrom 7.2lblin.z to7.36lblin.'z. We now extend the concept of partial derivative variables.
19.4.2Definition
to functions of n
x) be a point in R', and let f be a function of the n Let P(xt, x2, . variables x1r x2t . . . , xn. Then the partial derivative of f with respect to *2t
.
.
o
i
*nl
lv'\
1
I/Ltrr'L
IIr
rV
sraY
rvL
/
I
xk is that function, denoted by D rcf, such that its function value at any point P in the domain of f ,t given by Duf@t, x2, lim
.
.
,Xn):
(xr,
(x1, X2, . .
Lxrc
A'ro6
if this limit exists. In particular, If f is a function of the three variables r, y, and z, then the paftLalderivatives of f are given by D t f( x , U , z ) : l i m
f(x + L x , y , z )  f ( x , y , z ) Ar
and
if these limits exist. EXAMPtn 6:
Given
f (x, y, z) x'U * Yz' * zg verify that xf t(x, Y, z) * yfr(x,y, z) t ,ft(x,Y, z) : 3f(*,y,2).
soLUTroN:
Holding y and z constant, we get
f t(x, v, z) :2xY Hotding x artdz constant,we obtain fr(x,Y,z):f+z' Holding x and y constant,we get f"(x, y, z) :2Yz * 3z' Therefore, ) A ( * * z 2 *) z ( 2 y z * 3 z z ) x f { x , y , z )+ y f r ( x , y , z )+ z f r ( x , A , z ) : x ( 2 x 1 +
912
DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES
:' 2x2A* r'y * yz' * 2yz2* 3zg 3(x'y*yzr*zt) : 3f (x, y, z)
Exercises19.4 In ExercisesL through 6, apply Definition '!.g.4.'!, to find each of the partial derivatives. 1. f(x, A) :6x * 3y 7; Drf(x,y) Z . f ( x , A ) : 4 x z  3 x y ;D r f ( x , y ) 3. f (x, A) : 3xy* 6x  Ar;Drf(x, y) 4 . f ( x , A ) : r y z S y * G ; D r f( x , y ) 5.
6.f (x,a):#,
f(x,a): ffi';Drf(x,y)
D,f(x,y)
In Exercises7 through 10, apply Definition 1,9.4.2to find each of the partial derivatives. 7 . f ( x , y , z ) : f y  3 x y ,* 2 y z ;D r f ( x , y , z ) 8. f (x, y, z) : x2* 4y, * 9zz;Drf (x, y, z) 9. f(x, !, z, r, t)  xyr * yzt * yrt * zrt; Dnf@, z, r, t) !,
70.f (r, s, t, u, o, 7u): 3rzstI spa 2tup2 tow * 3uutz; Dsfe, s, t, u, o, w) LL. Givenf (x, y)  x!  9y2.Find Drf (2,1) by (a)apglyrng formula (3);(b) apptnng formula(S);(c)applyrngformula (t) and then replacingx and,y by 2 and 1, respectively. 12' For the tunction in Exercise11,find Drf (2,11by tul applyrr,g*formula (l); (b) applying formula(5);(c) appryingformula (2) and then replacingx and,y by2'and't,i"#"tii"iy. "
llrtil::'.Hl?i[::?l#
j1fi,1i,'*fted
13.f(x, A) :4yt + ffir;
partialderivatives bvholdinsaltbutoneorthevariables constant andappry
Drf (x,y)
L 4f.( x , A )=: &@, i ; D r f ( x , Y )
15.f @, Q) : sin 3d cos2g; Dzf@, O) 1 7 .z  s a t r :
0z
t" " v' uv
1,6.f (r, 0) : r2 cos0  2r tan 0; Drf (r, 0)
L8.r: eocos(0 + e); ! a0
21..f (x, A, z) : 4xyz * ln(2xyz); fs(x, y, z)
29.f (x, y, z) : suuz * tant!,
19.u  (x' + y, + rr1t,r.4
dz
22. f(x, y, z):
2A.u  tanr(xyzut), #
stu sinh 2z eru cosh2z;fr(x, A, z)
fr(x,y, z)
2 4 f ( r , 0 , 0 ) : 4 r z s i n 0 * s e 'c o s0 s i nr f  2 c o s @ ;f r ( r , 0 ,o ) In Exercises25 and 25, find fr(x,y) and fu(x,y). 25. f(x, A) :
fu
J, l n
s i nt d t
27.Givenu sin * h'. v"nfyt +,  ff #:
26.f (x, A) : fu e"*t dt Jr
o.
2 8 . G i v e n r p  * y + y , z I z z x . V e r i f yU + @ , d u ) , "y AxWnE:(r+y*z),
AND THE TOTAL DIFFERENTIAL 913 19.5 DIFFERENTIABILITY
It+ u' if (x'Y)+ (o'o) :1ffi (x, 29.Given f u) rf (x, y) : (0, 0) L0 Find (a\ f'(0, 0); (b)fr(O,0).
^r (\"x,y)+ ( 0,o) , \ :1 l,' +tyl if 30. Givenf(x, A) if (x, y) _ (0, 0) L0 Find(a)f'(0,y) if y * 0; (b)/,(0, 0).
31. For the tunction of Exercise30 find (a) fr(x,0) if r + 0; (b) fr(O,0).  9yz + 4z2* 36: 0 with the plane 32. Find the slope of the tangent line to the curve of intersection of the surface 36f 3). t/tZ, partial (t, as a derivative. this slope Interpret x:1 at the point 33. Find the slope of the tangentline to the curve of intersection of the surfacez: * * y2 with the planey: (2, 1,5). Draw a sketch. Interpret this slope as a partial derivative.
l atthepoint
34. Find equations of the tangent line to the curve of intersection of the surface * * y'* z2:9 with the plane y: 2 at the point (1,2,2).  4y2. lf. distance is measured in feet, 35. The temperature at any point (r, y) of a flat plate is T degrees and 7 : S+ W along the plate in the directions of the find the rate of change o1 the temperature with respect to the distance moved (3, 1). point positive r and y axes, respectively, at the 36. Use the ideal gas law for a confined gas (seeExample 5) to show that
v_ .{. e!_:r aT dP dV g7.lf.V dollars is the present value of an ordinary annuity of equal payments of $100 Per year for f years at an interest rate of 100i percent Per Year, then
11(1+t)'l V : 1 0 0l 
til result of part (a) (a) Find the instantaneous rate of change of V per unit change in i if f remains fixed at 8' (b) Use the time remains to 7vo and the 6vo from changes rate interest if the io find the approximate change in the lresent .raloe (d) Use the at 0'06'' fixed remains if I in f change per unit oIV fixed at 8 years. (c) Find the instantaneous rate of change and T.years 8 to from decreased is time the if value present tlie in change result of part (c) to find the approximate the interest rate remains faed at 6Vo.
19.5 DIFFERENTIABILITY AND THE TOTAL DIFFERENTIAL
In Sec.3.6 in the proof of the chain rule we showed that if I is a differentiable function Of the single variable r and y : f (x), then the increment Ay of the dependent variable can be expressedas Ay: f' (r) Ar * q Lx + 0' where 4 is a function of Ar and q + 0 as Ax From the above it follows that if the function / is differentiable at ro, the increment of.f at ro, denoted by Lf (xs)' is given by A / ( x o ) f ' ( x o ) A r * n L x
where r: O. ligo
For functions of two or more variables we use an equation colTesponding to Eq. (1) to define differentiability of.a function. And from the almition we determine criteria for a function to be differentiable at a
I
914
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
poinl we give the details for a function of two variables and begin by defining the increment of such a function. 19.5.1Definition If /isafunctionof twovariablesxand !,thentheincrementof fatthe point (ro, ye), denoted by Lf(xs, AJ, is given by (2)
Figure 19.5.1illustratesEq. (2) for a function which is continuous on an open disk containingthe points (xo, yo) and (16* Lx,ys+ Ay). The figure shows a portion of the surface,J f(*, y). Af(xo,iJ : ffn, is (16 the point * M,yo!.Ly, f(xo,'y))ana'i isihe point lvhere^Q ( x s * L x , A o *N A , f @ o +A x , y o + A y ) ) . z
(xo,yo,f(xo,VJ)
z: f(x,y) \
(ro * ax,uo* Lv,f@o* ax,ys+ av))
+ Ay)
o)
F i g u r e1 9 . 5 . 1
o rLLUsrBArrow1: For the function defined by /
f(x, y) :3x  xaz we find the increment of at any point (xn, ys). / Lf (xo,Ai : f (xr* M, yo+ Ail f Go, yo) :3(ro * Ar)  (ro * Lx)(yo+ Ay), _ (3ro _ xoAoz) : 316* 3 Ar  xeiloz_ AozLx _ 2*oyoAy _ 2yoAx AV  xo(Ay),  Ax(Ay)z _ 3xs * x6ysz : 3 Ax Ao2Ax ZxoyoLy 2ys Ax Ay _ xs(Ay), _ ArlAyy, . 19'5'2 Definition
rf f is afunction of two variables x and y and the increment of f at (xo, yo) can be written as Af (xo, y) : Drf(xo,Ui Ax * Drf (xs, A) Ly * e1Ar + ez Ly.
(3)
AND THE TOTAL DIFFERENTIAL 915 19.5 DIFFERENTIABILITY
where e1and e2are functions of Ar and Ay such that e1+ 0 and €z+ 0 as (M, Ly)  (0,0), then/is said tobe differentiable at (xs,ys). . rLLUsrRArrou 2: We use Definition 19.5.2 to Prove that the function of Illustration 1 is differentiable at all points in R2. We must show that for all points (xo, yo) in R2 we can find an e1 and an e2 such that
Af(xo,y)  Dtf(xo,U) Lx  D"f(xo,y) Ly: e1Ar I e2Ay and e1+ 0 and €z+ 0 as (Ar, Ay) + (0,0). Because f(x, y):3x rA2, Drf(xo,y) :3  voz and D"f(xo,ao): 2xoyo Using thesevaluesand the valueof Lf (xo'yd from IllustrationL, we have LxDrf (xo,yi AA:xo(N)'2ys Lx LV Lx(LV)' Lf (xo,yo)Dtf(xo,Uo) The right sideof the aboveequationcanbe written in the followingways: l2yo Ay  (AV)'l Ar * (ro Ay) LV or (2y, Ly) Ar * (Ar Ly  xo LY) LV or [(Ay)'] Lx* (2ysArre Ly) LV or 0'Ax*
f  Z y oA r  L x L y xs Lyl Ly
So we have four possible pairs of values for €1 efld €21 €1: 2yo Ly  (Lv)' and €2: xo LA or €r : 2yo Ly
and
G2: Lx Ly  xo LU
€1:  (Ly)'
and
€z: 2Yo Ax  xo LA
€r:0
and
€z: 2Yo Lx  Lx LY  xo LU
or or For each paut,we see that lim
(Lr,au)(o,o)
€r:0
and
lim
(ar,Lu)(o,o)
€z:0
It should be noted that it is only necessary to find one pair of values o for e1 and e2.
L9.5.3 Theorem
If a function f of two variables is differentiable at a point, it is continuous at that Point.
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
pRooF: If / is differentiable at the point (ro, a), it follows from Definition 19.5.2that f(xo+ Ax, yo+ Ay)  f (xo,yi : Drf(xo,y) Lx * Drf(xo,y) Ly *erArie2Ly where €r + 0 and e2+ 0 as (Lx, Ly) t (0,0). Therefore, f (xo+ Ax, Ao+ Ay) : f (xr, ys) * Dlf @s,U) Ax * Drf (xo,y) Ay *erAr*ezLA
Taking the limit on both sides of the above as (Lx, Ly) 
(0, 0), we obtain
l i m . , f ( x o * A x ,A o * A y ) : f ( x o , A o )
(4)
(Ar,ag)(o,o)'
If we let xo* Ax:, ?n!yoi AA:y,,'(Ax, Ay) (0,0)" is equivalent to "(x, y) t (xo, yo)." Thus,we have from (4)
y) : ,,,,1\T,*rf(x' f(x" yo) which proves that / is continuous at (xs, ys).
r
Theorem 19.5.3 states that for a function of two variables difierentiability implies continuity. However, the mere existence of the partial deriv.ativr D'! Drf at a point does not imply differentiability at that ya point. The following example illustrates this. ExAMpLE 1.: Given
f(x,y)
SOLUTION:

f  ( 0 ' 0 ) t i m 0  0 0
if (x, y) # (0,0)
D r f ( 0 , 0 ) t i m f k ' 0 )
if (x, y) : (0,0)
(0,a'r): D,f ; t*' t, l,l''g: ; &],
prove that Dl(0, 0) and Dd@, 0) exist but that f is not differentiable at (0, 0).
u'0
Y0
u0 Y
\
Therefore,both Drl(0, 0) and Drf(0,0) exist. In Example2 of Sec.L9.2we showedthat for this function lim f(x, a) (.r,g)(o,o)'
does not exist; hence,/ is not continuous at (0,0). Because/is not con_ tinuous at (0, 0), it follows from Theorem 19.5.3that is not differenti/ able there. ... P"f9f stating a theorem that gives conditions for which a tunction will be differentiable at a point, we Jonsider a theorem needed in its proof. It is the mean:value t*reoremfor a function of a single variaule appli"d to function of two variables. "
AND THE TOTAL DIFFERENTIAL 917 19.5 DIFFERENTIABILITY
19.5.4 Theorem
Let / be a function of two variables defined for all r in the closed interval la, bl and all y in the closedinterval lc, d). (i) If.D1f@,/o) existsfor some yoinlc, d] and for all xinfa, b], then there is a number f, in the open interval (a,b) such that
f(b,y)  f(a,AJ: (b a)D'.f(t',A)
(s
(ii) lt Drf (xo,y) existsfor somexoin la, bl and for all y in lc , dl , then there is a number f, in the open intewal (c, d) such that (5) : f (xo,d) f (xo,c) (d c)Drf(xo, tr) Beforeproving this theorem, we interpret it geometrically. For part (i) refer to fig. tS.S.i, which shows the portion of the surface z:f(x,y) above the rectangular region in the xy plane bounded by the lines x: a, x: b,y  c, and y : d: The plane y : yo intersects the surface in the curve represlnted by the two equations y: /o and z: f(x, y). The.slopeof the line through the points A(a, yo,f@, y)) and B(b, Ao,f(b, /o)) is
f (b,v)
 f(a' Yo)
ba Theorem 19.5.4(i)statesthat there is some point (tt, Ao,fGr, yJ ) on the curve between the points A and B where the tangent line is parallel to the secantline through A and B; that is, there is some number (1in (a, b) such  a), and this is illustratedin that D/(f1, yJ: lf (b, y)  f(a, y)ll(b the figure, for which Drf (€t, /o) < 0. Figure 19.5.3illustrates part (ii) of Theorem 19.5.4.The plane x: xo interseits the surface,: f (x, y) in the curve rePresentedby the two equaslope of the line through the points tions r:xo ;lfld z:f(x,y).The D(*o, d, f (ro, d))
z:
(tr,ao,fGr,Yo))
f(o,y)

(ro, tr, f(xo' tr)) 
f(b,vo)
z: f(x,Y B(b,Ao, f (b, yo))
Figurc 19.5.2
f(x,Y)
F i g u r e1 9 . 5 . 3
f ( x o ,c )
918
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
C(xo,c,f (xo,c)) and D (xs,d,f (xo,d) ) is [f (xo,d)  (xo,c)]tG c), and f Theorem 19.5.4(ii) states that there is some point (xo, tr, f(A, €")) on the curve between the points c and D where the tangent tine is parallel to thesecantline through c and D; that is, there is somenumber in {, 1c,d,1 such that Drf (xo, €r) : lf (xo,d)  f (xo,c)llG  c). PRooFoF THEoREM19.5.4(i): Let g be the function of one variable r defined by g(r): f(x,yo) Then 8'(r) : Drf(x, yo) BecauseDaf (x, y) exists for arl r in fa, bl, itfoflows thatg'(r) exists for alr x in fa, bf, and therefore g is coniinuous on [a, bl. so u]' ir," meanvarue theorem (4.7.2) for ordinary derivatives there exiits a number (, in (a, b) such that
8'(f') or, equivalently, (b' Yo) f (o' Yo) Drf (€r,Ao) f aa from which we obtain f(b, yr)  f(a, Ao): (b  a)Dl(€r, Ao) The proof of part (ii) is similar to the proof of part (i) and is left as an exercise(seeExercise11). I Equation (5) can be written in the form
f(xo+ h, Ui  fko, A) : yDJ(€r, Ur)
(7)
where f1 is befween xe and xs * h and h is either positive or negative (see Exercise 12). Equation (6) can be written in the form f (xo, yo+ I()  f(xo, yr) : lDrf (xo, €r) (8) is between yo and ao * 11andk is either positive or negative (see Yh"r" f, ExerciseL3). EXAMPIr 2:
f (x, y)
Given
SOLUTION:
By Theorem "l'95.4,there is a number f1 in the open interval
(2, 5) such that
f (s'4)  f (2,4) : (s z)Dl(€,,4)
ANDTHETOTALDIFFERENTIAL9 1 9 19.5DIFFERENTIABILITY find a fr reguired by Theorem 19.5.4if x is in 12,5l a n d y : 4 .
So Fr6,^24 c5:J
(3+fJ
972
5reF (3+ t)':40 Therefore,
3+ fr: +2\m sign and obtain
But because2
€r:z\rc  3
The following theorem states that a function having continuous partial derivatives at a point is necessarily differentiable at the point. 19.5.5Theorem
Let/be a function of two variablesx andy. SupposethatDtf NrdDrf exist on an open disk B (Po;r), where P6is the point (16,y6). Then if Dtf andD2f are continuous at Ps,f is differentiable at Pq. pRooFr Choose the point (xs* Lx, yo* Ly) so that it is in B(Po; r). Then Lf(xo, y) : f(x, * Lx, ys + LV)  f(xo, yo) Subtractingand adding f (xo+ Lx, lJ to the right side of the above equation, we get Lf (xo,y) : lf (xoi Lx, ys+ LV)  f (xo* A,x,yr)f
, + l f ( x o* A r, A )  f ( x o a)l
(e)
Because Drf and Drf exist on B(Po; r) and (ro * Lx, Ao* Ly) is in B (Po; r) , rt follows from (8) that f ( x o * A , x ,A o * A y ) f ( x o * L x , y ) :
( A V ) D r f( x 0+ L x , t r )
(10)
where fr is between lo and Ao* Ly. From (7) it follows that f (xo* Lx, ys) f (xo,Ao) (Ar)D'f (€', Ao)
(11)
where fr is betwe€n .rs and xo * A,x. Substituting from (10) and (11) in (9), w€ obtain L f ( x o , A o ) ( L y ) D r f ( r o * L x , t )
+ (Ar)D'f (€r,Uo)
(12)
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
Because(ro * Lx,Ao* Ly) is in B(Po;r),€zisbetweenyoandAo* Ly, and Drf is continuous at P0,it follows that lim.
,Drf (Ar,Au)(o,o)
(xo* Lx, (r) : Drf (xo, Uo)
(13)
and, becausef, is between xs and xo* Lx and Drf rs continuous at Po, it follows that l i m . , D r'f ( h , A ) (Ar,As) (o,o)
: D r f( x o , A o )
(14)
If we let er: Drf(tv y)  D,,f(xo,yr)
1rs)
it follows from Eq. (14)that lim €r ' :0 ta',lir[to,ot
(16)
and if we let e2: Dd(xs * Ax, {r)  Drf (xr, yJ
Gz)
it follows from Eq. (13) that (18)
,. li+...G2:0 (AJ',au)(o,o) Substituting from Eqs. (15) and (12) into (12), we get Lf (n, yd : LylDd@o, yi * erl * AxlDi@s, yo) * erl from which we obtain Lf (xo, yi : Drf (xo, Ao)Lx * D2fks, y) Ly * e1Ar * e2Ay
(19)
From Eqs. (15), (18), and (19) we see that Definition 19.5.2holds; so is differentiable at (xo, Ai. / I A function satisfying the hypothesis of Theorem 19.5.5is said to be continuouslydiferentiable at the point po. ExAMpu 3: Given f :x2u2  r
.t ir(x,y)*
f(x,A):1r'*y'if (x,y): t 0
(0,0) (0,0)
use Theorem 1.9.5.5to prove that f is differentiable at (0, 0).
solurroN: To find Drf we consider two cases: (x, (x,y) * (0,0). If (x,y): (0,0), w€ have Dl(0,0) lim
f(x,O)f(0,0)
: t T, .;  00  0
(0, 0) and
A
lf. (x, y) * (0, 0) , f (x, y) : *yrl(* * yr). To find Drf (x, y) we can use the theorem for the ordinary derivative of a quotient and consider y as a constant. We have Drf(*, r, 2rY'z(xz !!'z) :?x(fY2) (* * y')'
AND THE TOTAL DIFFERENTIAL 921 19.5 DIFFERENTIABILITY
2xyn
:
1x4 11'
The function Dtf is therefore defined by
2*Yn  D r f( x , A ) : l ( * ' * y ' ) ' Lo
+ ( 0,0) if ( x , y)  ( 0,o)
rir^ (x, y)
In the same manner we obtain the function Drf defined by
, t , ry:^, ir(x,y)+ (o,o)
Drf (x, y) : ](x2  y t
L0
1f.(x, y) : 10,0)
Both D/ and Drf exist on every open disk having its center at the origin. It remains to show that Dtf and D2f are continuous at (0, 0). BecauseDrf(O,0) :0, Drf will be continuousat (0,0) if lim
D'f(x, U):O
(e,s)(oo)
Therefore, we must show that for any e ) 0 there exists a 6 > 0 such that  
'r*ol4 ^nJ
I
l(* *'r1'l
. A o ' e whenever
 ^/
 , l , l yn, = n W f W ) ^ l, ?*!n (x'+Y')' lry " r r l @'* y')':
(20)
 2\W
Therefore,  I
7rt14 tnl

l(x'+fuI "u
o whenever
So if we take 6 : !e, we have (20).Hence, Dt;f is continuous at (0,0)' In the same way we show that D2f is continuous at (0' 0). It follows from Theorem 19.5.5that f is differentiable at (0,0). If we refer back to Eq. (3), the expression involving the first two terms on the right side, Dgf(ro, A) Lx * D2f(xs, yr) NA,is called the principal part of.Lf (xs, yq) or ttre "total differential" of the function/ at (n, y). We make this as a formal definition' 19.5.5Definition lf f isafunctionof twovariables xandy,and/isdifferentiableat(x,y), ttren the total differential of f is the function df having function values given by
(2r) Note that df is a function of the four variables x, y, Lx,and Ly.If
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
z: f(x, A), we sometimes write dz in place oI df(x, y, Lx, Ay), and then Eq. (21) is written as (22) If in particularf (x, y) : x, then z : x, D 1f(x, y) : 1, and D2f(x, y) : 0, and so Eq. Q2) gives dz: Ar. Becausez: /, we have for this function dx: Ax.In a similar fashion,if we take/( x,U) : y, then z: U,Drf (x,y) : 0, and Drf (x,A) : l, and so Eq. Q2) givesdz: Ay. Becausez: y, we have for this function dy : Ay. Hence, we define the differentials of the independent variables as dx: Ar and dy : Ly. Then Eq. e2) can be written as d z  D r f( x , y ) d x * D r f( x , y ) d y and at the point (xo, Uo), we have dz  Drf (xo, A) dx * Drf (xo, U) dy In Eq. (3),letting Lz: Lf (xo,U), dx: Lx, and dy:
(23)
(24)
Ly, we have : Az Drf (xo, Ao)dx + Ddjco, Ao)dy + e1dx * e2dy (25) Comparing Eqs. @a) and (25), we see that when dx (i.e., Ax) and dy (i.e., Ay) are close to zero, and becausethen e, and eralso will be close to zero, we can conclude that dz is an approximation to Az. Becausedz is often easier to calculate than az, we make use of the fact that ilz  Az incertain situations. Before showing this in an example, we write Eq. (23)with the notation 0zl0x and 0zlOyinsteadoLDlf(d, y) and Drf(x, yl, respectively: (26)
ExAMPrn 4: A closed metal can in the shape of a rightcircular rylinder is to have an inside height of 5 in., €ln inside radius of 2 in., and a thickness of 0.1.in. If the cost of the metal to be used is L0 cents per in.t, find by differentials the approximate cost of the metal to be used in the manufacture of the can.
solurroN: The formula for the volume of a rightcircular cylinder, where the volume is V in.8, the radius is r in., and the height is h in., is V  nrzh
(zz7
The exact volume of metal in the can is the difference between the volumes of two rightcircular rylinders for which r : 2.'!.,h: 6.2, and,r : 2, h: 5, respectively. AV would give us the exact volume of metal, but becausewe only want an approximate value, we find dV instead. Using (26), we have
dv: ff a,+{, an
(28)
From Eq. (27) it follows that av _ ^_.t _,_r av and ;:2nrh ffi:
q, Trrz
t
AND THE TOTAL DIFFERENTIAL 923 19.5 DIFFERENTIABILITY
Substituting these values into Eq. (28) gives dV :Znrh dr * nrz dh B e c a u s er  2 , h : 6 , d r 0 . L , a n d d h : 0 . 2 ,
w€ have
dV :2n(2) (5)(0.1)* n(2)'(0.2)
 3.2Tr Hence, AV : 3.2rr, and so there is approximately 3'2rr in'8 of metal in the ' : can. Becausethe cost of the metal is 10 centsper in.3and 1.0 3.2tt:32tr manufacture the in used 100.53,the approximate cost of the metal to be of the can is $1.
we conclude this section by extending the concepts of differentiabilitv and the total differential to a function of n variables. 19.5.7 Definition
L9.5.g Definition
point If l is a function of the n variables x1'11z" " ' xn' antdP is the by given P is at of (xi, xr, . . . 'r'), then the increment f (29) f(P) Af(P) : f(x'+ Lx1,ir* Lx2,. .  ' ?n* Lx) If f is a functio_n of the n variables x1, x2, as f at the Point P can be written
. . , xn, and the increment of
' . . +D"f(P) Axn Lf (P) : D'f (P) Ar, + Drf(P) Lx2* * er Lxt * e z L x 2 * " ' + e n L x n
w h e r e € 1+
(30)
0r€z)0, . . ' ,€n+Q'a;s
(Arr,Lx2,...
,Lxr) +(01 0,
"
,0)
then f ," said to be differentiableat P '
Analogously to Theorem 19.5.5,it can be proved that sufficient conditions foia functio n f of.n variables to be difflrentiable at a point P are that Df , Drf , . . . , bnf all exist on arropen ball B(P; r) and that D/' Dzf, .. . ,'Dnf are all continuous at P. As was the case for functions of" two variabies, it follows that for functions of n vatiables differentiability implies continuity. Howevet, the existence of the partial derivatives Drf, Drf, . . ., Dof ata point does not imply differentiability of the function at the Point. 1.9.5.9Definition
If f isafunctionof thenvariables xy,x2,. . . 'xnarrd/isdifferentiable values at,P, then the total difierential of / is the function df havtng fu nction. given bY df(P,Lxr,Lx2,""Aro):Dl(P)Lx11D2f@)Mz*"'+DJ(P)Ar"(31)
92/T
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
Letting w: f (xr, x2, . . . , xn), definingdx, : Lxt, dxr: L,x2, . dxn: Lx, and using the notation 6w 0x; instead of D1f(P), w€ can Eq. (31)as (32)
EXAMPTn5: The dimensions of a box are measured to be 1o ln., 12 in., and 15 in., and the measurements are correct to 0.02 in. Find approximately the greatest error if the volume of the box is calculated from the given measurements. Also find the approximate percent error.
solurroN: Letting V in.3 be the volume of a box whose dimensions are tr in., y in., and z in., we have the formula
xyz The exactvalue of the error would be found from LV; however, we use dV as an approximation to LV. Using Eq. (32) for three independent variables,we have
av,,av da*{a, dx dy v d z
dV:+ dxt+ and so
dV:yzdx*xzdy+xy
dz
(33)
From the given information lArl < 0.02,lAyl < 0.02, and lArl = 0.02.To find the greatest error in the volume we take the greatest error in the measurementsof the three dimensions. So taking dx:0.02, dy:0.02, dz: 0.02,and r : 10,! : 12,z:1.5, we have from Eq. (33) 6y: (12)(1s) (0.02)+ (10)(15)(0.02)+ (10)(12)(0.02) 9 so, AV  9, and therefore the greatest possible error in the calculation of the volume from the given measurements is approximately 9 ins. The relative error is found by dividing the error by the actual value. Hence, the relative error in computing the volume from the given measurementsis LVIV;= ilVlV:r*o:zfo:0.005. So the approximatepercent error is 0.57o.
Exercises 19.5 1'. If f(x, A) : 3x2* 2xy  A', Ar  0.03,and Ay : 0.02, find (a) the increment of ar ( 1 , 4) and (b) the total differential f off at (1,4). 2. It f(x, y) : xye"u, Ax: 0.1, and Ay : 0.2, find (a) the increment oI at (2,4) f (2,4). 3. If f(1, y, z): xy *ln(yz), Ax:0.A2, Ay:0.04, and &:0.03, total differential of f at (4, 7, S).
and (b) the total differential of f at
find (a) the incrementof / at (4, I , 5) and (b) the
AND THE TOTAL DIFFERENTIAL 925 19.5 DIFFERENTIABILITY
4. If f (x, y, z)  x'y * Zxyz z.3,Lx: 0;0L, Ay: 0.03, and the total differential of f at (3 , 0, 2) .
find (a) the increment of f ut (3 , 0, 2) and (b)
In Exercises5 through 8, prove that / is differentiable at all points in its domain by doing each of the following: (a) Find Lf (xo, y) for the given function; (b) find an e1and an €2so that Eq. (3) holds; (c) show that the et and the e2found in part (b) both approachzero as (Lx, Ay) , (0,0). 6. ( x ' y )  2x2i 3y' 5. (x, : x'Y 2xY
f
f
U)
8'f ( x , y ) vx
7.f (x,,) :i s. Given f (x,a) :
Y 2
{;.
;l:}'r;i;t.
,
and,D2f(1,1) exist, but / is not differentiable at (1, 1).
Prove that Dl[,l)
' (x,d . l3!I ir \" r ' + @,0\ 10. Given f (x, y) : 1f + Vn it (x, y\ : (0, 0) l0 Prove that DlQ, 0) and D2f(0,0) exist, but / is not differentiable at (0, 0). 11. Prove Theorem 19.5.4(ii). 12. Show that Eq. (5) may be written in the form (7) where {1 is between 16 and xs I h. 13. Show that Eq. (6) may be written in the form (8) where {2 is between yn and lo * k. In Exercises14 through 77, use Theorem19.5.4 to find either a f1 or a f2, whichever applies.
14.f(*, a) : x2+ 3xY Y';r is in [1, 3]; Y  4 16.f(x, A) :h,
y LSin l2, 2f;x : 4
15.f (x, y) : xs y'; x is in 12,61;y  3
tr. f (x,y):'ffi;
y LsLnlo,4l;x Z
\'' y) + (0, o) , \  ?:Y:i ir (x, L8.Givenf (x, A) : 1*' + y' rf (x, y) : (0,0) L0 This function is continuous at (0, 0) (see Example 3, Sec. 19.2, and Illustration L, Sec. L9.3). Prove that Drf (0,0) and Drf (0,0) exist but Dtf and Drf are not continuous at (0, 0).
?ven if: (x' Y) + (o'o)
rs.Givenf (x,a) : I # L0
tf (x, y) _ (0, 0)
Prove that f is differentiable at (0, 0) by using Theorem 19.5.5. In Exercises20 and 21, prove that / is differentiable at all points in R3by doing each of the following: (a) Find, Af (xo)ys, zn); G2,and e3,so that Eq. (30) holds; (c) show that the e1,e2,and e3found in (b) all approach zero as (Ax, Ly, Lz) (b) find drr G11 approaches(0, 0, 0). 21,.f (x, y, z) : )cy xz * z2 20. f (x, A, z) :2x22  3yz'
*"
i I(x'v'z)+ (0'0' 0)
L0
it (x, y, z) : (0, 0, 0)
22 .G ivefn( x , y , z ) : Vi fu
(a) Show that D1l(0, 0, O), D2f@, 0, 0) , and D"f (0 , 0, 0) exist; (b) make use of the fact that differentiability implies continuity to prove that / is not differentiable at (0, 0, 0).
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES (
xttzz
if (x'v'z) * (0,0,0) if (x,y, z): (0,0, 0)
23. G i v e n / ( r!,, 2 ) : l 7 T 7 I 7
.O
Prove that / is differentiable at (0, 0, 0). 24. Use the total differential to find approximately the greatest error in calculating the area of a right triangle from the lengths of the legs if they are measured to be 6 in. aid 8 in., respectively, with i possible error of 0.1 in. for each measurement. Also find the approximate percent error. 25. Find approximately, by using the total differential, the greatesterror in calculating the length of the hypotenuse of the right triangle from the measurementsof Exercise24. Alio find the approximate percent error. 26' If the ideal gas law (see Example 5, sec. 19.4) is used to find P when T and,v are given, but there is an error of 0.37o in measuring T and an error of 0.8voin measuring V, find approximately the greatlst percent ertor in p. 27. The specific gravity s of an object is given by the formula A S : AW where A is the number of pounds in the weight of the object in air and W is the number of pounds in the weight of the object in water. If the weight of an obiect inair is read is 20 lb with a possible error of 0.0i lb and its weighiin water is read as 12 lb with a possible enor of 0.02.1b,find approximately the ligest possible ertor in calculating ifrom these measurements.Also find the largest possible relative error. 28' A wooden box is to be made of lumber that is 3 in. thick. The inside length is to be 5 ft, the inside width is to be 3 ft, the inside depth is to be 4 ft, and the box is to have no top. use the total differential to find the approximate amount of lumber to be used in the box. 29' A company has contracted to manufacture 10,000closed wooden crates having dimensions g ft, 4 ft,and 5 ft. The cost of the wood to be used is 5c per square foot. If the machines that are used to cut the pieces of wood have a possible eror of 0'05 ft in each dimension,find approximately, by using the total differential, the greatest possible error in the estimate of the cost of the wood. In Exercises30 through 33, we show that a function may be differentiable at a point even though it is not continuously differentiable there' Hence, the conditions of Theorem 19.5.5are sufficient but not necessaryfor differentiability. The function / in these exercisesis defined bv
f (x,y)
.1 sin
u76,
if (x, Y) + (0,o) if (x, y) : (0, 0)
30. Find A/(0, 0).
31. Find Drf(x, y) and Drf (x, y).
32. Prove that / is differentiable at (0, 0) by using Definition 19.5.2and the results of Exercises 30 and 31. 33. Prove that D1f and.Drf arc not continuous at (0, 0).
19.6 THE CHAIN RULE
In Sec.3.6 we had the following chain rule (Theorem 3.d.1) for functions of a single variable: rf y is afunction of a, defined by y : (u) , and,Dzl exists; f and z is a function of x, definedby u: g(x), anaoru exists; their y is a function of x, and D"y exists and ii given by Dry : D;y Dru
1 9 . 6T H E C H A I NR U L E
or, equivalently, da "/ dx
dy du du dx
(1)
We now consider the chain rule for a function of two variables, where each of these variables is also a function of two variables. 19.6.1 Theorem The Chain RuIe
and lf. u is a differentiable function of x and y, defined by u:f(x,y)' x: F(r, s) , U : G(r, s), and \xlEr, \xl0s, 0yl0r, and Eyl0sall exist,then u is a function of r and s and
pRooF: We prove (2). The proof of (3) is similar. If s is held fixed and r is changed by an amount Ar, then r is changed by an amount Ar and y is changed by an amount Ay. So we have (4) Lx: F(r * Ar, s)  F(r, s) and L y : G ( r * L r ,s )  G ( r , s )
(5)
Because/ is differentiable, (6) Lf (x, y) : Drf (x, y) A'x* Drf (x, v) Ay * e1Ar * e2Ly where er and e2both approach zero as (Lx, Ly) approaches(0, 0). Furthermore, we require that e1:0 and €z:0 when Ax: Ly:0' We make this requirement so that e1and er, which are functions of Ax and Ly, will be continuousat (Ax, Ly\: (0,0). If in (6) we replaceLf (x, y) by Lu,Drf (x, V) by \ul6x, andDzf G, U) by \ulay and divide on both sides by Lr (Lr * O), we obtain Lu Lr
:
0u Lx 0x Lr
Ly A'x , 0u La  T € r   :  T 'Lr Cr.Lr 0Y Lr
Taking the limit on both sides of the above as Ar approaches zero, we get
li$fl vt li$# * ,1i*1.,) Hql # +(1iq.,) H # :#rin ^+,.H Becauseu is a function of r and y and both r and y are functions of r and s, a is a function of r and s. Becauses is held fixed and r is changed by an
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
amount Lr, we have lim
A,r  O
L u _ r : _ u ( r * A r ,s )  u ( r , i l _ O , 'm Lr or
(8)
Also, Lx _ r:_ F(r * Lr,s)  F(r,s)
Lr: ll$
: 0x a,
l i mr y : hAm ro
ay
lim
Ar  0
(e)
and Aro Lf
0r
(10)
Because 0xl0r and 0yl6r exist, F and G are each continuous with respect to the variable r. (rorr: The existence of the partial derivatives of a function does not imply continuity with respect to au of the variables simultaneously, as we saw in the preceding section, but as with functions of a single variable it does imply continuity of the function with respectto each variable separately.) Hence, we have from (4) lim tF(r + L,r,s)  F(r, s)]
lim Ar:
Ar0
Ar0
: F(r, s)  F(r, s)
0 and from (5) lim AY: lim lc(r + Lr, s)  G(r, s)f
Ar0
Ar0
: G(r, s)  G(r, s) 0 Therefore, as Ar approacheszero, both Ar and Ay approach zero. And becauseboth e1and e2approachzeroas (Lx, Ay) approaches (0,0), we can concludethat €z: 0 li^ €r : 0 and Alim r0
( 1 1)
Ar0
Now it is possible that for certain values of Lr, Lx: Ly:0. Becausewe required in such a case that er:€2:0, the limits in (11,)are still zero. Substituting from (8), (9), (10), and (1.1)into (Z), we obtain
.(,H(y) #:(,H(#)
which proves (2), EXAMPLE
].:
uln@
GiVCN
0u
ay
0x
__ a V
dr
p8
19.6THECHAINRULE x : T€r,and y : f or, find 6ul0r
and oulos.
929
0x E:"r From (2) we get 0ux
Ar:W From (3) we get 0u:ffi(rer)+ffi(rtr)W x U , ^\ / .. _c\ r(xes yer) ds
As mentioned earlier the symbols 0ul0r,0ul0s, Eul\x,0ul0y' and so forth must not be considered as fractions. The symbols 0u, 0x,,and so on have no meaning by themselves. For functions of one variable, the chain rule, given by Uq. (t), is easily remembered by thinking of an ordinary derivative asthe quotient of two differentials, but there is no similar interpretation for partial derivatives. Another trouLlesome notational problem arises when considering u as a function of x and y and then as a function of r and s.lf. u: f (x, y) , x: F(r, s), and y: G(r, s), then u: f(F(r, s), G(r, s)). [Note that it is incorrect to write u : f (r, s).1 1: In Example 1, o rLLUsTRArroN
u: f(x, y) :lnty'p +V x: F(r, s) : re" A : G ( r ,s ) : r e  " and so
L t  f G ? , s ) , G ( r ,s ) ) : l n f f i ff(r, s): 6tffi
+ u'f
If we let f(E(r, s), G(t, s)):h(r, written resPectivelYas
s), then Eqs. (2) and (3) can be
hr(r, s) : f ,(x, y)Er(r, s) + fr(x, Y)Glt, s) and hr(r, s) : fr(x, A)F"(t, s) + 1r@,Y)Gr(r' s) In the statement of Theorem 19.6.1 the independent variables are r antd.s, and,u is the dependent variable. The variables r and y can be :alled the intermediate variables. We now extend the chain rule to z intermediate variables and m independent variables. 19.6.2 Theorem Suppose that u is a differentiable function of the n variables.rr, x2t . . . , ChainRule xn,' ind each of these variables is in turn a function of the m variables TheGeneral
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
Ar, Uz, . . ., A*. Supposefurther that each of the partial derivatives A x iAl y j ( i : 1 , 2 , . . . , n ; j  1 , 2 , . , m) exists. Then u is a function
ffi(,H #,:mGH.(#: (,HW) )+
dz\ )xr/
,)
*
:
#(,Hffi).(,H(,H. (.9 The proof is an extension of the proof of Theorem 19.6.1.. Note that in the general chain rule, there are as many terms on the right side of each equation as there are intermediate variables.
EXAMPLE
Given
solurroN:
uxy+xz*yz X: r, A  r cos t, and Z: find 0ul0r and 0ul0t.
r stn t,
By applying the chain rule, we obtain
#:e;e).w)(,*).(Hff) : (y * z)(1) + (x I z) (cosf) * (r * y) (sin f) : y * z * r cos t * z cos f * r sin t i y sin t : / cosf * r sin f * r cosf * (r sin t)(cos t) * rsinf * (rcos f)(sin t) :2r(cos f * sin t) + r(2 sin f cos f) : 2r(cos f * sin t) * r sin 2t
.ffi)(y).ffie,) #:(rJ(#J : (y * z) (O) + (x *z) (r sin r) + (s * y) (r cos f) : (r* r sin f)(r sin t) + (r* r cosf)(r cosf) :r2 sin f  rz sinzt * 12 cost I rz coszt : r2(cosf  sin t) * rz(coszf  sin2f) : rz(cos I  sin t) + 12 cos2t
Now suppose that z is a differentiable function of the two variables r and y, and both r and y are differentiabre functions of the single variable f. Then z is a function of the single variable t, and so instead of the partial derivative of z with respect to f, we have the ordinary derivative
1 9 . 6T H E C H A I N R U L E
given by
of. u
(12) we call duldt given by Eq. (12) the total deiaatiae of u with respect to f. If u is a differentiable function of the n variables x1, x2, . . ., xn and each ri is a differentiable function of the single variable t, then z is a function of f and the totdl derivative of a with resPectto f is given by
EXAMPLE
Given
ux2+2xy*y' )c: f cos t, and y : t sin t, find duldt by two methods: (a) using the chain rule; (b) expressing u in terms of f before differentiating.
solurroN: (a) 6uldx: 2x * 2y, \ul0y  2x * dyldt: sin t + f cos f. So from (12)we have
dxldt: cos t t sin
( 2 x * 2 y ) ( c o s t  f s i n f ) + ( 2 x * 2 y ) ( s i n t + t c o st )
#:
2(r*
il(cos t f sint+ sint+ f cosf) : / ( t c o s t + f s i n f ) ( c o st  f s i n t + s i n t + f c o s f ) : 2 t ( c o s 2 t  f s i n f c o st + s i n f c o st + f c o s 2t + s i n f cos t  t s i n 2t + s i n 2t + f sin f cos f)  2 t [ 1 + 2 s i n f c o st + f ( c o s 2 t sin2t)] :2t(1 + sin 2t + t cos2t) (b) u : (f cos t)' + 2(t cos t) (t sin f ) + (t sin t)2  szcos2t + t'(2 sin f cos f ) + t2 sinz f : t2 + tz stnZt
4:2t dt
+ 2t stn2t + Ztz cos2t
4: lt f is a differEXAMPLE soLUTroN: Let u  tbxz  *ay'. We wish to show that z f (u) satisfies the given equation. By the chain rule we get entiable function and a and b z: that prove are constants, a n d *ay: + Ydu: ay f ' ( u )(  n y ' ) !*a'lE \f Gax' *ayt) satisfiesthe Partial 0x du 0x J' differential equation
Y:+Y:f'(u)(bx)
Therefore,
' 0 2* b x * : a'Y' dy ax
a y z l f ' ( u ) ( b x )*l b x l f ' ( u ) (  a y ' ) l: 0
which is what we wished to Prove.
gilI
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
EXAMPLE5:
Use the ideal gas law (see Example 5, Sec. "1,9.4) with k  10 to find the rate at which the temperature is changing at the instant when the volume of the gas is L20 in.3 and the gas is under a pressure of 8 lb/in.2 if the volume is increasing at the rate of 2 in.t/sec and the pressure is decreasin g at the rate of 0.1.lb/in.2 per sec.
soLUTroN: Let t  the number of seconds in the time that has elapsed 7Tr
T D
Tf 
PV:
since the volume of the gas started to increase; the number of degreesin the temperature at t sec; the number of pounds per squareinch in the pressure at t sec; the number of cubic inches in the volume of the gas at t sec. F^ PV
L0T and so
r:1o
At the given instant, P : 8, V  !20, dPldt: 0.1, and dvldt 2. Using the chain rule, we obtain
d: T  aTdP.aTdv dt aPift'avdt : y d P _' L u 70 itt t0 dt :1#(_0.1) + fo(Z) :1.2* 1.6 :0.4 Thereforethe temperatureis increasingat the rate of 0.4 degreeper second at the given instant.
Exercises19.6 In Exercises1 through 4, findthe indicated partial derivative by two methods: (a) Use the chain rule; (b) make the substitutions for r and y beforc differentiating.
1. u xz y'; x: 3r s;y : r * 2s;#ry, 3. u  sutc'x:2r
cost; A : 4r sint, !, U 0r' at
2. u  3x2 * )cy 2y,+ 3x y; x: 2r 3s;y : r * s;#, 4 . u  x 2+ y , ; x : c o s hr c o st ; A : s i n h r s i n t ; U = O u or' at
In Exercises 5 through 10, find the indicated partial derivative by using the chain rttle.
5. Lt: sinl(3r* il; x: rler;y : sin rs; #, X
5. u  xeo; )c tanr (rst); y : ln(3rs
7. u cosh * : 3r2s; y : 5se'; X, #r#
8 . u : , c y + x z * A z ;x :
9. u xzI y' + 22;x  rsin @ cos0; y :r 10. u  x'yz; )c!;
ou. ou V : res;z re'r; 0r' ds
' ott ou ou sin @sin o; z: r cos a;
*; ao,M
r s ;y : r z  s 2 ;
X
19.6 THE CHAIN RULE
(b) make the substitu
In Exercises L1 through 14, find the total derivative duldtby two methods: (a) Use the chain tions for x and y or fiorx, y, and z before differentiating' 12. u : l n x: cos f; y  sin f LL. u: ye' + )cea; 13.u
x:
W;
t*
In ExercisesL5 through 18, find the total derivative duldt by using the chain differentiating.
x  rnt; a  et 1s.u tanr e), L 7 u.  Hy t,
xy+y';x:et)y: er
1,4.u :  . t ; x : 3 ye'
t a n t ; y  c o sf ; z  s t n t ; 0 < t < L n
933
et
sin t; A:
lnf
do not expres s u as a function of t before
1 , 6u.  x y + x z +
f cos
f sin
18. u ln(xz* y'+ t2);x  f sin t; U: cos f
x : l n f ;y : t " 1
In Exercises1"9through 22, assumethat the given equation defin es z as a function of x and y. Dlfferentiate implicitly to find 0zl0x and 0zlAy. 19. 3x2+ y'+
z2  3xy * 4xz  15 0
20. z (x'*
yz) sin xz
Zg. If f is a differentiable function of the variable u,let u: a(azlAx) + b(azlay) : 0, where a and b ate constants.
21. ye*o" cos 3xz  5
bx  ay and prove that z:
24. If.f is a differentiable function of two variables u arrd.o,let u: satisfiesthe equation 0zl6x* Azl0y 0.
 ay) satisfies the equation f (b*
x  y and a : !  r and prove that z: f (x  y ' y ')
ZS.If f rs a differentiablefunction of x and y andu f (x,A), x r cos 0, and 0u dx
22. zeazI Zxe*'  4e'o : 3
r sin 0, show that
0u ^ 6usin0 costA0 r 0r
0u 0u ^ ducos0 srn0+ ae r a": ar 2 6 . f f a n d g a r e d i f f e r e n t i a b l e f u n c t i o n s oxfa n d y a n d u : f ( x , i l dvl}x, then if x: r cos0 and'y: r sin 0, show that 'l,.  0u 0o 6v du
ar:V N
anda: g(x,V),suchthat dulax:oldyandauly:
and ar:7 ao
27. Supposefis a differentiablefunction ofr and y and.u: f(x,y). Then ifx: 6ul0p and duldw in terms of 0ul0x and 6ul6y.
cosh 7)cosToandy:
sinh o sin at, exPress sin 0, and
r sin d 2 8 . S u p p o s e / i s a d i f f e r e n t i a b l e f u n c t i o nxo, yf , a n d z a n d u : f ( x , V , z ) . T h e n l f x : r s i n @ c o s 0 , and ul0z' \ul0y, of. ul6x, terms in arrd aulae 6ul0Q, ,:, {, expressdul0r, "o" the 29. At a given instant, the length of one leg of a right triangle is 10 ft and it is increasing at the rate of 1 ftlmin and lengti of the other leg of tie right triangle is 12 ft and it is decreasing at the rate of 2 ftlmin. Find the rate of change of the measure of the acute angle opposite the leg of length 12 ft at the given instant'
the 30. A vertical wall rnakes an angle of radian measure Sa with the ground. A ladder of length 20 ft is leaning against ladder, by the formed triangle the area of is the How fast ftlsec. rate of 3 at the wall doiyn tne wall and its top is sliding the wall, and ihe g"orttJ changing when the ladder makes an angle of ta'radians with the ground? which 31. A quantity of gas obeys the ideal gas law (see Example 5, Sec.19.4) with k: 12, and the gas is in a container lb/in'2 and is 5 pressure is is being heated at a raie of 3o per second. If at the instant when the temperature is 300o,the instant. that at the volume of decreasing at the rate of 0.1 lb/in.'z per second, find the rate of change
9:II
DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES
ZZ. l'tater is flowing into a tank in the form of a rightcircular cylinder at the rate of *z fd/min. The tank is stretching in such a way that even though it remains cylindrical, its radius is increasing at the rate of 0.002ftlmin. How fast is the surface of the water rising when the radius is 2 ft and the volume of water in the tank is 20zfp?
19'7 HIGHEROR_DER If f is a function of two variables, then in general Drf and.D4f are also PARTIAL DERIVATIVES functions of two variables. And if the pa*iil derivatives of these functions exist, they are called second partial derivatives of. In contrast, f. Dtf and D"f are called first partial derivatives of f. There are four second partial derivatives of a function of two variables. If f is a function of the two variables r and !, the notations
Dr(Dl)
Drrf
f ,,
f *o
Arf 0y 6x
all denote the second partial derivative of /, which we obtain by first partialdiffere_ntiating/ with respect to r and then partialdifferentiating the result with respect to y. This second partial derivative is defined bi ft(x,U):lim
r
(1)
Ly
AuO
if this limit exists. The notations
Dr(Drf )
Drrf
fn
f ,,
Arf 0xz
all denotethe secondpartial derivativeof /, which is obtainedby partialdifferentiatingtwice with respectto r. we have the definition r (x * Ax,y) fr(x, y)
fu@,y):lntr
e)
if this limit exists. we define the other two second partial derivatives in an analogous way and obtain frr(x, a) : lim
(3)
f'@' Y * LY) fr(x' Y) f ,r(x, a) : lim ay
(4)
and As0
if these limits exist. The definitions of higherorder partial derivatives are similar. Again we have various notations for a specific derivative. For example,
Drrrf
frr"
f rcu
atf av axax
tlay$
all stand for the third partial derivative of which is obtained by partial/, differentiating twice with respect to r and then once with respect to y.
PARTIALDERIVATIVES 935 19,7 HIGHERORDER
Note that in the subscript notation, the order of partial differentiation is from left to right; in the notation 03f l0y 6x 0r, the order is from right to left.
EXAMPLE
f (x,u) :
Given e* sin
find: (a) D'f (x, (c) atfli.x oyz.
SOLUTION:
y+lnxy (b) D 'rf (x,
Drf (x, Y)
.,1 s r n '/u t 
)cy
er stn V +*
So (a)Dr,;f(r,A) : e" siny  Ll*; and (b) Drrf (x, y) : s" cos y. (c) To find AsflaxAy2,we partialdifferentiate twice with respect to y and then once with respect to x. This gives us .L slnut'  ,
cos
y'
a'f 6x 0y2
sin y
Higherorder partial derivatives of a function of n variables have definitions which are analogous to the definitions of higherorder partial derivatives of a function of two variables. lf f is a function of n variables, there may be n2 second partial derivatives of f at a particular point. That is, for a function of three variables, if all the secondorder partial derivatives exist, there are nine of them: /rr, f ,r, fr", frr, frr, fr", f"r, 1"r, and flr.
EXAMPLE
Given
f (x, A, z)  sin(xy * 2z) findDrtf(ic,y,z).
SOLUTION:
D'f (x, A , z )  y c o s ( x y * 2 2 ) D"f (x, y , z)  2y sin (xy * 2z) D t r r f( x , y , z )
EXAMPLE
Given
f ( x , a ) : f Y  y cosh xy find: ( a) Dt, f (x, U); ( b ) D r , , (f * , y ) .
sin (xy * 2z)  2xy cos(xY * 2z)
SOLUTION:
(a) Drf (x,y)  3x2U Y2 srnh xY Drrf(x,Y) :3x2  2Y sinh xY  xY' coshxY (b) Drf (x,Y) : x3 cosh )cY xY sr'.h xY Dr,,f(x,y):3x2  y sinh xy  y sinh xy xy'cosh xy :3)c2  2y sinh xy  xy' cosh ry We see from the above results that for the function of ExamPle 3 the "mixed" partial derivatives Dtrf (x, y) and Drrf(x, V) are equal. So for with this particular functioh, when finding the second partial derivative This immaterial. is differentiation respect to x and then !, the order of examPle following the However, condition holds for many functions' shows that it is not alwaYs true.
93G
DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES
ExAMPLE4: Let f be the function defined by t.
\ x27rz
f ( x , a ) : l ( xfvf )i Lo
i r( x , Y ) ( 0 ,0 ) if (x,y)
( 0 ,0 )
Find frr(0, 0) and frr(0, 0).
solurroN: In Example3, Sec. 19.4,we showed that for this function  y for all y f '(0, Y) and for allx fr(x,O) : r From formula (1) we obtain ft (0, o) : lim Ag0
But from (5),f ,(0, Ly) Ay andf ,(0, 0) and so we have O{,, O: lim (1) : 1 frr(0,0): lim Auo
ay
Auo
From formula (3), we get
f'(o * Lx' o)  fr(o' o) frr(o,o) : lim A,x From (5),fr(Lx,0) : Ax and fr(0,0) :0. Therefore,
lim#: f r , , ( oo,) : Aal0 AI
hm L: l Atr_Q
For the function in Example 4 we see that the mixed partial derivatives/tr(r, y) and frr(x, y) are not equal at (0,0). A set of'conditions for yo):fu(xo, ye) is given by Theorem 1g.7.'!., which follows. thi+ ft(xo, The function of Example 4 does not satiify the hypothesis of this theorem becauseboth f' and fn are discontittuou" at (0, 0). It is left as an exercise to show this (see Exercise20). 19.7.1 Theorem
*qp::g that / is a funcrion of rwo variables x and y defined on an open disk B((16, yJ; r) and f*, fu, f,r, ^d fu, also are aefinea on B. Furthermore, suppose that f ,u and fu, are continuous on B. Then f ,u(h, uo): fo,(xo, yo1 pRooF: consider a square hgifg its center at (xs, yo) and the length of its side 2lhl such that 0 < frlhl < r. Then ail the ioi"t, in the interior 9f the square and on the sides of the square are irithe open disk B (see F i g . . l 9 . 7 . l _S)o. t h e p o i n t s ( x r * h , y o * h ) , (xr*h,yo), ind 1xo,yoih1 are in B. Let A be defined by A: f(xo* h,yo+ h)  f@o+ h, y) f ( x o , V o +h ) + f ( x o , y o ) ( z ) Consider the function G defined by
Figure
c(r)  f(x, Ao* h)  f(x, yr)
(8)
PARTIAIDERIVATIVES937 19.7HIGHER.ORDER Then
G(x+h) f(x+h,Ao+h) f(x+h,Ar) So (7) can be written as AG(ro+h)G(ro) From (8) we obtain (10) G' ( x) : f"(x, Uo+ h)  f"(x, yo) Now, becausefr(x, Ao*h) and fr(x, Ao) are defined on B, G'(r) exists if r is in the closed interval having endpoints at re and xn*h. Hence, G is continuous if r is in this closed interval. By the meanvalue theorem (4.7.2)there is a number c, between xq and xs1' h such that
G ( r o+ h )  G ( r o ) : h G ' ( c ' )
(11)
Substituting from (11) into (9), we get A  hG'(ct)
(r2)
From (12) and (10) we obtain * h)  f,(ct, U)f A  hlf " ( c r ,U o Now Lf I is the function defined bY
s @ ) f " ( c r , Y )
(13)
(14)
we can write (L3) as
A  hls@o+ h)  s(yo)l
(ls)
From (L4) we obtain 8'(Y)  f"o(h,a)
(15)
Becausef,u(cr, y) is defined on B,g'(y) exists if y is in the closed interval having endpoints at yq and AoI h; hence, g is continuous if y is in this closed interval. Therefore, by the meanvalue theorem there is a number d1 between Ao and Ao* h such that (J7) g\o + h)  sg) : h7'(d') Substituting from (1,7)into (15),we get A:h2{ lows that L:
hzf,o(ct,dt)
(dr); so from (16) it fol(18)
for some point (c1,d1)in the open disk B. We define a function { by (1e) QQ) : f (xo+ h,y) f ( x o ,V ) + h ) . T h e r e f o r cQ, ) c a n b e and so Q@+h):f(xo*h,y+h\f(xo,A written as
a  6 u o + h ) f u o )
(20)
DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES From (19) we get
Q ' @ ): f o ( r o* h , Y )  f o 7 o , Y )
(2r)
{'exists if y is in the closed interval having yo and Ao*h as end.points because by hypothesis each term on the right side of (21) exists on B. Therefore, Q is continuous on this closed interval. so by the meanvalue theorem there is a number d2 between ye and * h such that ao
Q Q o +h )  A Q J = h O ' @ , ) From (20),(2L),and (22)it follows that A  hlfr(ro * h, dr)  fr(xo, dr)f
(22)
(23)
Define the function 1 by
x(r) : fo@,k)
e4)
and write (23)as A : h l y ( x 6 +h )  x ( r o ) l From (24)we get x'(x) : fo,@, d")
(2s)
e6) and by the meanvaluetheoremwe concludethat there is a number c2 betweenxoand xs* h suchthat x ( r o+ h )  X @ ) : h x ' ( c r ) e7) From (25),(26),and e7) we obtain A: h2fnr(c2,dr)
(2S)
Equating the right sides of (1g) and (2g), we get hzf,n(cr, ilr) : hrfur(cz, ilz) and becauseh # 0, we can divide by hr, which gives fru(cr, d):
fnr(cr, dr)
eg) where (cr, dr) and (c2,d) arc in B. Becausec, and c, are each between ro and xs * h, we can twite c, : xs*e1h, where 0 ( e, < L, and cz:xo*e2h,wheie0 ( e, < 1. Similarly, becauseboth i\ and d2,arebetween youia!o*h,*".1n iit" d,r:yo ,"h, where 0 ( es 0 and f ,,(a, b) > 0, and we wish to prove that f(a, b) is a relative minimum function value. Becausefr, f ,n, and,fuo are continuous onB((a,b);r), it follows that f is alsocontinuous on B. Hence, there existsan open disk B' ((a, b); r' ), where r' = r,such that Q@,y) > 0 and f",(x, A) > g for every point (r, y) in B'. Let h and k be constants, not both zero, such that the point (a + h, b + k) is in B'. Then the two equations x:a*ht
and y:b+kt
0 f(a,b)
I
(iv) is included to cover all possible cases.
ExAMPLEL: Given that / is the function defined bY
f(x,A):2f*Y'x22Y determine the relative extrema of f it there are arry.
SOLUTION:
:2Y  2 f , ( x , y ) : 8 x B 2 x f u @ ,Y ) x:0, and x:*' Setting Settingfr(x,y):0, we get x:t, and Therefore y:1. git , f n bolh vanish at the Points f, fok, y):6, *" (+, L), (0, 1), and (t, 1). To apply the secondderivative test, we find the second partial derivatives of / and get
 Z f,,(x, Y) :24f f, , (  i , 1 ) : 4 > 0 and
foo(r,Y) :2
f,u(x'Y) :0
:4' 2 o:8 > o f . , ? E , l ) f o o ? t ,t ) f * o ' e + , 1 )
Hence , by Theorem z}.g.g(i) , f has a relative minimum
value at
r)
AND LINEINTEGRALS OF PABTIALDERIVATIVES, APPLICATIONS GRADIENTS, DERIVATIVES, DIRECTIONAL
 : 4 ( 0 : f ,,(0, 1)fou(0,L) f ,u'(0, 1) (2) (2) 0 and so by Theorem20.3.8(iiil,/ does not have a relative extremumat (0,1). and
>0
f ""(*' l)
t)  f,o'G, t1  4 ' 2  0 : 8 > 0 f ,*(t, L)foo(L, Therefore,by Theorem 20.3.8(i)/ has a relative minimum value at (+, 1). Hence, w€ concludethat / has a relative minimum value of  I at each of the points (i, t) and (t, L7.
ExAMP:.;.2. Determine the relative dimensions of a rectangular box, without a top and having a specific voluril€, if the least amount of material is to be used in its manufacture.
Let )c the number of units in the length of the base of the box; y  the number of units in the width of the base of the box; z  the number of units in the depth of the box; S  the number of square units in the surface area of the box; V  the number of cubic units in the volume of the box (V is constant). x, y, and z are in the interval (0, +o;. Hence, the absolute minimum value of S will be among the relative minimum values of S. We have the equations solurroN:
S:xy*2xz*2yz
and V:ryz
Solving the second equation fot zin terms of the variables r and y and the constant V, we get z : Vlry . And substituting this into the first equation gives us S : x1y*tx! + ! Differentiating, we get
as E:y
2v f
azs 4v
M:7
ds 2v ay:rT azs
ffi_l
A2S
4V
oy'
y'
Setting 6Sl0x:0 and 0Sl0y:0, and solving simultaneously,we get x 2 Y 2 V : 0 r y 2 2 V : 0 from which it follows that r : 9fr
, and y : W.
For these values of r
20.3 EXTREMAOF FUNCTIONSOF TWO VARIABLES
and A, we have
azs_ 4v dxz (W)3 a2S 0x2
_  4 v : ) '> n/ v 2v
a 2 S _ ( a r s \ r _:@'798ffr _4V L_l_3>o
@w)
From Theorem 20.3.8(i), it follows that s has a relative minimum value nd y:{N' and hence an absoluie minimum value when x:flTv From these values of r and Y we get
etrWe therefore conclude that the box should have a square base and a depth which is onehalf that of the length of a side of the base' Our discussion of the extrema of functions of two variables can be extended to functions of three or more variables. The definitions of relative extrema and critical point are easily made. For example, if.f is a function : : of the three variablis r, y, alrrdz, arrd f ,(xo, Yr, zo) fo(xo, yo' zo) (rs, ys, zo) is a critical point of /' Such a point is fr(xo, yo,zo):0, then 'oUtui""a by solving simultaneously three equations inthree unknowns. For a functio n of.n variables, the cltical points are found by setting all the n first partial derivatives equal to zero and solving simultaneously the n equati6ns in n unknowns. The extension of Theorem 20.3'8 to functions of tht"u or more variables is given in advanced calculus texts. In the solution of Example 2 we minimized the function having function values ry * 2xz * 2yz, iubject to the condition that r, A, ad:rdz satr;sfy Cbmpare this with Example 1, in which we found the equatioi ryr:V. :2f * y2 * 2V' the relative extrema of t