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= A (iii) A U AI = S (iv) If A
~
B, then Au B = B
(v) AnS=A
n 4> = 4> A n AI = 4>
(vi ) A (vii)
(viii) If A
3.1.3
~
B, then An B = A.
Cartesian product of sets
Def. 9 Cartesian product: The cartesian product between two sets A and B is defined as A x B = {(x, y) : x E A and y E B}. For example, let A = {1,2} and B = {a,b} then A x B = {(I, a), (1, b), (2, a), (2, b)}. Q. 2 Define cartesian product of sets and give one example of it. Ex. 3 (i) A = {1,2}, B = {3, 4, 5} and C (B n C) = (A x B) n (A x C). (x is the cartesian product).
= {4, 5, 6}
then show that A x
(ii) If A = {I, 2, 3} and B = {3,4}, then find AU B, An B, (A x B) and (B x A).
(iii) For the sets A
A xB (iv) If A
=1=
= {a,b,c}
and B
= {1,2}
and C
= {2, 3},
verify that
B x A.
= {a, b}, B = {I, 2}
find (A x B) U (A x C).
CH.3: ABS'fRACT ALGELRA •
3.1
SOLUTION:
(i) B n C = {4, 5}, A x (B n C) = {(I, 4), (1,5), (2,4), (2, 5n A x B = {(I, 3), (1,4), (1, 5), (2,3), (2,4), (2, 5n A x C = {(1,4),(1,5),(1,6),(2,4),(2,5),(2,6n . . '. (A x B) n (A x C) = {(I, 4), (1,5), (2,4), (2, 5n. Hence A x (B n C) = (A x B) n (A x C). (ii) A = {I, 2, 3} and B = {3,4}. AUB = {1,2,3,4},AnB = {3} A x B = {(I, 3), (1,4), (2,3), (2,4), (3, 3), (3, 4n and B x A= {(3,1),(3,2),(3,3),(4,1),(4,2),(4,3n. (iii) A x B = {(a, 1), (a, 2), (b, 1), (b, 2), (e, 1), (e, 2n and B x A = {(I, a), (1, b), (1, e), (2, a), (2, b), (2, en . . '. A x B of B x A. (iv) A x B = {(a, 1), (a, 2), (b, 1), (b, 2n and A x C = {(a, 2), (a, 3), (b, 2), (b, 3n. (A x B) u (A x C) = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3n.
Exercise 2 (i) Find the cartesian product A x B of the sets A and B where A = {1,3,5} and B = {2,4}. (ii) If A
= {I, 2}, B = {2, 3}, C
= {3,4}. Find
A x (B U C).
[Ans. (i) {(I, 2), (1, 4~.(3, 2), (3, 4), (5,2), (5, 4n, (ii) {(I, 2), (1,3), (1,4), (2, 2), (2,3), (2,4) }.]
Def. 10 Power set: The collection of all sub-sets of a set A including A, is called power set of A and it is, generally, denoted by P(A). Ex. 4
(i) Write down the sub-sets of the set {x, y, z}.
(ii) Write down all the subsets of the set S = {p, q, r}, what is the power set of the set S? •
SOLUTION:
(i) The subsets of {:r,y,z} are 1>,{x},{y},{z}, {x,y},{y,z}, {z,x} arid
{x,y,z}. (ii) The subsets of S are 1>,{p},{q},{r},{p,q},{q,r},{r,p} and {p,q,r}. The power set of S is
{1>, {p}, {q}, {r }, {p, q}, {q, r}, {r, p}, {p, q, r} }. Note 3 If a finite set has n elements, then its power set has 2n elements.
3.8
V.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Theorem 1 DeMorgan's laws For any two sets A, B
(i) (A U BY = A'
n B'
n B)' = A' U B', where 'A" denotes the complement of the set A.
(ii) (A
Proof: (i) Let x be any element of (A U B)' x E (AUB)' x¢(AUB)
x ¢ A and x ¢ B x E A' and x E B' x E (A' n B'). Hence (A U BY = A'
n B'.
(ii) Proof is similar to (i). Ex. 5
(i) By Venn diagram prove that A - (B U C) = (A - B)
n (A -
C).
(ii) Show by Venn diagram that An (B n C) = (A n B) n C . • SOLUTION:
(i) In 'figure (a), the vertical lines represent the set (B U C) and horizontal lines represent the set A. The region shaded only by horizontal lines represent the set A - (B U C). In figure (b), the horizontal lines represent the set (A - B) and vertical lines represent the set (A - C). The double shaded region represents the set (A - B) n (A - C). Both the regions are identical, hence A - (B U C) = (A - B)
n (A -
C).
A
(a) Figure: 3.1.2
(b)
(ii) In figure (a), the vertical lines represents the set A and horizontal lines represents the set B n C. The double shaded region represents the set An (B
n C).
CH.3: ABSTRACT _\LGEBRA
3.9
In figure (b), the vertical lines represents the set (AnB) and horizontal lines represent the set C. The double shaded region represents the set (A n B) n C. Both the regions are identical, hence An (B n C) = (A n B) n C. A
C B
of
B
Figure: 3.1.3
3.2
Relation
Def. 1 A relation between two sets A and B is a sub-set of Ax B and is denoted by R. Thus RcA x B. The notation xRy is used to represent (x, y) E R. xRy is read as 'x is related to y'. If A = B, then we say that R is a relation in the set A. Def.2 For the set A the relation R = {(x,y) : x E A,y E A,x = y} is called the identity relation in A and is denoted by fA. For example, if A = {a, b, e} then
3.2.1
fA
= {(a, a), (b, b), (e, en·
Types of relations
Def. 3 Reflexive relation: Let A be a set and R be the relation defined in it. R is said to be reflexive, if
(a, a) E R for all a E A i.e., every element of A is related to itself. A relation R in a set A is not reflexive if there be at least one element a E A such that (a, a) rt R. Def. 4 Symmetric relation: Let A be a set and R be the relation defined in it. R is said to be symmetric, if
(a, b) E R ::::} (b, a) E R i.e., aRb::::} bRa for all (a, b) E R. A relation R in a set A is not symmetric if (a, b) E R does not imply (b,a)ER.
U.G.
3.10
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Def. 5 Anti-symmetric relation: Let A be a set and R be the relation defined in it. R is said to be anti-symmetric, if (a, b) E Rand (b, a) E R
=?
a = b.
A relation R in A is not anti-symmetric, if there exist elements a, b E A, a b such that (a, b) E Rand (b, a) E R.
i-
Def. 6 Transitive relation: Let A be a set and R be the relation defined in it. R is said to be transitive, if (a,b) E Rand (b,c) E R
=?
(a,c) E R,a,b,c E A.
A relation R is not transitive if (a, b) E Rand (b, c) E R do not imply (a,c) E R.
Def. 7 Equivalence relation: A relation R, defined in A is said to be an equivalence relation if and only if (i) R is reflexive, i.e., aRa for all a E A. (ii) R is symmetric, i.e., aRb
=?
bRa.
(iii) R is transitive, i.e., aRb and bRc
=?
aRc.
Ex. 1 If R be a relation in the set of natural numbers N defined by the expression' (x - y) is divisible by 5', i.e., R = {(x, y) : x, yEN, (x - y) is divisible by 5},
prove that R is an equivalence relation . •
SOLUTION:
(i) Let x E N. Then x - x = 0 which is divisible by 5. Therefore, xRx for all x E N. Hence R is reflexive. (ii) Let xRy hold. i.e., x - y is divisible by 5 =? -(y - x) is divisible by 5 =? (y - x) is divisible by 5 =? yRx. Hence R is symmetric. (iii) Let xRy and yRz, i.e., (x - y) and (y - z) are divisible by 5 ::} [(x - y) + (y - z)] is divisible: by 5 =? (x - z) is divisible by 5 =? xRz. Hence R is transitive. Thus R is an equivalence relation.
CH.3: ABSTRACT ALGEBRA
3.11
Ex. 2 Let A be the set of integers. Show that the relation R defined by 'a ~ b', that is, R = {(a, b) : a,b E A,a ~ b},
is not an equivalence relation . •
SOLUTION:
(i) Let a E A. Then a ~ a holds. i.e., aRa holds. Hence R is reflexive.
(ii) Let aRb holds, i.e., a ~ b. This does not imply b ~ a. Hence R is not symmetric. Thus R is not an equivalence relation.
3.3
Mapping
Def. 1 Mapping: Let A and B be two non-empty sets. If there exists ·a correspondence, denoted by f, which associates to each element x of A a unique element y of B, then we say that f is a mapping of A into B. The mapping of A into B will be denoted by f : A -+ B. The set A and B are called respectively domain and codomain (or range) of the mapping f. The element y is denoted by f(x) and called the f image of x or the value of the function for x.
Def. 2 Range: Let f : A -+ B be a mapping. The range of f consists of those elements in B which appear as the image of at least one element of A. Def. 3 One-to-one mapping: Let f : A -+ B be a mapping. If for Xl i= X2 =} f(XI) i= j(X2) then the mapping f is called one-to-one or one-one or 1-1 mapping. e.g., the mapping f : R -+ R where f(x) = x is one-to-one mapping. Def. 4 Onto mapping: Let f : A -+ B be a mapping. If f(A) the mapping f is called onto mapping. e.g., the mapping f : R -+ R where f(x) = x i'l onto mapping.
=B
then
Def. 5 Identity mapping: If each element of a set is mapped on itself then it is called the identity mapping. The mapping f (x) = x for all x E A is an identity mapping.
U.G.
3.12
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Def. 6 Many-to-one mapping: If two or more elements of A correspond to the same elements of B through the mapping f then it is, called many-to-one m~pping.
e.g., the mapping f : R --+ R where f(x)
= x 2 is many-to-one mapping.
De£ (-a).b + a.b = 0 (by right distributive law) .'. (-a).b is the additive inverse of a.b. Hence (-a).b = -(a.b), for all a,b E F. (iii) Let e be the multiplicative identity of F. Then a.a- 1 = e where a-I E F is the inverse of a. Also we have (a- I )-1.a- 1 = e. ... (a- I )-1.a- 1 = a.a- 1 or, (a -1) -1 = a (by right cancellation law).
Ex. 10
(i) In a field F, prove that a2 = b2 => either Po
(ii) Prove that a ring R is commutative if (a a,bE R
+ b)2' =
= b or a = -b. a 2 + 2ab + b2 for
all
(iii) If a, b E F (a field) and a =I 0 then show that there exists a unique element x such that a.x = b.
V.G.
3.28 •
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
SOLUTION:
(i) Given a 2 = b2 . Adding (_b 2 ) on both sides we get, a 2 + (-b 2) = b2 + (_b 2) (_b 2 is the inverse of b2) 2 2 =} a + (_b ) = 0 =} a.a - b.b = 0 =} a.a + a.b - a.b - b.b + 0 = 0 =} a.(a + b) - (a + b).b = 0 (by distributive) =} a.(a + b) - b.(a + b) = 0 (by additive commutative) =} (a + b).(a - b) = 0 (by distributive) Since field is without zero divisor, therefore, either a - b = 0 or a + b = 0, i.e., either a = b or a == -b. (ii) We have (a+b)2 = a2 +2ab+b2 or, (a + b).(a + b) = a.a + 2a.b + b.b or, a.(a + b) + b.(a + b) = a.a + 2a.b + b.b (by distributive law) or, a.a + a.b + b.a + b.b = a.a + 2a.b + b.b (by distributive law) or, a.b + b.a + b.b = 2a.b + b.b (by left cancellation law) or, a.b + b.a = a.b + a.b (by right cancellation law) or, b.a = a.b (by left cancellation law). Hence R is commutative. (iii) Since a =f. O.... a-I exists and a-I E F. Left multiplying ax = b by a -1 we get a-l.ax = a-l.b or, e.x = a-l.b (left identity) or, x = a-l.b. Ex. 11 (i) If in a ring R, a2 = a for all a E R, prove that a+b = 0 for all a, b E R.
=}
a= b
(ii) If D be an integral domain and x be an element of D such that x 2 = x, then prove that x = 0 or 1.
(iii) If a, bE F, F is a field and b =f. 0, then prove that a = 1, when (ab)2 = ab2 + bab - b2.
(iv) If a, b, e, d are the members of a ~eld then show that ~ = ~ if ad = be . •
SOLUTION:
(i) Let a E R. By additive closure a+a = (a+a)2 = (a+a)(a+a)
CH.3: ABSTRACT ALGEBRA
= (a 2 + a 2)(a 2 + a2) (by distributive law) = (a + a)(a + a) (since a2 = a) or, (a + a) + 0 = (a + a) + (a + a) (by additive identity) or, 0 = a + a (by left cancellation) i.e., a + a = O. Now, we have a + b = O. . a + a = a + b.. By left cancellation, a = b.
(ii) Let us consider the relation x(x - 1) = x 2 - x (by distributive law) = 0 (since x 2
= x)
Since D is an integral domain, so it is without zero divisor. Thus, x = 0 or l. (iii) Given (ab)2 = ab2 + bab - b2 (ab)(ab) = (ab)b + bab - bb (aba)b = (ab)b + bab - bb aba = (ab) + ba - b (by right cancellation) aab = 2ab - b (as ab = ba for a field) aab = (2a - l)b aa = (2a - 1) (by right cancellation) ,*aa-a-a+1=O (a - l)(a - 1) = 0 a-I = 0 ,*a=l.
'* '* '* '* '* '* '*
'*
'*
(iv) We have ~ = :. ab- 1 = ed- 1 b d b-1a = ed- 1 b(b-1a) = bed- 1 a = bed- 1 ad = bed-1d ad = bc. Again ad = be add- 1 = bed- 1 a = bed- 1
'* '* '* '* '*
'*
'* '* b-1a = b-1bed'* b-1a = ed- 1 '* ab- 1 = ed- 1 a e '* b = "d'
1
Ex. 12 Prove that every field is without zero divisor.
3.29
3.30
U.G. MATHEMATICS
(~HORT
QUESTIONS AND ANSWERS)
• SOLUTION: Let F be a field and a, bE F. We shall prove that, if a.b = 0 then either a = 0 or b = 0 for all a, bE F. Suppose a.b = O. Let a ¥= 0, so a-I exists. Now, a-l(a.b) = a-I.O => (a-Ia).b = 0 => b = O. Similarly, if b ¥= 0 then a = O. Thus F has no zero-divisor.
Note 1 A field is an integral domain.
".
Ch.4 II Geometry Two Dimensions 4.1
Transformation of Axes
Change of origin without change of direction of axes (translation): If (x, y) and (x', y') be the old and new coordinates respectively of a point P then x = x' + a,y = y' + (3 where (a,(3) is the new origin w.r.t. to the old coordinates system. Rotation of rectangular axes without changing the origin (Rotation): If (x, y) and (x', y') be the coordinates of the same point P in old and new coordinates system then x = x' cos 0 - y' sin 0, y = x' sin 0 + y' cos 0, where 0 is the angle of rotation. Equation of the curve after rotation at an angle 0: If the axes are rotated at an angle 0 then the curve ax 2 + 2hxy + by2 + 2gx + 2fy + c = 0 becomes in the form ax,2 + by,2 + 2g'x' + 2f'y' + c' = 0, where 0 is given by
2h tan 20 = --b' aCombination of translation and rotation: If the origin is shifted to (a, (3) and the axes are rotated through an angle 0 in the anticlock wise direction (positive) then the changes the coordinates (x, y) of a point P are given by x = a + x'cosO - y'sinO, y = (3 + x' sinO + y'cosO, where (x',y') are the coordinates of P in new system of axes. Ex. 1 Find the equation of the line
~+~
= 2, when the origin is transformed
to the point (a, b). •
Putting x = x' + a and y = y' The transformed equation is SOLUTION:
~+a
-a-
~+b
+ -b- = 2 or,
~ ~
~
+ b.
+ b + 2 = 2 or,
~ ~
~
+ b = O.
Ex. 2 (i) Find the transformed from of the equation x 2 - y2 = 4 when the axes are turned through an angle of 45° keeping the origin fixed.
V.G.
4.2
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(ii) Find the translation which transforms the equation x 2 + y2 - 2x + 14y + 20 = 0 into X,2 + y,2 = 30. (iii) Find the angle through which the axes are to be rotated, so that the equation 17x 2 + 18xy -7y2 = 1 may be reduced to the form AX2 + By2 = 1.
(iv) The coordinates axes are rotated through an angle 60°. If the transformed coordinates of a point are (2V3, -6), find its original coordinates. (v) Show that the equation 4xy - 3x 2 = 1 is transformed to x,2 - 4y,2 = 1 by rotating the axes through an angle tan- 1 2. (vi) To what point the origin is to be transferred to get rid of the first degree terms from the equation 8x 2 + 10xy + 3y2 + 26x + 16y + 21 = O? (vii) Transform the equation 3(12x - 5y + 39)2 - 2(5x + 12y - 26)2 = 169, taking the lines 12x - 5y + 39 = 0 and 5x + 12y - 26 = 0 as the new axes of x and y respectively. (viii) What will be the form of the equation x 2 - y2 = 4, if the coordinates axes are rotated through an angle -
7r
2"'
(ix) Find the angle of rotation of the axes for which the equation x 2 - y2 = a 2 will reduce to xy = c2 . Determine c2 . (x) Find the equations of the following when ax+by+c = 0 and bx-ay+d = o are considered as axes of x and y respectively. (ax + by + c)2 = a 2 + b2 and (ax + by + c)(bx - ay + d) = a 2 + b2 . (xi) Show that if the origin is transferred to (0,1) and the axes are rotated through 45°, the equation 5x 2 - 2xy + 5y2 + 2x - 10y - 7 = 0 referred x,2
to new axes becomes
y,2
3 + ""2 =
1.
(xii) P is the point (2,4). If the axes are turned through an angle of 30° keeping the position of the origin unchanged. Find the new coordinates of P . •
SOLUTION:
(i) For rotation x = x' cosO-y' sinO = x' cos 45°-y' sin 45° = x' /V2-y' /V2 and y = x' sinO - y' cos 0 = x' /V2 + y' /V2. The transformed equation is
(
X' - y')2 _ (X'
V2
+ y')2 = 4
vI2
or, -4x'y' = 8 or, :r'y'
+2 =
O.
4.3
CH.4: GEOMETRY
(ii) Given equation is x 2 + y2 - 2x + 14y + 20 = 0 or, (x _1)2 + (y + 7)2 = 30. Substitute x-I
=
x' and y + 7
. '. the translation is x' (iii) The angle of rotation tan2e .
=
=
x-I, y'
y' we get x,2 + y,2
= 30 .
= y + 7.
e is given by
2h a-
= --b where a = 17,b = -7,2h = 18.
tan 2e
18
3
1
= 17 _ (-7) = 4" or, e = "2 tan
-1
3
4"'
(iv) Let (x, y) and (x', y') be respectively the old and new coordinates after transformation . . '. x
=
x' cose - y' sine and y
Here x' = 2V3, y' = -6 and
= x' sine +
e=
y' cose.
60°.
Then
. '. the original coordinates are (4V3, 0). (v) Let ,
e = tan- 1 2 or, tane = 2. . n
Slnu
Then x
y
=
2
{)
1
= J5' cos u = J5' ,
, .
x'
= x cos e- y sm e = -
J5
x' sin 0 + y' cos 0 =
2x'
2y'
- -
J5'
y'
J5 + J5'
Substituting the values of x and y to the equation 4xy - 3x 2 = 1 we get
~(x' -
J5
2y'). (2x'
+ y')
J5
-
~(x' -
2y')2 = 1
5
or, 4(2x,2 - 2y,2 - 3x'y') - 3(x,2 - 4x'y' or, 5x,2 - 20y,2 = 5 or, x,2 - 4y,2 = 1.
+ 4y'2)
= 5
(vi) Putting x = x' + 0' and y = y' + (3. Then 8(x' +0')2+ lO(x' +O')(y' + (3)+3(y' +(3)2 +26(x' +0') + 16(y' + (3) +21
=0
4.4
V.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
or, 8x /2 + 10x'y' + 3y/2 + x'(16a + 10/3 + 26) + y'(10a + 6/3 + 16) + 8a 2 + 10a/3 + 3/32 + 26a + 16/3 + 21 = O. To remove the first degree terms, putting 16a + 10/3 + 26 = 0 and lOa + 6/3 + 16 = O. Solving we get a
= -1, /3 = -1.
. '. the required point is (-1, -1).
12x - 5y + 39 12x - 5y + 39 = J144 + 25 13 _ 5x + 12y - 26 _ 5x + 12y - 26 Y . J25 +144 13
(vii) L~X=
~
d
The given equation reduces to 3(13X)2 - 2(13y)2 = 169 or, 3X2 - 2y 2 = 1, where X = 0 and Y = 0 are new axes. (viii) If (Xl, yl) be the new coordinates of a point (x, y) then x = Xl cos e yl sine, y = Xl sine + yl cose. Here the angle of rotation Then x
e=
-11'/2.
= Xl cos( -11' /2) - yl sin( -11' /2)
y = Xl sin( -11' /2)
+ yl cos( -11' /2)
= yl and
= _Xl.
The equation reduces to y/2 - x/2 = 4. (ix) Putting x = Xl cos e - yl ~in e and y = Xl sin e + yl cos e, we get (Xl cos e yl sine)2 - (Xl sine + yl cose)2 = a2 or, X/2 (cos 2 e - sin 2 e) + y/2(sin 2 e - cos 2 e) + xlyl (-4 sin e cos e) = a 2. To reduce to the form xy = c2, putting cos 2 e - sin2 e = 0 or, tan 2 e = 1 or, tane = ±1 or, () = ±11'/4, which is the angle of rotation. The equation reduces to =r=4x'y'(sin1l'/4.cos1l'/4) = a2 I I 1 1 2 I I 2/ or, =r=4x y . J2' J2 = a or, x y = ±a 2.
Obviously c2 = ±a2 /2. (x) Let >X
Y
=
ax + by + c and Y
~2+~
=
bx - ay
+ d.
J~+~
The given equations reduce to (a 2 + b2)X2 = a2 + b2 and (a 2 + b2)XY = 2 a + b2, or, X 2 = 1 and XY = 1 where X = 0 and Y = 0 are the equations of the new axes.
4.5
CH.4: GEOMETRY
(xi) If the origin is transferred to (a,j3) i.e., (0,1) and the axes are rotated through 45° then
x= a
+ x' cos 0 -
~
~
= v'2 - v'2
y' sin 0 = 0 + x' cos 45° - y' sin 45°
and y
~
Putting these values to the given equation we obtain
5 or, 2"(x' - y')2 -
2( x' ~y' ) - (x' -
y')(x'
+ y') + 5
+5(X'~Y')2 + lO(x'~Y') + 2(X'~Y')
5 +_(X'2 2 or , 4x,2
+ 2x'y' + y'2) X,2
12 = 0
y,2
+ 6y,2 = 12 or , - 3 + -2 = 1.
(xii) Let (x', y') be the new coordinates of the point (x, y). Then x = x' cos 0 - y' sinO, y = x' sinO + y' cos O. Here x
.
,V3,1
or, or
= 2, y = 4 and 0 = 30°.
2 = x . 2""
V3x' -
-
, S + 4V3
=
,I,V3
y . 2" and 4 = x . 2"
y' = 4 and
x'
y'
~
= 13 + x' sin 0 + y' cos 0 = 1 + v'2 + v'2
x' + V3y'
+ y . 2""
= S.
1 =--4 + SV3 3 + 1
4.6
U.G.
,
or , x =
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
8 + 4V3 In, -4 + 8V3 In = 2 + y 3' y 4 = = -1 + 2y 3. 4
. '. the new coordinates are (2
+ V3, -1 + 2V3).
Exercise 1 (i) Find the angle of rotation of the coordinates axes about origin which will transform the equation x 2 - y2 = 4 to xV = 2. (ii) Find the coordinates of the point (-2,4) referred to new axes obtained by rotating the old axes through an angle of 45° in the positive direction. (iii) Find the transformed equation of the line y = V3x, when the axes are rotated through an angle in the positive sense.
i
(iv) Find a new origin without rotating the axes in order that the equation y2 + 2y - 8x + 25 = 0 will turn out to be y2 = 4ax. Determine a. [Ans. (i) 37rj4, (ii) (V2, 3V2), (iii) y'
= 0, (iv)
(3, -1), a
= 2.]
Ex. 3 (i) If the expression ax + by changes to a' x' + b' y' by a rotation of the rectangular axes about the origin, prove that a2 + b2 = a,2 + b'2. (ii) By shifting the origin to the point (a, /3) without changing the directions of axes, each of the equation x - y + 3 = 0 and 2x - y + 1 = 0 is reduced to the form ax' + by' = O. Find a and /3 . •
SOLUTION:
(i) Let the transformation be x y = x' sinB + y' cos B. Then ax + by
= a( x' cos 8 - y' sin 8) + b( x' sin 8 + y' cos 8) = x' (a cos 8 + b sin B)
+ y' ( -a sin B + b cos B)
+ b'y' where a' = acosB + bsinB and b' = -asinB + bcosB. Now, a,2 + b,2 = (a cos B + bsinB)2 + (-asinB + bcosB)2 = a2(cos 2 B + sin2 B) + b2(cos 2 B + sin2 B) + 0 = a 2 + b2 • Putting x = x' + a, y = y' + /3. Then the given equation becomes (x' + a) - (y' + /3) + 3 = 0 and 2(x' + a) - (y' + /3) + 1 = 0 or, x' - y' + (a - /3 + 3) = 0 and 2x' - y' + (2a - /3 + 1) = O. According to the problem a - /3 + 3 = 0 and 2a - /3 + 1 = O. Solving we get a = 2, /3 = 5. = a'x'
(ii)
= x' cos B - y'sinB,
4.7
eR.4: GEOMETRY
Ex. 4 Show that the equation of the circle x 2 + y2 = a2 is invariant under the rotation of axes . • SOLUTION: Putting x = x' cos 0 - y' sinO, y = x' sinO + y' cos O. Then x 2 + y2 = a2 becomes (x' cos 0 - y' sinO)2 + (x' sinO + y' cos 0)2 = a 2 or, x'2(cos 2 0 + sin2 0) + y'2(sin 2 0 + cos 2 0) +2x'y'( - sinO. cos 0 + cos O. sinO) = a2 or, x,2 + y,2 = a2. Hence the given equation is inv>triant w.r.t. rotation of axes.
4.2
Pair of Straight Lines
Some useful formulae on pair of straight lines: (i) A second degree homogeneous equation of the form ax 2 + 2hxy + by2 represents a pair of straight lines through the origin iff h 2 ~ abo (ii) The angle between the lines ax 2 + 2hxy + by2 = 0 are
0= tan- 1
(
(iii) Condition of coincidence: If 0 hence h 2 = abo
±
=0
2)::
~ ab).
= 0 then the lines are coincident and
(iv) Condition of perpendicularity: If 0 coefficient of x 2+coefficient of y2 = O.
=
rr/2 then a
+b =
0 i.e.,
(v) The equation of bisectors of the angles between the lines ax 2 + 2hxy + by2 = 0 are
(vi) The equation ax 2 + 2hxy + by2 + 2gx + 2fy + c straight lines if a h 9 h b f = O. 9 f c
= 0 represents a pair of
Ex. 1 Find the equation of the pair of straight lines joining the origin to the points of intersection of the line x + 2y = 5 and the parabola y2 = 8x. • SOLUTION: Since the pair of lines passing through the origin and the intersection of x + 2y = 5 and y2 = 8x, so it is a homogeneous equation of . 2 x + 2y x and y. To make homogeneous we wnte y = 8x = 1.8x = --5-.8x or, 5y2 = 8x 2 + 16xy, which is the required pair of the straight lines.
4.8
V.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Ex. 2 (i) Find the value of k for which the equation x2_y2_2x+4y+k = 0 may represent a pair of lines. (ii) For what value of A, does the equation xy a pair of straight lines?
+ 5x + AY + 15 =
0 represent
(iii) Show that 6x 2 - 5xy - 6y2 + 14x + 5y + 4 = 0 represents a pair of straight lines which are perpendicular to each other. (iv) Show that the equation 4x 2 + 12xy + 9y2 -16x - 24y + 16 a pair of coincident straight lines.
= 0 represents
(v) Show that the equation x 2 + 2V3xy + 3y2 - 3x - 3V3y - 4 = 0 represents a pair of parallel lines, by factorising the left hand side. (vi) Test whether the equation x 2 + 6xy + 9y2 - 5x -15y + 6 = 0 represents a pair of parallel straight lines. (vii) For what value of A, 3x 2 + AXY - 5y2 straight lines ?
+ 2x + 2y
= 0 will be a pair of
(viii) Show that the equation x 2 + 6xy + 9y2 + 4x + 12y - 5 = 0 represent pair of parallel lines and find the distance between them. (ix) Find the angle between the pair of straight lines represented by the equation 3x 2 - 10xy + 3y2 = O. (x) Prove that the equation y3- x 3+3xy(y-x) = 0 represent"s three straight lines. (xi) Find the equation of the lines which pass through the origin and whose distance from (h, k) are equal to d. •
SOLUTION:
(i) Comparing the given equation with ax 2 + 2hxy + by2 + 29X + 21Y + c = 0 we have a = l,h = O,b = -1,9 = 1,1 = 2,c = k. If the curve represents a pair of straight lines then a h 9 h b I 9 I c or, 1( -k - 4)
(ii) Here a
1 0 1 0 -1 2 1 2 k
+ 1(0 + 1) = 0 or,
=0 -k - 3
= 0 or,
k = -3.
= 0, h = 1/2, b = 0,9 = 5/2, 1= A/2 and c = 15.
If the given curve represents a pair of straight lines then
o
1/2
1/2
0 A/2
5/2
5/2 A/2 15
= 0
4.9
eH.4: GEOMET'lY
or _!(15 _ 5),) +~(~-O) =Oor 5), = 15 or ).=3. '22424 '44' (iii) Here a = 6, h = -5/2, b = -6, 9 = 7, f = 5/2 and c = 4. h
h b
f
9
f
c
a
Now,
9
6 =
-5/2 7
-5/2 7 -6 5/2 5/2 4
= O.
Hence the given curve represents a pair of straight lines. Here the coefficient of x 2 + coefficient of y2 = 6 - 6 = o. . .. they are perpendicular to each other. (iv) The given equation is 4x 2 + 12xy + 9y2 - 16x - 24y + 16 = 0 or, (2x + 3y - 4)2 = 0 or, 2x + 3y - 4 = 0, 2x + 3y - 4 = O. Hence the given curve represents a pair of coincident straight lines. (v) The given equation can be written as x 2 + x(2V3y - 3) + (3y2 - 3V3y - 4) or x
=
0
= -(2V3y - 3) ± V(2V3Y - 3)2 - 4.l.(3y2 - 3V3y - 4)
,
2 =
-(2V3y - 3) 2
± =
V(12 y2 - 12V3y + 9) - (12y2 - 12V3y - 16) 2
-(2V3y - 3) ± 5 2
or, 2x = -2V3y + 3 ± 5 or, 2x + 2V3y = 8 and 2x or, x
+ 2V3y =
-2.
+ V3y = 4 and x + V3y = -2.
Hence the given equation represents a pair of parallel straight lines. (vi) The given equation can be written as x 2 + x(6y - 5) + (9y2 - 15y + 6) = 0
or, x = =
-(6y - 5) ± v'(6y - 5)2 - 4.1.(9y2 - 15y + 6) 2 -(6y - 5) ± 1 2
V.G.
4.10
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
or, 2x = -6y + 5 ± 1 or, 2x + 6y = 6 and 2x or, x + 3y = 3 and x + 3y = 2.
+ 6y =
4
Hence the given equation represents a pair of parallel straight lines. (vii) Here a'= 3, h = >"/2, b = -5, 9 = 1, f = 1, c = O.
Now, D. =
a h 9
h 9 b f f c
3 >../2 1 >../2 -5 1 1 1 0
=
= 2 + >...
If the equation represents a pair of straight lines then D. = 0 or, 2 + >.. = 0 or, >.. = -2. -
(viii) The given equation can be written as x 2 + x(6y + 4) + (9y2 + 12y - 5) = 0
or, x
=
-(6y
+ 4) ± V(6y + 4)2 2
= -(6y ~ 4)
±6 =
4(9y2
-(3y + 2)
the straight lines are x
+ 12y - 5)
± 3.
+ 3y = -5 and x + 3y = 1.
Distance from (0,0) to the first line is
~= ~ 1+9
yl0
and that of from
· . -1 second 1me IS r;n' yl0 Hence the distance between the lines is
Iyl0 ~ - ylO -;,; I= yl0 !n.
(ix) We have 3x 2 - lOxy + 3y2 = 0 or, (3x - y)(x - 3y) = 0 or, 3x - !J = 0, x - 3y = 0 or, y = 3x and y = ix. Hence the angle (9) between these lines is
tan 9 =
3 _1 31
1 + 3'3
4
4
= -3 or, 9 = tan -1 -3'
Alternative method: The equation is of the form ax 2 + 2hxy + by2 = O. ... a = 3, 2h = -10, b = 3. The angle between these lines is
CH.4:
tan()
=
2../h 2 - ab a+b
or, () = tan
_1
=
2J25 - 9
3+3
GEO~1ETRY
4.11
4 3
=-
4
3.
(x) Given that y3 - x 3 + 3xy(y - x) = 0 or, (y - x)(y2 + xy + x 2 ) + 3xy(y - x) = 0 or, (y - x)(y2 + 4xy + x 2) = 0 or, (y - x)(y + 2x ± ../3x) = 0 or, y - x = 0, y + 2x ± ../3x = o. The three straight lines are y
=
x, y
= (../3 -
2)x,y
= (-../3 -
2)x.
(xi) Let y = mx be the equation of the line passing through the origin. Distance from (h, k) to the line is
k - mh 1= d (given) IJl+m2 or, (k - mh)2 = d2 (1 + m 2) or, m 2(h 2 - d 2) - 2khm + (k 2 - d 2) = 0 2kh
±
J4k 2h 2 - 4(h 2 - d 2 )(k 2 - d 2 )
or, m = ----'-----2-'-----'-'----'= kh
±
Jk 2h 2 - (h 2 - d 2)(k 2 - d 2)
Hence the required lines are y = (kh
± dJh2 + k 2 -
d 2)x.
Exercise 1 (i) Find the value of k for which the equation x 2 -y2+2x+k = o may represent a pair of straight lines. (ii) Examine whether the equation 2x 2+5xy+3y2+5x+6y+3 = 0 represents a pair of straight lines.
(iii) Find a value of k so that the equation
z2
+ kxy + y2 - 5x - 7y + 6 = 0
may represent a pair of straight lines. (iv) Test whether the equation 4x 2 - 4xy + y2 -12x + 6y + 8 = 0 represents a pair of parallel straight lines. (v) Find the lines represented by the equation x 2 + 2xy sec () + y2 = O. (vi) Find the angle between the pair of lines x 2 + 2xycosec a
+ y2 =
O.
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
4.12
(vii) Show that the equation 4x 2 + 4xy + y2 + 4x + 2y + 1 = 0 represent pair of parallel lines and find the distance between them. (viii) Show that the equation of the lines passing through the origin and perpendicular to 5x 2 - txy - 3y2 = 0 is 3x 2 - txy - 5y2 = o. [Ans. (i) k = 1, (ii) yes, (iii) -5/2,10/3, (iv) yes, (v) x (vi) 7r/2 - a, (vii) 0.]
= (-secO±tanO)y,
Ex. 3 If p be the perpendicular drawn from the origin upon a line whose ·
1
mtercepts on the axes are a and b, show that 2 a
+ b12
1 = 2. p
• SOLUTION: Since the intercepts on the axis are a, b then the equation of Y 1 · . x t h e 1me IS ;; + b = . Distance from the origin to this line is
-1
IJ~+b
1 or, 2 a
I = p (given) .
1
1
+ b2 = p2·
Ex. 4 (i) Show that origin and the point (1, -2) lie on opposite sides of the line 3x - 4y - 5 = O. (ii) Find the locus of a point which moves so that its distance from the axis of y is double the distance from the point (2, 2). • SOLUTION: (i) Distance from the origin to the given line
-5
(d1 )
= J9+16 =
(d )
= 3.1 - 4.( -2) V9 + 16
9 + 16
-1
< 0 and that from 5
=~
(1, -2)
O.
5> ,Since d1 and d2 have opposite signs then the origin anq the point (1, -2) are in opposite side of the given line. 2
(ii) Let (a, (3) be the point. Distance from (a,(3) to y-axis is a. Also distance between (a, (3) and (2, 2) is J(a - 2)2 + ((3 - 2)2. According to the problem a = 2J(a - 2)2 + ((3 - 2)2 or, a 2 = 4(a 2 + (32 - 4a - 4(3 + 8) or, 3a2 + 4(32 - 16a - 16(3 + 32 = O. Hence the locus of the point is 3x 2 + 4y2 - 16x - 16y + 32 =
o.
CR.4: GEOMETRY
4.13
Ex. 5 (i) Show that the condition for one of the straight lines given by 2 ax + 2hxy + by2 = 0 may coincide with one of the straight lines giyen by a'x 2 + 2h'xy + b'y2 = 0 is 4(ah' - a'h)(hb' - h'b) = (ba' - b'a)2. (ii) The gradient of one of the straight lines represented byax 2+2hxy+by2 =
o is twice that of the other.
Show that 8h 2
= 9ab .
• SOLUTION: (i) Let the common line be Ix + my = O. Let (lx + my)(llx + mlY) = ax 2 + 2hxy + by2. Comparing we have lh = a, mml = band lml + 11m = 2h. Again let (Ix + my)(l2x + m2Y) = a'x 2 + 2h'xy + b'y2, where 112 = a', mm2 = b' and 1m2 + 12m = 2h'. Now, 2(ah' - a'h) = lh(lm2
+ 12m) -1l2(lml + 11m)
= 12(hm2 -12ml).
Similarly, 2(hb' - h'b) = m2(ltm2 - mI12). Again, a'b - b'a
= 112mml -
mm21lt
= lm(ml12
-11m2).
Hence 4(ah' - a'h)(hb' - h'b)
(ii) Let
= 12m 2(m112 -11m2)2 = (a'b (llx + mly)(l2x + m2Y) = ax 2 + 2hxy + by2.
b'a)2.
On comparison we have lt12 = a, mlm2 = b and 11m2
+ 12m 1 = 2h. The lines are itx + mlY = 0 and 12x + m2Y = 0
11 12 or, y = - - x and y = --x. ml
m2
According to the problem - 11- = - 212- or, ltm2 = 212ml ml
(1)
m2
Now, ab = 1112mlm2 = (212ml)(l2mt} = 2(l2mt)2 [using (1)] and 2h = ltm2 + 12ml = 312ml·
Ex. 6 Find the vertex and latus-rectum of the curve 3x 2 + 12x - 8y • SOLUTION: We have 3x2 + 12x - 8y = 0 or, x 2 + 4x
+4 =
iy + 4
= O.
U.G.
4.14
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
8(3)
8
or, (x + 2) 2 = 3 y + 2" or, X 2 = 3Y, where X = x + 2 and Y = y + 3/2. The curve represents a parabola with vertex at (-2, -3/2) and latus rectum is 8/3 (as 4a = 8/3).
Ex. 7
(i) Find the pole ofthe line 5x-6y = 10 with respect to the parabola y2 = 5x.
(ii) Find the pole of the focal chord of the parabola y2 through (9a,6a) . •
=
4ax, passing
SOLUTION:
(i) Let (a,iJ) be the pole. The polar of it w.r.t. theparabolaisyiJ= ~(a+x) or, 5x - 2yiJ + 5a = O. -5a Comparing it with the given line, we have, -55 = -2iJ
-6
10'
or, iJ = 3, a = -2 .. '. the pole is (-2,3). (ii) The equation of the chord passing through the focus (a, 0) and (9a, 6a) is 6a y = -9- - ( x - a) or, 3x - 4y - 3a = O. (1) a-a Let (a, iJ) be the pole. The equation of the polar of (a, iJ) is yiJ = 2a(x + a) or, 2ax - iJy + 2aa = O. It is identical with (1). Comparing we get
2a iJ 2aa 8a - = - = - - or, a = -a,iJ =-. 3 4 -3a 3 .'. the required pole is (-a, 8a/3).
Ex. 8 Is the line 2x + 4y = 9 normal to y2 = 8x ? Find the foot of the normal. Let (a, iJ) be a point on the parabola y2 = 8x . ... iJ2 = 8a (1) The equation of tangent at (a, iJ) is yiJ = 4(x + a) or, 4x - yiJ + 4a = O. (2) The equation of normal at (a, iJ) is a line perpendicular to (2) . . '. its equation is iJx + 4y + k = O. It is identical with the given line. Then iJ 4 -k - = - = - or, iJ = 2, k = -9. 2 4 9 From (1), 8a = 4 or, a = 1/2 . . '. the given line is normal to the parabola and the foot of the normal is (1/2,2).
•
SOLUTION:
4.15
CH.4: GEOMETRY
(i) Show that 3x+4y+7 = otouches the circle'x 2+ y 2-4x-6y-12 o and find the point of contact. (ii) Find the tangent to x 2 + y2 - 6x - 6y + 14 = 0 from (5,5) .
Ex. 9
•
=
SOLUTION:
(1·) "tliT vve h ave 3x
+ 4 y + 7 = 0 or, x =
-4y3 - 7
Putting this value to the equation of the circle.
(-4~ -
7) 2 + y2 _ 4(-4 Y - 7) _ 6y -
0
12 =
3
or, (4y + 7)2 + 9y2 + 12{4y + 7) - 54y - 108 = O. or, 25y2 + 50y + 25 = 0 or, y2 + 2y + 1 = 0 or, {y + 1)2 = 0 or, y = -1,-l. Now, x = {4 - 7)/3 = -l. . .. the point of intersection between the line and the circle is (-I, -1). Since the number of point of intersection is one so the line touches the circle.
(ii) The equation of straight lines passes through (5, 5) are y - 5 = m(x - 5) which is a tangent of the given circle. The center and radius of the circle are respectively (3,3) and V9 + 9 - 14 = 2. Now the distance from mx - y + 5 - 5m = 0 to the point (".3,3) is 3m - 3 + 5 - 5m I = 2 1
2
or,
Vl+m 2
1
-
2m
VI +m2 I = 2
or, (1- m)2 = 1 + m 2 or, 1 + m 2 - 2m = 1 + m 2 or, m = O. . . . the required tangent is y = 5. Exercise 2 (i) Find the pole of the line x 2 x + y2 - 2x + 5 = O.
+ 2y + 3
= 0 w.r.t the circle
(ii) Find the equation of the tangent and the normal at (1, -1) to the conic y2 _ xy - 2x2 - 5y + x - 6 = O. [Ans. (i) (2, 2), (ii) x
+ 4y + 3 = 0 and 4x -
y - 5 = 0.]
U.G.
4.16
4.3
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Polar Equation
(i) Relation between cartesian and polar coordinates: If (x, y) and (r,8) are respectively cartesian and polar coordinates of a point P then x = r cos 8, y = r sin 8 and r = J x 2 + y2, 8 = tan -1 Jt. x (ii) Distance between two points: If A(rl. ( 1 ) and B(r2' ( 2) be two points then the distance between A and B is AB =
Jr~ + r~ - 2r1r2 COS(82 - ( 1 ).
(iii) Area of a triangle: Let A(rl. ( 1 ), B(r2' ( 2), C(r3, ( 3 ) be the vertices of a triangle AB C then the area is
(iv) Polar equation of a straight line: The general form of a straight line is
~ r
=
A cos 0 + B sin 0, where A and B are constants. The slop of the
line is -AlB. (v) Polar equation of the line }:Jassing through two points: If the line passing through the points (rl' 8d and (r2' ( 2) then its equation is
(vi) Polar equation of a circle: (a) Let (c, a) be the polar coordinates of the centre and a be the radius then the equation of the circle is r2 - 2cr cos(8 - a) + c2 - a 2 = O. (b) If the circle passes through the pole with centre at (c, a) and radius a (in this case c = a) then the equation ofthe circle is r = 2a cos(8a). This equation is of the form r = A cos 0 + B sin O. (c) If the pole is on the circle and the initial line passes through the centre then the equation of the circle is r = 2a cos 8, where a is the radius and (a,O) is the centre. I
(vii) folar equation of a conic with the focus as the pole: If the focus and the pole of the conic are same then the equation of the conic is
~r = 1 -
e cos 0, where 1 and e are respectively semi-latus rectum and eccentricity. e < 1, conic is an ellipse If e = 1, conic is a parabola { e> 1, conic is a hyperbola
CH.4: GEOMETRY
4.17
Ex. 1 (i) 'Transform the cartesian coordinates (-1, -1) to the polar coordinates. (ii) Find the polar coordinators of the point whose cartesian coordinates are (1, -y'3).
• SOLUTION: (i) Let -1 = r cos 0, -1 = r sin O. . '. r2 = 1 + 1 = 2 or, r = J2 and tanO = =~ = 1. Since the point is on the third quadrant and tan 0 = 1. Then 0 = 7r/4 + 7r = 57r/4 . . '. the required polar coordinates is (v2, 57r/4) or (v2, -37r /4). (ii) Let 1 = r cos 0, -y'3 = r sinO. r2 = 1 + 3 = 4 or, r = 2. Also, tan 0 = -y'3 or, 0 = -7r/3.
... the required polar coordinates is (2, -7r /3). Ex. 2 (i) Find the cartesian coordinates of the point whose polar coordinates is (2,7r/2). (ii) Find the cartesian coordinates of the point whose polar coordinates is (2,150°).
• SOLUTION: (i) Here r = 2 and 0 = 7r/2. Then x rsin7r/2 = 2.1 = 2.
=
rcosO
=
2cos7r/2
=
0 and y =
Hence the corresponding cartesian coordinates is (0,2).
= 2 and 0 = 150°. Then x = r cos 0 = 2 cos 150° = 2 cos (90° + 60°) = -2 sin 60° = -2yt = -y'3 and y = rsinO = 2sin1500 = 2 cos 60° = 2.~ = 1.
(ii) Here r
Hence the required cartesian coordinates is (-y'3, 1). Ex. 3 Find the distance between the points whose polar coordinates are (-12,7r/3) and (9,-7r/6). • SOLUTION: The distance between the points (-12,7r/3) and (9, -7r /6) is
J( -12)2 + 92 -
2( -12)(9) cos( 7r /3 + 7r /6) = J225 + 216 cos 7r /2 = J225 = 15.
Ex. 4 If (8,7r/4) and (6,-7r/12) are the polar coordinates of two adjacent vertices of a square, find its area.
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
4.18
• SOLUTION: The distance between the given points is y'(8)2 + 62 - 2.8.6 cos(1I"/4 + 7r /12) = y'100 - 96. cos 7r /3 = y'100 - 96/2 = v'52. the area of the square is (v'52)2 = 52.
Ex. 5 Find the polar equation of the straight line joining the points whose polar coordinated are (1, 7r /2) and (2, 7r). • SOLUTION: The equation of line joining the given points is
~ sin(7r or,
7r /2) +
~ sin 7r /2 + ~ cos 8 -
~ sin(7r /2 -
sin 8
8) +
l
sin(8 - 7r) = 0
=0
1 . 8 1 8 or, ;- = sm - 2cos .
Ex. 6 . (i) Transform to cartesian coordinates the equation
r~ cos ~ = at.
(ii) Transform the polar equation r cos(8 - a) = p to cartesian equation. (iii) Transform polar to cartesian equation r3 = a cos2 ~. • SOLUTION: (i) We have r 1/ 2 cos
~2 =
a 1/ 2.
Squaring both sides r cos2 8/2
= a or,
i(1 + cos 8)
=a
or, r + r cos 8 = 2a or, r = 2a - r cos 8 or, r2 = (2a - r cos 8)2 or, x 2 + y2 = (2a - x)2 or, x 2 + y2 = x 2 - 4ax + 4a 2 or, y2 + 4ax - 4a 2 = 0, which is the required equation. (ii) Given that rcos(8-a) = p or, r(cos8cosa+sin8sina) y sin a = p, which is the cartesian equation. (iii) The equation can be written as r3 = ~(1 + cos 8) or, or, or, or,
2r 4 = a(r + rcos8) or, 2(x 2 + y2)2 = ar + ax {2(x 2 + y2)2 - ax}2 = a2(x 2 + y2) (as x 2 + y2 = r2) 4(x 2 + y2)4 + a 2x 2 _ 4ax(x 2 + y2)2 = (l2X 2 + a2y2 4(x 2 + y2)4 _ 4ax(x 2 + y2)2 = a 2y2.
= p or, xcosa+
4.19
CH.4: GEOMETRY
Exercise 1 (i) What is the cartesian equation of the polar equation 2 a cos 20 ?
r2
=
. (1'1') Trans£orm the poI ar equation -[ = 1 + cos 0 to a cartesian equation. r [Ans. (i) (x 2 + y2)2 = a2(x 2 - y2), (ii) y2 + 21x _[2 = 0.] Ex. 7 (i) Transform the cartesian equation x 2 + y2 - 2ax = 0 to the polar form.
(ii) Transforms the cartesian equation x 3 = y2(2a - x) to polar equation. (iii) Transform to polar coordinates the equation (x 2 + y2)2 = a 2(x 2 _ y2) . •
SOLUTION:
(i) Putting x = r cos 0, y = r sin 0 to the given equation we get r2 cos 2 0 + r2 sin2 0 - 2ar cos 0 = 0 or, r2 = 2ar cos 0 or, r = 2a cos 0, which is the required polar equation. (ii) We have x 3 + xy2 = 2ay2 or, x(x 2 + y2) = 2ay2 or, r cos O.r2 = 2ar 2 sin2 0 (since x = r cos 0, y = r sinO) or, r cos 0 = 2a sin2 O.
(iii) Putting x = r cos 0, y = r sinO to the given equation. Then (r2)2 = a2(r2 cos 2 0 - r2 sin2 0) or, r 4 = a2r2 cos 20 or, r2 == a2 cos 20. Exercise 2
(i) Transform the equation (x 2 + y2)2 = ax 2 to polar equation..
(ii) Transform the cartesian equation xy = a 2 to polar equation. (iii) Change the cartesian equation y2 = 4x + 3 to polar equation. [Ans. (i)
r2 =
a cos 2 0, (ii)
r2 sin 20 =
2a 2, (iii)
r2 sin2 0 = 4r cos 0
+ 3.]
Ex. 8 (i) On the parabola 3r(1- cos 0) = 1 find the point with the smallest radius vector.
(ii) Find a point on the conic
~r =
(iii) Find the points on the conic •
1 - cos 0 having the least radius vector.
~r = 1 + 2 cos 0 whose radius vector is 5.
SOLUTION:
(i) We have 3r(1 - cos 0) = 1 or, r = 3(
1 0)' 1- cos The radius vector r will be minimum when cos 0 is minimum. minimum value of cos 0 is -1 at 0 = 71'. 1 1 1 Hence the minimum value of r is ( ) = 3.2 = -6' 3 1 - COS7r
The
V.G.
4.20
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(ii) Here r =
i . The radius vector r will be least if 1 - cos 0 is maxI - cosO imum. Its maximum value is 2 at 0 = 11'.
Hence the least value of r is
~.
The required point is (1/2,11').
(iii) When radius vector r = 5 then we have from the curve
5
"5
= 1 + 2 cos 0 or, 1 + 2 cos 0 = 1 or, cos 0 = 0 or, 0 =
11' 311'
2"' 2
Hence the required points are (5,11'/2) and 5,311'/2). Exercise 3
(i) On the curve r
=
21 0 find the point with the small5 - 2 cos
est radius vector. (ii) On the conic r =
6
1 - cos
0' find the point with the smallest radius vector.
(iii) Find the point on the conic
~ = 1r
cos 0 whose radius vector is 4.
(iv) Find the point on the conic 12 = 1 - 4 cos 0 whose radius vector is 4. r
[Ans. (i) (3,11'), (ii) (3,11'), (iii) (4, cos- 1
i), (iv) (4,120°).]
~=
3 - 3 cos 0. r (ii) Determine the nature and latus rectum of the conic whose polar equation . 8 IS r = . 4 - 5 cos 0
Ex. 9
(i) Find the nature of the conic
(iii) Find the nature of the conic r
=4
(iv) Determine the nature of the conic
1 0 - 5 cos
~ r
= 3-
v'3 cos 0.
(v) Find the latus rectum and eccentricity of the conic •
~= r
SOLUTION:
(i) The given equation can be written as 2/3 = 1 - cos O. r
. .It WIt 'h -i = 1 - e cos u, II Comparmg r
we have 1 = 2/3 and e = 1. Since e
= 1,
the given equation represents a parabola.
3 - 4 cos O.
4.21
CH.4: GEOMETRY
(ii) The equation r
=
8
4 - 5 cos
l} can be written as
8 2 5 - = 4 - 5 cos 0 or, - = 1 - - cos O. r r 4
Comparing it with ~ = 1 - e cos 0, we have I = 2 and e = 5/4. Since r e > 1 the equation represents a hyperbola whose semi-latus rectum is 2 i.e., latus rectum is 4. 1 1 (iii) We have r = 4 0 or, - = 4 - 5 cos 0 - 5 cos r 1/4 or, r
=1-
5 - cos O. Here I 4
= 1/4 and e = 5/4 > l.
Hence the conic is a hyperbola (iv) Here
~= r
semi-latus rectum is 1/4.
3 - V3cosO or, 4/3 = 1r
Comparing it with e=
~nd
1/v'3 < l.
~r = 1 -
~ cosO.
v3
= 4/3 and
e cos 0 we get I
Hence the given conic is an ellipse. (v) The given equation can be written as Comparing with
~
,r
5~3
= 1-
~ cos O.
= 1 - e cos 0 we get I = 5/3 and e = 4/3.
Hence the latus rectum is 21 = 10/3 and the eccentricity is 4/3.
Exercise 4
(i) Find the nature of the conic
~= r
4 - 5 cos O.
(ii) Determine the nature and latus rectum of the conic whose polar equation . 3 Isr=34 Ll' COS!7
(iii) Determine the nature of the conic r = 2
+ ; cos O'
(iv) Determine the nature of the conic ~ = 2 + 4 cos (J and also find its length r of the latus rectum. [Ana. (i) hyperbola, (ii) hyperbola, 2, (iii) hyperbola, (iv) hyperbola, 3.] (i) The latus rectum of a conic is 10 and its. eccentricity is ~. Find the length of the focal chord inclined at an angle 60° with the major axes.
Ex. 10
U.G.
4.22
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(ii) Find the length of the focal chord of the parabola is inclined to the axis at an angle 7r /3 . •
~r =
1 - cos 0 which
SOLUTION:
(i) The equation of a conic with focus as pole is
~ =1-
e cos O. Here 21 r and e = 1/2. Let S be the focus and PSP' be a focal chord.
SP=
I 1 - ecosO 5
20 5 = 1 1 =-. 0 1 - 2 cos 60 1 - 2' 2 3 I 5 5 SP = 1 = 1 1 = 4. 1cos(-1200) 1,2 2' (--) 2 =
1
Hence the length of the focal chord is
SP + SpI
= 10
P600
~ S
-120 0
pI Figure: 4.3.1
= 20 + 4 = 32. 3
3
(ii) Here the equation of a conic is
~=
1 - cos O. r Let S be the focus and PBP' be a focal chord. 6 6 Then SP = 1 /3 = - - 1 = 12 and -COS7r 1- 2
SpI =
6 = 6 = 6 - 4 1 - cos ( -27r /3) 1 - cos 1200 1 + 1/2 - .
Hence the length of the focal chord is
SP + BP' = 12 + 4 = 16. (i) Find the centre and radius of the circle r = 3 cos 0 + 4 sin O.
Ex. 11
(ii) Find the centre and radius of the circle r = acosO + bsin O.
(iii) Find the polar coordinates of the centre of the circle r = 8 cos O. •
SOLUTION:
(i) Let 3 = a cos 0', 4 or,
0'
= tan
_1
4
= a sin 0'. Then 9 + 16 = a2 or, a = 5 and tan 0' = 4/3
3'
.. . r = 3 cos 0 + 4 sin 0 = a cos () cos 0' + a sin () sin 0'
4.23
CH.4: GEOMETRY
= acos(O - G) = 5cos(O - G).
Hence the polar coordinates of the centre are
5/2. (ii) Let a =
ACOSG,
b = AsinG. Then
A
=
(~; tan -1 ~)
and radius is
Ja 2 + b2 and
b -. a Thus r = a cos 0 + b sin 0 = A (cos 0 cos G + sin 0 sin G)
G = tan
-1
= Acos(O - G) =
A 2.2 cos(O 2
the radius of the circle is centre are
(
G). 2
Ja + b
Ja 22+b2 ' tan- 1 ~b) .
2
and the polar coordinates of the
(iii) The given equation is of the form r = 2a cos O. . a = 4. Hence the polar coordinates of the centre are (4, 0) and radius is 4.
Ex. 12 Find the polar equation of the circle which passes through the pole and two points whose polar coordinates are (d,O) and (2d, 7r /3). Find also the radius and the centre of the circle. • SOLUTION: Let the equation of the circle be r = a cos 0 + bsinO. Since it passes through (d,O) and (2d,7r/3) then d = a and
(1)
. .. the required equation of the circle is
(~
~
r = d cos 0 + V3d sin 0 = 2d cos 0 + sin 8 ) = 2d(cos 8 cos 7r/3 + sinO sin 7r/3) = 2d cos(O - 7r/3). Hence the radius is d and centre is (d, 7r /3).
Ex. 13 Show that the equation of a parabola in polar coordinates can be expressed in the form r 1 / 2 sin(8/2) = (1/2)1/2, where I is the semi latus rectum of the parabola. • SOLUTION: The equation of the parabola i~ polar coordinates is or, -I r
= 2'Slll2 -(}2
~= r
1 - cos (}
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
4.24
or rl/2 sin -() = ( -l ) 1/2 , 2 2
4.4
General Equation of Second Degree
The general equation of second degree is ax 2 + 2hxy + by2 + 2gx + 2fy + c = O. The curve represented by this equation is a conic section or conic. Let D = a h 9 2 ab - h and Do = h b f '. The nature of the conic is represented by these 9
f
c
two quantities. Type
Elliptic
Parabolic
Do D= ab - h 2 D>O Do < 0 D >0 Do < 0 D>O Do> 0
Canonical form
Name
~+~=1
D>O
Do = 0
a2x 2 + b2y2 ::;: 0
D=O
=1=
D 0 and a h 9 1 -1 -2 ~ = h b / = -1 2 -3 = -26 9 / e -2 -3 3
< O.
Hence the given conic represents an ellipse. Ex. 2
(i) Find the centre of the ellipse 7x 2 - 2xy + 7y2 + 22x -lOy + 7 = O.
(ii) Show that the curve 4x 2 + 4xy + y2 + 4x + 2y + 20 = 0 has infinitely many centres.
4.27
CH.'t: GEOMETP' •
SOLUTION:
(i) Comparing the given equation with ax2 + 2hxy + by2 + 2gx + 2/y + c = 0 we have . a = 7,h = -l,b = 7,g = 11,/ = -5,c = 7. The centre (a, (3) is obtain from the equations aO' + h(3 + 9 = 0 and hO' + b(3 + / = o. . '.7O'-(3+11=Oand-O'+7(3-5=O. Solving we get 0'= -3/2, (3 ::;; 1/2. Hence the required centre of the conic is (-3/2,1/2). (ii) Comparing the given equation with ax 2 + 2hxy + by2 + 2gx + 2/y we have a = 4, h = 2, b = 1, 9 = 2, / = 1, c = 20.
+c = 0
The (a, (3) be the coordinates of the centre and it is obtain from the equations aO' + h(3 + 9 = 0 and hO' + b(3 + / = O. or, 40' + 2(3 + 2 = 0 and 20' + (3 + 1 = O. These two equations are identical and they are 20' + (3 + 1 = O. This equation has infinite solutions. Hence the given conic has infinite many centres.
Three Dimensions 4.5
Direction Cosines and Ratios
Some important formulae on three dimensions (i) Distance between two points: Let P(XI, Yb Zl) and Q(X2, Y2, Z2) be two points. The distance between P and Q is V(X2 - Xl)2
+ (Y2 -
Yl)2
+ (Z2 -
zt)2.
(ii) Internal point: If R divides P(XI, Yb zt) and Q(X2, Y2, Z2) internally in the ratio m : n then the coordinates of R are (
mX2 + nXl mY2 + nYl mZ2 + nZl) m+n ' m+n '. m+n .
(iii) Mid-point: If R is the mid.point of P(XI, Yll zt} and Q(X2, Y2, Z2) then the coordinates of R are (
X2
+ Xl 2
Y2
'
+ Yl 2. '
Z2
+ Zl) 2'
.
4.28
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iv) Direction cosines: If a directed line makes angles Ct., /3, 'Y with the positive directions of x, y, Z axes respectively the cos Ct., cos /3, cos'y are called the direction cosines of this line. Direction cosines (d.cs) are generally denoted by 1, m, n. The direction cosines 1, m, n satisfy the relation 12 + m 2 + n 2 = 1. (v) Direction cosines of a line joining two points: Let P(XI,YI,Zl) and Q(X2,Y2,Z2) be two points. Directio/Il ratios (d.rs) of this line are X2 - Xl, Y2 - Yl, Z2 - ZI and the direction cosines are X2 - Xl Y2 - Yl Z2 - Zl . . . PQ ' PQ ' PQ ,where PQ IS the dlstance between the pomts
P and Q. ~
(vi) Angle between two lines: If h, ml, nl and h, m2, na are the d.cs of two lines then the angle () between them is
If ab bb C! and a2, b2 , C2 are the d.rs of two lines then the angle () between them is £I -1 ± ala2 + bl b2 + ClC2
u=cos
Ja~ + b~ + C~ Ja~ + b~ + C~
.
(vii) Condition for perpendicularity: If iI, mb nl and 12, m2, n2 are d.cs or d.rs of two lines then they are perpendicular to each other if and only if h12 + mlm2 + nln2 = o. (viii) Condition for parallel lines: If iI, ml, nl and 12, m2, n2 are d.cs of two lines then they are parallel if and only if
11 12
= ml = nl = 1. m2
n2
Ex. 1 (i) Find the direction cosines of the straight line passing through the points (1,2,4) and (3,1,3).
(ii) Can 1, 2, 3 be the direction cosines of a straight line? Can they be direction ratios ? Justify your answer. (iii) Can the numbers
)a, ~, - ~ be the direction cosines of any line?
Justify your answer. (iv) A directed straight line makes angles 60°,45° with the axes of x and Y respectively. What angle does it make with the z-axes ? (v) Find the direction cosines of a line equally inclined to the coordinates . axes.
CH.4: GEOMETRY
4.29
• SOLUTION: (i) The d.rs of the line joining the given points are {I - 3,2 - 1,4 - 3} or, {-2, 1, I}. The d.cs are
-2 { vi4 + 1 + 1 '
(ii)
II}
{ -2
II}
vi4 + 1 + 1 ' vi4 + 1 + 1 or, J6' J6' J6 . {I, 2, 3} do not form d.cs, because 12 + 22 + 32 = 14 =1= 1. But they may be d.rs of a straight line.
1)2 (1)2 (iii) ( v'3 + J2
(
III}
Hence { v'3'
1)2
+ - J2
J2' - J2
=
1
1
1
3" + "2 + "2
=1=
1.
can not be a d.cs of any line.
(iv) Let a = 60°,13 = 45°. We know cos2 a + cos2 13 + cos2 , or, cos 2 60° + cos 2 45° + cos 2 , = 1 or, cos 2 , = 1 - cos 2 60° - cos 2 45° = 1 - ~ - ~ = ~. or, cos,
=~
or, ,
=,.
= 60°. Hence the line makes 60°
=1
with the z-axis.
Then the d.cs are {cos a, cos 13, cos,} (v) Here a ~ 13 , {cosa cos a cosa} or, {cosq, cos a, cos a} or, .f ' .f '-.Ir==::;:= 2 2 v 3 cos a v 3 cos a v 3 cos2 a
III}
or, { v'3'
v'3' v'3 .
Ex. 2 The length of the projections of a segment of a straight line on the axes of coordinates are 3, 4, 5 units respectively. Find the length of the segment. • SOLUTION: Let the line be PQ and make angles with coordinates axes be a,f3" then we know that P'Q' = PQcos9, where P'Q' is the projection of PQ on a line AB and 9 is the angle between PQ and AB. Taking AB as x-axis, y-axis and z-axis successively we have 3 = PQ cos a, 4 = PQ cos 13, 5 = PQcos,. Squaring and adding we get 9 + 16 + 25 = (PQ)2(cos 2 a + cos 2 13 + cos 2 ,) or, (PQ)2 = 50 or, PQ = 5J2, which is the length of the line segment.
Ex. 3 Find the projection of the curve x 2 + 2y2 + 3z 2 = 12, 3x + 1 = 0 on the yz-plane. • SOLUTION: Putting x = -1/3 in the equation x 2 + 2y2 1 107 (-1/3)2 + 2y2 + 3z 2 = 12 or, 2y2 + 3z 2 = 12 - 9 = 9 y2 z2 2 or, 18y2 + 27z = 107 or, 107/18 + 107/27 = 1.
+ 3z 2 = 12 we get
4.30
U.G. MATHEMATICS (SliURT QUESTIONS AND ANSWERS)
This equation represents a curve surface (elliptic cylinder) projecting the given curve on the yz-plane. The projection is an ellipse with centre at (0,0) .h and Wit axes.
V{ITff 18 and V{ITff 27
. . and seml-mmor .. as t h e Ieng th 0 f·Its semi-maJor
Ex. 4 Find the locus of a point P which is equidistance from the points (2, -2, 1) and (0,2,3). • SOLUTION: Let the coordinates of P be (x, y, z). Distance from (x, y, z) to (2, -2, 1) is V(x - 2)2 + (y + 2)2 + (z - 1)2 and that of from P to (0,2,3) is V(x - 0)2 + (y - 2)2 + (z - 3)2. According to the condition, V(x - 2)2 + (y + 2)2 + (z - 1)2 = vx2 + (y - 2)2 + (z - 3)2. Squaring we get (x - 2)2 + (y + 2)2 + (z - 1)2 = x 2 + (y - 2)2 + (z - 3)2 Of, -4x + 8y + 4z = 4 or, x - 2y - z + 1 = 0.
Ex. 5 Find the coordinates of the points which divide the distance between two points (2,0,1) and (4,-2,5) internally and externally in the ratio 3:2. • SOLUTION: The coordinates of the internal point are
(
3.4+2.2 3.(-2)+2.0 3.5+2.1) r (16 -6 17) 3+2' 3+2 ' 3+2 a 5' 5 ' 5
and the coordinates of the external point are
(
(8 -6 1 ) 3.4 - 2.2 3.(-2) - 2.0 3.5 - 2.1) 3 - 2' 3- 2 ' 3_ 2 or , , 3.
Ex. 6 Show that the points (0,7,10), (-1,6,6) and (-4,9,6) form an isosceles right-angled triangle. • SOLUTION: Let the coordinates of A, B, C be respectively (0, 7, 10), (-1,6,6) and (-4,9,6). (AB)2 = (0 + 1)2 + (7 - 6)2 + (10 - 6)2 = 1 + 1 + 16 = 18 (BC)2 = (-1 + 4)2 + (6 - 9)2 + (6 - 6)2 = 9 + 9 + = 18 and (CA)2 = (-4 - 0)2 + (9 - 7)2 + (6 - 10)2 = 16 + 4 + 16 = 36. Since AB2 + BC2 = CA2 and AB = BC, ABC is an isosceles right-angled triangle.
°
Ex. 7 Show that the points (1,1,1), (-2,4,1), (-1,5,5) and (2,2,5) form a , square.
4.31
CH.4: GEOMETRY
• SOLUTION: Let A(l, 1, 1), B( -2,4,1), C( -1,5, S) and L!(2, 2, 5) be the corner points of the square ABCD. AB2 = (1 + 2)2 + (1 - 4)2 + (1 - 1)2 = 9 + 9 + = 18 BC2 = (-2 + 1)2 + (4 - 5)2 + (1- 5)2 = 1 + 1 + 16 = 18 CD 2 = (-1- 2)2 + (5 - 2)2 + (5 - 5)2 = 9 + 9 + = 18 and DA2 = (2 - 1)2 + (2 -1)2 + (5 - 1)2 = 1 + 1 + 16 = 18 .
° °
. AB = BC = CD = DA. Again AC2 = (1 + 1)2 + (1 - 5)2 + (1 - 5)2 = 4 + 16 + 16 = 36 and BD2 = (-2 _2)2 + (4 - 2)2 + (1- 5)2 = 16 +4+ 16 = 36. i.e., the ~iagonal AC and BD are equal. Hence ABCD is a square.
Ex. 8 If a, f3, 'Y be the direction angles of a line then show that sin2 a+sin2 f3+ sin2 'Y = 2. • SOLUTION: We know, if a, f3, 'Y be the direction angles of a line then cos2 a+ cos 2 f3 + cos 2 'Y = 1 or, (1 - sin2 a) + (1 - sin2 (3) + (1 - sin2 'Y) = 1 or, 3 - (sin~ a + sin2 f3 + sin2 'Y) = 1 or, sin2 a + sin2 f3 + sin2 'Y = 2. Exercise 1 (i) What are the direction cosines of the straight line joining the points (3,4, -1) and (1,7, -I)? (ii) Show that 1, 1, 1 can not be the d.c's of any directed line.
[Ans. (i) {2/vU, -3/Vl3,0}.]
4.6
Plane
°
(i) General equation of plane: ax+by+cz+d = is the general equation of the plane, where a, b, c are d.rs of the plane. (ii) Equation of the plane passing through a point: The equation of the plane passing through a point (Xl, YI, Zl) and having d.rs a, b, c is a(x - xt} + bey - yt} + c(z - zt} = 0. (iii) Equation of the plane in normal form: Ix + my + nz = p is the normal form of a plane, where I, m, n are the d.cs of the plane and pis the perpendicular distance from origin to the plane. (iv) Intercept form of a plane: If the plane cuts the coordinates axes at (0, 0, 0), (0, b, 0) and (0,0, c) then its equation is :. + !b + :. ::;: 1. a
c
(v) Perpendicular distance of a point from a plane: Let ax + by + cz + d = be the equation of the plane and (Xl! yl! zt) be a given point. Then the distance is
°
ax 1 +
l va
byl + CZI + 2 +h2+c2
dl.
4.32
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(vi) Angle between two planes: The angle between two planes alx+bly+ ClZ + dl = 0 and a2x + b 2 y + C2Z + d 2 = 0 is given by -1
±
cos·
ala2 + bl b2 + ClC2 v'a~ + b~ + c~v'a~ + b~ + C~·
(vii) Condition for perpendicularity: If the planes alx+blY+ClZ+d l = 0 and a2x+b2Y+C2Z+d2 = 0 are perpendicular then ala2+blb2+CIC2 = o. (viii) Condition for parallel planes: If the planes and a2x + b2Y + C2Z + d 2 = 0 are parallel then
al
bl
a2
b2
alX
+ bly + CIZ + dl
= 0
Cl C2
(ix) Planes bisecting the angles between two planes: Let the equations of the planes be alX + blY + ClZ + dl = 0 and a2X + b2 y + C2Z + d2 = o. The equations of the planes are alx
+ bly + ClZ + dl v'ai + bi + ci
= ± a2x
+ b2y + C2Z + d 2 v'a~ + b~ + c~ .
(x) Coplanarity of four points: If the four points' (Xl, Yb zI), (X2,Y2,Z2), (X3,Y3,Z3), (X4,Y4,Z4) are coplanar then X2 - Xl
Y2 -
Xl :1:4 -j'>Xl
Y3 -
X3 -
Y4 -
Yl Yl YI
z2 - zl Z3 z4 -
Zl = O. Zl
Ex. 1 (i) Find the equation of the plane which passes through the point (1, -1, 2) and is parallel to the plane 2x + 3~ - Z + 1 = o. (ii) Find the equation of the plane passing through the line of intersection of the planes 3x + Y - 5z + 7 = 0 and X - 2y + 4z + 3 = 0 and through the point (-3,2, -4). (iii) Find the equation of the plane passing through P(a, b, c) and perpendicular to OP, where 0 is the origin. (iv) Find the equation of the plane so that the foot of the perpendicular from the origin to it is (2,3, -1). (v) Find the equation of the plane that bisects perpendicularly the line segment joining the points (2, -1,3) and (-4,2,2). (vi) Find the equation of the plane passing through the point (4, -3,5) and containing the y-axis.
CH.4: GEOMETRY
4.33
(vii) Find the equation of the plane passing through the points (I, 0, 0), (O, 1,0) and (O, 0,1) . •
SOLUTION:
(i) Let the equation of the plane passing through (I, -1,2) be a{x - 1) + b{y + 1) + c{z - 2) = O.
(I)
Since it is parallel to 2x + 3y - z + 1 = 0, the d.rs are proportional, i.e., abc - = - = - =k. 2 3 -1 Putting the values of a, b, c to (I) we get the required equation of the plane. Hence the required equation of the plane is 2{x - 1) + 3(y + 1) - l(z - 2) = 0 or, 2x + 3y - z
+ 3 = O.
(ii) Let the equation of the plane passing through the given planes be (3x + y - 5z + 7) + k{x - 2y + 4z + 3) = O.
If iJ passes through the point (-3, 2, -4) then (-9 + 2 + 20 + 7) + k{ -3 - 4 - 16 + 3) = 0 or, 20 - 20k
= 0 or, k = 1.
Hence the required equation of the plane is (3x + y - 5z + 7) + l{x - 2y + 4z + 3) = 0 or, 4x - y - z
+ 10 =
O.
(iii) Let the equation of the plane passing through P{a, b, c) be l{x - a) m{y - b) + n{z - c) = O.
+
The d.rs of the line OP are {a - 0, b - 0, c - O} or, {a, b, c}. If the plane is perpendicular to the line 0 P then
~a =
Hence the equation of the plane is + b{y - b) + c{z - c) = 0 or, ax
mb =
~. c
a2 + b2 + c2 . (iv) The d.rs of the line passing through origin (0) and P(2, 3, -1) are {20,3 - 0, -1 - O} or, {2, 3, -I}. a{x - a)
+ by + cz =
(2,3, -1) is the foot of the perpendicular on the required plane. Let the equation of the plane be a(x - 2) + b(y - 3) + c(z
+ 1) =
Since OP is perpendicular to the plane (1), Hence (1) becomes 2(x - 2) + 3(y - 3) - l(z
(1)
O.
~ = ~ = ~1
+ 1) = 0 or, 2x + 3y -
z = 14.
(v) The requi~ed plane passing through the middle point of A(2, -1,3) and B{ -4,2,2) i.e., through the point (-1,1/2,5/2).
V.G.
4.34
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
The d.rs of AB are {6, -3, 1}. These d.rs are proportional to the d.rs of the line perpendicular to the plane. Hence the equation of the plane is 6(x + 1) - 3(y - 1/2) + l(z 5/2) = or, 6x - 3y + z + 5 = 0.
°
(vi) Let the equation of the plane passing through (4, -3,5) be a(x - 4) + b(y + 3) + c(z - 5) = 0. (1) The plane (1) containing y-axis, so all the points on y-axis are the points on the plane (1). Let (0,1,0) be a point on y-axis. Equation (1) passes through it, so -4a + 4b - 5c = 0. (2) Again (1) containing y-axis, so a.O + b.1 of y-axis) i.e., b = 0 .
+ c.O =
0 ({O, 1, O} are the d.cs
. '. (1) and (2) become a(x - 4) + c(z - 5) = 0 and 4a + 5c = O. x-4 z-5 On dividing we have -4- = -5- or, 5x-20 = 4z-20 or, 5x-4z (vii) Let the equation of the plane through (1,0,0) be a(x - 1) + b(y - 0) + c(z - 0) = O. It also pass through (0,1,0) and (0,0,1), a(O - 1) + b(l - 0) + c(O - 0) = O. a(O - 1) + b(O - 0) + c(l - 0) = O. Eliminating a, b, c from (1), (2) and (3) we get x -1 .:...1 -1
y z 1 0 0 1
or, (x - 1) - y (-1 - 0) + z (0 + 1) = Ex. 2 z
(1)
(2) (3)
= O.
°
or, x + y + z = 1.
(i) Find the distance of the point (1, 2,-3) from the plane 5x - 3y + 0.
+5=
(ii) Find the distance between the planes x + 2y - 3z + 5 = 3z - 7 = 0 . •
= O.
°
and x + 2y -
SOLUTION:
(i) The distance from the point (1,2, -3) to the plane 5x - 3y + z + 5 = 0 is
5.1- 3.2 + 1.(-3) + 51 __1_ y'25+9+1 - y'35' 1
4.35
CH.4: GEOMETRY
(ii) The distance from the origin to the plane x 5 5 ~v'~1=+=;=4=+:::;:::9 = vT4'
+ 2y
Again the distance from the origin to the plane x
-7
~v'~i=+=;=4=+=='9 = -
7
- 3z
+ 2y -
+5
= 0 is
3z - 7
=
0 is
vT4'
Hence the required distance between the given planes is 5 12
1vT4 -
-71
vT4
=
vT4'
Ex. 3 Show that the points (1,1,1) and (2,1,1) lie on the opposite sides of the plane 2x + y + 3z - 7 = o. • SOLUTION: At (1,1,1), dl = 2x + y + 3z - 7 = 2.1 + 1 + 3.1 - 7 = -1 < 0 and at (2,1,1), d2 = 2x + y + 3z - 7 = 2.2 + 1 + 3.1 - 7 = 1 > O. Since dl and d2 are opposite in sign, the points lie opposite sides of the given plane.
Ex. 4 (i) Reduce the equation x + 5y - 7z + 10 = 0 to the normal form. Hence find the length of the perpendicular from origin to the plane.
(ii) Write down the intercept form of the equation of the plane 5x
+ 2y +
3z -17 = O. •
SOLUTION:
(i) Dividing the given equation by
~+~-~+~=O 5V3
J1 2 + 52 + (_7)2 = J75 =
5V3 we get
5V3
x or, 5V3
+
5V3 5V3 y 7z 2 V3 - 5V3 = - V3'
Which is the normal form of the given plane and the perpendicular distance from the origin is
1- ~ 1= ~.
(ii) The equation can be rewritten in the form 5x + 2y + 3z = 17 or, x y z 17/5 + 17/2 + 17/3 = 1, which is the required intercept form of the given plane.
= 9 and 2i+y+z = 7. Determine the value of h for which the planes 3x - 2y + hz - 1 = 0 and x + hy + 5z + 2 = 0 may be perpendicular to each other.
Ex. 5 (ii)
(i) Find the angle between the planes x-y+2z
p.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
4.36 • SbLUTlON:
(i~:,,,·The_~;v.gle
() between the planes is given by
cos ()
=
Jli
1.2 + (-1).1 + 2.1 = -1 = + (-1)2 + 22v'22 + 12 + 12 2
cos
600
or, () = 60°. (ii, :;Since the planes are perpendicular i.e., the angle between them is ~, 7r 3.1 + (-2).h + h.5 cos 2" = -..;r=9=+=4=+=======h"""2";-'-;=1=+:::::h=;;:2=+=2==='5
;'
or, 3 - 2h + 5h = 0 or, 3h + 3 = 0 or, h = -1.
Ex. 6 Show that the point (3,9,4), (4, 5,1), (-4,4,4) and (0, -1, -1) are coplanar.
•
SOLUTION: Here
4-3 5-9 1-4 -4-3 4-9 4-4 0-3 -1-9 -1-4
-4 -3 -7 -5 0 -3 -10 -5 1
=
=0.
Hence the given points are coplanar.
Ex. 7 (i) Find the equation of the plane through the point (1,2, -3) and normal to the straight line joining the points (-1,3,4) and (5,2, -1).
(ii) Find the equation of the plane which is perpendicular to the plane x
+
2y - z + 1 = 0 and which contains the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y + z + 2 = O. (iii) Find the equation of the plane passing through the point (2,5, -8) and perpendicular to each of the planes 2x - 3y + 4z + 1 = 0 and 4x + y 2z + 6 = O. (iv) Find the equation of the plane passing through (4,1,0) and parallel to the plane 3x - 4y + 7z + 10 = O. (v) Find the equation of the plane passing through (2,1, -1) and perpendicular to the line whose d.rs are 3,4, -2. (vi) Find the equation of the plane passing through (2,3,4) and parallel to the zx-plane. (vii) Find the equations of the planes bisecting the angle between the planes 3x - 4y + 12z = 26, x + 2y - 2z = 9.
4.37'
CH.4: GEOMETRY
(viii) Show that 2x2 - 6y2 - 12z2 planes . •
+ 18yz + 2zx + xy =
0 represents a pair of
SOLUTION:
(i) The d.rs of the line joining the points (-1,3,4) and (5,2, -1) are { -6,1, 5}. Since the plane normal to this line {-6, 1, 5} are also d.rs of the required plane. Hence the equation of the plane passing through the point (1,2, -3) is -6(x - 1) + 1(y - 2) + 5(z + 3) = 0 or, 6x - y - 5z = 19. (ii) Let the equation of the plane contains the line of intersection of the given planes be x + 2y + 3z - 4 + k(2x + y + z + 2) = 0 or, (1 + 2k)x + (2 + k)y + (3 + k)z + 2k - 4 = O. {1) If it is perpendicular to the plane x + 2y - z 2(2 + k) - 1.(3 + k) = 0 or, k = -2/3.
Then (1) becomes (1 - 4/3)x or, x - 4y - 7z + 16 = O.
+ (2 -
+1= 0
then 1.(1 + 2k)
+
2/3)y + (3 - 2/3)z - 4/3 - 4 = 0
(iii) Let the equation of the plane passing through the point (2,5, -8) be a(x - 2) + b(y - 5) + c(z + 8) = O. (1) If it is perpendicular to the given planes then 2a - 3b+ 4c = 0 and 4a + b - 2c = 0
(2) (3)
Solving (2) and (3) we get abc 6- 4
= 4 + 16 = 2 + 12
abc
or
(1) becomes 1.(x-2)+1O(y-5)+7(z+8) which is the equation of the plane.
i = 10 = "7 = 0 or, x+10y+'lz+4 = 0,.
(iv) Since the required plane is parallel to the given plane, the d.rs of the required plane are {3, -4, 7}. Hence the equation of the plane is 3(x 4) - 4(y - 1) + 7(z - 0) = 0 or, 3x - 4y + 7z - 8 = O. (v) The plane is perpendicular to the line with d.rs 3,4, -2, therefore the d.rs of the plane are also 3,4, -2. Hence the equation of the plane passing through the point (2,1, -1) is 3(x - 2) + 4(y - 1) - 2(z + 1) = 0 or, 3x + 4y - 2z = 12. (vi) The plane is parallel to the zx-plane and passes through the point (2,3,4). Then its equation is 1.{x - 2) + O.(y - 3) + l.(z - 4) = 0 or, x + z = 6.
4.38
U.G. MATHEMATICS (SHORT:QUESTIONS AND ANSWERS)
(vii) The required equations of the planes are 3x - 4y + 12z - 26 = J9 + 16 + 144 3x - 4y
+ 12z -
26
x
+ 2y -
± x + 2y -
J1
2z - 9 +4 +4
2z - 9
or, 13 =± 3 or, 9x - 12y + 36z - 78 = ±(13x + 26y - 26z - 117) or, -4x - 38y + 62z + 39 = 0 and 22x + 14y + lOz -195 Hence the required planes are + 38y - 62z - 39 = 0 and 22x
+ 14y + 10z 12z2 + 18yz + 2zx + xy =
4x
= O.
195 = O.
(viii) We have 2x2 - 6y2 0 or, 2x2 + x(2z + y) + (18yz - 6y2 - 12z2) = O. x= = =
-(2z
+ y) ±
-(2z
+ y) ± (7y -
J(2z
+ y)2 -
4.2.(18yz - 6y2 - 12z2) 4 -(2z + y) ± J49 y2 + 100z 2 - 140yz 4
10z)
4
or, 4x = -(2z + y) ± (7y - 10z) or, 4x = -(2z + y) + (7y - lOz) and 4x = -(2z + y) - (7y - lOz) or; 4x = 6y - 12z and 4x = 8z - 8y .
. '. the equation of the planes are 2x - 3y + 6z = 0 and x + 2y - 2z = O. Exercise 1 3y
(i) Find the distance of the point (1, 2, 0) from the plane 4x +
+ 12z + 16 =
O.
(ii) Find the equation of the plane passing through the point (3, -2,6) and containing the x-axis. (iii) Find the equation of the plane passing through the intersection of the planes 2x + y + 2z = 9 and 4x - 5y - 4z = 1 and the point (3,2,-1). (iv) Find the equation of the plane passing through the point (-1,0,1) and the line of intersection of the planes 4x - 3y + 1 = 0 and y - 4z + 13 = O. (v) Express the following in the normal form 2x - 3y + 4z - 5 = O. (vi) Find the equation of the plane passing through (3,5,1), (2,3,0) and (0,6,0). (vii) Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y - z + 5 = 0 and which is perpendicular to the plane 5x + 3y + 6z + 8 = O.
4.39
CH.4: GEOMETRY
(viii) Show that the four points (0, -1,0), (2, 1, -1), (1, 1, 1) and (3,3,0) are coplanar. (ix) Find the angle between the planes 2x + Y + Z
= 6,
x - Y + 2z
= 3.
(x) Find the distances of the points (2,0,1) and (3, -3,2) from the plane x - 2y + z - 6 = 0. Do the points lie on the same side or opposite sides of the plane ? (xi) Find the distance between the parallel planes x - 4y + 8z - 90 x - 4y + 8z + 20 = 0.
°
= and
[Ans. (i) 2, (ii) 3y + z = 0, (iii) 11x - 5y - z = 24, (iii) 12x - 8y - 4z + 16 (iv) 12x-8y-4z+16 = 0, (v) Jg- ~+ ~ = ~, (vi) 3x+2y-7z-12 (vii) 51x + 27y - 22z (xi) 110/9.]
4.7
+ 96 =
= 0, = 0,
0, (ix) 60°, (x) -3/.;6,5/.;6, opposite side,
Straight Lines
(i) Equations of straight lines: (a) If the line passes through the point then the equations of the line are x -
Y - YI
Xl
m
(Xl,
=
YI, Zl) and its d.rs I, m, n
Zl
Z -
n
This is known as the symmetric form or the canonical form or the standard form.
(b) If the line passes through two given points then its equations are X -
Xl
=
=
(Xl,
Z -
YI, zt) and
(X2' Y2, Z2)
Zl
(c) Parametric form: If the line passes through the point (Xl, Yl, Zl) and d.rs be I, m, n then its equations are X = lr+xl, Y = mr+YI, Z = nr + Zl, where r is the parameter. (d) Plane intercept form: The planes alx + b1y + CIZ + dl = 0 and a2x + b2 y + C2Z + d2 = 0 together represent the straight line of intersection of the planes.
(ii) Coplanarity of two straight lines: The lines X - Xl Y - Yl Z - Zl X - X2 Y - Y2 ---,---"=- =- and - = -- =
It
ml
nl
12
m2
Z -
Z2
-n2
4.40
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
are coplanar if and only if X2 - Xl
Y2 - Yl
h
ml m2
l2
=0.
(iii) Perpendicular distance of a point from a line: The perpendicular distance from the point (ex, 13, "() to the line X -
Y-
Xl
Yl
Zl .
Z -
--=--=--IS 1 m n
[(ex -
Xl)2
_ {l(ex -
+ (13 - yd 2 + ("( - zd 2
xd + m(j3 - Yl) + n("( - Zd}2] 1/2 [2 + m 2 + n 2
(iv) Shortest distance between two skew lines: The shortest distance between the lines X - Xl
Y - YI
h
ml
Z -
Zl
--=--=--and
X - X2
nl
Y - Y2
Z - Z2
m2
n2
=--=--
l2
is
±
X2 - Xl
Y2 - YI
Z2 - Zl
h
ml m2
n2
l2
[
2 ml
nl
m2
n2
+
nl 2
h
nl
l2
n2
+
h
ml
l2
m2
r'
Ex. 1 Find the angle between the two straight lines whose direction ratios are respectively 5, -12, 13 and -3,4,5 . •
SOLUTION:
The angle B between the lines is given by
cosB =
5.(-3) + (-12).4 + 13.5 = 2 =..!... v'25 + 144 + 169v'9 + 16 + 25 13\1"2.5\1"2 65
or, B = cos
-1
1 65 .
Ex. 2 Obtain the direction cosines of the line formed by the planes 3x + 2y + Z - 5 = 0 and x + y - 2z - 3 = O.
CH.4:
4.41
GEOMETRY
• SOLUTION: Let I, m, n be the d.rs of the line . . '. 31 + 2m + n = 0 and 1 + m - 2n = O. Solving these equations m
1+6
-4 - 1
=
n m n = = 3 - 2 -5 7 1
Also [2 + m 2 + n 2 = 25 + 49 + 1 = 75. the d.rs of the line are {-5, 7, 1} and the corresponding d.cs are -5 { 05'
7
1}
v'15' V75
{ -1 or
7
1}
V3' 5V3' 5V3 .
Ex. 3 Obtain in the symmetrical form, the equation of the straight line x 2y + 3z = 4, 2x - 3y + 4z = 5. • SOLUTION: Let l, m, n be the d.rs of the line. · '. 1 - 2m + 3n = 0 and 2l - 3m + 4n = O. Solving n m n m -8+9=6-4=-3+41=2='1' Putting.z = 0 to the given equations we get x - 2y = 4 and 2x - 3y = 5. Solving these equation we obtain x = -2, y = -3. Hence the equation of the line in symmetrical form is
x+2_y+3_:. 1
-
2
-1'
Ex. 4 Are the two lines
x-2 y-3 z+4 x+l y-2 z+3 -3- = -2- = -4- and -5- = ~ = -2perpendicular to each other ? • SOLUTION: The d.rs of the given lines are {3, 2, 4} and {5, -6, 2}. Now, 3.5 + 2.( -6) + 4.2 = 11 =1= 0, therefore, the given lines are not perpendicular. Ex. 5
(i) Examine whether the straight lines x-I = y - 2 = z - 3 and x-I = y - 1 = z - 1
2 are coplanar.
3
4
1
2
3
U.G.
4.42
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
x-I (ii) Does the line -1-
y-2
z-3
lie on the plane 2x + 4y - 3z
= 1?
(iii) Find the values of I and a for which the line x-2 y+3 .::-6. . -Z- = -4- = ---=2 IS perpendIcular of the plane 3x-2y+az+1O
= O.
= -1- = -2-
(iv) Find the values of a and d for which the line Y+8 :; - 3. . x - 10 ---1- = -2- = -1- lIes In the plane ax + 2y - 3z + d •
= O.
SOLUTION:
(i) Here
1 -1 2-1 3-1 2 3 4 1 2 3
=
0 1 2 2 3 4 1 2 3
=0.
Hence the lines are coplanar. x-1 y-2 z-3 (ii) Let - - = - - = - - =r. 1 1 2 ... x = r + 1, y = l' + 2, z = 2r + 3. Putting these values in 2x + 4y - 3z = 1 we get 2(r + 1) + 4(r + 2) - 3(2r + 3) = lor, 1 = 1 Le., every point on the line is on the plane. Hence the line lie on the plane. (iii) Since the given line and plane are perpendicular, so their d.rs are proportional, 1 4 -2 1 - = - = - or, - = -2 or, 1 =-6 3 -2 a 3
-2
and -
a
= -2 or, a = 1.
... the required values of 1 and a are respectively -6 and 1. (iv) The point (1O, -8, 3) on the straight line is also on the given plane . ... lOa + 2.( -8) - 3.3 + d = 0 or, lOa + d = 25 (1) Since the line is on the plane, (-l).a
+ 2.2 + 1.(-3) =
0 or, a = 1.
From (1), d = 25 - lOa = 25 - 10 = 15.
Ex. 6
(i) Find the distance of the point (3, 2,1) from the line
x-1 y z-2 -3- = '4 = -1-·
CH.4: GEOMETRY
4.43
(ii) Find the shortest distance between the lines
x-1 y-2 z-3 x-3 y-3 z-4 -2- = -3- = -4-' -3- = -4- = -5-
• SOLUTION: (i) The distance from the point (3, 2, 1) to the given line is [(
3 _ 1)2
22 +
+
(1 _ 2)2 _ {3.(3 - 1) + 4.2 + 1.(1 - 2)}2] 1/2 32 + 42 + 12
= [4 + 4 + 1 -
(6+8-1)2]1/2 26
=
f5 V"2;
(ii) The required shortest distance is 1-3 2-3 3-4 2 3 4 3 4 5
[ ! : '+ i : '+ ; ! ']'/2 -2 -1 -1 234 345
= [(15 -
16)2 + (10 - 12)2 + (8 - 9)2]1/2
=
1 yI6'
Ex. 7 Two lines have d.c.'s {1,a,a}, {a,a,1}. Find the d.c.'s of a line perpendicular to them . • SOLUTION: Let the d.cs of the requires line be {l,m,n}. Since this line is perpendicular to the lines whose d.cs {1, a, a} and {a, a, 1}, I + a + a = a and a + a + n = a or, I = a, n = a. Hence the required d.cs are {a, 1, a}.
Ex. 8
(i) Find the coordinates of the point where the line
x-a
y-(3 z-, =--=--
m
n
meets zx plane.
(ii) Find the coordinates of the point where the straight line joining the points (5, -2, 3) and (3, a, 1) cuts the xy-plane.
U.G.
4.44 •
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
SOLUTION:
(i) The given line is x - a y .- (3 z- , =--=-m n
x-a -(3 z-, On the zx-plane y = O. Then - - = = -I m n x - ex -(3 1(3 or, -1- = -:;;; or, x = a - ;:;;:. n(3. · ·1ar1y, z = , - SIIDI m
. h . d . . ( 1(3 0" - -n(3) . .. t e reqUIre pomt IS a - -, m m (ii) The equations of the line joining the given points is
x-5 5- 3
--=
On the xy-plane, z
y+2 z-3 x-5 y+2 z-3 = - - or - - = - - = - - . -2 - 0 3- 1 2 -2 2
=0
x- 5 y+2 0- 3 x- 5 y+2 -3 - - = - - = - - or - - = - - = 2 -2 2 1 -1 1 or, x = 5 - 3 = 2 and y + 2 = 3 or, y = 1. the required point is (2,1,0).
. 0 f mtersectlon . . 0 f t he 1·me -2 x - 2 y - 1 = -2z- 3 E x. 9 (I.) F·m d t he pomt - = ---=-3" and the plane 2x + y - z = 3. (ii) If the lines x-I = y - 4 = z - 5 and x - 2 = y - 8 = z - 11 2 1 2 -1 k 4 are intersecting, find the value of k. (iii) Find the value of m for which the line to the plane x - 3y + 6z •
+7=
x; 1 = y: 2 = z~23 parallel
O.
SOLUTION:
x-2 y-l z-3 (i) Let -2- = ---=-3" = -2- = r or, x = 2r + 2, y = -3r + 1, z·= 2r + 3. Putting these values in 2x + y - z = 3 we get 2(2r + 2) + (-3r + 1) - (2r + 3) = 3 or, -r - 1 = 0 or, r . .. x = 0, y = 4, Z = 1. Thus the required point of intersection is (0,4,1).
= -1.
CH.4: GEOMETRY
4.45
(ii) If two line3 intersect then they should be coplanar. 2-1 2 -1
i.e.,
8-4 11-5 1 2 k 4
(4 - 2k) - 4(8 k = 3.
or, or,
= 0 or,
1 2
4 6 1 2
= 0
-1 k 4
+ 2) + 6(2k + 1) = 0 Of,
10k - 30
(iii) If the line is parallel to the given plane then 3.1 + m.( -3) or, -3m - 9 = 0 or, m = -3.
=0
+ (-2).6
= 0
Ex. 10 Find the perpendicular distance of the point P(x, y, z) from the yaxis. . 0 f t h e y-axIs . .IS 0 x z • S OLUTION: The equatIon = iy = O· The distance from the point P(x, y, z) to the y-axis is
Ex. 11 Find the equation of the straight line passing through the point (1, 2, 3) and parallel to the line
~ = ~ = ~.
• SOLUTION: Since the required line is parallel to the given line, the d.rs of the required line are {2, 3, 4}. Hence the equations of the line passing through the point (1, 2, 3) are x-I y-2 z-3 -2- = -3- = -4-' Exercise 1 (i) Find the direction cosines of the lines represented by the equations 3x - 2y + z + 4 = 0 and 2x + 2y - z - 3 = O. (ii) Find the direction cosines of the line, represented by the equations x y - z + 1 = 0 and 4x + y - 2z + 2 = O.
+
(iii) Examine if the straight lines
x-I
y-2
z-3
-2- = -3- = -4- and
x-2 ----,.. 3
y-3
z-4
= -4- = -5-
are cpplanar. (iv) ,
Prov~
that the perpendicular distance from the point (-1,3,9) to the . x - 13 Y+8 z - 1. I1. = 1/3
and 5 + 4
= 3/ A or,
Hence the equation of the generator passes through (3, 4, 5) is z - y = tx,z + y = 3x. Ex. 3 Show that the straight line x-I = y - 2 surface x 2 - xy + 2x - 3y + 2z + 7 = O.
= z + 1 lies entirely on the
• SOLUTION: Let x-I = y - 2 = z + 1 = r . ... x = r + 1, y = r + 2 and z = r - 1. Putting these values to the given surface (r + 1)2 - (r + l)(r + 2) + 2(r + 1) - 3(r + 2) + 2(r - 1) + 7 = 0 or, (r2 + 2r + 1) - (r2 + 3r + 2) + (2r + 2) - (3r + 6) + (2r - 2) + 7 = 0 or, 0 = o. ... (r + 1, r + 2, r - 1) is a point on the surface, for all values of r. Hence the given line lies entirely on the given surface. Ex. 4 Find the equation of the right circular cone whose vertex is origin, axis 7r the z-axis and semi-vertical angle "4.
· fh . x Y z • S OLUTION: E quatlOn 0 t e z-aXIS are 0 = 0 = 1·
4.52
U.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Let P(x, y, z) be any point on the cone. The d.rs of the generator OP are {x,y,z}. Now semi-vertical angle of cone is ~ which is angle between OP and axis z. Thus
7r
cos
'4 =
x.O + y.O
+ z.l
-vr:O~2=+=O=:=;2;;=+=1~2-v't=x:::;. when
a = 5i-3J+8k, b= >.i+3J+2k and IO:-bl = 9.
(v) Two vectors 5 and iJ are obtained by joining the origin to the points (1,0, -1) and (-1, I, 1). Find the value of 15 + iJl . •
SOLUTION:
(i) We have 0: = 4i + nJ - 3k and 10,1 = 13 . . '. V16 + n 2 + 9 = 13 or, 25 + n 2 = 169 or, n 2 = 169 - 25 = 144 or, n = ±12.
(ii) 0: - b= (2i + J + 3k) - (i + 3J + k) = i - 2J + 2k.
10: - bl
=
VI + 4 + 4 =
3.
Hence the unit vector in the direction of 0: -
0: -
i - 2J + 2k
b
10: - bl
=
3
bis
5.3
eR.S: VECTOR ALGEBRA
(iii) The vector AB = position vector of B - position vector of A = (9i + 2] +
k) - (Si + S] + k) = 4i - 3]. '. the distance between A and B is IABI = )16
+9=
5.
(iv) ii - b= (Si - 3] + 8k) - (>.i + 3] + 2k) = (5 - >.)i - 6] + 6k . . '. Iii - bl = )(S - >.)2 + 36 + 36. Given Iii - bl = 9.. '. )(5 - >.)2 + 72 = 9 or, (5 - >.)2 + 72 = 81 or, (5 - >.)2 = 81 - 72 = 9 or, 5 - >. = ±3 or, >. = 5 ± 3 or, >. = 8,2.
i + 0] - k and iJ = -i +] + k. . '. 5. + iJ = (i - k) + (-i + ] + k) = ]. Hence 15. + iJl = 1]1 = 1.
(v) Here the vectors 5. =
(i) Show that the points A, B, C whose position vectors are respectively -2i + 2] + 3k, 2i + 3] + 3k and -i - 2] + 3k form an isosceles triangle.
Ex. 2
(ii) Prove by vector method that the three points A(2, 3, 4), B(I, 2,3) and C(4, 2,3) form a right-angled triangle. (iii) Show that in a triangle ABC, point of BC . •
AB + A.C = 2AM, where M
is the mid-
SOLUTION:
(i) AB=p.v. of B - p.v. of A = (2i + 3] + 3k) - (-2i+ Similarly, and
eA =
EG = (-i -
2] + 3k) - (2i +- 3] + 3k) = -3i (-2i + 2] + 3k) - (-i - 2] + 3k) = -i + 4;'
m, Jf+16 = m.
Now, IABI = )16
leAl = As IABI =
2] + 3k)
+1=
IEGI
= 4i +;.
5]
= )9 + 25 = J34,
leAl, the triangle ABC is isosceles.
(ii) AB = p.v. of F1 - p.v. of A = (1,2,3) - (2,3,4) = (-1, -1, -l)i BC = (4,2,3) - (1,2,3) = (3,0,0) and CA = (2,3,4) - (4,2,3) =
(-2,1,1).
°
Now, IABI = )1 + 1 + 1 = )3, IBCI = )9 + + 0= 3, ICAI = )4 + 1 + 1 = )6. Since IABI2 +'ICAI 2 = IBCI2, the triangle ABC is right angled.
\.
D.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
5.4
(iii) Let the p.v. of A, B, C be respectively --;--:t
Now , A Ali =
b~ - a~ A--;-± ~ ~ AM) = , AU = c - a ,
a, b, c. The p.v. of Mis (b+C)/2. b+ c ~ b+ c - 2a . -- - a = 2
2
Exercise 1 Show that the points A, B, C whose position vectors are respectively 2i + 4) - 3k, 4i + 5) + k and 3i + 6) - 3k form a right angled triangle. Ex. 3 D, E, F are the mid points of the sides BC, CA, AB respectively of a
~riangle ABC.
Show that
AD + BE + CF = o.
• SOLUTION: Let the p.v. of A, B, C be respectively
.
a, b, c.
Now, p.v. of
b+c a+c a+b
D,E,F are respectIvely -2-' -2-' -2-'
AD = b+ c _ a = b+ c - 2a BE = a + c _ b= a + c - 2b and 22' + b _ c= + b- 2c.
cF = a
2
a
2
Th
2
2
AD + BE + cF -_ b+ c2- 2a + a + c2- 2b + a + b2en,
Ex. 4 For what value of c, the length of the vector unit?
ex = c(3i -
2c _ ~ - O.
6) + k) is of 3
• SOLUTION: lal = v'9c 2 + 36c2 + c2 = cv'46. Since lal = 3,cv'46 = 3 or c = 3/v'46. (i) For what values of A and /-L, the vectors -3i + 4) + Ak, and /-Li + 8) + 6k are collinear?
Ex. 5
(ii) Show that the three points collinear.
-i - 2) + 7k, 3i + k and 5i + ) - 2k
are
• SOLUTION: (i) If two vectors are collinear then the coefficients of i,) and tional. -3 4 A -3 1 i.e., -;; = 8" = "6 or -;; = "2 or, /-L = -6 and 1
A
2
6
- = - or A = 3.
k are propor-
5.5
. :l.5: VECTOR ALGEBRA
2J + 7k, 3£+ k and 5£ + J - 2k . AB = (3T + k) - (-T - 2J + 7k) = 4T + 2J - 6k and AC = (57 + J - 2k) - (-£ - 2J + if) = 67 + 3J - 9k.
(ii) Let p.v. of A, B, C be respectively -£ -
. '. .'.
2i.e., AB = 3AC.
Hence the vectors are collinear. (i) If
Ex. 6
oF = T+ 3J - 7k and oQ = 5T - 2J + 4k, then find the unit
vector parallel to
PQ.
(ii) Find a unit vector parallel to the vector •
-3T + 6J - 2k.
SOLUTION:
PQ = oQ-oF = (57-2J+4k)-(T+3J-7k) = 4T-5J+11k,
(i) The vector and
IPQI
= V16
+ 25 + 121 = JI62.
The unit vector parallel to
PQ is
±PQ
--=±
5J + 11k .
4T -
IPQI
v162
(ii) The unit vector parallel to -3£ + 6J -
if is
-3T + 6J - 2k
-3T + 6J - 2k
± I - 3T + 67.- 2kl = ± V9 + 36 + 4
5.2
-3T + 6J - 2k ±
7
.
Scalar Product or Dot Product
a
Def. 1 Scalar product: The scalar or dot product between two vectors and b is denoted by a.b and defined as a.b = lallbl cosO, where 0 is the angle between a and b. The value of a.b is a scalar quantity.
Q. 1 Define dot product of two vectors. Properties:
a.b = b.a (ii) a.a = lal 2 = a2
(i)
(iii) The angle 0 between
a and bis given by
o= (iv) Two vectors
cos
-1 (
a.b)
± Iall bl .
a and bare perpendicular if and only if a.b = o.
5.6
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Q. 2 Give the geometrical interpretation of a.b = O. Ex. 1
(i) Show that a.a =
lal 2 .
(ii) Find the vector ii collinear with;3 = (4,2, -2) and satisfying ii.;3 = 6. • SOLUTION:
(i) Let a = ali + a2] + a3k . ... a.a = (ali + a2] + a3k).(ali + a2] + a3k) =
ar + a~ + a~.
lal = Jar + a~ + a~. Hence a.a = lal 2 . Again,
(ii) The vector collinear with ;3 is
c;3, for some constant
c.
i.e., ii = c;3 = (4c, 2c, -2c). Since 5..;3 = 6, or, (4c, 2c, -2c).(4, 2, -2) = 6 or, 16c + 4c + 4c = 6 or, 24c = 6 or c = 1/4. He~ce
ii = (1,1/2, -1/2).
Def. 2 Coplanar vectors: Three vectors a.(b x C) = O.
a, b, c are
said to be coplanar, if
Ex. 2 If ii,;3, 1 are three non-coplanar vectors and 8.ii = 8.;3 = 6.1 = 0 then show that 6 is a null vector. • SOLUTION: Since 8.ii = 6.;3 = 8.1 = 0, the vector ii,;3 and 1 are perpendicular to 8. This indicates that the vectors ii,;3 and 1 are coplanar, but it is given that they are non coplanar, hence 6 = 0, i.e., a null vector.
Ex. 3 (i) Find (b - a).(3a + b) where a = i - 2] + 5k, b = 2i +] - 3k and find the angle between b - a and 3a + b. (ii) If a = i + 2] - 3k and b = 3i - ] a+2b.
+ 2k, .find the angle between 2a + band
k and 3i+ 4] - k. Find the value of A for which two vectors 2i +] - k and i + 4] + Ak are
(iii) Find the sine of the angle between the vectors - 2i - ] (iv)
perpendicular to each other. • SOLUTION: (i) b - a = (2i + ] - 3k) - (i - 2] + 5k) = i + 3] - 8k and 3a + b = 3(i - 2] + 5k) + (2i + ] - 3k) = 5i - 5] + 12k. Now, (b - a).(3a + b) = (i + 3] - 8k).(5i - 5] + 12k) = 5 - 15 - 96 = -106.
L'-"" 5:
VECTOR
5.7
J. LGEBRA
Ib - 51 = VI + 9 + 64 = J74 and 135 + bl = V25 + 25 + 144 = V194. Then the angle
0= cos-
1 (
ffi~) = cos-
±
1
~~)
±
(
(ii) 25 + b = 2(i + 2J - 3k) + (3i - J + 2k) = 5i + 3J - 4k and 5 + 2b = (i + 2; - 3k) + 2(3i - J+ 2k) = 7i + k.
Now, (25 + b).(5 + 2b)
= 35 - 4 = 31. 125 + bl = V25 + 9 + 16 = 5V2 and 15 + 2bl = V49 + 1 = 5V2.
Then the angle is
(iii)
= cos
-1( ±
_1(±31)
(25+b).(5+2b)) ~ ~
= cos 12a + blla + 2bl 50 (-2i - J- k).(3i + 4; - k) = -6 - 4 + 1 = -9. I - 2i - J - kl = vi4 + 1 + 1 = v'6 and 13i + 4J - kl = V9 + 16 + 1 = J26.
LI
u
,
Then the cosine of angle is cos 0 = .'. sinO
0 Jl-
= Vl- cos 2 =
1 1856
.
-9
v'6J26' 6 26
=
J
5 1: 6
(iv) If the given vectors are perpendicular then (27 + J- k).(i + 4; + >"k) = 0 or, 2 + 4 - >..
= 2~'
= 0 or, >.. = 6.
Exercise 1 (i) Find the angle between the vectors 5 = 2i + 2J b = 3i +4k.
k and
(ii) Show that the vectors corresponding to the positions of the points (3, -2, 1) and (2,3,0) are at right angle. (iii) Show that the vect.ors 5i - 4; + 3k and 2i + ; each other.
-
(iv) Find the value of >.. for which two vectors '5 = 3i + >..; - if are perpendicular to each other. (Ans. (i) cos- 1
2 15 ,
(iv) >.. =
-6.J
2k are perpendicular to
>..i -
4; + 3k and
t=
D.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
5.8
• SOLUTION: Let a = ali + a2) + a3k and b = bli + b2) + b3k. Then a2 = lal 2 = + a§ + a~ and b2 = Ibl 2 = b.i:+ b§ + b~. Again a.b = (ali + a2) + a3k).(bli + b2) + b3k) ~ alb l + a2b2 + a3b3· Also a.b = ab cos e. Therefore, alb l + a2b2 + a3b3 = abcos e or, (alb l + a2b2 + a3b3)2 = a2b2cos 2 e Since cos 2 e :S 1, we have
ar
(albl + a2b2 + a3b3)2 :s a2b2 i.e., (alb l + a2b2 + a3b3)2 :S (ar + a§ + a~)(br + b§ + b~).
Ex. 5 Justify the following statements: In case of three non-null vectors a, band e, a.b = a.e does not always imply that b= e. • SOLUTION: Let a = i + ), b = 2i + ) and e = -i + 4]. Then a.b = 2 + 1 = 3 and a.e= -1 + 4 = 3, i.e., a.b = a.ebut b i- e. Thus the given statement is correct.
Ex. 6 (i) Show that the vector {(5.1)/J - (5./J)1} is perpendicular to the vector 5. (ii) If two vectors a and b be such that la + bl = la - bl find the angle between them. • SOLUTION:
(i) 5. {(5.1)/J - (5./J)1} = (5.1)(5./J) - (5./J)(5.1) = O. Hence {(5.1)/J - (a:./3)1} is perpendicular to 5.
(ii) Given that la + bl = la - bl· _ or, la + ~2 = la..:- bl 2 or, or, or, or,
... the angle between
5.3
_
(a + b).(a + b) = (a - b).(a - b) a.a + a.b + b.a + b.b = a.a - a.b - b.a + b.b 4a.b = 0 [as a.b = b.a] a.b = o.
a and b is 7r /2.
Vector Product or Cross Product
Def. 1 Vector product: The vector product or cross product between two vectors a and b is denoted by a x b and is defined by a x b = lallbl sin eii, where is the angle (acute) between and band ii is a unit vector perpendicular to both a and b.
e
a
Q. 1 Define vector product of two vectors.
eR.5: VECTOR ALGEBRA
5.9
Properties:
(i) (ii) (iii) (iv)
a x b= -b x a. Ifax b= 0 then a and bare parallel. a x a = O. If a and b represent the adjacent sides is la x bl·
(v) If a and (vi) (vii)
of a parallelogram then its area
bbe any two sides of a triangle then its area is ~Ia x bl.
Ifax b= c then c is perpendicular to both a and b. a x (b xC) = (a.C)b - (a.b)c.
Ex. 1 Give the geometrical interpretation of ax
b= O.
• SOLUTION: If the vector product of two vectors then the vectors a and b are collinear or parallel. Ex. 2
(i) If a =
zero, i.e.,
a x b= 0
ii - 2J - k and b= i + J + k find a x b.
iJ
(ii) If 5. and are two vectors such that the value of 15. x iJl. (iii) If
a, b is
lal = 4, Ibl =
3 and
15.1 = 8, liJl = 6 and &iJ = 0,
find
a.b = 5, find la x bl·
• SOLUTION:
i
J k
(i)
a x b=
(ii)
= i( -2 + 1) - J(2 + 1) + k(2 + 2) = -i - 3J + 4k. Since &iJ = 0, 5. and iJ are perpendicular, i.e.,
2 -2 -1 111
the angle
. 7r IS
"2'
15. x iJl = 15.lliJl sin e = (8)(6) sin 7r /2 = 48. Here a.b = 5 or, lallbl cos e = 5 Now,
(iii)
or, (4)(3) cos e = 5 or, cos e = 5/12 or sine ,
la x bl-
=
=
VI - cos 2 e = )1- 144 25 =
lallbl- sine =
J119. 12
JTI9~
(4)(3)~ = v 119.
e between them
U.G.
5.10
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Exercise 1 (i) If 5 and jj are two vectors such that 151 = 4,ljjl = 5 and &jj = 0, find the value of 15 x jjl. (ii) If a and bbe two vectors such that lal = 10, Ibl = 1 and value of la x bl· (iii) If a =
2i If it = 2i -
2] -
(iv)
2] -
•
-i -
3] + 4k, (iv) vI26.]
a x (b + C) + b x (e + a) + e x (a + b) (a x i) +] x (a x ]) + k x (a x k) = 2a .
(i) Prove that
(ii) Prove that
i
x
6, find the
k and b= i + ] + k, find a x b. k and v = i +] + k, find lit x vi.
[Ans. (i) 20, (ii) 8, (iii)
Ex. 3
a.b =
=
O.
SOLUTION:
(b+C)+bx (e+a)+ex (a+b) =axb+axe+bxe+bxa+exa+exb =axb-exa+bxe-axb+exa-bxe = 0 = R.H.S. [since a x b = -b x a] L.H.S.= i x (a x i) + ] x (a x ]) + k x (a x k) = (i.i)a - (ta)i + (;'])a - (;'a)] + (k.k)a - (k.a)k = (a + a + a) - (ali + a2] + a3k) where a = ali + a2] + a3k = 3a - a = 2a = R.H.S.
(i) L.H.S.=ax
(ii)
Ex. 4
a
(i) If and b are two mutually perpendicular vectors, show that I(a + b) x (a - b)1 = 2lallbl·
(ii) Show that •
(a - b)
x
(a + b)
=
2(a x b) .
SOLUTION:
(i) I(a + b) x (a - b)1 = la x a - a x b+ b x a - b x bl = I - a x b+ b x al = Ib x a + b x al = 21 b x al = 21bllal sin 7r /2 [since the vectors are perpendicular] = 2lallbl· (ii)
(a - b)
Ex. 5 Ifax
(b -
C).
x
(a + b)
=
a x a + a x b- bx a - bx b= 2(a x b).
b= eX J and a x e= bx d,
then show that
(a - d)
is parallel to
eR.5: VECTOR ALGEBRA
• SOLUTION: We have Subtracting we get
5.11
a x b= 2 x land a x 2 = b x d~
axb-ax2=2xl-bxl or, a x (b - C) = (2 - b) x l or, a x (€ - C) + (§ - 0 x l =
0 x (b - C) - d x (b - C) = 0
or, a or, (a - d) x (b - C) = O. Hence the vectors a - land
b-
2 are parallel.
Ex. 6 (i) If a, b, 2be respectively the position vectors of the vertices A, B, C of the triangle ABC, then show that the area of the triangle ABC is
!lbx2+2xa+axbj. (ii) Find the area of the triangle two of whose sides are given by the vectors 2i +] - 3k and 3i - ] + 2k. (iii) Find the area of the parallelogram formed by two vectors 3i + 2] and
2] - 4k. •
SOLUTION:
1- (i) The area of a triangle ABC is 2"IAB x ACI.
Now, AB = b- a and AC = 2 - a. ... !IAB x ACI = !I(b - a) x (2 - 0,)1
= ! Ib x 2 - b x a - a x 2 + a x 0,1
a b+ bx 2 + 2 x 0,1.
= !I x
(ii) The area of the triangle is !1(2i +; - 3k) x (37 - ] + 2k)l. Now, (2i +] - 3k) x (37 -] + 2k)
=
7 i
k
2 1-3 3 -1 2
= 7( 2 - 3) - ;(4 + 9)
+ k(- 2 -
-7 - 13] - 5k. Its magnitude is VI + 169 + 25 =
3)
=
VI95.
Hence the required area is !VI95. (iii) The area of parallelogram formed by the given vectors is
1(3i + 2]) x (2] - 4k)l. Now, (37+2;) x (2J - 4k)
D.G.
5.12
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Z
-
o
2 -4
J 3 2
k o
7( -8 - 0) -)( -12 - 0)
+ k(6 -
0)
-87 + 12) + 6k. magnitude is )64 + 144 + 36 =
)244 =
2J6I.
= =
Its Hence the required area is
b = ii x c then b= C? Justify your answer. It is given ii x b= ii x c and b#- c. What conclusion you can draw? If ii #- 0 and ii x b = ii x c, does it imply that b = c? If it does not, what
Ex. 7 (ii) (iii)
2J6I.
(i) If ii x
does it imply? •
SOLUTION:
(i) If ii x b = ii x c then ii x b- ii x c = (5 or, ii x (b - C) = (5. This implies either ii = (5 or b= c, or ii and b- c are parallel.
b= ii x c then ii x (b - C) = (5. If b#- c then either ii = (5 or ii and b- c are parallel.
(ii) If ii x
(iii) If ii #- (5 and ii x b= ii x c then ii x or ii and b- c are parallel.
(b -
C)
= (5 implies either b=
c
/
Ex. 8 If 5 •
+ iJ + 1 = (5 show that
SOLUTION:
Given 5
+ iJ + 1 =
5 x
iJ = iJ x 1 = 1 x a.
(5.
.'. 5 x (5 + iJ + 1) = 0 or, 5 x 5 + 5 x iJ + 5 x 1 = (5 or, 5 x iJ = -5 x 1 [since 5 x 5 = 0] or, 5 x iJ = 1 x 5 Again iJ x (5 + iJ + 1) = 0 or, iJ x 5 + iJ x iJ + iJ x 1 = 0 or, -5 x iJ + iJ x 1 = (5 or, iJ x 1 = 5 x iJ From (1) and (2), 5 x iJ = iJ x 1 = 1 x 5
(1)
(2)
Ex. 9 (i) Find the unit vector perpendicular to each of the vectors ii = and b=) + k. (ii) Find a vector of magnitude 5 perpendicular to both the vectors and i - 2) + k.
7-2)
2i+) - 3k
eR.5: VECTOR ALGEBRA
5.13
• SOLUTION: ' vector perpend'lCUI ar to a- an d umt (1.) Teh
--b IS •
ii x b . ±--_ Iii x bl
Now, ii x
Iii x
i
b=
; k
1 -2 0 o 1 1
= -2i - ; + k. bl = y!4 + 1 + 1 = V6.
. the required vector is (ii) Let ii
1 (- 2i- - j- + k). ± V6
= 2i + J - 3k and h = i - 2J + k.
Now, ii x
i
b= =
J k
2 1-3 1 -2 1
-5i - 5J - 5k.
Iii x hi = y!25 + 25 + 25 = 5V3. The vector perpendicular to ii and
± 5(ii x
!)
=
bof magnitude 5 is ± 5( -5i - 5J - 5k) = ± -5(i +; + k)
Iii x bl
5V3
V3
Exercise 2 Find the unit vector perpendicular to each of ii = 6i + 2; + 3k and
b= 3i - 6J - 2k.
[Ans. ±(2i+ 3J - 6k)/7.]
Ex. 10 If 1i5..(iJ x 1)1
a, iJ =
and 1 be three mutually perpendicular vectors, show that lalliJlI11·
• SOLUTION: Since the vectors are mutually perpendicular, the angle between any two vectors is 7r /2. Now, 1i5..(iJ x 1)1 = 1i5..(liJI111 sin7r/2ii)1 where ii is a unit vector perpendicular to both iJ and 1, i.e., ii is parallel to i5.. or, 1i5..(iJ x 1)1 = liJI111 (i5..ii) = liJl 111 lalliil cos 0 = lalliJl111· Ex. 11 Show that la x iJI 2 + (i5..iJ)2 = lal 2liJl2 . • SOLUTION: 1& x iJ12.+ (a.iJ)2 = 1&121iJ12 sin2 (J + lal 2 1iJI 2 cos 2 (J = l 1 l.e.,
Xn > Xn+l.
n n Hence the sequence is monotone decreasing. Xn+l
Also, the lower bound of the sequence is 0 (when n -+ 00). Hence the sequence is convergent. Ex. 3 (i) If the sequence {an}n converges, show that the sequence {Ianl}n is also convergent. (ii) If a sequence {xn}n converges to 1 show that the sequence {Ixnl}n converges to Ill . •
SOLUTION:
(i) Let c: be any given positive number. Let the sequence {an} converges to the limit l . ... there exists m such that Ian - II < c: for all n Now
Ilanl-llll ::;:
Hence {Ianl}
> m.
-ll < c: for all n > m. converges to Ill· Ian
(ii) Let c: be any given positive number. Since Xn -+ 1 as n -+ 00, there exists m such that IX n - II n>m.
Ilxnl- Illl ::;: IX n -ll < c: for all n > m. This shows that {Ixnl} converges to Ill.
Now,
Ex. 4 Find the value of lim {vn + 1- Vri}. n-+oo
•
SOLUTION:
(vn + 1 - Vri) = =
[since
01
(vn+l - y'n) (V'n+l + y'n) (v'n+T + y'n) 1
(Vn + 1 + v'n)
v'nTI > y'n)
1 2y'n·
.
3n + 2 3n-1 3 2 n+ n+ (3n + 2)(n + 2) - (3n - l)(n + 3) = (n+3)(n+2) 7 (n + 3) (n + 2) > 0 for all n ;::: L
=
Xn ,
i.e.,
{xn}
is monotone increasing.
6.9
U.G. MATHEMATICS (SHORT QUESTIONS 'AND ANSWERS)
6.10
(ii) Here . x ..
2n-1 3n + 4 . 2(n + 1) - 1 2n + 1 - 3(n + 1) + 4 - 3n + 7·
=
Xn
n+1
Now,
Xn+1 - Xn =
=
2n + 1 2n -1 3n + 7 - 3n + 4
+ 1)(3n + 4) - (2n - 1)(3n + 7) (3n + 4)(3n + 7) 2 (6n + 11n + 4) - (6n 2 + 11n - 7) (3n + 4)(3n + 7) (2n
11 (3n+4 )(3n+7)
Thus (iii) Let .
Xn+1
Xn
> X n , i.e.,
{xn}
°
> for all n > 1. -
is monotone increasing.
n+1
= --. n
n+2 = n + l'
.. Xn+1
Now , x n +1
- X
n+2
n
n+1
= - - -nn+1
n(n+2)-(n+1)2 n(n + 1) 2 (n + 2n) - (n 2 + 2n
+ 1)
n(n + 1) -1
n (n+ 1) Thus
Xn+1
1. -
is monotone decreasing.
Exercise 2 (i) Show that the sequence {2:: 1} is monotone increasing and bounded. (ii) Show that the sequence
{n: 1} is monotone decreasing.
Ex. 6 Give an example of a sequence which is neither monotonic increasing nor monotonic decreasing.
,
• SOLUTION: The sequence {(_1)n + 1} i.e., {O,2,O,2, ... } is neither monotone increasing nor monotone decreasing. The sequence is oscillating. E x. 7 If
Xn =
such that IX n
-
2n + 5 were h . a pOSItive '" mteger, fi n d t h I ·mteger m n IS e east 6n -11 ~ I < 10- 3 whenever n > m.
CH.6: DIFFERENTIAL CALOULUS
• SOLUTION:
I
Xn -
3"11
2n
5 - 3"11 = 1 6n -+11
= 118n2~ 331 < € or, 26 or, n
< 18n€ 26
>
6.11
33€ or, 18n€
+ 33€ 26 18€ = 18€
where
€
= 10-
3
.
> 26 + 33€
33
+ 18 = 1.4444 x
3
10
+ 1.8333
or, n > 1446.2333 ... the least positive integer m = 1447.
n 1 Theorem 3 If {xn} be a sequence such that lim Ix + 1 = l, where 0 :S l < 1, then lim
Xn
n-+oo
n-+oo
= O.
Xn
Xn
Ex. 8 Show that -, = 0 for any real x. n. Xn
• SOLUTION: Let an = -, . n. . x n +1 .. an+! = (n + I)! .
I I xn+!
an+I Now, - =
I an
xn! -
I
+ I)! xn = In : 11 = n: 11 X I -+ 0 as (n
n -+
00
for all real x.
Hence lim an = 0 n-+oo n'
or, lim ~, = 0 for all real x. n-+oo n .
. Theorem 4 (Sandwich Theorem): If {an}, {b n }, {en} are three sequences such that (i) an :S bn :S en for all n, and (ii) lim an = lim en = l n-+oo n-+oo then lim bn = l. n-+oo
Proof. Since {an} and {en} are converge to l, therefore,' there exists positive integers mI, m2, such that Ian - II < € for all n ~ m I, (1) and len -ll < € for all n ~ m2. (2) Let m = max(mb m2)' Then for n ~ m, we have by (1), (2) and an :S bn :S en l - € < an :S bn :S en < 1 + €
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
6.12 ¢:} ¢:}
< bn < 1 + €, for all n 2: m, Ibn - II < €, for all n 2: m.
1-
€
Hence lim bn = l. n-+oo
Ex. 9 Show that the sequence {b n }, where 1 bn = [ (n+l)2
1
1
+ (n+2)2 + ... + (n+n)2
]
'
converges to zero. • SOLUTION: It is obvious that, for all k, 1
~
_1_ < . 1 (2n)2 - (n + k)2
~
k
~
n
\ ' for all n. n
n n Then (2n)2.~ bn ~ n 2 ' for all n. 1 1 Now, let an = - and Cn = - be the nth terms of two sequences such that 4n n (i) an ~ bn ~ Cn for all n, and (ii) lim an = lim en = o. n-+oo
n-+oo
Then by Sandwich theorem lim bn n-+oo
= O.
Theorem 5 (Cauchy's first theorem on limits): If lim an = l, then n-+oo
lim (a 1 + a2 n-+oo
+ ... + an)
= l.
n
Ex. 10 Show that . [1 11m v'n 2 + 1
+
n-+oo
• SOLUTION: Let Therefore,
ak =
.J
1 v'n 2 + 2
n n 2 +k
· 11ma n
n-+oo
'
+ ... +
1 ] =l. v'n 2 + n
k = 1,2, ... , n.
1·1m p1= l . = n-+oo 1+1 n
Thus by Cauchy's first theorem on limits, we have
. al -hm n-+oo
+ al + ... + an n
..
=1
6.13
CH.6: DIFFERENTIAL CALCULUS
or
. 1[ n . [1 vn 2 +
hm -
n~oo n vn 2 + 1
or
hm
n~oo
1
+
+
n vn 2 + 2
1
vn 2 + 2
+ ... +
+ ... +
n J= 1 vn 2 + n
1 ]=
vn 2 + n
1.
Ex. 11 Find the limit of the sequence {an} defined by
•
SOLUTION:
Let lim an = l. n~oo
Now lim an+! = lim
n~oo
or ,
n~oo
~2 (an + ~) an
l = ~2 (l + ~) l
or, l2 = 9 or, l = ±3. But, l i= -3 as an > 0 for all n. Hence l = 3.
Exercise 3 Find the limits of the following sequences (i) an+! =
~(an +
:J,k >
1,a1
>
0,
n (1'1') an+1 =34 + 3a 2 ,n~1,a1=1.
+
[Ans. (i)
6.3
V
an
k : l' (ii)
J2.]
Infinite Series
Def. 1 Series: A series is the sum of the terms of a sequence. Thus if aI, a2, a3,··· is a sequence then the sum a1 + a2 + a3 + ... of all the terms is n
called an infinite series and is denoted by
L an or simply by L: an· n=l
Def. 2 Partial sum: Let 8 n = a1 + a2 + ... + an for all n. Then 8 1 = aI, 8 2 = a1 + a2, 8 3 = a1 + a2 + a3, and so on. The sequence {8 n } is called the sequence of partial sums of the series. lim 8n is said to be the sum of the series. n~oo
V.G. MATHEMATICS (SHORT QUESTIONS AND'ANSWERS)
6.14
Def. 3 Convergence of series: An infinite series is said to be converge, diverge or oscillate according as its partial sums {Sn} converges, diverges or oscillates. Theorem 1 A necessary condition for convergence of an infinite series is that lim an = O.
L
an
n---+oo
Ex. 1 Show that the series 123
-+-+-+". 2 3 4 is not convergent. • SOLUTION: Let an
=
n --1 n+
Now, lim an
= n---+oo lim ~1 = n+
Since lim an
=1=
n---+oo
n---+oo
1
=1=
O.
0, therefore, the series is not convergent.
Cauchy's general principle of convergence for series A necessary and sufficient condition for the convergence of an infinite series 00
L
an is that the sequence of its partial sums {Sn} is convergent. That is, a
n=l
series that
L
an converges iff for each Ian+!
€
> 0, there exists a positive integer m such
+ a n+2 + ... + an+pl < €,
for all n ~ m and p ~ 1.
Ex. 2 Vse Cauchy's general principle to show that the series converge .
L.!.n does not
• SOLUTION: If possible, let the series be convergent. Then for any given € (say, 0.25), there exists a positive integer m, such that 1- + -1- + " . + -1- I < € In+1 n+2 n+p'
for all n
> m and p > L -
-
In particular, if n = m and p = m we get
Thus there is a contradiction. Hence the given series does not converge.
CH.6: DIFFERENTIAL CALCULUS
6.3.1
6.15
Series of positive terms
Theorem 2 The positive term geometric series 1 + r for r < 1 and diverges to +00 for r ;::: 1.
+ r2 + r3 + .. " converges
Comparison test Theorem 3 A positive term series
L nP1 is convergent if and only if p > 1.
Theorem 4 A. If L: an and L: bn are two positive term series, and k is any positive fixed real number (independent of n) and there exists a positive integer m such that an :::; k bn for aU n ;::: m, then (i) L: an is convergent, if L: bn is convergent, and (ii) L: bn is divergent, if L: an is divergent. B. (Limit form) If E an and L: bn are two positive term series such that lim (abn ) = 1, where n-too
n
1 is a non-zero finite number, then the two series converge or diverge together.
Ex. 3 Show that the series
L sin -n1 is divergent.
. 1
1
• SOLUTION: Let an = sm - and bn = -. n n Now, l' sin lin =1. · an 11m-=lm n-too bn n-too 1I n Therefore, two series either converge or diverge together .. But, the series L: bn diverges, therefore, by comparison test diverges.
Ex. 4 Test the convergence of the series 1
L
L: sin lin
also
nili/n'
1
• SOLUTION: Let an = n1+ 1 / n and bn = ~. Now, an l'1m -1/ 1 1 · -b 11m = n-too n-too n n n = . Hence the two series either converge or diverge together. But, the series L: bn is a divergent series, therefore, by comparison test the series E an is also divergent.
6.16
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Some standard tests for positive term series: (i) Cauchy's root test:
1f:E Un
be a series of positive terms, and let
lim u 1 / n = l.
n-too
l < 1, l > 1, l = 1,
If If If
n
then :E Un is convergent, then :E' Un is divergent, then the test fails.
1f:E Un
(ii) D' Alembert's ratio test: let
be a series of positive terms, and
Un+l l --=. 1I· m
n-+oo
If If If
l l l
< 1, > 1, = 1,
then :E Un is convergent, then :E Un is divergent, then the test fails.
(iii) Raabe's test:
1f:E Un
be a series of positive terms, and let lim
n-too
If l > 1, If l < 1, If l = 1,
Un
{n(~ -I)} = Un+l
l.
then :E Un is convergent, then :E Un is divergent, then the test fails.
(iv) Gauss's test:
1f:E Un
be a series of positive terms, and ~ be exUn+l
pressed in the form
then
Ex. 5
:E Un
is convergent if f3
< 1,
. 1 (i) Show that the senes -
and divergent if f3 2: 1. 2
3
n
3
n
2+ -22 + -2 + ... + -2
(1'1') Show t h at t h e series - 1 1.3
+ -1 + -1 + ... 3.5
5.7
+ ... IS. convergent .
is convergent.
(iii) Prove that the series
1 (2)2 + (3)3 "7 + ... + (n)n 2n + 1 + ...
3+ 5 converges.
6.17
CH.6: DIFFERENTIAL CALCULUS
(iv) Prove that the series
2}+~.(~)2 +~.(~)3 + ... + n+1.(~)n + ... 3 3
8
3
27
3
n
3
is convergent. 00
(v) Prove that
L;
n
diverges. n=l n. (vi) Show that the series 1 1 1 -+-+-+ ..... . 1.2 3.4 5.6
is convergent.
(vii) Test whether the series •
00
1
1
n
L - is convergent or divergent.
SOLUTION:
(i) Here Un = ~. 2n . n+1 .. U n+l =21 . n+Now, . Un+l 11m - n-+oo Un
=
n +1 2n 1im -1 - x n n-+oo 2 + n . n +1 11m - n-+oo 2n
~2n-+00 lim (1 + .!. ) n 1
- 1. n-+oo
1)
Hence by Raabe's test the series is convergent. (iii) Let
Un
=
If·
(2n:
Now, lim u~/n = lim ( n-+oo
n-+oo
.
= 11m n-+oo
n
2n + 1
)
1 2 + lin
1
=- 1, n
> U n +l.
Hence {un} is a monotonic decreasing sequence. Also, lim Un n--+oo
= n--+oo lim .! = O. n
Hence by Leibnitz test the series is convergent. 1 . 1 (ii) Here Un = n 2 ' .. un+! = (n + 1)2' Now,
~
=
(n+ 1)2 = (1 + .!)2 > 1,
U n +l
i.e." Un
n
n
> U n +l.
Hence {un} is a monotonic decreasing sequence. . l'1m Un = l'1m "2' 1 = O. Agam, n--+oo n--+oo n Hence by Leibnitz test the series is convergent . . Def. 4 Absolutely convergent: An alternating series absolutely convergent if the series z:::: Ian I is convergent.
L: an is said to be
CH.6:
1..'.
'FERENTIAL ,'ALCULUS
6.21
Def. 5 Conditional convergent: An alternating series L an is said to be conditional convergent if the series L an is convergent, but the series L Ian I is not convergent. Ex. 7 Show that the series 111 1 - - + -2- - +2 · · · · · · 22 3 4 is absolutely convergent. • SOLUTION: The absolute series, which is obtained on taking every term of the given series with a positive sign 111 1 + 22 + 32 + 42 + ..... . 1 nP' where p = 2 is convergent, because, the series is of the form Hence the given series is absolutely convergent.
L
> 1.
Ex. 8 Show that the series 111 1--+---+······ 234 is conditional convergent. 1 • SOLUTION: Let an = -. The given series is convergent by Leibnitz test as n
lim an = 0 and lan+ll < lanl for all n. n~oo But, the absolute series 111 1+-+-+-+ ...... 234 is not convergent, because, this series is of the form Hence the given series is conditional convergent.
L
1
nP' where p = 1.
Theorem 6 Every absolutely convergent series is convergent. Proof. Let L an be an absolutely convergent series. So the series L lanl is convergent. Hence, for any c > 0, by Cauchy's general principle of convergence, there exists a positive integer m such that lan+ll +
lan+21 +
Also for all nand p
... + lan+pl
< c;
for all n ~ m and p ~ 1.
> 1,
Ian+! + an+2 + ... + an+pl
< lan+ll + la n+21
+ ... + lan+pl
< c,
for all n ~ m and p ~ 1. Hence, by Cauchy's general principle of convergence the series verges.
L
an con-
6.22
U.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Function
6.4
Def. 1 Function: Let A and B be two non-empty sets. A correspondence which associates with each element x E A an unique element y E B, is called a mapping or a function. It is generally denoted by f : A -+ B. Def. 2 Co-domain: Let f : A -+ B be a function. Then A is called domain and B is called co-domain of the function f. Def. 3 Range: Let f : A -+ B be a function. The subset of B containing those y E B for which there exist x E 11 such that f(x) = y will be called the range or image set of f, denoted by f (A). '
x-a
x-b
= b b _ a + a a _ b' f(a) + f(b) = f(a + b). (ii) If f(x) = x 2 - 3x + 7, show that
Ex. 1
(i) If f(x)
{f(x
+ h) -
then show that
f(x)}
h •
= 2x - 3 + h .
SOLUTION:
x-a x-b (i) Wehavef(x)=bb_a +aa_b . ... f(a) = a, f(b) = band b a f(a+b)=bb_a +aa_b b2 a2 b2 - a 2 ----b-a b-a b-a = b + a = f(b) + f(a). Hence f(a) + f(b) = f(a + b). (ii) f(x) = x 2 - 3x + 7, ... f(x + h) = (x + h)2 - 3(x + h) + 7 = x 2 + 2hx + h 2 - 3x - 3h + 7
= x 2 + (2h - 3)x + h 2 - 3h + 7. f(x + h) - f(x) Now, h {x 2 + (2h - 3)x + h2 - 3h + 7} - (x 2 - 3x + 7) = h 2 2hx + h - 3h = h = 2x + h - 3 = 2x - 3 + h.
CH.6:
Ex. 2
D.:- }'ERENTIAL
6.23
I' \LCULUS
(i) Find the domain of definition of the function f(x) =
JX=l +
V5=X. (ii) Find the domain of definition of the function log(x 2
-
5x
(iii) Find the domain of definition of f where f(x) = VlOg 5x
+ 6).
~x
2 .
(iv) Find the domain of definition of f where f(x) = J(3x - 1)(7 - x). (v) Find the domain and range of the function f(x) = (vi) Show that vx 2 •
-
5x
+ 6 is not
defined for 2 < x
1:.1. x
< 3.
SOLUTION:
(i) The function is defined if x-I 2: 0 and 5 - x 2: 0 or, x 2: 1 and x ::; 5. Combining we get 1 ::; x ::; 5. Which is the domain of definition of f(x). (ii) The function is defined if x 2 - 5x + 6 > 0 or, (x - 2)(x - 3) > O. Either x - 2> 0, x - 3> 0 or, x - 2 < 0, x - 3 < O. The inequalities x - 2> 0 and x - 3> 0 imply x > 3 and the inequality x - 2 < 0 and x - 3 < 0 imply x < 2. Hence the domain is x < 2 and x > 3, i.e., (-00,2) and (3,00) . (iii) The function
.Ilog 5x -4 x V
2
is defined when
5x - x 2 4 2: 1
(as log a < 0 when 0 < a < 1) or, 5x - x 2 2: 4 or, x 2 - 5x + 4 ::; 0 or, (x - 1)(x - 4) ::; O. i.e., when x-I < 0, x - 4 > 0 or x-I> 0, x - 4 < o. But x < 1, x > 4 does not give any real value of x, so when x the function is defined.
> 1, x < 4
Hence 1 < x < 4 is the domain of definition. (iv) The function is defined when (3x - 1)(7 - x) 2: 0, i.e., when 3x - 1 2: 0,7 - x 2: 0 or 3x - 1 ::; 0,7- x ::; O. The relation 3x - 1 2: 0,7 - x 2: 0 imply x 2: ~,x ::; 7 or, ~ ::; x ::; 7. The relations 3x - 1 ::; 0,7 - x ::; 0, i.e., x ::; ~, 7 ::; x does not give any valid x. Hence the domain of definition is ~ ::; x ::; 7. (v) The function is defined for all real values of x except O. Therefore the domain of definition is R - {O}, R is a set of real numbers.
D.G.
6.24
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Also,
f(x) =
{-I, 1,
x
O.
f is {-I, I}. (vi) When x - 5x + 6 > 0 or, (x - 2)(x - 3) > 0 then the function is defined. i.e., when x - 2 > 0, x - 3 > 0 and x - 2 < 0, x - 3 < O. i.e., x > 2, x > 3 and x < 2, x < 3. Hence the range of 2
Combining these relations we get x > 3, x < 2 . . '. the function is defined when x not defined when 2 < x < 3.
3 and consequently f(x) is
(i) Find the domain of definition of f(x) where
Exercise 1
f(x) =
loge
4x - x 2 3 ' x and f(x) are real numbers.
(ii) Find the domain of definition of the function f(x) [Ans. (i) 1 :::; x :::; 3, (ii) 2 Ex. 3
(i) If f(x) = x
(ii) If f(x) =
+ lxi, find whether
f(2) and f( -2) are equal?
Elx and c (~ 0) be any real number show that
SOLUTION:
= x -t-Ixl . . '. f(2) = 2 + 121 = 2 + 2 = 4 and f( -2) = -2 + 1- 21 = -2 + 2 = O.
(i) Here f(x)
f(2) and f( -2) are not equal. (ii) Here f(x) =
2 + )4 - x.
< x < 4.]
If(c) - f( -c)1 = 2 . •
= )x -
El. x
.'. If(c) - f(-c)1 = I~
- ~I·
When c
> 0 then this is equal to I~
When c
< 0 then this is equal to I~c - ~ I = 2.
-
~c I = 2.
Hence If(c) - f( -c)1 = 2. Ex. 4 Draw the graphs of the following functions:
CH.6: DIFFERENTIAL CALCULUS
6.25
(i) f(x) = x - [x], where [xl denotes the greatest integer not greater than x.
(ii)
y~{
1, 0, -1,
(iii)
y~ {
0, 1 +x, 1- x,
•
when x> 0 when x = 0 when x < o. when Ixl > 1 when - 1 :S x :S 0 when 0 < x :S 1.
SOLUTION:
(i) The following figure is the graph of the given function. y
_({j!;+x Figure: 6.4.1
(ii) The graph of the function
IS
shown below.
y 1
CJ)-----
----+----- X
o
---~-1
Figure: 6.4.2
(iii) The graph of this function is shown below.
y
1 --¥'---+---4--- X
-1
o
Figure: 6.4.3
1
6.26
6.5
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Limit and Continuity
Def. 1 Limit: The limit of a function f(x) for a given value, a, of the independent variable x is the constant I for which the function f(x) can be made as small as we please by making x to approach sufficiently nearer to its assigned value a (denoted as x -+ a). It may be expressed as lim f(x) = I.
x-+a
That is, a function f(x) is said to have a limit I as x approaches a, if for any positive number c, there exists a number 8 depending on c can be found out, such that If(x) -II < c for all values of x for which 0 < Ix -
al < 8.
This definition is called Cauchy's definition of a limit of a function. Ex. 1 Define the symbol: lim f(x) = I. x-+a
lim f(x) = I means, given any pre-assigned positive number €, however small, we can determine another positive number 8, depending on c, such that If(x) -II < c for all values of x satisfying 0 < Ix - al ::; 8. •
SOLUTION:
x-+a
Ex. 2 Define the statement 'lim f (x) = x-+a
00'
If f(x) tends to infinity with the same sign when x tends to a either from the right or left, then w(:) say that as x tends to a, f(x) tends to infinity. In other words, if corresponding to any pre-assigned po~itive number M, we can determine a positive number 8, such that f(x) > M whenever 0 < Ix - al ::; 8, we say lim f(x) = 00. •
SOLUTION:
x-+a
Ex. 3 What are the differences between lim f (x) and x-+a
f (a).
f(a)isthevalueofthefunctionf(x)whenx=a. But limf(x) x-+a is the value of the function f(x) when x is very close to a but not equal to a. This is called the limiting value of the function f(x). •
SOLUTION:
Theorem 1 If lim f (x) = I and lim c, whenever IXI - X2L< Hence the function f(x) is not uniformly continuous on (0,1].
a.
CH.6: .clFFERENTIAT CALCULUS
6.37
Ex. 12 Prove that sinx is uniformly continuous on [0,00) . • SOLUTION: Let c IsinXI - sinx21
> 0 be given. Let XI,X2 =
0 such that If(xd - f(X2)·1 ~ c, where Xl,X2 are any two points of I for which IXI - x21 < o. Let X E I and X2 = x. Then for c > 0 there exists 0 > 0 such that If{x) - f{xdl < c whenever Ix - xII < o. Hence the function is continuous at every point Xl E I. Thus the function f (x) is continuous on I.
6.7
Differentiation
Def. 1 Derivative: The differential coefficient of y = f (x) with respect to is dy 1· f(x + h) - f(x) f '() X or -d = 1m h X
X
h~O
provided this limit exists.
Ex. 1 Show by a diagram the difference between fl.y and dy. • SOLUTION: fl.y is the increment of y as the limiting value of fl.y as fl.x -+ O.
X
Figure: 6.7.1
is incremented by fl.x and dy is
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
6.38
Ex. 2 State the geometrical interpretation of :: . • SOLUTION: For the curve y = J(x), :: is the gradient of tangent at any point (x, y).
Differential coefficient in some standard cases
d
d
dx (c) = 0
n 1
dx (xn) = nx -
d
dx (aX) =
d~(sinx) =
aX loge a
d~ (cos x) =
cos x
- sin x
d
dx (cot x) = -cosec 2x
d
d
dx (cosec x) =
dx (sec x) = sec x tan x
-cosec x cot x
d 1 -(log x) = dx
-
x
d (Sin . -1) x =
1 '
v'f="P
dx
d
-(tan
. dx
v"f=X2
l
-1
1 x) = - 1 + x2
d -1 -d (sec x) = x
d~ (sinh x) =
d 1 1 -(cot- x) = - - -2
1+ x
dx
1
JX2=1' x x -1
cosh x
d~ (~anhx) = seth 2x d
dx (sechx) = -sechx tanh x
Ex. 3
_d (cos- 1 x) = _--==1= dx
d (coshx) ="'sinhx dx '
d"
d
dx (cosechx) = -cosechxcothx
(i) Differ~ntiate sin- 1 J(x) w.r.t. x.
(ii) Differentiatei loglo x w.r.t. x 8 .
.
dx (cothx) = -cosech 2x
CH'c
un Fi:RENTIAL CAl CULUS
6.39
(iii) Differentiate x cos- 1 x w.r.t. cos- 1 x. Lr . ) D'IuerentIate ' tan- 1 (IV
[
COS .X ] w.r.to x 1+smx (V) Differentiate lOiogcou w.r.t. x.
(vi) Differentiate logsinx x w.r.t. x. (vii) Differentiate log(.Jx •
a+ .Jx -
b) w.r.t.
x.
SOLUTION:
.
d { .
(1) dx sm
-I} f(x)
f'(x)
d
1
= V1- {f(x)}2' dx {f(x)} = V1 -
{f(x)}2'
(ii) Let y = log10 x and z = x 8 • The derivative of log10 x w.r.t. x 8 is
Now, y = loge x.:loglO e. dy 1 -', dz 7 .. -d = - log10 e and -d = 8x . x x x Hell,ce dy =' dy . dx = log10 e . _1_ = 1 dz dx dz X 8x 7 8x8 loge 10 (iii) Let y = x cos- 1 x and z = cos- 1 X. dy -1 1 dz 1 .. -d = 1. cos x-x. ~2 and -d = - .J 2 X 1- x X 1- X -1
x
dy dy dx cos x - v'1-x2 Hence - = - . - = 1 dz dx dz - r.;===if vl-xM
=
X -
VI - x 2 cos- 1 x.
(iv) Let Y =
2 2 -1 [COS x/2 - sin X/2] -1 [ cos X] tan 1+sinx =tan (cosx/2+sinx/2)2
-1 [COSX/2 -
sinx/2] . cosx/2 + smx/2
=
t~
=
1 tan- [tan
(~- ~)]
=
=t~
_I [1- tanx/2]
~ -~.
dy 1 dx = -2' (v) Let y = lOiog cou . Taking logarithm (with base e) 'both sides we get, loge Y = log cos x.loge 10.
Differentiating w.r.t. x 1 dy 1. --d = loge 10.--( - smx) 'lL.!?cos X
= -loge 10. tanx.
1 + tanx/2
~~.
6.40
V.G. MATHEMATICS (SHORT: QUESTIONS AND ANSWERS) dy or, dx
= -loge 1O.ytanx = -loge 10. tanx.lO 1ogcosx
(vi) Let y = logsinx X = loge x logsinx e = I
loge x . oge smx
Differentiating w.r.t. x we get, dy
x1 l oge' smx -
dx
(loge sin x)2 1 cot x.loge x x loge sin x (loge sin x)2 1 cot x.logsinx X x loge sin x loge sin x 1 - x cot x logsinx x x loge sin x
= = =
1 cos X. Ioge X SiiiZ
(vii) Let y = loge vx - a + vx - b). Differentiating w.r.t. x we get, dy
=
dx
=
1
2' VX =
{!(x _ a)-1/2
1
+ !(x _
rx=a + v'"X-=b 2 2 1 1{ 1 1} rx=a + Vx - b . 2 vx=a + v'"X-=b 1
2'
vx=a
1 + v'"X-=b a + Vx - b' vx - a.v'"X-=b
1 v(x - a)(x - b)'
Exercise 1 (i) Differentiate loglO x with respect to x 3 • 1 (ii) Find the differential coefficient of x sin - x w.r.t sin- 1 x. (iii) Differentiate 1010gsinx w.r.t. x. (iv) Differentiate sec(tan- 1 x) w.r.t. x. .) lOglO e (") [A ns. (1 ~,11
(1ogx+ V1 - x2sin X) xsm
·-1
X
1
"') 1 10 1010gsinx (. ) xsec(tan- x) ] (111 oge . cot x. ,IV 1 + x2 •
Ex. 4 (i) For the function J(x) = Ixl find 1'(0). (ii) If 1 + X, x> 0 J(x) = { 1- x, X $ 0, examine whether 1'(0) exists.
X,
b)-1/2}
6.41
CH.6: DIFFERENTIAL CALCULUS
(iii) Suppose
f(x) =
X2 when x is rational 0,' when x is irrational
{
Show ~hat 1'(0) = O.
+ 1)x+2, findJ'(O). 21xl + Ix - 21 fikd f'(1) .
(iv) If f(x,} = (x (v) If f(:f) = •
SOLUT~ON:
(i) L.H.D.= lim f(h) - ~/(O) = lim Ihl-IOI =' lim h~O+ h h~O+ h h~O+ h
1M
h !' 1 =1. = Iim-=!,-lm h~O+ h h~O+
RH.D.= lim
f(h\~ f(O) = lim Ihl-IOI
h~O-
h
h~O-
lim
h~O-
1Mh
-h
= lim -1 = -1. hi h~OThus L.H.D.:;i: RH.D. Hence f'(O) does not exist. = lim
h~O-
-..:..:J.
") L .H.D - l' f(h) -h f(O) -_ l'1m 1 + hh - 1 -(11 . - 1m h~O+
h~O+
l'1m -h h -- 1.
h~O+
RH.D.= lim f(h) - f(O) = lim 1 - h - 1 h~Oh h~Oh
-h
= lim -h =-1. h~O-
Thus L.H.D.:;i: RH.D. Hence f'(O) does not exist. (iii) We know 1'(0) = lim f(h) h~O
~ f(O).
When h tends to rational numbers
2
then f'(O) = lim h - 0 = lim h = O. h~O h~O h (f(0) = 02 = 0, since 0 is a rational number) Similarly, when h tends to irrational numbers then
1'(0)
= lim 0 _h 0 h~O
= lim 0 h~O
= O.
In both cases, 1'(0) = O. Hence f'(O) = O. (iv) Taking logarithm both sides we get log f (x) = (x + 2) log (x Differentiating w.r.t. x we obtain 1 x+2 f(x/'(x) = log(x + 1) + x + 1
+ 1).
U.G.
6.42
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
x+2}
or, f'(x) = f(x) { log(~ +x) + x + 1
or, 1'(0) = f(O) { log(l + 0) + 0+2} 0 + 1 =12{0 + 2} = 2. (v) 1'(1)
= lim
f(1
h~O
+ h) -
f(l)
h
~ lim (211 +
hi + Ih -
h~O
11) - (2.1 + 11 - 21) h
= lim {2(1 + h) - (h - In - 3 = lim!: = lim 1 = l. h~O . h h~O h h~O
Ex. 5
(i) If x = e3t + 5 and y = sin3 t find
(ii) Find
~~
(iii) Find
~~, if x =
~~.
when x = a cos ¢, y = bsin¢: .
a(e - sine),y = a(1 + case).
. 2t. 2t dy (IV) Iftany= I_t 2 ,smx= l+t 2 ,showthat dx =l. (V) Find •
~~
when x = a(2 cos t
+ cos 2t), Y =
a(2 sin t - sin2t) .
SOLUTION:
(i) We have x = e3t + 5 and y = sin3 t. Differentiating w.r.t. t we get
~: = 3e3t , ~~ = 3 sin2 t cos t. , dy dy dt 3' 2 1. 2 3t " dx = dt' dx = sm tcost x 3e3t = sm tcoste- . (ii) Here x = acos¢,y = bsin¢. Differentiating w.r.t. ¢ we get dx . dy d¢ = -asm¢, d¢ = bcos¢. , dy dy d¢ 1 b ,. -d = d' A.·-d = bcos¢ . A. = --cot¢. x 'I' x -asm'l' a (iii) ,Given that x = a(e - sine), y = a(1 + cos e). Differentiating w.r.t. e we get dx dy. dB = a(l- cos B), dB = -asme.
dy = dy. de = -asinB 1 dx dB dx a(1 - cos B) 2 . 8 8 B = sm 2' cos 2 = _ cot _. 2sin 2 !l.2
2
6.43
CH.6: DIFFERENTIAL CALCULUS
•
(iv) Putting t = tan8. . 2 tan 8 .. tan y = 1 -tan2 8
= tan 28.
or, y = 28. .. 2 tan 8 . Ll Agam, smx = 1 + tan2 8 = sm2u. or, x = 28. Thus y
= x = 28.
Differentiating y
=x
w.r.t x we get
:~ = 1.
(v) Here x = a(2 cos t + cos 2t), y = a(2 sint - sin2t). Differentiating w.r.t. t we get
.~~ = a( -2 sin t dy
dx
=
2 sin 2t) and
~~ = a(2 cos t -
2 cos 2t)
dy dt
dt ·dx
= a(2 cos t - 2cos2t) x
1
( 2 . 2. ) a - smt - sm2t _ eos 2t - cos t _ 2 sin( -t/2) sin(3t/2) - sin2t+sint - 2sin(3t/2)cos(t/2) = -tant/2.
Exercise 2
(i) Find
~~
when x =' at2 , y = 2at.
(ii) Find
~~
where x = acos 2 8,y = asin2 8.
(iii) Find
~~
when x = a(cost + tsint),y = a(sint - teost).
[Ans. (i) l/t, (ii) -1, (iii) tan t.]
Ex. 6
y (i) Find dd when y = sin2 2x. x I .
(ii) Find
~~
when x 3 + y3 = 3axy.
(iii) Find
~~
when x 2/ 3 + y2/3 = a2/ 3.
(iv) Find
~~
if x = y log(xy).
(v) Find
~:
when xY.yX = 1.
(vi) Find
~~
when xY = yX.
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
6.44
(vii) Find dy when e XY dx (viii) If x
Y
=e
x-y
(ix) If log(xy) •
-
4xy = 2.
dy ,show that -d X
= x 2 + y2
log x
= (1 + 1ogx )2'
dy prove that dx
y(2x2 -1) 2 2)' y
= x(1 _
SOLUTION:
(i) We have y = sin2 2x. Differentiating w.r.t. x we get dy = 2sin2xcos2x.2 = 2sin4x. dx ' (ii) Give that x 3 + y3 = 3axy. Differentiating w.r.t. x we get 2 3x + 3y2 = 3a (y + x )
~~
~~
or, (3y2 - 3ax) ~: = 3ay - 3x 2 dy ay - x 2 or, - = . dx y2 - ax (iii) Given that x 2/ 3 + y2/3 = a 2/ 3. Differentiating w.r.t. x we get ~x-l/3 + ~y-l/3 dy = 0 3 3 dx
1/3=_(~)1/3.
or, dy =_xdx y-l/3
x
(iv) Here x = ylog(xy) or, x = y(logx + logy) Differentiating w.r.t. x we get dy 1 dy 1 dy 1 = -d log x + y.- + -d logy + y'-'-d x x x y x dy y or, 1 - - = -d (log x + log y + 1) x x or, x - y = (log x x
= ylogx + ylogy.
+ log y + log e) ~
dy x -y or, dx = xlog(xye)'
(v) We have xY.yX = 1. Taking logarithm both sides we get y log x Differentiating w.r.t. x we get
+ x log y =
O.
6.45
CH.6: DIFFERENTIAL CALCULUS
dy 1 -logx + y.- + logy + dx
or,
x
~~ ( log x + ~)
1 dy y dx
X.-.-
= - ( ; + log
=0
y)
y
or, dy
= _ ;- + log: = _ Y.. (y + x log y) .
log x + _ . x x + y log x Y (vi) Taking logarithm both sides we get y log x = x log y. Differentiating w.r.t. x we get dy 1 1 dy -d log x + y.- = 1. logy + x'-'x x y dx dx
or, dy ( log x dx
dy 'dx
or - =
=) = log y _ y
logy-~
log x - ~
Y.. x
y(xlogy-y) x(ylogx - x)'
=..:......;.,-..::...:.-~
(vii) Given that e XY - 4xy = O. Differentiating w.r.t. x we get y y xy e x· + y.1 } _ 4 x. dx + y.1 } -_ 0
{d
{d
dx
dy
or, dx (xe xy
-
4x) = 4y - ye
xy
dy 4y - ye xy y(e xy = = 'dx xe XY - 4x x (e xy
or -
(viii) Given that x Y =
eX-Yo
-
4) Y =-4) x.
Taking logarithm both sides we get
ylogx=(x-y)loge=x-y
(1)
Differentiating w.r.t. x we get dy 1 dy dy y - . log x + y. - = 1 - or, - (log x + 1) = 1 - dx
dy 'dx
or -
=
x dx dx x - y ylogx = ---,-:::....-~--,... x(log x + 1) x(log x + 1)
From (1), y(logx + 1)
=x
y or, x
x
1
= 1ogx+ 1
dy log x .. dx = (1 + logx)2'
(ix) Given that logxy = x 2 + y2 or, log x + logy = x 2 + Differentiating w.r.t. x we get .! + .! dy = 2x + 2y dy or, dy (~ - 2y) = 2x _ .! x
y dx
dx
dx y
x
y2.
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
6.46
dy 2x2 - 1 Y y(2x 2 - 1) or - = - - - ---"---;::. dx x ' l - 2y2 - x(l - 2y2) .
Ex. 7 If I(x) is an even function and 1'(0) exists, show that /,(0) = O. If I(x) is an even function then I(x) = I( -x). Differentiating w.r.t. x we get /,(x) = - /'( -x). At x = 0, /,(0) = -1'(0) or, 2/,(0) = 0 or, 1'(0) = O. •
SOLUTION:
(i) If x
Exercise 3
(ii) If x
= sin2 e, y = tan e find ~~.
= cos 2 e, y = tan2 e then find ~~.
(iii) Find
~~, where x = a + 2 cos e, y = b + 2 sin e.
(iv) Find
~~
when
(v) Find
~~,
when y = (sinx)logx.
yx.
xY =
[Ans. (i) ~ see 3 e.cosec e, (ii) - sec4 e, (iii) - cot e, . y(xlogy-y) ,(v) (sinx . 1 (IV) - - +eosx.logx )(smx)ogx.] x(y log x - x) x
Ex. 8 (i) If the side of an equilateral triangle increases at the rate V3 ft/sec and its area increases at the rate of 12 sq. ft./sec, find the sides of the triangle. (ii) If the area of a circle increases at a uniform rate, show that the rate of increases of the perimeter varies inversely as the radius . •
SOLUTION:
(i) Let a be the side of the equilateral triangle. Therefore its area A = Given that and
dA
dt
V;
a2
(1)
~~ = V3
(2) (3)
= 12
Differentiating (1) w.r.t. t we get dA = dt
V3 2a da = V3 a.V3 = ~a 4
or, 12 = ~a or, a = 8.
dt
2
[using (3)]
2
[by (2)]
CH.6: DIFFERENTIAL CALCULUS
(ii) Let r be the radius of the circle. Then its area A = 271"r2 . dA GIVen that dt =constant = k (say). From (1),
dA
dt
=
6.47
(1)
dr 271"r dt
dr dr k or,271"r- = k or, -d = -2-· t 7I"r dt Again perimeter P = 271"r. dP dr k k ... dt = 271" dt = 271", 271"r = -:;:. Hence the rate of increase of perimeter varies inversely as the radius. Def. 2 Monotone increasing function: A function is said to be monotonic increasing (non-decreasing) if f(XI) 2: nX2) for Xl > X2· e.g., the function f(x) = x 3 is a monotonic increasing function. Def. 3 Monotone decreasing function: A function is said to be monotonE decreasing (non-increasing) if f(XI) S f(X2) for Xl > X2· e.g., the function f (x) = 1 - X is a monotonic decreasing function. Formula 1 If f'(x) > 0 in a S X S b then f(x) is monotone increasing in S X S b. If f'(x) < 0 in a S x S b then f(x) is monotone decreasing in a S x S b.
a
Ex. 9 (i) Examine whether 2x 3 ~ 12x2 + 24x + 6 is increasing or decreasing on the real line. (ii) Show that x 3 + x 2 - 5x + 3 is monotonic increasing for x > 1. (iii) Show that the function f(x) = -2x 3 +15x 2 -36x+6 is strictly increasing in the interval 2 < x < 3. (iv) For 0 •
< e < 71"/2, Si;e
decreases as
e increases.
Is it true or false?
SOLUTION:
12x2 + 24x + 6. Differentiating w.r.t. x we get f'(x) = 6x 2 :- 24x + 24 = 6(x 2 - 4x
(i) Let f(x) = 2x 3
Hence
-
=
6(x - 2)2 > 0 for all real x.
f (x) is increasing for all real x.
(ii) Let f(x) = x 3 + x 2 - 5x + 3 . ... f'(x) = 3x 2 + 2x - 5 = 3x 2 - 3x (x - 1) (3x + 5) > 0 if x > l. Hence
+ 4)
f (x)
is increasing for x
> 1.
+ 5x - 5 = 3x(x - 1) + 5(x - 1)
=
6..18
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iii) Here J(x) = -2x 3 + 15x 2 - 36x + 6 . . '. J'(x) = -6x 2 + 30x - 36 = -6(x 2 - 5
+ 6)
= -6(x - 2)(x - 3) ,=
6(x - 4(3 - x). This snows that J'(x)
> 0 if x > 2 and x < 3,
i.e., if 2
< x < 3.
sinO (iv) Let J(O) = -0-'
. J'(O) = OcosO -l.sinO. .. 02 Let (0) = OcosO - sinO . . '. (0) = O. Now, ¢'(O) = -OcosO < 0 in 0
< 0 < 7r/2 . . '. ¢(O) is monotonic decreasing in 0 < 0 < 7r/2. So ¢(O) > ¢(O) or, OcosO - sinO < O. Hence J'(O) < 0 in 0 < 0 < 7r/2. sinO . Thus -0- is decreasmg for 0
6.8
< 0 < 7r /2.
Successive Differentiation
Ex. 1
(i) If Y = xn then show that Yn
= n!.
(ii) If Y = eax then show that Yn = ane ax .
(1'1'1') If Y = -1- t h en s how t h at Yn = ( (-1)nn! ) +1' x+a x+a n (iv) Find Y3, if Y = x 2 10gx.
3x +4 (v) If Y = (x - 2 )( 2x + 1)' find (Y5)O. (vi) If Y
= sin2 x
(vii) If Y =
~1' x+
show that YlO(O)
= 29.
show that Y5(O) = 5!.
= 2cosx(sinx - cos x) show that (YlO)O = 210. Prove that Y4 + 4y = 0 when Y = e- x cos x.
(viii) Ify (ix)
(x) If Y = x 2n , where n is a positive integer, show that Yn = 2n {1.3.5 ..... (2n -1)}xn. (xi) If Y = A sin mx
+ B cos mx,
then show that Y2
+ m 2y =
(xii) If Y = sin(msin- 1 x), then show that (1- X2)Y2 - XYI (xiii) If Y
= easin-l x
show that (1 - X2)Y2 - XYI - a2y
= O.
O.
+ m 2y =
O.
CH.6: DIFFERENTIAL CALCULUS
(xiv) If y
= (sin- I X)2, find the value of k so that d2y dy (1 - x ) dx 2 - x dx 2
•
+k =
O.
SOLUTION:
(i) We have y = xn. Differentiating w.r.t. x we get YI = nxn- I . Again differentiating w.r.t. x we get Y2 = n(n - l)xn-2. Similarly, Y3 = n(n - 1)(n - 2)xn-3. In this way, Yn = n(n - 1)(n - 2)···2· lx n- n = n!.x o = n!. (ii) Here y = eax . Differentiating w.r.t. x we get Yl = ae ax . Again differentiating we get Y2 = a2eax . Similarly, Y3 = a3eax , Y4 = a4eax and so on. In this way we get Yn = ane ax . 1 (iii) We have Y = - - = (x + a)-I. x+a Differentiating w.r.t. x we get YI = -(x + a)-2. Again differentiating Y2 = (-1)( -2)(x + a)-3 = (_1)2 .2!(x + a)-3. Similarly, Y3 = (-1)3.3!(x + a)-4, Y4 = (-1)4.4!(x
_ (_)n I( Hence Yn 1 .n. x
+a
+ a)-5.
)-(n+1) _ (-I)nn! - (x + a)n+1 .
(iv) Given Y = x 2 10g x. Differentiating w.r.t. x we get Yl = 2x log x + x 2. ~ = 2x log x + x. Again differentiating Y2 = (2logx + 2x.~) + 1 = 210gx + 3. 2
Hence Y3 = -. x (v) We have Y
=
3x + 4 (x _ 2)(2x + 1)
= 2(x - 2)-1 -
2
=x- 2(2x + 1)-1.
Differentiating w.r.t. x we get Yl = 2(-I)(x - 2)-2 - (-I)(2x + 1)-2.2 Again differentiating we get Y2 = 2( -1)( -2)(x - 2)-3 - (-1)( -2)(2x = 2( -1)2.2!(x - 2)-3 - (-1)2.2!(2x
1
2x
+1
+ 1)-3.2 2
+ 1)-3.22
6.49
V.G.
6.50
~ATHEMATICS (SHORT QUESTIONS AND ANSWERS)
_ (_ )2 ,[ 2 _ 22 ] 1 .2. (x _ 2)3 (2x + 1)3 .
_ (_ )5
In thIS way, Y5 -
(Y5)O
2
,[
1 .5. (x _ 2)6
_
2
5
(2x + 1)6
= -240 [ (x
1 16] _ 2)6 - (2x + 1)6
1
15345
16]
= -240 [ 26 -1
]
= -4-'
(vi) Here Y = sin2 x. Differentiating w.r.t. x we get Yl :...= 2 sin x cos x = sin 2x. Again differentiating we get 'Y2 = 2 cos 2x = 2 sin (~ + 2x) and Y3 = 22 sin (2.~ + 2x) In this way YlO Hence YIO(O)
(Vll.. )
= 29 sin (9.~ + 2x)
= 29 sin (9.~)
= 29 sin ( 47r + ~) = 29 sin ~ = 29 . H ere Y = -x- = x + 1 - 1 = 1 - -1- = 1 - ( 1 +) x - 1.
x+1 x+1 x+1 Differentiating w.r.t. x we get Yl = 0+ (1 +x)-2. Again differentiating we get Y2 = -2(1 + X)-3. Similarly, Y3 = 3!(1 + x)-4, Y4 = -4!(1 +.x)-5 and Y5 = 5!(1 + x)-6. Hence Y5(0) = 5!.
(viii) Given that Y = 2cosx(sinx - cos x) = 2 cos x sinx - 2 cos 2 X = sin2x - 2 cos 2 x. Differentiating w.r.t. x we get Yl = 2 cos 2x + 2.2 cos x sin x = 2 (cos 2x + sin 2x). Again differentiating we get
+ 2 cos 2x) = 22(-sin2x + cos2x). 22 { cos (~ + 2x) + sin (~ + 2x) }.
Y2 = 2(-2sin2x =
3 Similarly, Y3 = 2 { cos
(2.~ + 2x) + sin (2.~ + 2x) }.
10 In this way, YIO = 2 { cos (9.~ + 2x)
+ sin (9.~ + 2x)}.
Hence (YlO)O = 210{ cos (9.~) + sin (9.~)}
6.51
CH.6: DIFFERENTIAL CALCULUS
=210(sin~+cos~) =210. (ix) Given that Y = e- x cos x. Differentiating we get Yl = -e- x cos x - e- x sinx. Again differentiating Y2 = (e- X cos x + e- X sinx) - (_e- X sinx + e- x cos x)
2e- x sinx. Y3 = 2(-e- X sinx + e- x cos x). Y4 = 2{(e- X sinx - e- x cos x) + (-e- X cos x - e- x sinx)} =
= 2( -2e- x cos x) = -4y. Hence Y4 + 4y = O. (x) Here y = x2n. Differentiating we get Yl = 2nx2n-l. Again differentiating Y2 = 2n(2n - 1)x2n-2. Similarly, Y3 = 2n(2n - 1)(2n - 2)x 2n - 3, Y4 = 2n(2n - 1)(2n - 2)(2n - 3)x 2n - 4. In this way, Yn = 2n(2n - 1)(2n - 2)(2n - 3) ... (n + 1)x 2n - n
= =
2n(2n - 1)(2n - 2)(2n - 3) ... (n n!
+ 1)n(n -
1) .. ·2· 1
~
{1.3.5 .... (2n -1)}{2.4.6 .... (2n - 2)2n} n
x
n!
= ___ {1_.3_.5_._ .. _.(~2_n_-_1.:...:.)}_2n_._n! n = 2n{1 3 -
,
n. (xi) Given that Y = Asinmx + B cos mx . ... Yl = mAcosmx - Bmsinmx.
X
•• 5....
Again differentiating we obtain Y2 = -m 2 Asinmx - m 2B cos mx = -m 2(Asinmx + B cos mx) = _m 2y. or, Y2 + m 2y = O. (xii) We have y = sin(m sin- 1 x). Differentiating we get Yl =.cos(msin-1.x). ~. v1- x1 or, V'f=X2Yl = m cos(msin- x). Squaring we get (1 - x2)y~ = m 2 cos 2 (msin- 1 x) = m 2 {1 - sin2 (msin- 1 x)} = m 2 (1- y2).
( 2n_ 1)} x n .
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
6.52
Again differentiating we get (1 - x 2 )2YIY2 - 2xYr = m 2 ( -2YYl) or, (1 - X2)Y2 - XYI + m 2y = O. .
(xiii) Here Y = easm
-1
x.
a
.
= Vf=X2easm
Differentiating we get
Yl
or, (1 - x 2)yi
(by squaring)
= a 2y2
ay
-1
x
= VI _ x 2
Again differentiating we get (1 - x 2 )2YIY2 - 2xyi = a 22YYl or, (1 - X2)Y2 - XYI = a 2y. (xiv) Given that Y = (sin- 1 x)2. Differentiating we get
Yl
1
= 2(sin- 1 x) ~. vI- x 2
or, VI - X 2 Yl = 2 sin- 1 x or, {I - x 2)yr = 4(sin- 1 x)2 (by squaring) or, (1 - x 2)yi = 4y.
Again differentiating we get (1 - x 2 )2YIY2 - 2xYt = 4Yl (1 - X2 )Y2 - XYI - 2 = O. Comparing with the given equation we get k = -2.
Exercise 1
(i) If Y
= cos x( sin x -
(ii) Prove that (1 - X2)Y2 - XYI
=
cos x), show that (YlO)O
2 when yY
= 29.
= cos- 1 x.
(iii) If x = acos(},y = bsin(}, then show that Y2 = - b2cosec3(}. a
(iv) If Y =
Ae mx
+
Be- mx ,
then show that Y2 - m 2y = O.
d2 y Ex. 2 If sinx + cos y = 1, then find dx 2 ' • SOLUTION: We have sin x + cos Y = 1. Differentiating w.r.t. x we get . dy dy cos x cos x - sm Y - = 0 or, -d = -.-. x smy dx Again differentiating w.r.t. x - sin x sin y - cos x cos y 2
sin Y
=
cos(x -y) sin2 y
Theorem 1 Leibnitz's theorem: If U and v are two functions of x, then the nth derivative of their product, that is, (uv)n ;::= UnV +n C1Un-lVl +n C2 U n -2 V2 + ... +n Crun-rv r + ... + UV n , where the suffixes denote the order of differentiation with respect to x.
6.53
CH.6: DIFFERENTIAL CALCULUS
Q. 1 State Leibnitz's theorem. Ex. 3 (i) Differentiate n times the equation (1 + X2)Y2 + (2x - l)Yl = o. (ii) If (1 - X2)Y2 - XYI + m 2y = 0, then show that (1 - x2)Yn+2 - (2n + l)xYn+1 + (m 2 - n 2)Yn = •
o.
SOLUTION:
(i) We have (1 + X 2 )Y2 + (2x - 1)Yl = o. Differentiating n times using Leibnitz's theorem we get {(I + X2)Yn+2 +n Cl(2x)Yn+1 +n C2(2)Yn} +{(2x - l)Yn+1 +n C 1 (2)Yn} = 0 2 (1 + X )Yn+2 + 2nxYn+1 + n(~-l) .2Yn + (2x - l)Yn+1 + 2nYn = 0 (1 + X 2)Yn+2 + (2nx + 2x - l)Yn+l + (n 2 -Tl- + 2n)Yn = 0
+ X2)Yn+2 + {2x(n + 1) -l}yn+1 + n(n + l)Yn (ii) Given that (1 - X2 )Y2 - XYI + m 2y = O. (1
= 0
Differentiating n times using Leibnitz's theorem we get {(1- x2)Yn+2 +n C 1(-2x)Yn+l +n C 2(-2)Yn} -{xYn+1 +n C 1 (1)Yn} + m 2Yn = 0 . (1- X2)Yn+2 - 2nxYn+1 - n(n2-1).2Yn - XYn+1 - nYn + m 2Yn = 0 (1- X2)Yn+2 - 2nxYn+1 - (n 2 - n)Yn - xYn+1 - nYn + m 2Yn = 0 (1 - X2)Yn+2 - (2n + l)xYn+1 - (n 2 - n + n - m 2)Yn = 0 (1 - X2)Yn+2 - (2n
6.9
+ l)xYn+1 + (m 2 -
n 2)Yn = 0
Partial Differentiation
au The first order partial derivative of u = u(x, y) w.r.t. x is denoted by ax or U x ·
a
2u The second order partial derivative of u = u(x, y) is denoted by ax 2 or U xx . au The first order partial derivative of u = u(x, y) w.r.t. Y is ay or u y . 2
. IS . a . I d· The second ord er partla envatlve ay2u or
U yy .
2
. I d· . 0 f U w.r.t. Y IS . ayax a u or u yx · The partla envatlve x The partial derivative of u y w.r.t. x is
::;Y
or u xy ·
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
6.54
Ex. 1 If x = r eos 0, y = r sin 0, does 8x __1_7 80 - 80 . 8x
• SOLUTION: We have x = reosO, y = rsinO. Div-iding we get tan 0 = JL or 0 = tan -1 JL. x x Differentiating x = reosO partially w.r.t. 0 we get
~:
= -rsinO.
Differentiating 0 = tan- 1 JL partially w.r.t. x we get x 80 8x
1 r 80 = - sin 0
8x 80·
=1=
8x
(i) Find fx, fy if f(x, y) = sin- 1
Ex. 2
(~). 82u
82 u
(ii) If u =
JXY, find the value of 8x 2 + 8 y 2.
(iii) Let U
= x Y , prove that Uxy = Uyx
(iv) If u = x log y, (y
(v ) If f( x, y )
at any point (x, y).
> 0) show that u xy
= x 3 y + exy2 ,show t hat
= u yx . 2
8 f 8x8y
• SOLUTION: (i) Here f(x, y) = sin-1(y/x). Differentiating partially w.r.t. x we get
fx = J1-
~Y7x)2 ( - :2)
x = J x 2 - y2 I
(
Y)
Y
- x 2 = - x J x 2 _ y2 .
Again differentiating partially w.r.t. y we get
fy
=
1
(1)
J1- (y/x)2 -;
=
1
Jx2 _ y2·
2
8 f = 8y8x·
Ch.O: DIFFERE"'''T'IAL CALCULUS
Hence 2 2 8 u 8 u __ ! ( -3/2 1/2 8x2 + 8y2 4 x Y
2 1/2 -3/2) __ ! x + y2 +x Y 4 (xy)3/2'
(iii) Given that U = xY. Differentiating partially w.r.t. x we have Ux = yx y - 1 . Again differentiating partially w.r.t. y we get Uyx = x y- 1 + y.xy-1logx = xy-1(1 + ylogx). Similarly, Uy = x Y log x. Uxy = yxy-1logx + xy~ Hence Uxy
= xy-1(1 + ylogx).
= Uyx ,
(iv) We have u = x log y. Differentiating partially w.r.t. x we get u x = log y.
1 Again differentiating partially w.r.t. y we get u yx = -. y Similarly, u y
=~ y
and u xy
= !'. y
Hence u xy = u yx . (v) Here f(x, y) = x 3 y + e xy2 . Differentiating partially w.r.t. x and y we get
8f
2
2
8x = 3x y + eXY .y
2
~f 2 xy 2 8y8x = 3x + (e .2xy)y2 = 3x 2 + e
Similarly,
xy2
+ eXY
2
.2y
(2 xy3 + 2y).
:~ = x 3 + exy2 .2xy
82 f 2 xy 2 2 XY 2 8x8y = 3x + (e .y )2xy + e .2y = 3x 2 + e
xy2
(2 xy 3
+ 2y).
6.55
U.G.
6.56
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Def. 1 Homogeneous function: A function u = f(x, y) is said to be homogeneous function in x and y if it can be expressed as one of H J following form: (i) f(x, y) = xn - 1 +a > -+c + b2'
I.e., - 12
Therefore, 1 tan- 1 b - tan- 1 a --< I + b2 b- a
1
< - -2 1+a
'
i.e.,
b- a
1 + b2
'-1
< tan
-1
b - tan
a
b- a
< 1 + a2 •
6.67
CH.6: DIFFERENTIAL CALCULUS
Theorem 5 General mean value theorem: Taylor's theorem: If f(x) is a real valued function of x such that its derivatives up to r-1(x) are all continuous in a :::; x :::; a + hand fn(x) exists in a < x < a + h, then
. h2 hn- 1 hn f(a + h) = f(a) + hj'(a) + 2f fl/(a) + ... + (n _ 1)! r-l(a) + n! fn(a + (}h), where
0 < (}"< 1.
Q. 5 State Taylor's theorem. Note 3 The Taylor's theorem can be written in the interval [0, xl as
x2 x n- 1 xn f(x) = f(O)+xj'(O)+2f 1"(0)+·· .+ (n _ 1)! r-1(0)+ n! r((}x),
0 < () < 1.
This finite series is obtain by sUbstituting a = 0 and h = x. This series is also known as Maclaurin's finite series and the last term is ca:led remainder term (due to Lagrange). Note 4 If the remainder term vanishes for a particular values of x E I then the above series becomes
f(x) = f(O) + x!,(O) +
x2
2f 1"(0) +
xn -
1
... + (n _ 1)! r-l(O) + . ",
0 < () < 1.
This infinite series is known as Maclaurin's infinite series and it is valid for the values of x E I. This series is also used to expand a function as an infinite series in powers of x. Ex. 12 Find an infinite series for the function eX. •
SOLUTION:
f(x) = eX I'(x) = eX I"'(x) = eX fiv(x) = eX
Let f(x) = eX. Then 1'(0) = 1 1"(0) = 1 f,I/(O) = 1 fiv(O) = 1 and so on.
Then by Maclaurin's infinite series we have 2
/(x)
=
/(0) + x/'(O) +
x2
=
1 + x.l + 21. 1 +
x2 =
x3
3
~! 1"(0) + ~! /111(0) x3
31. 1 +
4
+
~! /iv(O)
+ ...
x4 4!.1 + ".
x4
l+x+2f+3!+4f+'"
Exercise 1 Find the infinite series for the functions (i) log(1 and (iii) cos x.
+ x), (ii)
sinx
V.G.
6.68
6.11
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Maximum and Minimum
Formula 1 If c be a point within the domain of definition of f (x), and if f'{c) = 0 and 1"{c) i= 0, then f{c) is a maximum if 1"{c) < 0 and a minimum if 1"{c) > o. (i) Show that the function 2x 3 - 21x2 + 36x + 125 is maximum at x=1. (ii) Show that the function x 3 - 3x 2 + 6x + 3 does not possess any maximum or minimum value. (iii) Show that f(x) = x 3 - 6x 2 + 24x + 4 has neither a maximum nor a
Ex. 1
minimum. (iv) Show that f(x)
= Ixl has minimum at x = o.
(v) Show that the maximum value of x
+ .!.
is less than its minimum value. x (vi) Show that the largest rectangle with a given perimeter is a square. (vii) Show that the maximum rectangle inscribable in a circle is a square . •
SOLUTION: - 21x2 + 36x + 125 . ... f'{x) = 6x 2 - 42x + 36 = 6{x 2 - 7x + 6). At extrema f'{x) = 0, giving 6{x 2 - 7x + 6) = 0 or, x 2 - 7x + 6 = 0 or, (x - 6)(x - 1) = 0 or, x = 1,6.
(i) Let f(x) = 2x 3
Again 1"{x) = 6{2x - 7) . ... 1"(1) = 6{ -5) = -30 < o. Hence f (x) is maximum at x = 1. (ii) Let f{x) = x 3 - 3x 2 + 6x + 3. .. f'{x) = 3x 2 - 6x + 6 = 3{x 2 - 2x + 2). At extrema f'{x) = 0, giving 3{x 2 - 2x + 2)
=0
_ 2 ± v'4=8 _ 2 ± 2i _ 1 . or, x 2 2 - ± t.
This shows that l' (x) i= 0 for any real x. Hence f (x) does not possess any maximum or minimum value. (iii) Let f{x) = x 3 - 6x 2 + 24x + 4 . ... 1'{x) = 3x 2 - 12x + 24 = 3{x 2 - 4x + 8). At extrema f'{x) = 0, giving 3{x 2 - 4x + 8) = 0 or, x =
4 ± }16 - 32 2 = 2 ± 2i.
Thus x has no real value. Hence f{x) has neither maximum value nor minimum value.
CH.6: DIFFERKFTIAL CALCULUS
6.69
(iv) The function f(x) = Ixl is always positive for all real values of x. The minimum value of Ixl is 0 when x = o. Hence f(x) is minimum at x (v) Let f(x)
= x +! .. '. J'(x) = 1 - ~.
x At extrema J'(x) = 0 1 or, 1 - 2 = 0 or, x = x Now,
= o. x
± l.
2 f I/() x = 3.
x 1"(1) = 2 > o. ... f(x) is minimum at x = 1 and minimum value is f(l) = 2.
1"( -1) = -2 < O. Thus f(x) is maximum at x = -1 and maximum value is f( -1) = -2. Hence the maximum value is less than its minimum value. (vi) Let a, b be the sides of a rectangle. Then its perimeter is 2(a (say), k is constant, and its area is A = abo A = ab = a(k - a) = ak - a2.
+ b)
= 2k
dA Now, da = k - 2a. At maxima dA = 0 or, k - 2a = 0 or, a = k/2. da 2 d A Also, da 2 = -2 < O. Hence A is maximum at a = k /2 and b = k /2. As a
= b = k/2,
the largest rectangle is a square.
(vii) Let a, b be the sides of a rectangle which is inscribed in a circle of radius r. The diagonal of the rectangle is J a 2 + b2 which is equal to the diameter. Ja 2 + b2 = 2r or, a2 + b2 = 4r2. Area of rectangle is A = abo or A2 = a 2b2 = a 2(4r2 _ a 2) = 4r 2a 2 - a 4 .
dA2 2 Now, da = 8r a - 4a 3 . dA At extrema - = 0 or, 8r 2 a - 4a 3 = 0 da or, a 2 = 2r2 or, a = r..j2 . .. b2 = 4r2 - a 2 = 4r2 - 2r2 = 2r2 or, b = rV2. That is, a
= b = r..j2.
Henr.:e the largest rectangle is a square.
6.70
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Ex. 2 Let 1 be defined in the closed interval [a, b] and a < c < b. Does I'(c) = 0 always imply the existence of an extrema of 1 at x = c? Justify your answer . • SOLUTION: f'(c) = 0 does not always imply,. is an extreme point. For example, if I(x) = x 3 - 6x 2 + 12x - 3, then I'(x) = 3x 2 -12x + 12 at x = 2. But I"(x) = 6x - 12 = 0 at x = 2 and 1"(2) = 0, 1"'(2) = 6 -=I O. Hence 1'(2) = 0 but x = 2 is not an extreme point.
=0
Ex. 3 Show by an example that a maximum value of a function at a point may be even less than the minimum value at another point . • SOLUTION: Let I(x) = x
+.!.x .... J'(x) =
1 - -\-. x
At extrema f'(x) = 0 1 or 1 - "2 = 0 or, x = ± l. x 2 Now, 1" (x) = 3' x ... 1"(1) = 2 > O. Thus I(x) is minimum at x = 1 and minimum value is 1(1) = 2. 1"( -1) = -2 < O. So I(x) is maximum at x = -1 and maximum value is 1(-1) = -2. Hence the maximum value is less than its minimum value.
6.12
Tangent and Normal
(i) Equation of the tangent of the curve I(x, Y) = 0 at the point (Xl> YI) is
Formula 1
(ii) Equation of the normal of the curve I(x, Y) = 0 at the point (Xl, YI) is
Ex. 1 X
=
(i) Find the equation of the tangent at the point (J .J3 sin (J, y = 2 cos (J ,
(ii) Find the equation of tangent at 6 cos (J,
(J
=~
=
to the curve
1r
X
/3 of the curve
= 5 sin (J, y =
6.71
CH.6: DIFFERENTIAL CALCULUS
(iii) Show that the abscissae of the points on the curve y where tangents are parallel to the axis of
= x(x - 2)(x -
x are given by x =
4}
2
2 ± v'3'
(iv) Find the equation of the tangents at origin of the curve (x 2 + y2)2 = 4(x2 _ y2) . •
SOLUTION:
(i) The given curve is x = v'3sinO,y = 2cosO .
. dx r.; dy . . . dO = v 3 cos 0 and dO = =2 8m O. Thus dy dx
= dy. dO = _ . 2 sin 0 = _ ~ tan O. dO dx
v'3 cos 0
r.; .
7r
v'3
r.;
7r
v'3 3
3'x = V3S1U 3 = v3' 2 = 2' 7r 1 Y = 2 cos 3' = 2. 2 = 1 and dy = -~tanO = -~tan~ = -~.V3 = -2. At 0 =
dx
v'3
v'3
3
v'3
Hence the equation of the tangent at 0 =
= -2(x - 3/2)
y -1
(ii) Here x
i
is
or, y - 1 = -2x + 3 or, y + 2x = 4.
= 5 sin 0, y = 6 cos O.
Differentiatin~
w.r.t. 0 we get dx dy dO = 5cosO and dO = -6sinO.
. dy . . dx At 0
dy dx =
= _ 6 sin 0 = _ ~ tan O. 5cosO
7r
5
7r
7r
= -2' x = 5 sin -2 = 5, y = 6 cos -2 = 0 and 00
dx or dy = O.
The equation of tangent at ~ is
or, x - 5
= O.
(iii) We have y = x(x - 2)(x - 4) . .'.
:~ = (x -
2)(x - 4)
+ x(x - 4) + x(x - 2).
U.G.
6.72
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
If the tangents are parallel to x-axis,
+ x(x 6x + 8 + x 2 -
(x - 2)(x - 4)
or, x
2
-
or, x =
~~ = 0 giving
+ x(x - 2) = 0, 4x + x 2 - 2x = 0 or, 4)
3x 2
-
12x
+8 =
0
12 ± )144 - 4.3.8 12 ± 4V3 2 6 = 6 = 2 ± V3'
(iv) Let Y = mx be the tangents at the origin. Putting Y = mx to the given curve we get (x 2
+ m 2 x 2 )2 =
4(x 2
-
m 2 x 2 ) or, x 2 (1
+ m 2 )2 = 4(1 -
m 2 ).
Since y = mx touches the curve, the values of x should be equal. Hence 1 - m 2 = 0 or, m = ±1. Thus the required equation of the tangents are y = ±x. y
Figure: 6.12.1 Formula 2
(i) Cartesian subtangent= TM = ycot1/J = Y/Yl'
(ii) Cartesian subnormal= M N = Y tan 1/J = Y.Yl. (iii) Length of normal = PN = ysec1/J
~ yJ1 + Yr.
(iv) Length of tangent = PT = ycosec 1/J = (Y J 1 +
yi) /Yl·
de (v) Polar subtangent = r2 dr' dr (vi) Polar subnormal = de'
Ex. 2 (i) For the curve r = eO, show that the polar subtangent=the polar subnormal. (ii) Find the length of the polar subtangent for the curve r = a(1 + cos e) at
e=7r/2. (iii) Show that the length of the cartesian subtangent of the curve Y = e- x / 2 is 2 units.
6.73
CH.6: DIFFERENTIAL CALCULUS
•
SOLUTION:
(1·)"D ror t h e curve r = e(J , dr dO = e(J = r. dO polar subtangent = r2 dr Hence they are equal. (ii) Here r
1
dr
= r 2 .;: = r and polar subnormal = dO = r.
= a(l + cos 0) .. ·. ~; = -a sin O.
The polar subtangent dO 1 a2 (1 + 0)2 = r2 -dr = a2(1 +- cos 0)2 = --''-----'-(at 0 = 7r/2) ' -asinO -a.1 = -a. dy = _~e-X/2 . (iii) Given that y = e- x / 2. . . . dx 2 e- x / 2 y The cartesian subtangent = - = 1 _ /2 = -2. Yl -2 e x Hence the length of cartesian subtangent is 2 units.
Ex. 3 Write the condition that the two curve, f(x, y) = 0 and ¢(x, y) = 0 cut orthogonally. •
SOLUTION·
The gradient of the curve I(x, y) = 0 is
dy = - Ix ml = dx Iy and that 0 f t he curve ¢ ( x, y) =. 0 IS dy ¢x m2= _. = - - .
dx
¢y
If they cut orthogonally, mlm2 = -1 or, -
or, Ix¢x
+ Iy¢y
~:. -
:: = -1
= 0, which is the required condition.
Formula 3 If ¢ be the angle between the tangent and radius vector at any point on the curve r = I (0). then
(i) p=rsin¢. 1 dr (ii) cot ¢ = ;: dO.
x
o
Figure: 6.12.2
6.74
U.G:
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Ex. 4 With usual notations prove that p = r sin ¢. •
SOLUTION:
From the above figure we see that
~~
= sin¢ or,
~=
p = rsin¢.
Ex. 5 Show that for log r = aO + b, p ex r. •
Differentiating log r = aO + b w.r.t. 0 we get I dr [ . I dr ] _;: dO = a or, cot ¢ = a smce;: dO = cot ¢ I SOLUTION:
. '" or, sln¥,=
... P =
1 v'1+a 2.
r sin
0)
= - cos x
f sin mx dx = - c0"mmx
f sin x dx
f cos mx dx = sin:x
f cos x dx = sin x
f sec 2 mx dx =
f sec 2 x dx = tan x
tanmmx
f cosec 2mx dx = _
co~mx
f cosec 2 x dx
J sec x tan x
se~mx
f sec mx tan mx dx =
f cosec mx cot mx dx = _
cos~ mx
= - cot x
dx = sec x
f cosec x cot x dx = -cosec x
f sinh mx dx = cos~mx
f sinh x dx = cosh x
f cosh mx dx = sin~mx
f cosh x dx = sinhx
f sech 2mx dx = tan~mx
f sech 2x dx = tanh x
f cosech 2mx dx = _ cot~mx
f cosech 2x dx = - cothx
f sech mx tanh mx dx
= - se~mx
fsecx dx = logltan(i
/
+ ~)I
cosech mx coth mx dx = -
f sech x tanh x dx = -sech x f cosec x dx = log I tan ~I
cosechmx m
/ cosech x coth x dx = -cosech x dx = sin- 1 x or - cos- 1 x, Ixl / V1-x2
1
= tanh- 1 x, Ixl < 1 or coth- 1 x, Ixl > 1
~ 2
!
x
! ! !
2dx 2=~10glx-al=-~coth-l::,lxl>lal x - a 2a x +a a a
= sec- 1 :1' or - cosec -lx, Ixl > 1 x -1 dx 1 -1 ;c x 2 + a 2 = ~ t.an ~
2dx 2 a - x
! vi ! vi ! !
=~lOgla+xl=~tanh-l::,lxl lal x2 - a a dx
vla 2 -
= sin- l ::, Ixl > lal x a 2
x
a2
2
2
Jx 2 +a 2 dx= -Jx2+a 2 +-loglx+Jx2+a21 2
x./2 +a 2 +-smh-a. IX =-yx 2
!
! f f
2
x Jx2 - a2 dx = -Jx 2 - a 2 2
a
2
a -log Ix + Jx 2 - a 2 1 2 x x a2 = -J.r2 - a 2 - - cosh- l 2 2 a x a2 x Ja 2 - x 2 dx = -Ja 2 - x 2 + -sin- l 2 2 a ax b d ax a cos bx + b sin bx e cos x x = e . a2 + b2
e
ax . 8m
-
b d ax a sin bx - b cos bx x x = e. a 2 + b2
Ex. 1 Obtain the formula for tions of x.
J uv dx,
where u, v are both differential func-
7.3
CH.7: INTEGRAL CALCULUS • SOLUTION: We know ~(uV) = du V dx dx dx and integrating we get,
or,
I
d(uV) =
I
d' d:V dx
+
I
+ u ddV ' x
Multiplying both sides by
dV u dx dx
or, UV=ldUVdx+ludV dx dx dx Putting V =
u
I
v dx =
[As V
J v dx, the above equation becomes
I (:: I
+
I
uv dx
= J v dx or, v = ~~ J
I I J ++ (. .)I + 11 + + or,
v dX) dx
uvdx = u
J(~: J
vdx -
vdx) dx.
Ex. 2 Integrate the following:
C) I
III
(v)
dx J3x 4 J3x 2 x 1 (x 2 _ 1)2 dx,
+ 10
d
(. )I .I
cos x dx, smx + cos x dx x(x + 1)'
11.
x,
(IV)
~ sin x cos x dx . smx cos x
• SOLUTION:
(i) dx
=
I I I
=
-~
=
_! [! (3x + 4)3/2 _! (3x + 10)3/2]
=
J3x
+ 4 + J3x + 10 J3x + 4 -
(J3x
J3x
+ 10
d
+ 4 + J3x + 10)( J3x + 4 -
J3x
J3x + 4 - J3x + 10 d (3:r + 4) - (3x + 10) x
I
{(3x
6 3
+ 4)1/2 3/2
(3x
3
+ 1O)1/2} dx 3/2
C
+
=
-~~ [(3x + 4)3/2 -
(3x
+ 10)3/2] + C
=
1 - 27 [(3x
(3x
+ 10)3/2] + c,
+ 4)3/2 -
+ 10)
x
V.G.
7.4
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
c is constant of integration. (ii) d cos x . x / smx + cos:!'
! / (sinx + cos x) + (cos x - sinx) dx 2 sinx + cos x
= =
! / (1 + c~s x - sin X) dx 2 smx + cos x
=
! [ / dx + / d(~in x + cos x)] 2 smx + cos x
~[x + log(sinx + cos x)] + c, c is constant of integration. 2
2
... / x +1 / x +1 (m) (x 2 _ 1)2 dx = (x _ 1):l(x + 1)2 dx
_! / {(x -1 1)2 + (x +1 1)2 } dx
- 2
= ~ / {(x - 1)-2 + (x + 1)-2} dx _!{(X-1)-1 - 2 -1
+
(X+1)-1} -1
+c
= _!{_1_'+ _1_} +c
2 x-I
x+1
x
= -1--2 +c,
-x
c is constant of integration. d (iv) / X(X :1) = X:1]dX=logX-lo g(1+X)+C
/[tx
+c, x+ c is constant of integration. = log--1
(v) / 1 ~ 2 sin x cos x dx = / (si~x + cosx)2 dx smx + cos x smx + cos x = J(sinx + cos x) dx
= - cos x + sin x + c, c is constant of integration. Ex. 3 Integrate the following: (i) J x 2ex3 dx, (iii) J sin3 x cos 2 x dx, (v) J xa x2 dx,
(ii) J cos 3 BdB, (iv) J sin 2x cos 2 2x dx,
7.5
CH.7: INTEGRAL CALCULUS
!
(vi)
(. . )! VB}
!
(x)
tan log x dx,
. (u)
X
dx
eX
+ e- X
dx
VB
(;;,
X+yX
/
dx
(1
+ x 2 )vtan- 1 x + c
e2x
1 + c~sx dx, x+smx
.. ) / (XB •
(")!
2 cos x dx, cos2x
(xi)
- - 1 dx,
.. !
,
+
eX
/
X
e (1 +x) 2() dx. cos xe X
(XlII)
SOLUTION:
(i) Putting x 3 = z. 3x 2 dx = dz. 3 d 1 1 13 J x 2 eX dx = J eZ 3z = - J eZ dz = - eZ + c = - eX 333 c is constant of integration. (ii) J cos3 0 dO
+ c,
= J cos 2 0 cos 0 dO = J(1 -
sin2 0) cos 0 dO
= J(1- z2) dz [where z = sin 0 and dz
= cos 0 dO]
sin3
0 -· 3 3 c is constant of integration. z3
= z - -
+ c = sin 0 -
+ c,
(iii) J sin3 x cos 2 x dx = J sin2 x cos 2 x sin x dx = J (1 - cos 2 x) cos 2 x sin x dx z, ,', sinxdx = dz.
Putting cos x = Then the given integrand becomes J(1 - z2)z2( -dz)
= J(z4 -
z5
z2) dz
z3
= 5" - 3 + c
sin3 x
cos5 x
= - 5 - - -3-+ c ,
c is constant of integration. (iv) Putting cos2x = z, Then -2 sin2x dx = dz. 1 1 z3 J sin 2x cos 2 2x dx = J - - z2 dz = - - - + c 2
1
2 3
3
= - 6" cos 2x c is constant of integration. (v) Putting x 2 = Z • . '. 2xdx = dz.
.
J xa
x2
dx
1
= -2 J a
Z
dz
+ c,
1 aZ
1 aX
2
= -2 - + c = -2 - + c, log a log a
c is constant of integration.
,
U.G.
7.6 (vi)
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
J
J
2 cos x dx = 2 cos x dx cos 2x 1 - 2 sin2 x Putting sinx = z. '. cosxdx = dz. Then the given integral becomes
J(1
J
1)
2dz d 1 - J2z + 1 + J2z z 1 - 2z2 = 1 1 - J2 log 11 - v'2zl + J2log 11 + v'2zl + C
11
- 1 1og + J2zl +C J2 1- J2z
C
+ J2sinxl +C 1- J2sinx
- 1 1og \1
=
J2
is constant of integration.
(vii) Putting 1 + .
.I
..
vx = z. Then 2yx 1;;;;- dx = dz.
2 dz vx = Jvx(1dx+ vx) = J-z= 2 log Izl + = 2log(1 + vx) +.c,
dx
x+
C
C
is constant of integration.
(viii) Putting log x = z. Then .!.dx = dz. x tan log x x dx = I tanz dz = log sec z
J
+c =
log sec (log x)
c is constant of integration.
(ix) Putting tan- 1 x
!
+c =
z2. . _1_2 dx = 2z dz.
.. l+x
dx
(1
+ x 2)vtan- f .r + c
=
!
2z dz = 2 I dz = 2z
z
= 2vtan- 1 x
(x) Putting x
+ sinx =
!
z . . '. (1
+ cos x) dx =
1 + cos x dz - - . - dx = - = log Izl x + SlllX Z is constant of integration.
!
(xi) Putting eX
!
+ c + CI,
is constant of integration.
Cl
C
+ Cl
+1=
~ dx = eX
+1
!
+C=
log Ix
z . . '. eX dx = dz. x
eo" .e dx eX + 1
=
!
(z - 1) dz z
dz. . + slllxl + c,
+ c,
7.7
CH.7: INTEGRAL CALCULUS
(1 - ~) dz = z -
= !
= (eX + 1) - log( eX c is constant of integration. (xii)
! eX eX)
log Iz I + c
+ 1) + c,
dx X =! 2 eX 1 dx (multiplying numerator and denominator by + ee X+
Putting eX = z . . '. eX dx = dz. dz Then! dx =! _ 2 = tan- 1 z + c = tan- 1 (e X ) eX + e- X z +1 c is constant of integration. (xiii) Putting xe x = z. Then (eX + xe X) dx = dz or, eX(x + 1) dx = dz. . ! eX (1 + x) dx = ! ~ = .. cos 2 (xe X ) cos 2 z
+ c,
f sec2 z dz
= tanz + c = tan(xe X ) c is constant of integration.
+ c,
Ex. 4 Integrate the following: (i)
f
(iii) •
log x dx,
(ii)
f
f
(iv)
f tan- 1 x
xe x dx,
x log x dx, dx .
SOLUTION:
1x
(i)
f
log x dx = f 1. log x dx = log x f 1.dx - f{ (log x). f 1.dx }dx = logx.x - f ~.xdx = xlogx - f dx = xlogx - x + c, c is constant of integration.
(ii)
f
x log x dx = log x f x dx -
(iv)
x} dx
f x' 1 2" d x = 2" Iog x - f 2" dx 2 = ~2 log x - x4 + c, c is constant of integration. f xe x dx = x f eX dx - f{1. f eX dx} dx = xe x - f eX dx = :re x - eX + c, c is constant of integration. f tan- 1 x dx = f 1. tan- 1 xdx x2 = Iog x. 2" -
(iii)
f {1x (log x) f x2
x2
X
1
1
= tan·- x ! 1.dx - ! {:x (tan- x) ! 1.dx }dx =
x tan -1 x - / _1_x dx = x tan -1 x _ ~! 2x dx 1 + x2 2 1 + x2
=
x tan- 1 x _
~! d(1 + x ) 2 1 + x2 2
7.8
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS) 1
= x tan-l x - "2log(1
+ x 2 ) + c,
c is constant of integration.
Ex. 5 If f'(x) = 0 for every point in the closed interval [a, bJ then show that f (x) is constant throughout the interval. • SOLUTION: Integrating f'(x) or, f(x) = c within [a, b].
7.1.2
= 0 we get
I f'(x)dx
= IO.dx
within [a,b].
Definite integration
Theorem 1 Fundamental Theorem of Integral Calculus: If (i) I: f(x) dx exists and (ii) there exists a function 4>( x) such that 4>' (x) = f (x) in a ::; x ::; b, then
I: f(x) dx = 4>(b) - 4>(a). Q. 1 State the fundamental Theorem of Integral Calculus. Ex. 1 Give the geometrical meaning of I: f(x) dx .. • SOLUTION: I: f(x) dx is the area of the region bounded by x-axis, the curve Y = f (x) and the straight lines x = a, x = b. Ex. 2 Does Ii f(x) dx answer.
= Ii g(x) dx
always imply f(x)
= g(x) ?
Justify your
• SOLUTION: I: f(x) dx = I: g(x) dx does not always imply f(x) = g(x). For example, let a = 0, b = 1, f(x) = 2x and g(x) = 3x 2 • 1 1 Now, Il f(x) dx = 10 2xdx = 1 and 10 g(x) dx = Il3x 2 dx = 1. 1 1 i.e., 10 f(x) dx = 10 g(x) dx but f(x) = 2x -:F 3x 2 = g(x).
Some usefull formulae on definite integrals I. I: f(x) dx
= I: f(z) d;;. a
II. I: f (x) dx = - Ib f (x) dx. III. I: f(x) dx = O.
IV. loa f(x) dx = I; f(a - x) dx.
V. loa f(x)dx
a
= 2 Ioa/ 2 f(a - x)dx,
if f(a - x)
VI. Ii f(x) dx = 2 loa f(x) dx, if f(a
= f(x).
+ x) = f(x).
CR. 7: INTEGRAL CALCULUS
7.9
VII. I~a f(x) dx = 2 I; f(x) dx, if f( -x) = f(x) i.e., f(x) is even function. VIII. I~a f(x) dx = 0, if f( -x) = - f(x) i.e., f(x) is odd function.
Ex. 3 (i) What do you mean by an odd function of x ? If f(x) is an odd function of x, prove that f(x) dx = O.
ra
(ii) Prove that I~af(x)dx (iii) If f(x) = f(a
+ x)
= 2 loa f(x)dx, if f(x) = f(-x).
then prove that
f3a
10 •
fa f(x) dx = 3 10 f(x) dx .
SOLUTION:
(i) A function f(x) is called odd function if f( -x) = - f(x). Now, I~a f(x) dx = I~a f(x) dx + I; f(x) dx Putting x. = -y in the first integral.
= -dy and when x = 0 then y = 0 and when x = -a then y = a . I~af(x)dx = I~ f(-y) (-dy) + loa f(x)dx = - I; f(y) dy + loa f(x) dx (since f(x) is odd function) = - I; f(x) dx + I; f(x) dx (changing the variable y to x)
.'. dx .".
= O.
(ii) I~a f(x) dx = I~a f(x) dx
+ I; f(x) dx. Putting x = -y in the first integral.
= -dy and when x = 0 then y = 0 and when x = -a then y = a. I~a f(x) dx = I~ f( -y) (-dy) + loa f(x) dx = I; f( -y) dy + I; f(x) dx = loa f( -x) dx + loa f(x) dx (changing the variable y to x) = I; f(x) dx + I; f(x) dx [since f( -x) = f(x)]
.'. dx
= 2 loa f (x) dx.
(iii) L.H.S. = I~a f(x) dx = loa f(x) dx + I;a f(x) dx Now we consider the integral I;a f (x) dx.
Putting x = a + y. Then dx x = 2a then y = a .
+ 123aa f(x) dx.
= dy and when x = a then y = 0 and when
+ y) dy = I; f(y) dy [since f (x) = f (a + x)] Similarly, Iiaa f(x) dx = loa f(y + 2a) dy = loa f(y) dy . '. I;a f(x) dx = I; f(a
7.10
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS) =
[since f(y Hence
I;a f(x) dx
It f(x) dx
+ 2a) = f(y + a + a) = f(a + y) = f(y)} = It f(x) dx + It f(x) dx + loa f(x) dx =
3 loa f(x) dx.
Ex. 4 Prove that
(i)
loa f(x) dx = loa f(a -
(ii)
I: f(a + b - x) dx = I: f(x) dx,
x) dx,
(iii) I~a ¢(x 2) dx = 2 loa ¢(x 2) dx, (iv) I~a x¢(x 2) dx (v) 1;/2 sin2 x dx •
= 0,
= 1;/2 cos 2 X dx .
SOLUTION:
(i) Putting x = a - y. Then dx = -dy and when x = a, y x = O,y = a . ... loa f(x) dx = I~ f(a - y)( -dy) = It f(a - y) dy = It f(a - x) dx (changing y to x)
= 0 and
when
(ii) Putting a + b - x = z. Then -dx = dz and when x = a then z = band when x = b then z = a .
... I: f(a+b-x) dx = fb f(z)( -dz) = I: f(z) dz = I: f(x) dx (changing z to x). (iii) Here ¢( x 2) is an even function as ¢{( - x )2} = ¢( x 2). Hence I~a ¢( x 2) dx = 2 It ¢( x2) dx. (iv) The integrand x¢(x 2) is an odd function as
(-x)¢{( -x)2} = -x¢(x 2). Hence I~a x¢(x 2) dx = O.
(v) 1;/2 sin2 xdx = 1011"/2 sin2(7r/2 - x) dx [Using the prperty It f(x) dx = It f(a - x) dx] = 1011"/2 cos 2 X dx
Ex. 5 Show that (i) x¢(sinx) dx = ~ 1; ¢(sinx) dx,
I;
1 ... ) 11/2 (III -1/2 cos X og
I-x l+x
dx = 0,
(v) J~ll e1xl dx = 2(e - 1),
(ii) I01l"/21ogtanx dx = 0, (iv)
I:~~2 sinx dx
= 0,
CR,7: INTEGRAL CALCULUS
•
SOLUTION:
(i) Let 1
= 10'11" x¢(sinx) dx = Io'll"(7r [using
I; f(x) dx = loa
x)¢(sin(7r - x)) dx f(a - x) dx]
= Io'll"(7r - x)¢(sinx) dx = 7r 10'11"
¢(sin.T)dx - I; x¢(sinx)dx
I; ¢( sin ;r) dx or, 2I = 7r I; ¢(sin x) dx or, 1 = ~ I; ¢(sinx) dx. = 7r
(ii) Let 1
1
= 1;/2 log tan x dx = 1;/2 log tan(7r /2 [using
I; f(x) dx = loa f(a -
x) dx
x) dx]
I; /2 log cot x dx = - 10'11"/2 log tan x dx = -1 or, 21 = 0 or, 1 = 0 or, 10'11"/2 log tan x dx = O. =
(iii) Let f(x) = cos'X log 1- x. , l+x , ( ) Now, f -x = cos( -~ );log 1l+x _ x I-x) , +x
I-x l+x
= cos.r ( - log ; - - 1 = - cos x log - = - f(x).
i.e., f(x, is an odd function. 1/2 . 1- x Hence 'cos x log - - dx = O. / -1/2 1+x (iv) Let f(x) = sinx . . '. f(-x) = sin(-x) = -sinx = -f(x). i.e., f(x) is an odd function. Hence
/'11"/2 -'11"/2
sin x dx = O.
(v) Here the function e lxl is an even function . . '. I~l e lxl dx = 2 Ii e 1xl dx = 2 Ii eX dx = 2[e X]fi = 2(e - 1). (vi) Io'll"/2(a 2 cos 2 x + b2 sin2 x) dx = Io'll"/2(a 2 - a2 sin2 x + b2 sin2 x) dx
= 10'11"/2 {a 2 + (b 2 - a2) sin2 x} dx
H(1- cos 2x)} dx
= 10'11"/2 {a 2 + (b 2 - a2
= 10'11"/2 {a 2 + !(b 2 - a2) - !(b2 - a2) cos 2x} dx
7.11
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
7.12
[~(b2 + a2)x _ ~(b2
=
_ a 2)Sin22X ]:/2
= [~(b2 + a2)I - ~(b2 - a2)Si~7I"]
= i(a 2 + b2).
Ex. 6 Evaluate the following: (i) fo271" I cos xl dx, (iii) (v)
71"/2
sinx
.
10o
(ii)
SIll X
+ cos x
10
r/
ax,
(iv) 10
r loge x dx, la b
(vi)
X
•
71" /2 sin x - cos x . dx, o 1 + SIll X cos X
f
4
loge(1
a
-a
xe
x
+ tanO) dO,
4
dx.
SOLUTION:
(i) The function cos x is positive in 0 to 7r /2 and 37r /2 to 27r and negative in 7r /2 to 37r /2.
r2
, " 10
71"
Icos x I dx =
=
71"/2
10o
Icos x Idx +
1371"/2
Icos x Idx +
71"/2 3 IT /2 1 71" /2 cos x dx + (- cos x) dx o 71"/2
1
1271" 371"/2
+ 1271"
371"/2
Icos x Idx cos x dx
+ [sin x] 271" 71"/2 371"/2 = sin 7r /2 - (sin 37r /2 - sin 7r /2) + (sin 27r - sin 37r /2) = [sin x] 71" /2 _ [sin x] 371" /2
o
= 1 - (-1 - 1) + (0 - (-1)) = 4. (ii) Let 1
= =
71" /2 sin x - cos x dx o 1 + sin x cos x 71"/2 sin(7r/2 - x) - cos(7r/2 - x) dx o 1 + sin(7r/2 - x) cos(7r/2 - x)
In
1 1
71" /2 cos.r - sin x dx o 1 + sin x cos x 71" /2 sin x - cos x = dx =-1 o 1 + sin x cos x or, 21 = 0 or 1 = 0, 71"/2 sinx (iii) Let 1 = , dx o SIll X + cos x 71" /2 s'in( 7r /2 - x) =
10
=
1 1 o
sin( 7r /2 - x)
+ cos( 7r /2 -
x)
dx
CH.7: INTEGRAL CALCULUS
7.13
7r/2
cos x dx 1no sinx + cos x 7r /2 sin x 1n7r/2 cos x 21=1+1= . dx+ . dx . . 0 sm x + cos x 0 sm x' + cos X 1n 7r /2 sin x + cos X dx = 1n7r/2 dx = 7r= o sin x + cos x 0 2 1n 7r or, 1 = 4'
=
(iv) Let 1 = =
r/4
10
10r/
log(l + tanB)dB
4
log{1+tan(7r/4-0)}dB
r/ 1 - tanB) = 10 log 1 + 1 + tan B dB = r/ {(I + tan B) + (1 - tan B)} dB 10 g 1 + tanB 4
(
4
10
4
2 ) dB 1 + tanB 10o r/ 4 r/4 = log 2 10 dB - 10 log(l + tan B) dB 7r /
=
log (
= log2'
4-
or, 21 =
7r
7r
1 7r
4 log 2 or, 1 = 8" log 2.
(v) Putting log x = z. .'. ~ dx = dz and when x = a, z = log a and when x = b, Z = log b. 10gb log x [Z2]IOgb --dx= zdz=.. a x. loga 2 loga 1 1 = 2'(log2 b -log2 a) = 2'(log b - log a) (log b + log a)
. lb
=
1
~ log (~) log(a.b)
(vi) Let f(x) = xe x4 • Then f(-x) = _xe x4 = -f(x). Hence f(x) is an odd function. Thus
Ex. 7
I:a xe
x4
dx = O.
(i) Evaluate J~l f(x) dx where f(x) = x + lxi,
(ii) If f(x) = Ix - 11, evaluate
Kf f(x) dx,
4 (iii) Evaluate J2 f(x) dx where f(x) = Ix - 21 + Ix - 31,
U.G.
7.14
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iv) Find the value of Ii 11 •
xl dx .
SOLUTION:
(i) I~1 f(x) dx = I~1 (;r + Ixl) dx = I~1 X dx + I~1 Ixl dx. Here x is an odd function and Ixl is an even function. I ~ 1 f (x) dx = 0 + 2 IOI Ix Idx = 2 Ii x dx X2] 1
1
=2[20=2'2=1. (ii) I02 f{x) dx = Ii
Ix -
11 dx = I011x - 11 dx + I121x - 11 dx
= Ii -(.r - 1) dx 2
= [- x2 +x]:
+ I12(x -
+ [x22
1) dx
-x]:
[-! + 1] + [(! - 2) - (! - 1)] = ! + (0 + !) = 1. =
(iii) I24 f(x) dx
= I24(lx - 21 + Ix - 31) dx = I24 Ix - 21 dx + I24 Ix - 31 dx 4 = I24(x - 2) dx + Ii Ix - 31 dx + I3 1x - 31 dx 3 = I24(x - 2) dx + I 2 -(x - 3) dx + I34(X - 3) dx x2 ] 4 [X2 ] 3, x 2 4 [ = 2" - 2x 2 + - 2" + 3x 2 + [2" - 3x] 3 =
(iv) Ii 11 -
2+ (~- 4) - (4 - ~) = 3.
xl dx =
IOI 11 -
xl dx + I1211 - xl dx
= Ii(1- x) dx + If -(1- x) dx x2 2 x2 1 =[x- 2 ]0+[-x+ 2
L
=
[1-~] +
[(-2 + 2) - ( -1 +
Ex. 8
(i) Evaluate lim [_1_+_1_+ ... n--+oo n + 1 n +2 5 5 5 1 + 2 + 3 + ... + n 5 ] (ii) Evaluate lim [ 6 . n-+oo n
(iii) Evaluate lim n--+oo
[~+ _1_ + n n+1
...
+~]. 2n
~)]
+~]. 2n
= 1.
7.15
CH.7: INTEGRAL CALCULUS
•
SOLUTION:
[_1_ + _1_ + ... + ~] n-+oo n + 1 n +2 2n
(i) lim
· = 11m
1~ n r=1 n
?; 00
= limh h-+O
= [log(l
(ii) lim
5
+ x)16
L:
l'Im1 n r=11 1+x
0
= log 2. 5 ]
= lim
h-+O
n
00
0
1·
=
n ~ n r=1
+1
0
1] ... +2n
1~ n n r=O n + r
=, l'1m
n-+oo
l'
n-+oo
L: 1n +r
r=O
1~ 1 n r=O 1 + r In
[1 _1_ dx = [log(l + x)16
io
.!.~(~)5
n
=lm-~--=lm-~
n-+oo
n-+oo
x 1 1 L:(rhl = 101 x 5 dx = [-] =-. 6 6
r=1
n
= lim
n6
6
l'1m [1 1· -+--+
n-+oo
~r5
n-+oo ~ r=1
n6
lim h
1
+ rln
10 1 - -1 d x
[1 +2 +3 + ... +n
=
r=1
n-+oo
5
n-+oo
(1'1'1')
n
+r
-1- = 1 + rh
5
= n-+oo~ lim ' " 'n +-r n
-~--=
n-+oo
1
n
1+x
l'
h~
=lm~
h-+O
r=O
= log 2.
= e·r(sinx - cosx) and f(O) = 1, find f(x), (ii) If f'(x) = eX (sin x + cos x) and f(O) = 0 then find f(x) .
Ex. 9
•
(i) If f'(x)
SOLUTION:
(i) We have f'(x) = eX(sinx - cos x). Integrating we get
f (x) = =
f f
= eX
eX (sin x - cos x) dx eX sinxdx -
f
sin x dx -
f
= _ex cos x
+
= _ex cos x
+ c,
f f f
eX cosxdx
(eX
sin x dx) dx -
eX cos x dx -
f
f
eX cos x dx
eX cos x dx
where c is arbitrary constant. Given that f(O) = 1. .'. 1 = -eo cos 0 + c or, 1 = -1 + c or, c = 2. Hence f(x) = 2 - eX cos x.
1 1 + rh
V.G.
7.16
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
+ cos x).
(ii) Here f'ex) = eX(sinx Integrating we get f (x) =
J J
eX (sin x
= - eX cos x X
-e cos x
= _ex cos x
+2
=
J
eX
+ 2.e
sin x dx) dx
(eX
eX
X
.
J +J
sin x dx
J J J
sin x dx -
= eX
=
+ cos x) dx
+
cos x dx
eX
eX
cos x dx
cos x dx
cos x
+ sin x 2
+ c,
+ eX (cos x + sinx) + c =
eX
sinx + c,
where c is constant. Given that f(O) = O.... 0 = 0 + cor, c = O. Thus f(x) = eX sinx.
Exercise 1 Evaluate the following integrals: (i)
Ixl dx,
(ii)
joo ~, 1+x
(iv)
[21
(iii)
[11
Ixl dx, e 1xI dx,
[11
-00
j7r/2 sin -7r/2 ( .. ) j7r /2
(v)
(vi) f0211- xl dx,
0dO,
io
cos x d x, 2 sin x + 3 cos 2 X
Vll
.) (IX
7
j
... ) (Vlll
X
2
-7r/2 2 2·
-2
7r / 2
10o
~ ysmx d x, Jsinx + Jcosx
x smx d 6 12 x, x +
(X) If f(x) is an odd function show that J~3 f(x) dx = O. [Ans. (i) 5/2, (ii) 1, (iii) 71", (iv) 2(e - 1), (v) 0, (vi) 1, (vii) 0, (viii) 71"/4, (ix) 0.]
Exercise 2
.
.
(1) Evaluate: hm
1k
n
n~oo
(ii) Evaluate: lim
110
n~oo
[Ans. (i) k~l' (ii)
A·]
+ 2k + 3k + ... + n k k+l
+ 210 + 310 + ... + n lO n
11
,k
> O.
CR.7: INTEGRAL CALCUI,US
7.2
7.17
Beta and Gamma Functions
Def. 1 Gamma Function: The gamma function is denoted by r(n) and is defined as r(n) = Iooo e- x x n- 1 dx, n > O. Q. 1 Define gamma function. Def. 2 Beta Function: The beta function is denoted by B(m, n) and is defined as B(m, n) = Io1 xm-1(1 - x)n-1 dx, m, n > O.
Q. 2 Define beta function. Some usefull properties: I. r(n + 1) = nr(n), n>0 II. r(l) = 1 III. r(n + 1) = n!, n being a positive integer m,n > 0 IV. B(m,n) = B(n,m), 1 2m V. B(m, n) = 2 I 01l'/2 sin - e cos 2n- 1 e de, m, n > 0
VI.
r/2. 1 (p+1 q+1) 1r(~)r(~) q smPBcos Bde="2 B -2-' -2- ="2 r(p+i+ 2 ) ,p,q >-1
io
VII. B(1/2, 1/2) = 7r VIII. B(m, n)
=
~~:):(:j
IX. r(1/2) = .fi 7r X. r(n)r(l - n) = -.- , 0 o.
Xl, X2
(ii) Determine the feasible space of the following LPP
Maximize z = 2XI + X2 Subject to Xl X2
Xl
+ X2
XI,X2
< 2 < 3 > 1 > O.
(iii) Determine the feasible region of the following LPP
+ 5X2 + 5X2 < 2Xl + X2
O.
(iv) Find graphically the feasible space, if any, for the following: Xl
+ 2X2
O. 0, -2XI
+ X2
~ 2, Xl ~
SOLUTION:
(i) The constraints are treated as equations along with the nonnegativity relations. Then 3XI + 5X2 = 10, 5XI + 3X2 = 15, Xl = 0,X2 = O. or, 1~73 +¥ = 1, ¥+~ = 1, Xl = 0,X2 = O.
-!~~~....--lXl
o Figure: 9.1.1 The shaded portion is the required feasible region.
0, X2
~
0
CH.9: LINEAR PROGRAMMING PROBLEM
(ii) The constraints are treated as equations along with the non-negativity rela2, tions. Then Xl X2 = 3, Xl + X2 = 1, Xl = 0,X2 = 0.
X2
9.3
=3
-1---"*""-+---
Xl
o Xl +xi= 1 Figure: 9.1.2 The shaded portion is the required feasible region. (iii) The constraints are treated as equations along with the nonnegativity relations. Then Xl + 5X2 = 15, 2Xl + X2 = 8, Xl = 0,X2 = 0. or '15 ~ + ~ ~ + ~ Figure: 9.1.3 3 - 1' 4 8 - 1, Xl = 0,X2 = 0.
The shaded portion is the required feasible region. X2
(iv) The constraints are treated as equations along with the nonnegativity relations. Then Xl Xl
+ 2X2 =
Xl -
= 0,X2 = 0.
or ' 7 ~ Xl
7,
=
+~ 3.5 -
0,X2
1' 4 ~
X2
= 4,
+~ -4 -
+ 2x
X
-7 .. Xl - X2
= 4
~~~~+--Xl
1,
= 0.
Figure:
9.1.4
The shaded portion is the required feasible region. (v) The constraints are treated as equations along with the nonnegativity relations. Then Xl - X2 = 0, - 2XI + X2 = 2, 'VI = 0, X2 = 0. or, Xl - X2 = 0, ~ + = 1, Xl = 0,X2 = 0.
X2
-2Xl +X2 1 -
X2
=
=
°
2
-+--+~~-'Xl
¥
Figure:
9.1.5
The shaded portion is the required feasible region.
Exercise 1 (i) Find graphically the feasible space, if any, for the following system of equation and inequations: Xl
+ 2X2
4XI
6 3
> o.
Xl,X2
(ii) Determine the feasible space of the following LPP
Maximize z = 3Xl + 2X2 Subject to Xl - X2 < 1 Xl
+ X2 >
3
>
o.
Xl, X2
(iii) Determine the feasible space of the following LPP
Minimize z = 9Xl when Xl
+ 7X2 + 2X2
o.
(iv) Find graphically the feasible space for the following LPP:
+ 3X2 3Xl + 6X2 < 8 5Xl + 2X2 < 10 XI,X2 > o.
Maximize Z =
Xl
(v) Find graphically the feasible space, if any, for the
+ 3X2 .::; 2Xl + 3X2 > 2.1:1
Xl,X2
follow~ng:
6
6
> o.
(vi) Find graphically the feasible space, if any, for the following:
+ X2 < 5Xl + 3X2 > 2Xl
Xl, X2
6 15
> o.
(vii) Find graphically the feasible space, if any, for the following:
+ 2X2 < Xl + 3X2 >
4
>
o.
Xl
Xl,X2
6
CR.g: LINEAR PROGRAMMING PROBLEM
9.5
(viii) Sketch graphically the feasible region, if any, for the following:
> 0 2XI + 3X2 < 6 Xl,X2 > O. Xl -
x2
(ix) Find graphically the feasible space for the following LPP:
Maximize Z =
+ 2X2 Xl + X2 < Xl + X2 >
Xl
XI,X2
9.2
2
1 > O.
Mathematical Preliminaries
Ex. 1 (i) Express (7, 11) as a linear combination of (3,5).
0
= (2,3) and /3 =
(ii) Express (5, 2, 1) as a linear combination of (1, 1,0) and (3, 0, 1). (iii) Express the vector (1, 1, 1) as a linear combination of the vectors (1, 2, 3), (4,2, 1), (2,4,2) in E3 . • SOLUTION: (i) Let (7,11) = a(2, 3) + b(3, 5) where a, b are scalers. or, (7,11) = (2a + 3b, 3a + 5b) Comparing we get 2a + 3b = 7, 3a + 5b = II. Solving we get a = 2, b = I. .'. the required relation is (7,11)
= 2(2,3) + (3,5) = 20 + /3.
(ii) Let (5,2,1) = a(l, 1,0) + b(3, 0,1) = (a + 3b, a, b) where a, b are scalers. Comparing we get a + 3b = 5, a = 2, b = I. Solving we get a = 2, b = I. .'. the required relation is'f, 1) = 2(1,1,0) (iii) Let (1,1,1) = a(l, 2, 3) + b(4, 2,1) + c(2, 4, 2) 4c, 3a + b + 2c) where a, b, c are scalers. Comparing we get a + 4b + 2c = 1
+ (3,0,1). = (a + 4b + 2c, 2a + 2b + (1)
2a + 2b + 4c = 1
(2)
3a + b + 2c = 1
(3)
9.6
U.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Using (1) x2 - (2) we get 6b = 1 or, b = 1/6 . . '. (2) and (3) become 2a + 4c = 1 - 2/6 = 2/3 and 3a + 2c = 1 - 1/6 = 5/6.
' Ib 1 1 Sl o vmg we get a = 4' = 6' c = 24' . '. the required relation is 1 1 (1,1,1) = 4(1,2,3) + 6(4,2,1)
1
+ 24 (2, 4, 2).
Exercise 1 (i) Express b = (5,4) as a linear combination of al = (3,2), ail = (2,1). (ii) In the Euclidean space E2, show that the vector (1,3) can be expressed as a linear combination of (1,2) and (2,3). (iii) Express the vectors x = (5,9) as a linear combination of the vectors a = (1,2),b = (3,5). (iv) Express x = (4,5) as a linear combination of a = (1,3), b = (2,2). (v) In Euclidean space E2, show that the vector (10,8) can be expressed as a linear combination of the vectors (2,3) and (4,-1). (vi) Express (5, 2, 1) as a linear combination of (1, 1, 0) and (3,0,1). (vii) In three dimen~ional Euclidean space E3 express the vector (2,1, -8) as linear combination of the vectors (3,0,2), (7,0,9) and (4,1,2). [Ans. (i) b = 3al - 2a2, (ii) (1,3) = 3(1,2) - (2,3), (iii) x = 2a + b, (iv) x = ~a + ~b, (v) (10,8) = 3(2,3) + (4, -1), (vi) (5,2,1) = 2(1,1,0) + (3,0, 1), (vii) (2,1, -8) = 4(3,0,2) - 2(7,0,9) + (4, 1,2).] Def. 1 Linearly Dependent: A set of vectors {Xl, X2, ... , xn} of En is said to be linearly dependent if there exists a set of scalars Ci, i = 1,2, ... , n not all of which are zero, such that
where 0 is the null vector in En. Def. 2 Linearly Independent: A set of vectors {Xl, X2, ... , xn} of En is said to be linearly independent if the only set of Ci for which
holds, be
Ci
= 0, i = 1,2, ... , n.
Q. 1 Define linearly dependent and linearly independent set of vectors.
9.7
CH.9: LINEAR PROGRAMMING PROBLEM
ai bi Ci Formula 1 Let ~ = a2 ~ C2 a3 b3 C3 If ~ = then the vectors (ai, bi , Ci), (a2, b2, C2) and (a3, 03) C3) are linearly dependent and if ~ i= then they are linearly independent.
°
°
Ex. 2 (i) Examine whether the set of vectors (1, 1, 0), (1, 1, 1) and (2, 1, 3) is linearly independent in E3. (ii) Prove that vectors (2,2,8), (1,0,4) and (1,2,4) are linearly dependent. (iii) Examine if the vectors (1,2, -3), (2, -3, 1) and (-3,1,2) are linearly independent or not. (iv) Show that the vectors (2, 1, 2), (8, 4, 8) are linearly dependent . •
SOLUTION:
110 1 1 1 = 1(3 - 1) - 1(3 - 2) 213 vectors are linearly independent.
(i) Here
228 (ii) Here 1 4 = 2(0 - 8) - 2(4 - 4) 124 vectors are linearly dependent.
°
°
+ =
+ 8(2 -
1
i= 0.
Hence the given
0) = 0. Hence the given
1 2-3 2 -3 1 = 1(-6 -1) - 2(4 + 3) - 3(2 - 9) = 0. Hence the -3 1 2 given vectors are linearly dependent.
(iii) Here
(iv) Let a = (2,1,2), b = (8,4,8). Let us consider the relation cia + C2b = 0. or, Ci (2,1,2) + c2(8, 4, 8) = (0,0,0) Comparing we get 2Ci +8C2 = O,Ci +4C2 = and 2Ci +8C2 = 0. Solving we get Ci = -4C2 = k (say) ... Ci = k, C2 = -k/4 for any real k. As Ci i= 0, C2 i= the given vectors are linearly dependent.
°
°
Exercise 2 (i) Prove that vectors a = (3,0, -3), b = (-1,1,2) and (4,2, -2) are linearly dependent.
C
=
(ii) Prove that vectors (-1,2,1), (3,0,-1) and (-5,4,3) are linearly dependent. (iii) Prove that vectors (1,1,0), (3,0,1) and (5,2,1) are linearly independent.
9.8
V.C. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iv) Prove_that vectors (1, 2, ~4), (3,2, -1) and (6,8, -13) are linearly dependent. (v) Show that (1, 2, 0), (0, 3, 1), (1, 0, 1) are linearly independent. (vi) Show that the vectors (4, 3, 2), (2, 1, 4) and (2, 3, -8) are linearly dependent.
Ex. 3 Does a set of vectors containing the null vector is linearly independent ? Give reason.
• SOLUTION: Let Xl, X2, ... ,Xr be a set of vectors, they may be linearly dependent or independent. Then for the scalars Cl, C2, •.• , Cr, CIXI
+ C2X2 + ... + CrX r
= 0
(1)
Now let () be the null vector . . '. CIXI + C2 X 2 + ... + CrXr + Cr+l() = 0 or, cr +!() = 0 [using (1)] This is true for any non zero value of Cr+!. Hence a set of vectors containing null vector is linearly dependent.
Def. 3 Spanning set: A finite set of vectors SeEn (Euclidean space) is said to span En if every vector of En can be expressed as a linear combination of the vectors of the set S.
Q. 2 Define spanning set. Def. 4 Basis: Any linearly independent set of vectors SeEn which spans En is called a basis of En.
Q. 3 Define a basis in En. Def. 5 Orthogonal Basis: A basis {Ol' 02, ... , a r } is said to be an orthogonal basis if for all i, j, i f. j, 0i.Oj = O.
Q. 4 Define orthogonal basis in En.
Ex. 4 Prove that S = {(I, 0, 0), (1, 1,0), (1, 1, I)} spans E3 . • SOLUTION: Let (aI, a2, a3) E E3 be any vector. If the given vectors span E3 then we can find a, b, C such that (aI, a2, a3) = a(l, 0, 0) + b(l, 1,0) + c(l, 1, 1). or, al = a + b + c, a2 = b + c, a3 = c. i.e., C = a3,b = a2 - a3,a = al - a2 - a3. From these equations, real value of a, b, C can be found uniquely. Hence the given vectors span E3.
CH.9: LINEAR PROGRAMMING PROBLEM
9.9
(i) Prove that the set of vectors (1, 1, 0), (0, 1, 1) and (1, 0, 1) from a basis of E3.
Ex. 5
(ii) Show that the vectors a = (1,2,1), b = (2,3,0) and- c = (1,2,2) form a basis in E3. (iii) Can (1,0,0), (0,1,0) and (1,1,1) form an orthogonal basis in E3 ? (iv) Show that the vectors (1,0,0), (0,2,0) and (0,0,3) form an orthogonal basis . •
SOLUTION:
(i) The determinant formed by the given vectors is 110 1 1 which is equal to 1(1 - 0) - 1(0 - 1) = 2 -=I 0. 101
°
Thus the given vectors are linearly independent. If the given vectors span E 3, then for any vectors (aI, a2, a3) we can find a, b, c such that (aI, a2, a3) = a(l, 1,0) + b(O, 1, 1) + c(l, 0,1). or, al = a + c, a2 = a + b, a3 = b + c. From these equations, real values of a, b, c can be found. Hence the given vectors form a basis for E3. (ii) The determinant formed by the given vectors is 1 2 1 2 3 =1(6 - 0) - 2(4 - 0) + 1(4 - 3) = -1 -=I 0. 122
°
Thus the given vectors are linearly independent. Since the number of vectors is equal to the number of components of each vector and they also linearly independent so the given vectors span E3. Hence the given vectors is a basis for E3.
(iii) The determinant formed by the given vectors is
o1
°1 °°
= 1 -=I 0.
111
Thus the given vectors are linearly independent. Since the number of vectors is equal to the number of components of each vector and they also linearly independent so the given vectors span E3. Hence the given vectors form a basis for E3.
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
9.10
On
Now, {(I, 0, 0).(0,1, = 0 + 0 + 0 = 0, but {(I, 0, 0).(1,1, In= 1 i= O. . '. the vectors (1,0,0), (0,1,0) and (1,1,1) are not orthogonal. Hence the given vectors does not form an orthogonal basis. (iv) The determinant formed by the given vectors is 100
o 2 0 003
= 6
i= o.
Thus the given vectors are linearly independent. If the given vectors span E 3, then for any vectors (aI,a2,a3) we can find a, b, c such that (aI, a2, a3) = a(l, 0, 0) + b(O, 2, 0) + c(O, 0, 3). or, al = a, a2 = 2b, a3 = 3c. From these equations, real values of a, b, c can be found. Hence the given vectors span E3. Again, {(I, 0, 0).(0, 2, {(O, 2, 0).(0, 0, 3)}
On = 0, {(I, 0, 0).(0, 0, 3n = 0 and
= O.
Hence the given vectors form an orthogonal basis.
Exercise 3
(i) Prove that (1, 1,0), (1, -1, 0), (0,0,1) span E3.
(ii) Prove that the set of vectors (1, 1, 0), (1, -1, 0) and (0, 0, 1) from a basis of E3. . (iii) Do the vectors 01 = (1,0,7),02 = (4,0,6) and 03 = (1,0,0) from a basis . for E3 ? (iv) Do the vectors (1,1,0), (0,1,1) and (1,0,1) form a basis for E3 ? (v) Examine whether the following vectors a = (1,2,3), b = (3, -2, 1) and c = (4,2,1) form a basis for three dimensional space E3 ? (vi) Show that the vectors (-1,1,0), (1, 1,0) and (0,0,1) form a basis in E3. (vii) Does the vectors (1, 2, 3), (4, 5, 6), (3, 6, 9) form a basis in E3 ? [Ans. (iii) no, (iv) yes, (v) yes, (vi) yes, (vii) no.]
Def. 6 Feasible Solution (F.S.): In a linear programming problem a solution satisfying the constraints and non-negativity conditions is called a feasible solution. Q. 5 Define feasible solution.
CH.9: LINEAR PROGRAMMING PROBLEM
9.11
Def. 7 Basic Solution (B.S.): Let Ax = b be a m linear simultaneous equations of n (n > m) variables. We assume that rank of A and rank of the augmented matrix are equal to m. If any m x m non-singular sub-matrix be chosen from A and if all the remaing (n - m) varibales those are not associated with this sub-matrix be set equal to zero, the solution to the reducing system of equations is a basic solution.
Q. 6 Define basic solution. Def. 8 Basic Feasible Solution (B.F.S.): In a linear programming problem a solution satisfying the constraints, non-negativity conditions and basic is called a 'basic feasible solution.
Q. 7 DeHne basic feasible solution. Def. 9 Degenerate Basic Solution: In a basic solution, at least (n - m) variables must be zero. If the number of non-zero basic variables be less than m or if anyone of the basic variables be zero, the solution is called degenerate basic solution.
Q. 8 Define degenerate basic solution. Ex. 6 Explain the difference between a basic solution and a degenerate basic solution . • SOLUTION: In the basic solution, the values of the basic variables are nonzero, on the other hand, in degenerate basic solution the values of at least one basic variable is zero. Ex. 7 How many basic solutions are possible in a system of m-equations and n-unknowns? • SOLUTION: The maximum number of basic solutions in a system of m equations in n unknowns is
nc _ m -
Ex. 8
n! m!(n - m)!'
(i) Find a basic solution of 2Xl + X2 + 4:t3 = 11, 3XI + X2 + 5X3 = 14.
(ii) In the given equations find the basic solutions with variable: Xl
+ 4X2 -
X3
= 3, 5Xl
+ 2X2 + 3X3 =
4.
X3
as the non-basic
U.G.
9.12
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iii) Show that Xl = 5, X2 = 0, X3 = -1 is a basic solution of the system of equations
+ 2X2 + X3 2XI + X2 + 5X3 Xl
•
=
4
=
5.
SOLUTION:
(i) Let Xl = 0. Then the equations become X2 + 4X3 = 11 and X2 + 5X3 = 14. or, BX = b, where B = Now, IBI = . "" B-
1
=
~
:
= 1
[~
: ] , X = [ :: ] ,b = [
¥ 0,
adj.(B) =
~! ].
[_~ -~] .
af~f = [_~ -~].
Hence X = B-Ib =
[_~ -~] [ ~! ]=
[
-! ].
""" (0, -1,3) is a basic solution. (ii) Since
X3
is a non-basic variable, we put
X3
= 0.
. "" the equations become BX = b, where
B =
[!
~], X = [ :~ ] ,b =
Now, IBI =
I! ~ I
= -18
" B-1 = adj.B = """ IBI
_~
[
18
¥
!l
[
0, adj.(B) = [
Xl
X2
1
.
1
= [
a, ~~, 0) is a basic solution with
+ X3 =
-4]
2 -4]. -~ ] [! ] ~! l
-5
HenceX=B _ lb=-ls [ -52
Hence (iii) When
_~
X3
as a non-basic variable.
= 0, then the equations become 4 and 2XI + 5X3 = 5.
or, BX = b, where B = Now, IBI =
[~
!],
X = [ :: ] ,b = [ : ].
I ~ ! 1= 5 - 2 =3 ¥ 0.
Xl = 5, X2 = 0, X3 = -1 satisfy the given equations and the matrix B is
non-singular so it is a B.S.
9.13
CH.9: LINEAR PROGRAMMING PROBLEM
Ex. 9
(i) Find a basic feasible solution of the system: 1
(ii) Find a basic feasible solution of the following equations: Xl
+ 3X2 + 2X3 + 4X4
10
+ 4X3 + 6X4
16.
2XI -
X2
(iii) Find two basic feasible solutions of the following equations:
•
SOLUTION:
(i) Let
X3
= 0. Then the equations become
Xl
= 1, X2 = 4.
Hence (1, 4, 0) is a basic feasible solution. (ii) Let Xl and X2 be the non-basic variables. Setting Xl = equations become 2X3 + 4X4 = 10, 4X3 + 6X4 = 16. Solving we get
X3
X3
= 0.
Then
= 0. Then the
= 1, X4 = 2, both are positive.
Hence (0,0,1,2) is a basic feasible (iii) Let
X2
Xl
sol~tion.
= 1, X2 = 4.
... (1,4,0) is a basic feasible solution.
Again, let Xl
= O.
Then
X3
= 1/2 and X2 = 4 -
1/2 = 7/2.
So (0,7/2,1/2) is another basic feasible solution. Exercise 4 tions: Xl
(i) Find the basic solutions of the following system of equa-
+ X2 + X3
= 4, 2XI
+ 2X2 + 3X3 =
10.
(ii) Find a basic solution of the following system of equations: Xl
2XI
+ X2 + X3
+ 5X2 -
2X3
=
4
=
3.
(iii) Find a basic feasible solution of the following equations:
+ X3 -2XI + X2 + 2X3 2XI -
3X2
=
4
=
5.
9.14
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iv) Find a basic feasible solution of the system
+ 2X3 = X2 + X3
Xl
Xl, X2,
X3
1 4
> 0.
(v) Find the basic solutions of the following system of equations: Xl
+ X2 + X3 3XI + 2X2 XI,X2,X3
8
>
18 0.
[Ans. (i) (0,2,2),(2,0,2), (ii) (\1,0, i), (iii) (!,0,3), (iv) (1,4,0), (v) (6,0,2), (0,9,-1), (2,6,0).] Def. 10 Point Sets: The point sets are the sets whose elements are points or vectors in n-dimensional Euclidean space En. The set X = {(XI, X2) : x~ + x~ < 4}, i.e., the set of all points lying inside a circle of radius 2 with centre at the origin, is a point set in E2.
Q. 9 Define point sets. Def. 11 Hyperplane: A set of points in n-dimensional space whose coordinates satisfy the linear equation of the form qXI
+ C2X2 + ... + CnXn =
Z
is called a hyperplane for fixed values of z and Ci, i = 1,2, ... , n. 2XI - X2 + X3 + 8X4 = 5 is a hyperplane in 4-dimension.
Q. 10 Define hyperplane. Def. 12 Open Half Spaces: The sets Xl = {x: ex X2 = {x : ex > z} are- called open half spaces.
< z}
and
Q. 11 Define open half spaces. Def. 13 Closed Half Spaces: The sets Xl {x : ex 2': z} are called closed half spaces.
{x : ex ::; z} and X 2 =
Q. 12 Define closed half spaces. Def. 14 Line: Let Xl a.nd X2 be two points in Euclidean space En. A line in En passing through the points XI, X2 is defined to be the set of points X = {x : X = >'X2 + (1 - >')Xl, >. is real}
CH.9: LINEAR PROGRAMMING PROBLEM
9.15
Q. 13 Define line.
Def. 15 Interior Point: A point x is said to be an interior point of the set X if an c:-neighbourhood about x contain only points of the set X. Q. 14 Define interior point.
Def. 16 Boundary Point: A point x is said to be a boundary point if every c:-neighbourhood about x contains points of the set and also points outside the set. Q. 15 Define boundary point.
Def. 17 Convex Combination: Let {Xl, x2, ... ,, xr} , be a set of points in En. The linear combination x = AIXI + A2X2 + ... + ArXr is said to be convex a combination if Ai 2': 0, i = 1,2, ... , r and Al + A2 + ... + Ar = 1. Q. 16 Define convex combinat:on.
Def. 18 Convex set: A set X is said to be convex set if for any two points Xl, x2 in the set, all the points on the line segment joining these two points are the member of the set X. In other words, a set X is said to be a convex set if any convex combination of any two points of the set X belongs to the set X. The set X = {(Xl, X2) : + X~ ::; 1} is a convex set.
xi
Q. 17 Define a convex set.
Def. 19 Extreme Point: A point x in a convex set is said to be an extreme point, if it does not lie on the line segment joining any two points other than I X in the set. In other words, a point x in a convex set is said to be an extreme point if x cannot be expressed as a convex combination of two other points. Every point on the circumference of the circle
xr + x~ ::; 1 is an extreme point.
Q. 18 Define extreme point of a convex set.
Def. 20 Convex hull: Let X be a given set of points. The set of all convex combinations of the set of points from X constitutes a convex hull of the given set X.
9.16
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
If X be the set of points lying on the boundary of a circle then the circle with
its interior is the convex hull of the circle. Q. 19 Define convex hull.
Def. 21 Convex Polyhedron: The set of all convex combinations of a finite number of points is called the convex polyhedron generated by these points. The convex hull of the eight vertices of a cube is a convex polyhedron. Q. 20 Define polyhedron.
Def. 22 Simplex: An n dimensional convex polyhedron having exactly (n + 1) vertices is called a simplex. A tetrahedron is a simplex in three dimensions. Q. 21 Define simplex.
Ex. 10 Give examples of (i) convex hull in in E 2 , (iii) simplex in E2 .
E2
and
E3,
(ii) convex polyhedron
• SOLUTION: (i) The points lying on the boundary of a circle with its interior is its convex hull in E 2 . The whole cube is the convex hull of the set of points consisting of only the eight vertices of the cube in E3. (ii) The convex hull of the three vertices of a triangle is a convex polyhedron in E 2 . (iii) A triangle is a simplex in two dimensions. Ex. 11 (i) Is the set of points (0,0), (0,1), (1,0) and (1,1) on the xy-plane a convex set ? (ii) Show that the set S =
{(Xl, X2) : Xl
+ X2
;:::
2} is convex.
(iii) Discuss whether the following set is convex or not X = {x: Ixl::; 2}. (iv) Examine whether the following set is convex S = {(Xl, X2) : X2 ::; 4, Xl ;::: 2}. (v) Examine whether the set X = {(Xl. x2)12xI + X2 ::; 14,2xI
+ 3X2
::;
22, Xl. X2
;:::
a} is a convex.
(vi) Discuss whether the following set is convex or not X = {(Xl, X2) : X~ + X~ = 25}. (vii) Show that the set of points on the line y = mx + c is convex.
CH.9: LINEAR PROGRAMMING PROBLEM
9.17
• SOLUTION: (i) Here the number of points of the set is four, a finite number. A finite set can not be a convex set. Hence the given points does not form a convex set. (ii) Let YI = (XI, X2) and Y2 = (X3, X4) be any two points of X. Then Xl + X2 ~ 2 and X3 + X4 ~ 2. Now, Y = )..YI + (1- )..)Y2,0:S)..:S 1 = )"(XI, X2) = ()..XI
+ (1 -
+ (1 -
)..)(X3, X4)
)..)X3, )..X2
+ (1 -
)..)X4).
Let us consider {>'XI + (1 - )..)X3} + {>.X2 + (1 - )..)xd = )..(XI + X2) + (1 - >')(X3 + X4) ~ 2>' + 2(1 - )..)
= 2.
Hence the given set is a convex set. (iii) Let Xl, X2 EX. The we have IXII :S 2, IX21 :S 2. Let us consider the convex combination of Xl and X2 which is >'XI + (1- )..)X2,0:S)..:S 1. Then I)..XI
+ (1 -
)..)x21
:S )..IXII + (1 -
)..)IX21
:S ),,2 + (1 ~ )..)2 = 2.
Thus the point determined by the convex combination of Xl, x2 is also in X. Hence the given set is a convex set. (iv) Let YI
= (Xl, X2)
Then Xl
~
and Y2
2, X2 :S 4, X3
= (X3, X4) ~
be two points in S. (1)
2, X4 :S 4
+ (1 - )..)Y2, 0 :S ).. :S 1 = >'(XI,X2) + (1- )..)(Xa,X4) = (>'XI + (1 - )..)xa, )..X2 + (1 - )..)X4). Now, >'XI + (1 - )..)xa ~ 2)" + 2(1 - )..) = 2 and )..X2 + (1 - )..)X4 :S 4).. + 4(1 - )..) = 4 [using (1)] Let Y = >'YI
... yES. Hence S is a convex set.
(v) Let YI = (XI,X2) and Y2 = (X3,X4) be two points in X. Then 2XI + X2 :S 14,2xI + 3X2 :S 22, Xl ~ 0, X2 ~ 0 and 2X3 + X4 :S 14, 2xa + 3X4 :S 22, Xa ~ 0, X4 ~ 0
+ (1 - )..)Y2, 0 :S >. :S 1 )..(XI, X2) + (1 - )..)(xa, X4) (>'XI + (1 - )..)xa, )..X2 + (1 -
Let Y = )..YI =
=
)..)X4).
(1)
9.18
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
2{'\Xl +(1-'\)x3}+{Ax2+(1-'\)xd = '\(2Xl +x2)+(1-,\)(2x3+x4) ~ 14,\ + 14(1 - ,\) = 14. Similarly, 2{AXl
+ (1 -
'\)X3}
+ {AX2 + (1 -
'\)xd
~
22 .
. '. Y EX. Hence X is a convex set.
(vi) The set X, i.e., the points on the perimeter of a circle is not a convex set, because, the line segment joining any two points on the perimeter, does not lie on the boundary. Hence X is not a convex set. (vii) Let X be the set of points on the line y = mx + c. Let Yl = (Xl, mXl + c) and Y2 = (X2' mX2 + c) (since Y = mx + c) be two points in X. Let Y = )..Yl
)..)Y2, a ~
+ (1 -
= )..(Xl' mXl
+ c) + (1 -
+ (1 = ()..Xl + (1 -
= ()..Xl
).. ~ 1 )..)(X2' mX2
+ c)
+ c) + (1 - )..)(mx2 + c)) )..)X2' m{)..xl + (1 - ,\)X2} + c) EX. )..)X2' )..(mxl
Hence X is a convex set. (viii) In the figure, the shaded region represents the set X. A portion of the line segment AB lies outside the shaded region. Hence the set X is not a convex set.
Exercise 5
o Figure: 9.2.1
(i) Prove that the set X = {(x, y)!x
+ 2y ~ 5}
is a convex set.
(ii) If Xl, X2 be real, show that the set X = {(Xl, X2)!Xl +X2 ~ 50, Xl +2X2 ~ 80, 2Xl + X2 ~ 20, Xl, X2 ~ a} is convex set. (iii) Examine whether the following set is convex S = {(Xl,X2)!Xl ~ 2,X2 ~ 3} (iv) Examine whether the following set is convex X = {(Xl, X2)!2xl + X2 ~ 20, Xl + 2X2 ~ 80, Xl
+ X2
~ 50, Xl, X2 ~
[Ans. (iii) yes, (iv) yes.]
Ex. 12 (i) Find the extreme points, if any, of the following sets: (a) S = {(x,y) : x2 + y2 ~ 25} (b) S = {(x, y) : !xl ~ 1, !Y! ~ 1}
a}.
9.19
CH.9: LINEAR PROGRAMMING PROBLEM
(ii) Does the set X = {(x, y)lx2 •
+ y2
:::; 9} contain any extreme point?
SOLUTION:
(i) (a) All the points on the circumference of the circle x 2 + y2 extreme points.
= 25 are the
(b) We have S = {(x,y) : Ixl :::; 1, Iyl :::; I} or, S = {(x, y) : -1 :::; x :::; 1, -1 :::; y :::; I}. The four corner points (-1, -1), (-1,1), (1, -1), (1, 1) are extreme points. (ii) All the points on the circumference of the circle x 2 + y2 = 9 are the extreme points of the set X. Ex. 13 (i) In which of the half spaces determined by the hyperplane 3XI + 2X2 + 4X3 + 6X4 = 7, does the point (-6,1,7,2) lie? (ii) A hyperplane is given by the equation 3XI + 2X2 + 4X3 + 7X4 = 8. Find in which half space divided by the hyperplane the points (-6,1,7,2) and (1,2,-4,1) lie. (iii) Determine the position of the point (1, -2,3,4) relative to the hyperplane 4XI + 6X2 + 2X3 + X4 = 2. •
SOLUTION:
(i) We have P == 3XI + 2X2 + 4X3 + 6X4 - 7 = 0 At the origin, Po = -7 < O. At (-6,1,7,2), PI = -18 + 2 + 28 + 12 - 7 = 17> O. Since Po and PI are in opposite in sign, so the given point lies in the open half space not containing the origin.
(ii) The given hyperplane P == 3XI + 2X2 + 4X3 + 7X4 - 8 = 0 At the origin, Po = -8 < O. At (-6,1,7,2), PI = -18 + 2 + 28 + 14 - 8 ~ 18 > O. Since Po and PI are in opposite in sign, so the given point lies in the open half space not containing the origin, i.e., in 3XI + 2X2 + 4X3 + 7X4 > 8. Again, at (1,2, -4, 1), P 2 = 3 + 4 - 16 + 7 - 8 = -10 < O. Since Po and P2 have same sign, so the given point lies in the open half space containing the origin, i.e., in 3XI + 2X2 + 4X3 + 7X4 < 8. (iii) Here 4(1) + 6( -2) + 2(3) + 4 - 2 = O. Hence the point (1, -2,3,4) lies on the hyperplane. Exercise 6 A hyperplane is given by the equation 2XI + 3X2 + 4X3 + 5X4 = 7. Find in which half space defined by the hyperplane the points (1,2,3,4) and (1,2,3,-4) lie.
U.G.
9.20
Ex. 14
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(i) Show that a hyperplane is a convex set.
(ii) Show that the intersection of two convex sets is also a convex set.
(iii) Show that the union of two convex sets is not necessary a convex set . •
SOLUTION:
(i) Let us consider the hyperplane X = {x: ex = z}. Let Xl and X2 be two points in X. Then eXI = z and eX2
= z.
Let Xa = AXI + (1 - A}X2, 0 ~ A ~ 1. Then eXa = Aexi + (1 - A)ex2 = AZ + (1 - A}Z = z . . '. Xa, a convex combination of Xl and X2, is in X. Hence, X is a convex set. (ii) Let Xl and X2 be two convex sets and X = Xl n X2. Let Xl, X2 E Xl n X2· Implies XI,X2 E Xl and XI,X2 E X 2. Then AXI + (1 - A)X2 E Xl. 0 ~ A ~ 1 and AXI + (1 - A}X2 E X 2 , 0 ~ A ~ 1. Hence AXI
+ (1 -
A}X2 E Xl n X 2 = X .
. '. the convex combination of Xl, x2 E X belongs to X. Hence X is a convex set. (iii) The triangular regions Figure (a) and Figure (b) are convex sets. But, their union, shown in Figure (c) is not a convex set, as the line segment AB does not lie wholly in the set.
~B (a)
·9.3
(b) Figure: 9.2.2
(c)
Simplex Method
Theorem 1 Fundamental theorem of linear programming: If the L.P.P. has an optimal solution, then the optimal solution will coincide with at least one basic feasible solution of the problem. Q. 1 State the fundamental theorem of linear programming.
CH.9: LINEAR PROGRAMMING PROBLEM
Ex. 1
9.21
(i) Reduce the L.P.P.:
Maximize Z = 2Xl + X2 Subject to 3Xl
6X3
~ 2Xl + 3X3
Xl, X2, X3
0
into standard form. (ii) Reduce the following L.P.P. to its standard form:
Maximize Z = 3Xl Subject to 3Xl + 2X2 Xl Xl
X2
+ 2X2 Xl,X2
+
3X2
< 6 > -1 < 1 > O.
(iii) Rewrite the following inequations in the form of equations: Xl -
2X2
2Xl
+ X2
+ 5X3 > -
7X3
Xl, X3
6
< 5 > 0,
X2
is unrestrict ion in sign.
(iv) Express the following LPP in the matrix form:
+ 3X2 + X2 + X3 Xl + 2X2 5Xl - 2X2 + X3
Maximize Z = Subject to Xl
2Xl
Xl, X2, X3
+
4X3
> 5 = 4 < 9 > O.
(v) Write down the following LPP in its standard form introducing slack and surplus variables.
Minimize Z = -3Xl Subject to Xl -3Xl
+ 2X2 - 4X2
+ 2X2 Xl, X2
< -14 < 6 > O.
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
9.22 •
SOLUTION:
(i) The constraint 2 30,2xI + 3X3 ~ 2.
~
2XI
+ 3X3
o.
(iv) Introducing surplus variable X4 to the first constraint and b.ack variable X5 to the third constraint respectively. The standard form of the given L.P.P. is
= 2Xl + 3X2 + Subj'ect to Xl + X2 + Xa - X4 = Xl + 2X2 5Xl - 2X2 + Xa + X5 = Maximize
Z
Xl,X2,Xa,X4,X5
4xa
+ 0.X4 + 0.X5
5
4
9
> o.
The matrix form is AX = b, where A = [
~ ~
1 -1
00 0
5 -2 1 Xl X2
X=
Xa X4 X5
(v) We first rewritten the given problem as Z* = 3Xl Subject to - Xl
Maximize
- 2X2
+ 4X2 > -3Xl + 2X2 < Xl,X2
14 6
> o.
Introducing surplus variable Xa to the first constraint and slack variable X4 to the second constraint respectively.
9.24
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
The standard form of the given L.P.P. is
= 3XI - 2X2 + Subject to - Xl + 4X2 - X3 = -3XI + 2X2 + X4 = Maximize Z·
Xl, X2,
Exercise 1
X3, X4
0.X3
+ 0.X4
14
6
> O.
(i) Reduce the following L.P.P. to its standard form:
Maximize Z = 3XI
2X2
+ X2 < -Xl + 3X2 >
Subject to 2XI
XI,X2
1 4
> O.
(ii) Rewrite the following problem in equality form by introducing slack and surplus variables and identify them:
Maximize Z = 2XI
+
+ 2X2 < -3XI + 5X2 >
Subject to Xl
XI,X2
3X2
5
8
> O.
(iii) Introduce slack and surplus variables convert system into standard form
Maximize Z = 2XI
+ X2
Subject to Xl 2XI
+ X2
XI,X2
< 4 > 1 > O.
(iv) Express the following LPP in standard form with the help of slack and surplus variables:
Maximize Z = 9XI
+ 2X2
Subject to 4XI -
Xl
< >
X2
O.
CH.9: ~INEAR PROGRAMMING PROBLEM
(v) Convert the following
9.25
LPP:
+ 7X2 Subject to 2X1 + 3X2 < Xl + 7X2 >
4
>
O.
Maximize Z =
6X1
X1,X2
6
into standard form by introducing slack and surplus variables. (vi) Express the following LPP in normal (standard) form with the help of slack and surplus variables:
Maximize Z =
4X1 - 5X2
Subject to
Xl
+ X2 < 13 X2
X1,X2
> 10 > O.
(vii) Express the following LPP in normal (standard) form with the help of slack and surplus variables:
+ 2X2 - 2X1 + 3X2 < -Xl + 8X2 >
Maximize Z = SUbject to
7X1
Xl X2 X1,X2
5 6
< 3 < 4 > O.
(viii) Introduce slack and surplus variables to convert the system into standard form: Zmax = 3X1 - 2X2 Xl
~
-5
3X1 - 5X2
~
3 and
Subject to [Ans. (i)
Max Z =
3X1 - 2X2 '
+ X2 + X3 -Xl + 3X2 - X4
Subject to
2X1
Xl,
(ii)
Xl, X2 ~
X2, X3, X4
=
1
=
4
> O.
+ 3X2 Subject to Xl + 2X2 + X3 = -3X1 + 5X2 - X4 = X1,X2,X3,X4 > Max Z =
2X1
5 8 0
O.
9.26
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
is slack variable and X4 is surplus variable. Max Z = 2XI + X2 (iii) 4 Subject to Xl + X3 2XI + X2 - X4 = 1 Xl, X2, X3, X4 > o.
X3
Max Z = 9XI + 2X2 Subject to 4XI - X2 + X3 Xl - X4 X2 +xs Xl, X2, X3, X4, Xs
(iv)
= = =
> o.
Max Z = 6XI + 7X2 Subject to 2XI + 3X2 + X3 Xl + 7X2 - X4 Xl, X2, X3, X4
(v)
Max Z = 4XI - 5X2 Subject to Xl + X2 + X3
(vi)
X2 -X4
Xl, X2,
X3, X4
= =
= = >
X2 +X6 XI,X2,X3,X4,XS,X6
Max Z = 3XI - 2X2 Subject to - Xl + X3 3XI - 5X2 - X4 Xl, X2, X3, X4
(viii)
9.4
6 4
> o.
Max Z = 7XI + 2X2 Subject to - 2XI + 3X2 + X3 -Xl + 8X2 - X4 Xl + Xs
(vii)
120 10 15
= = >
13 10
o. = = =
5 6 3 4
> o. 5 3
o.
Duality
Ex. 1 (i) Construct the dual problem from a given standard linear programming problem.
CH.9: LINEAR PROGRAMMING PROBLEM
9.27
(ii) Find the dual of the L.P.P
+
Maximize Z = 3Xl
Subject to
2X2
+ 4X2 < 3Xl + 2X3 < 3Xl
(iii) The dual of the problem is
22 16
> O.
XI,X2
+ 13x2 Subject to 3XI + 7X2 >
Minimize z = 7Xl
4 4Xl - 2X2 > 4 XI,X2 > O.
Find the primal problem. (iv) Write down the dual of the following L.P.P: Maximize Z = 6XI
Subject to
Xl
+ 8X2
+ X2 -
10X3
X3
> 2
2XI - X3
> 1 > O.
xl,X2, X3
(v) Find the dual of the following L.P.P:
+ X2 Subject to 2Xl + X2 < Xl + 3X2 =
10 3
>
O•
Maximize Z = Xl
Xl, X2
• SOLUTION: (i) Let us consider the L.P.P. Maximize z = ex subject to Ax ~ b x~O
Here A is an m x n matrix, x and e are n-component column and rm" matrices ann b is an m-component column matrix. The dual problem of this primal problem is Minimize w = btv subject to A tv ~ e t v~O
where v is an m-component column matrix.
9.28
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Note 1 If the primal problem be Minimize z = ex subject to Ax ~ b x~o
then its dual problem is Maximize w = btv subject to A tv ~ e t v ~ o. (ii) It is a maximization problem and the constraints are of dual problem is
+ 16v2 Subject to 3VI + 3V2 > 4VI + 2V2 >
Minimize w =
'~'
type. So its
'~'
type. So its
22vI
VI, V2
3
2
> 0
or,
+ 16v2 Subject to VI + V2 > 2VI + V2 >
1
>
o.
Minimize w =
22vI
VI,V2
1
(iii) It is a minimization problem and the constraints are of dual problem is
+ 4V2 Subject to 3VI + 4V2
o.
(iv) We first rewrite the problem in the following form
lOxg + 8X2 Xl - X2 + Xg < -2 -2XI + Xg ~ -1
Maximize Z = Subject to -
6XI
> o.
CH.9: LINEAl. .?ROGRAMMTNG PROBLEM
9.29
It is a maximization problem and the constraints are of ':::;' type. So its dual problem is
Minimize w = Subject to
-2VI - V2 - VI - 2V2 -VI
+ V2
VI
VI,V2
> 6 > 8 > -10 > 0
or,
Maximize w* = Subject to -
2VI
+ V2
VI -
2V2
-VI -VI - V2 V!, V2
> 6 > 8 < 10 > O.
(v) The given problem is rewrite as
+ X2 + X2 < Xl + 3X2
O.
It is a maximization problem and the constraints are of ':::;' type. So its dual problem is
Minimize w = 10VI + 3V2 Subject to 2Vl + V2 - V3 VI
+ 3V2 -
3V3
v!, V2, V3
3V3
> > >
1 1 0
or,
Minimize w = 10VI + 3v~ Subject to 2VI + v~ > 1 VI
+ 3v~ > VI
1 > 0, v~ =
V2 - V3
unrestricted in sign.
9.30
Exercise 1
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(i) Find the dual of the L.P.P
+ 2X2 Subject to 3XI + 4X2 < 7XI - 2X2 < Zmaz = 4XI
7 13 > O.
Xl, X2
(ii) Write down the dual of the following primal problem: Maximize
Z =
Subject to
Xl
+ 3X2 < 3 < 5 > O.
2XI
+ 2X2 Xl + X2 XI,X2
(iii) Write down the dual of the following L.P.P:
Z = lOXI + 15x2 Subject to Xl + 4X2
Maximize
4XI
< 20 < ·35 > O.
+ X2
XI,X2
(iv) Find the dual of the L.P.P Zmaz = 3XI - 7X2
Subject to xl - 7X3 > 1 Xl
+ 3X3
O.
XbX2,X3
(v) Find the dual of the following problem:
+ 5X2 Subject to 3XI + 4X2 < Xl + 3X2 :::;
Minimize Z = Xl
Xl, X2
~
6 2 O.
(vi) Find the dual of the primal L.P.P Maximize
Z=
Subject to
5XI Xl
+ 3XI 2X2 + X3
4X3
2XI
:::;
4
+ X2 + 4X3 < Xl! X2, X3 >
7 O.
CH.9: LINEAl\,
[Ans. (i)
PROGRAM~TNG
= 7Vl + 13v2 Subject to 3Vl + 7V2 > 4VI - 2V2 :>
(v)
> O.
= 3Vl +5V2 Subject to VI + V2 > 2VI + v2 >
= 20VI + 35v2 Subject to VI + 4V2 > 4Vl + V2 >
10 15
> O.
= -VI + 5V2 Subject to - VI + V2 > Min w
Min w
-7Vl - 3V2
3 < 7
Vl,V2
> O.
= -6Vl -
2V2
< 1 < 5 > O.
Vl,V2
= 4VI + 7V2 Subject to 5Vl + V2 > -2Vl + V2 > VI + 4V2 >
Min w
Vb V 2
9.5
3
> O.
Subject to - 3Vl - V2 -4VI - 3V2
(vi)
2
Min w
Vl,V2
(iv)
4 2
Min w
VI,V2
(iii)
9.31
Min w
Vl,V2
(ii)
PROBLEM
2
3 -4
> O.
Transportation Problem
Def. 1 Transportation Problem: A transportation problem is a particular type of linear programming problem. Here a particular commodity which stored at different origins is to be transported to different destinations in such a way that the transportation cost is minimum. Let there be m origins 0 1 , 02, ... , Om and the quantity available at origin Oi be ai, i = 1,2, ... , m and let there be n destinations Db D2,"" Dn and the demand at the destination Dj be bj, j = 1,2, ... , n.
·9.. 32
U.G. MATHEMATICS (SHoaT QUESTIONS AND ANSWERS) m --..............
Now, if
n
L ai '= L i=l
bj then the transportation problem is called a balanced
j=l
transportation problem.
Q. 1
(i) What do you mean by a transportation problem?
(ii) Define a balanced transportation problem.
Ex. 1 Express a general transportation problem in mathematical form. or Write down the general transportation problem involving m origins and n destinations as a linear programming problem . • SOLUTION: A general transportation problem involving m origins and n destination is stated as below: Minimize z = ex Subject to Ax = h, x 2: O. Where e = (CU,CI2, ... ,Cij, ... ,Cmn) is an mn-components row vector, x = [xu, X12,··., Xij, ... ,xmnl is an mn-components column vector, h = [aI, a2,"" am, bl , b2,· .. , bnfis an (m+n)-components column vector and A = [an, aI2, ... , aij, .. · , amnl is the coefficient matrix in which aij is the column vector associated with the variable Xij. Ex. 2 (i) Using north west corner method find the initial basic feasible solution to the following transportation problem:
8 9
3
(ii) Find the initial basic feasible solution of the following transportation problem by matrix minima method:
I 5~
5 10
4
(iii) Find the initial basic feasible solution by row minima method for the following transportation problem:
CH.9: LINEAR PROGRAMMING PROBLEM
9.33
D1 D2-' D3 01 O2 03
6 4 1
8 9 2
6
10
14 12 6 5 15 4 3
(iv) Find the initial basic feasibl~ solution by column minima method for the following transportation problem:
I II III
c
A 50 90 270
B 30 45 200
220 170 50
4
2
2
1 3 4
(v) Find a feasible solution of the following transportation problem by VAM
D3 2
1 130 2 40 30
3
20
(vi) Using north west corner method find the initial basic feasible solution to the following transpo~tation problem:
D1 D2 D3 D4 01
O2
3 9
03
10
10
7 4 2 5
1 11 3 20 3 35 30
2 7 8
21
Is this solution non-degenerate ? • SOLUTION: (i) Applying north-west corner method to find initial B.F.S.
D1
8 3
6
5
9.34
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS) The initial B.F.S. is
Xu
= 6, XI2 = 3, XI3 = 1, X23 = 5.
(ii) Minimum'value among all elements is 5. Selecting this value firstly. DI
D2
..!J
6
D3
5
8
~
5
6 7
4
5
8
6
The next minimum element among the elements offirst and third columns is 5. The reduce matrix is
4 7
5 The initial B.F.S. is
X12
6
= 4, Xl3 = 4, X21 = 5, X23 = 2.
(iii) Applying row minima method to find an initial B.F.S.
Dl
D
D 14
12 03
The initial B.F.S. is
XI3
5
= 14, X21
= 6, X22
= 5, X23 =
1, X32 = 5.
(iv) Applying column minima method to find an initial B.F.S.
ABC I
1 3
II
4
2
9.35
CH.9:, LINEAR PROGRAMMING PROBLEM
= 1, X21 = 3, X32 = 2, X33 = 2.
The initial B.F.S. is Xu
(v) Applying VAM to find an initial B.F.S.
D2 01
1--.::...r-r=-t--r=--l30 (0) 10 (1) IC..--_..I....-_..l...-_..J
x
40(1) 40 (1) 40
x
20 The initial B.F.S. is Xu
20
= 20, X13 = 10, X22 = 20, X23 = 20.
(vi) Applying north-west corner method to find an initial B.F.S.
11 20
35
The initial B.F.S. is Xu 30.
= 10, Xu = 1, X22 = 4, X23=
16, X33
= 5, X34 =
All the values are non-zero, so the solution is non-degenerate. Exercise 1 (i) Using north west corner method find the initial basic feasible solution to the following transportation problem:
01 02
03 bj
Dl D2 D3
ai
2 1 3 16
6 10
3 9 4 9
5 2 1 3
12
(ii) Using north west corner method find the initial basic feasible solution to the following transportation problem:
9.36
U.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
8 9
3 (iii) Find the initial basic feasible solution of the following transportation problem by North-West corner m~thod:
1~ ~ ~ 1
4
6
(iv) Using North-West corner method, find the initial basic feasible solution to the following transportation problem:
20 10 1500. 15 25 500 300 400 (v) Using North-West corner method, find the initial basic feasible solution to the following transportation problem: 8 9
3 (vi) Using north west corner method find the initial basic feasible solution to the following transportation problem:
8 5 5 (vii) Find the initial basic feasible solution by North-West corner method for the following transportation problem: 8 9 2
4 3
14
4 1
6
5
6
10
15
6
12
CR.!): LTNEAR PROGRAMMING PROBLEM
9.37
(viii) Find an initial B.F.S. of the following T.P. by North-'West corner rule: 8 3 11 50
7 8 3 80
3 9 5 80
60 70 80
[Ans. (i) Xu = 6, X21 = 3, X22 = 7, X32 = 9, X33 = 3, (ii) Xu == 5, X12 = 3, X13 = 2, X23 = 5, (iii) Xu = 5, X12 = 3, X22 = 1, X23 = 6, (iv) Xu = 300, X12 = 200, X22 = 100, X23 = 400, (v) Xu = 6, X12 = 3, X13 = 1, X23 = 5,
= 6, X12 = 4, X22 = 1, X23 = 4, (vii) Xu = 6, X12 = 8, X22 = 2, X23 = = 5, (viii) Xu = 50, X12 = 10, X22 = 70, X33 = 80.]
(vi) Xu 10, X33
9.6
Assignment Problem
Def. 1 Assignment Problem: The assignment problem is a particular type of transportation problem where all the origins (or jobs) are to be assigned to an equal number of destinations (or machines) in one-to-one basis such that the assignment cost is minimum (or maximum).
Q. 1 Define an assignment problem. Ex. 1 State why an assignment problem is not a LPP ? • SOLUTION: Let Cij is the cost of assigning the ith job to the jth facility. The assignment problem is Determine Xij ~ 0; i, j = 1,2, ... ,m which optimizes the total cost m
Z
=L
m
m
L
CijXij
subject to
i=l j=l
L j=l
1,2, ... ,m. The required X,.
~J
=
{I, 0,
Xij ~
m
Xij
= 1, i =
1,2, ... , m and
L
Xij
= 1,
=
0 in the problem has explicit form
if the ith job be assigned to the jth facility otherwise
and as such the assignment problem is not a L.P.P. as the variables assume only 0 and 1.
Ex. 2
j
i=l
Xij
can
(i) The following assignment problem is a maximization problem
9.38
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS) convert it to a minimization problem.
(ii) Solve the following assignment problem: Salesman A B C D
I 0 9 23 9
II III IV territory 14 20 0 12
9 0 3 14
3 22 0 0
(iii) Solve the following assignment problem:
I A B C
6 16 19
II III 14 2 10
17 12 9
(iv) Solve the following assignment problem:
I
•
A
11
B C
22 29
II III 23 25 13
16 19 27
SOLUTION:
(i) To convert the given problem a minimization problem we subtract all the elements from a large number 9. The reduce problem is:
(ii) Here each row and each column contains at least one zero. Now, we draw minimum number of horizontal and vertical lines to cover all the zeros.
I
II
A
q
1~
B
~
,
C 2~ D
~
III IV
~
29
~ 22
1~ 1~
q
q q 3 q
Cf-I.!): LINEAR PROGRAMMING PROBLEM
9.39
Here the number of lines is equal to the order of the mn.trix. For optimal assignment we consider only the zero elements.
I A
B
II
III
IV
~--+----r--~--~
C
D
r---~--r---4-~~
Hence the optimal assignment is A --+ I, B --+ II I, C --+ I I, D --+ IV and the minimum cost is 0 + 0 + 0 + 0 = 0 units. (iii) We subtract the minimum element of each row from all elements of the corresponding row. The reduce cost matrix becomes:
I 0
II
III
A
8
11
B C
14 10
o
10
1
o
We draw the minimum number of lines to cover all the zeros.
ABC I
..... (}..........-8.. .···11·· ..
II ···14···· ·····0 ..... ··-lB····
III ···10···· ·····1····· ·····0······
Here the number of lines is equal to the order of the matrix. For optimal assignment we consider only the zero elements.
I
II
III
~~ Hence the optimal assignment is A --+ I, B --+ I I, C --+ I I I and the minimum cost is 6+2+9=17 units. (iv) We subtract the minimum element of each row from all elements of the corresponding row. The reduce cost matrix becomes:
U.G.
9.40
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
A
I 0
II 12
III 5
B C
3
6
0
16
0
14
We draw the minimum number of lines to cover all the zeros.
ABC I
····0···· ····12········-5······
II ·····3···········6···..······0······ III ···t6····· ·····0······ ···14····
Here the number of lines is equal to the order of the matrix. For optimal assignment we consider only the zero elements. I
II
III
H~ Hence the optimal assignment is A -+ I, B -+ Ill, C -+ I I and the minimum cost is 11+19+13=43 units.
Exercise 1 lem
(i) The following assignment problem is a maximization probI 16 6 11
A B C
II 25 22 22
III 35 2 19
convert it to a minimization problem. (ii) Solve the following assignment problem: I
A B C D
0
9 3 5
II 1 6
III 3
0
2 1
2
0
(iii) Solve the following assignment problem:
IV 2 3 1 0
CH.9: LINEAR PROGRAMMING PROBLEM I
A B C
II
III
16 22 26 2 1 19
23 19
7
(iv) Solve the following assignment problem: I
A B C
1 2 2
II
III
3 5 3
11
1 10
9.41
(h.lD II Dynamics of Particles 10.1
Principle of Dynamics
Formula 1
(i) If x is the displacement of a particle at time t then its
velocity (v)=
~:.
(ii) If v be the velocity of a particle at time t then its acceleration
(f)
=
dv dt
=
d 2x dt2
dv = v dx '
where x is the displacement of the particle at time t. (iii) If a particle is moving in a plane with displacement in x and y directions are respectively x and y then its velocity is dX)2 ( dt
+ (dY)2 dt
and its acceleration is
Ex. 1
(i) The law of motion of a body moving along a straight line is x = !vt; prove that its acceleration is constant.
(ii) The velocity of a particle moving in a straight line is given by the relation v 2 = ax 2 + c, (a i- 0). Prove that its acceleration varies as the distance from the origin in the line. (iii) If a particle moves in a straight line according to the law x = 4 sin( 2t +a), prove that v 2 = 4(16 - x 2 ), where v and x are respectively the velocity and distance traversed in time t. (iv) A particle moves along a straight line so that after t sees, its distance 8 from fixed point 0 on the line is given by 8 = (t - 1)2(t - 2). Find its distance from 0 when its velocity is zero. (v) A particle moves along a straight line according to the law 4t + 3. Prove that the' acceleration varies as 1/8 3.
8
2
= 6t 2
+
(vi) The displacement x at time t of a particle moving along the x-axis is given by t = ax 2 + bx, where a, b are constants .. Express the velocity v of the particle at any moment as a function of x and prove that the retardation of the particle is 2av 3 . (vii) A particle moves in a straight line according to the law v 2 = 6a(xsinx+ cos x), where x is the distance from a fixed point on the line at any instant and v is the velocity there. Find its acceleration.
U.G.
10.2
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(viii) If a particle moves in a straight line according to the law x = a sin(j.£t +c), prove that v 2 = (a 2 - x 2 )j.£2, where v and x are respectively the velocity and distance traversed in time t. (ix) The coordinates of moving point at time t are x = a(2t a (1 - cos 2t). Prove that its acceleration is constant.
+ sin2t), y
=
(x) The acceleration of a particle starting from rest and moving along a line is given by the relation f = 15 - 0.4t 2 . Determine the velocity and distance described after time t. (xi) The acceleration of a particle t secs. after it starts from rest is ~e-2at ftj(sec)2. Find the maximum velocity. (xii) If the position of a moving point is given by x = acost and y = bsin2t, find the path. (xiii) A wheel makes 400 revolution per hour. What is its angular velocity per second? •
SOLUTION:
1 (i) H ere x = '2vt. Differentiating w.r.t t, we have dx 1 dt = '2 [v
dv
+ t dt 1or,
1 1 dv -v = -t2 2 dt dv dv dt or, v = t dt or, -; = t'
Integrating we get log v = log t + log c or, log(vjc) = logt or, v = ct . dv dt = c =constant. Thus the acceleration is constant. (ii) Here v 2 = ax 2 + c. Differentiating w.r.t x we get dv dv 2v dx = 2ax or, v dx
<X X.
the acceleratioll is proportional to distance.
CIi.!o:
DYNAMIC~ OF PARTICLES
(iii) We have x = 4sin(2t + a). . dx .. dt = 8cos(2t + a) or, v 2 = 64cos 2(2t + a) = 64{1 - sin2(2t + a)} or, v 2 = 64(1 - x 2/16) = 4(16 - x 2). (iv) Here s = (t - 1)2(t - 2). ds 2 ... dt = 2(t - 1)(t - 2) + (t - 1) = (t - 1)(3t - 5). ds Hence dt = 0 when t = 1 and t = 5/3. When t = s = (5/3 When t = Hence the
5/3, 1)2(5/3 - 2) = 4/9( -1/3) = -4/27. 1, s = o. required distances from 0 are 4/27 and O. (v) Given s2 = 6t 2 + 4t + 3. Differentiating w.r.t t we get ds ds 6t + 2 2 1/2 2s- =12t+4,or,-=--=(6t+2)(6t +4t+3)- . dt dt s Again differentiating w.r.t t, we obtain
d2 s
-
dt 2
= 6(6t 2
+ 4t + 3)-1/2
+(6t + 2)( -1/2)(6t2 + 4t + 3)-3/2(12t + 4) 6(6t 2 + 4t + 3) - (6t + 2)2 18 - 4 14 (6t 2 + 4t + 3)3/2
= (s2)3/2 -
-,;3.
·· · as "3' 1 Hence acce IeratlOn IS vanes s
(vi) Here t
=
ax 2 + bx.
'a .. DIllerentlatmg w.r.t x we get -dt = 2ax + b
dx 1 . d' dx or, v = - = , is the velocIty at any Istance x. dt 2ax + b Again differentiating w.r.t x we get dv -dx =-
2a dv or v (2ax + b)2 , dx
=-
2a (2ax + b)3
= -2av 3 .
Hence the retardation of the particle is 2av 3. (vii) Given v 2 = 6a(xsinx + cos x). Differentiating w.r.t x we obtain 2v ~~
= 6a(sinx + xcosx - sinx) = 6axcosx.
Hence the acceleration is v ~~ = 3ax cos x.
10.3
V.G.
10.4
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(viii) We have x = asinC{Lt+e). Differentiating w.r.t t we get dx v = dt = a{Lcos({Lt + e). or, v 2 = a 2{L2 cos 2({Lt + e) = a2{L2{1 - sin2({Lt + e)} = a2{L2(1 - x 21a 2) = {L2(a 2 - x 2). (ix) Here x = a(2t + sin2t), y = a(l - cos 2t). Differentiating w.r.t t, we obtain x = a(2 + 2 cos 2t) and iJ = 2a sin 2t. Again differentiating w.r.t t we get fi = -4a sin 2t and jj = 4a cos 2t. Hence the acceleration is Vfi 2 + jj2 = V16a 2 sin2 2t + 16a 2 cos 2 2t = 4a, which is constant.
d2 x 2 (x) Here f = - 2 = 15 - O.4t . dt Integrating w.r. t t we get dx = 15t _ 0.4 t 3 + c dt 3 Since v = 0, t = 0, c = O. . . 0.4 3 the velocIty IS 11 = 15t - Tt . Again integrating we get t2
0.4 4 2 12 Initially, x = 0, t = 0, so d = O. Hence t h e reqmre . d d'Istance x = 2"t 15 2 - 12t 0.4 4 . x = 15- - - t +d
d2 x 1 (xi) Given - 2 = _e- 2at . dt 2
Integrating
.. 11 I mba y, t
~;
=
_~e-2at + c.
dx = 0'dt =.O'
c
= 4"1
dx 1 2 t Hence dt = 4'(1 - e- a ) ftjsec.
~;
will be maximum when e- 2at is zero.
When t -+ 00, e- 2af -+ O. Hence the maximum velocity is
i ft/sec.
(xii) We have x = a cos t, y = b sin 2t. Now, y = bsin2t = 2bcos tsint = 2b(xja)V1- cos 2 t
CH.10 =
:)YNAMICS OF PARTICLES
10,5
2xb./ -;;:v 1- x2 /a 2
2
= 4b ;2 (1 _ x2 /a 2 ) which is the required path. a (xiii) The wheel moves 400 x 211" radian per hour. So its angular velocity is 400 x 211" 60 x 60 rad/sec= rad/sec=40 degree/sec. or,
y2
2;
Exercise 1 (i) The distance of a particle moving along a straight line from a fixed point at time t is given by 8 2 = at 2 + 2bt + c, where a, b, care constants, prove that the acceleration ex: ~. (ii) The velocity of a particle moving in a straight line is given by the relation v 2 = ax 2 + 2bx + c. Prove that its acceleration varies as the distance from a fixed point in the plane. (iii) A particle moves along a straight line according to the law 8 2 = t 2 +2t+3. Show that its acceleration varies inversely as s3. (iv) The velocity v of a particle moving along a straight line given by the relation v 2 = as 2 + b where 8 is the distance travelled from a fixed point on the line and a, b are constants. Prove that its acceleration varies as the distance from the fixed point on the line. (v) A wheel makes 200 revolution per hour. What is its angular velocity in radians per second? (vi) The distance in ems. described by a particle in t sees., is given by 2 8 = at + bt + c. Find the velocity at the end of 3 sees. and acceleration at the end of 8 sees. [Ans. (v) 11"/9 rad/sec., (vi) (6a
10.2
+ b)
em/sec, 2a cm/sec2 .]
Work, Power and Energy
Def. 1 Work: If the force is constant, the work done by it is defined as the product of the magnitude of the force and the distance through which the point of application of the force moves in the direction of the force. That is, Work = Force x displacement along the action of the force.
Q, 1 Define work. Def. 2 Power: Power is the rate of work done. That is, Power= ~s = Fv, where F is the force, s is displacement, t is the time duration and v is the uniform velocity.
10.6
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Q. 2 Define power. Def. 3 Energy: Energy is defined as the capacity of doing work by a man or machine. The energy are mainly classified as kinetic energy (K.E.) and potential energy (P.E.). Kinetic Energy: Kinetic energy of a body is the capacity of its doing work by virtue of its motion and is measured by the amount of work that the applied force will do in bringing the body to rest. The K.E. of a particle of mass m moving with velocity v at time t is ~mv2. Potential Energy: Potential energy of a body is its capacity for doing work by virtue of its position and is measured by the amount of work that the forces acting on the body will do when the body moves from its present position to some fixed position. The P.E. of a particle of mass m placed at a height h above the ground is mgh.
Q. 3 Define energy, kinetic energy and potential energy. Ex. 1 State the principle of conservation of energy. • SOLUTION: The change in kinetic energy of a body is equal to the work done by the acting force. Ex. 2 (i) How much work is done in rising a mass of 2 kg. to a height of 5 meters? (ii) The velocities of a particle of mass m moving with uniform acceleration at two instances are VI ft/sec and V2 ft/sec. Find the work done by the force during this period.
(iii) For uniform acceleration motion of a particle of unit mass its velocities at two instances are 3ft/sec and 5ft/sec. Find the work done by its force during this interval time. (iv) What is the work by gravity second of its fall ?
on-a_~tone
of mass mlb during the t-th
(v) What is the work by gravity on a stone of mass 70 gms, during the 10 th second of its fall ? • SOLUTION: (i) Weight of the mass = 2 x 1000 x 980 dynes and displacement = 5 x 100 c.m . ... work done = 2 x 1000 x 980 x 5 x 100 ergs = 98 x 107 ergs = 98 joules.
CH.10:
10.7
DYNAMICS OF PARTICLES
(ii) Let VI and V2 be the velocities of a particle at the points A (at a distance a) and B (at a distance b). The work done by the force F is b b r r dv r Ja F ds = Ja mv ds ds = J
2
mv dv
VI
(iii) Here m = 1, VI = 3 ft/sec and V2 . the work done
1
1
= "2mv~ - "2mv~
poundals.
= 5 ft/sec.
1212122
= "2mV2
- "2mVl = "2 m (V2 - VI)
= !.1.(25 =
9)
= 8 poundals = 382 ft-Ibs
i ft-Ibs.
(iv) The displacement after t second is s time
=
1
"2gt2 and the work done at this
= mg.~gt2 = ~m(gt)2.
The velocity at (t - 1)th and tth secs are respectively
1
1
"2mg2 (t - 1)2 and "2m g2 t 2.
.the work done at tth sec is
1 1 2 "2m g2 {t - (t - 1)2} = "2mg2(2t -
1) poundals.
(v) We know that the work done by a mass m at tth sec. is Here m
1).
= 70 gm, 9 = 980 cm/sec2 , t = 10 sec.
the work done is
Ex. 3
~mg2(2t -
!
x 70 x 9802 (2.10 - 1) = 6.39
X
108 erg.
(i) Prove that the K.E. of a body of mass m and moving with velocity
v is ~mv2.
(ii) Prove that the change in kinetic energy of a body of mass m is equal to the work done by acting forces. (iii) A particle is falling freely under gravity, show that the sum of K.E. and the P.E. is constant during the motion. (iv) Show that the rate of change of K.E. of a particle is equal to its power. (v) At what height would the K.E. of a body falling from a height h be equal to half its P.E. ?
V.G.
10.8 •
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
SOLUTION:
(i) Let a particle of mass m moving along the line OX. Also let V be the velocity at A (OA = a) on OX., A force F acting along X 0 opposite to the direction of motion brings the particle to rest at B (0 B = b). Let P be any point on OX (OP = x).
a
1__---_+
X
0
A B Figure: 10.2.1
The velocity of the particle at P be v. Then F = -mv dv, dx Hence the total work done by the force to brings the particle from A to B is dv Fdx = _ mv dx = _ mvdv = !mV 2. Ja Ja dx Jv 2
b r
rb
Thus the KE. is
rO
!mV 2 .
(ii) Let u and v be the velocities of a particle of mass m at A (at a distance a) and at B (at a distance b). The total work done by the force F is
b lb l a
Fds =
a
dv ds = mVdS
l
u
v
2 - -mu 2. 1 1 mvdv = -mv 2
2
Thus the change in KE. is equal to the work done by the force. (iii) Let a particle of mass m falling freely under gravity from a point A at a height h above the earth's surface. Let P be a point at height x from A. If v be x the velocity at P then v 2 = 2gx. The KE. at P = ~mv2 = ~m.2gx = mgx. h The P.E. at P = mg(h - x). The sum of KE. and P.E. is mgx + mg( h - x) = mgh, which is independent of x. Hence sum of KE. and P.E. is constant throughFigure: 10.2.2 out the motion. (iv) If the acting force force be constant then the rate of change of KE. is
~ (!mv2) dt 2
=
~ (!mv2) ds
ds 2 dv = mv ds v
dt
= Fv
dv (since mv ds
= F)
= Power
Hence the rate of change of KE. is equal to its power.
CH.10: DYNAMICS OF PARTICLES
10.9
(v) Let h be the total height and v be the velocity at a height x (from bottom) where the KE. is equal to half the P.E. ... v 2 = 2g( h - x). 121 The KE. is "2mv = "2m.2g(h - x) = mg(h - x) and the P.E. is mgx. . 1 1 Smce KE.= "2 P.E., mg(h - x) = "2 mgx
or, h =
3
"2 x
or, x =
2
"3 h.
Ex. 4 (i) Show that the H.P. of an engine which can project vertically upwards 10,000 lb of water per minute with a velocity of 80 ftjsec is 30~~ . (ii) Find the horse power of an engine which draws a train at a uniform rate of v ftjsec against a resistance of P Ib-wt . •
SOLUTION:
(i) Work done per minute =
~mv2 = ~
x 10000
X
(80)2 poundals
= 10000 x 80 x 80 ft.-lbs. 2 x 32 10000 x 80 x 80 Power = 2 x 32 x 60 ft.-lbsjsec. . 10000 x 80 x 80 1000 10 H.P. of the eng me = 2 x 32 x 60 x 550 = ~ = 30 33 . (ii) Resistance = P lb-wt. Power = Pv ft.-lbs.
H.P. of the engine is
:s~.
Exercise 1 (i) Find the H.P of an engine which draws a train at a uniform rate of 45 milesjhr. against a resistance of 900 lb-wt. (ii) Find the horse power of an engine which draws a train at a uniform rate of 60 mjh against a resistance of 4480 lb-wt. (iii) What is the work by gravity on a stone of mass 80 gms, during the 8 th second of its fall ? (iv) Find the horse power of a pumping machine which can project 10,000 lbs. of water per minute with the speed of 44 ft. per sec. [Ans. (i) 108, (ii) 716.8, (iii) 5.76
r
lOB erg, (iv)
91.]
10.10 _
10.3
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Impulse and Impulsive Forces
Def. 1 Impulse: If the force be constant in magnitude and in direction then the impulse of the force is the product of the force and the time during which it acts. Impulse=Ft = mft = m(v - u) [since v = u + ft] . '. Impulse=change in momentum.
Q. 1 Define impulse. Q. 2 From the definition of impulse, show that impulse is equal to the change of momentum. Def. 2 Impulsive Forces: An impulsive force is a very large force which acting on a body for an infinitesimal small time produces in it a finite change of momentum but the displacement of the body during this time is negligibl~ small. The momentum equation in this case can be written as (Pr)T-+o = m(v - U) Here r is infinitesimally small while the force P is very large and they cannot be measured separately but the change of momentum can be measured experimentally.
Q. 3 Define an impulsive force and explain how it is measured.
Ex. 1 State the principle of conservation of linear momentum. • SOLUTION: If the sum of the external forces acting on a system of particles be zero in any direction then the total momentum of the system in that direction remains constant during the motion.
Ex. 2 (i) A force acting on a body of mass 10lb changes its velocity from 45 miles/hr. to 60 miles/hr. Find the impulse of the force. (ii) A cricket ball moving with a velocity 25 m/s is struck by a bat which causes it to move in the same straight line but in opposite direction with a velocity of 15 m/s. If the impulse generated by the bat be 600 dyne-sec, find the mass of the ball. (iii) A cricket ball of 5~ oz. has a velcity 60 ft./sec., before the batman strikes it. It is hit back in the same direction with a velocity of 80 ft./sec. What impulse has the ball received? (iv) Two spheres of masses 6 and 4 oz. collide in the line joining their centres with velocities 25 and 35 ft./sec. respectively in the same sense; if they do not separate after collision, find their common velocity.
CH.10: DYNAMICS OF PARTICLES
10.11
(v) An inelastic particle of mass 20 lbs. impinges on another particle of mass 2 lbs. coming from the opposite direction with an equal veloclty. If the common velocity after impact be 9 ft./sec., find their velocities before impact. (vi) A gun fires a shot weighing 800 lbs with a muzzle velocity of 2,000 ft/sec. The recoil of the gun is checked by a constant .force in ~ th of a second. Find the force. (vii) For rectilinear motion of a particle if an impulse I change its velocity from u to v and E is the change in K.E., show that E = I(ut V ) • •
SOLUTION:
(i) Here m = 10 lbs., u = 45 miles/hr., v = 60 miles/hr. . 1760 x 3 lmpulse = m(v - u) = 10 x (60 - 45) x -6-0-x-6-0 = 220 poundals sec.
(ii) Here u = 15 m/s = 1500 c.m./s, v Impulse I = 600 dyne-sec.
= 25 m/s = 2500 c.m./s.
Since 1= m(v - u), 600 = m(2500 + 1500) 600 3 or, m = 4000 = 20 gm. 51 11 (iii) Here m = 1~ = 32 lbs. u = 60 ft/sec, v = 80 ft/sec. impulse
= m(v =
u)
11 32 x 140
= ~~ (80 + 60) .
= 14.13 poundals-sec.
(iv) Let V be the common velocity of the two spheres after impact. From conservation of linear momentum, we know momentum before impact = momentum after impact or, 6 x 25 + 4 x 35 = (6 + 4) V or, 10V = 290 or, V = 29 ft/sec. (v) Let v be the velocity before impact. From conservation of linear momentum, momentum before impact = momentum after impact or, 20v - 2v = (20 + 2) x 9 or, 18v = 22 x 9 or, v = 11 ft/sec. (vi) Let m be the mass of the gun and the recoil velocity of the gun be v. Then Mv = 800 x 2000.
10.12
V.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Since the recoil is checked in . .. 0
~th sec
1
= v - 5 x f or, f = 5v, where f is the retardation induced by the force
P. . '. P = mf = 5vm = 5 x 800 x 2000 poundals=111.6 tons-wt.
(vii) Impulse 1= m(v - u) and change in K.E. (E) is 1 2 1 2 1 1 E = 2mv - 2mu = 2m(v - u)(v + u) = 2I(v + u).
Exercise 1 (i) A force acting on a body of mass 5lb changes its velocity from 30 miles/hr. to 45 miles/hr. Find the impulse of the force. (ii) A force acting on a body of mass 10lb changes its velocity from 60 miles/hr. to 90 miles/hr. Find the impulse of the force. (iii) A force acting on a body of mass 5lb changes its velocity from 60 miles/hr. to 90 miles/hr. Find the impulse of the force. [Ans. (i) 110 poundal, (ii) 440 poundal, (iii) 220 poundal.]
10.4
Collision of Elastic Bodies
Note 1 (i) The impact is said to be direct for two bodies if the relative direction of motion of each other, just before impact, be along the common normal called the line of impact, at the point of contact. Otherwise the impact is said to be oblique.
(ii) Newton's Experimental Law: If UI and
U2 be the component velocities of two bodies of mass mI and m2 respectively before impact along their common normal and VI and V2 be the velocities in the same direction after impact, then VI - V2 = -e(ul - U2). The constant e is called the coefficient of elasticity or restitution.
(iii) If e = 0 the bodies are said to be inelastic. (iv) If e = 1 the bodies are said to be perfectly elastic. (v) If u and v be the velocities before and after direct impact of a particle on a fixed horizontal plane then v = eu. (vi) If u is the initial velocity of a particle then the velocities after second, . third, fourth etc. direct impact on a fixed horizontal plane are respectively e2 u, e3 u, e4 u.
CH.10: DYNAMICS OF PARTICLES
10.13
(vii) If a particle falls from a height h above the plane then the height H to which the particle moves up after first rebound is given by 2
H
= -v2g = e2 h.
(viii) If a particle falls from a height h above the plane then the height after second, third, fourth etc. impact are respectively e4 h, e6 h, e8 h. Ex. 1 (i) A sphere falling from a height h impinges directly on a floor and rises to a height hI' Show that hI = e2 h where e is the coefficient of restitution. (ii) A ball is dropped vertically on a fixed horizontal plane from a height of 100 ft. If the coefficient of restitution be 0.4, find the height of the first rebound. (iii) From what height must a heavy elastic ball be dropped on a floor so that after rebounding once it will reach a height of 25 metres? (e = 0.5). (iv) A heavy elastic ball is dropped on a smooth floor from a height hand after rebounding thrice, it rises to a height of 64 cm. Find, h if e = 0.5. (v) A ball falls from a height of 64 ft on a fixed horizontal floor and rises 4 ft after two rebounds. Find the coefficient of elasticity. (vi) A perfectly elastic sphere falling from a height h impinges on a fixed horizontal plane and rebo.unds to a height hI after impact. Show that hI = h.
(vii) When two equal perfectly elastic balls come into collision, prove, that their velocities are interchanged after collision. ' (viii) A sphere of mass 1 lb. moving with a velocity of 2 ft./sec. impinges directly on a sphere of mass 2 lbs. at rest. If the first sphere comes to rest after the impact, find the velocity of the second sphere and the coefficient of restitution. (ix) Two balls of masses 2 lbs. and 3 lbs. are moving with velocities 6 ft./sec. and 3 ft.fsec. respectively in the same direction along the same straight line and collide with one another. Find the K.E. lost by impact if the coefficient of restitution be 2/3 . •
SOLUTION:
(i) Let u be the velocity of the sphere with which it strikes the plane. ,', u 2 = 2gh. The velocity after first rebound = eu = ev'29h . . '. the height of the first rebound
10.14
U.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
2 hI = (eu)2 = e (2gh) = e2h. 2g 2g (ii) Here e = 0.4 and initial height h = 100 ft. Then the height of the first rebound is e2 h = (0.4)2 X 100 ft.=16 ft. (iii) Here e = 0.5 and height after first rebound is H = 25 meters. Let h be the initial height. Then H = e2h. 25 or, 25 = (0.5)2h or h = - ( )2 = 100 metres. 0.5 (iv) The height of the third rebound is e 6 h = 64 c.m. (given) and e = 0.5. 64 64 . h = 6 = -( )6 = 64 x 64 c.m.= 4096 c.m. e 0.5 (v) Here h = 64 ft, H
=
4 ft.
We know H = e4 h or e4 = H = ..! = ~ or e = ~ , h 64 16' 2 (vi) Let u be the velocity of the sphere with which it strikes the plane . ... u 2 = 2gh. Since the sphere is perfectly elastic, e = 1. The velocity after first rebound = eu = u = y'2gh. u2 2gh the height of the first rebound hI = 2g = 2g = h. (vii) Let UI, U2 be the velocities before impact and VI, v2 that of after impact. The balls are perfectly elastic, e = 1. Then by Newton's experimental law V2 - VI = -e(u2 - UI) (1) From the principle of conservation of linear momentum, mV2 + mVI = mU2 + mUI where m is the mass of the balls. or, Solving (1) and (2) we get V2 =
UI
(2)
and VI = U2.
Hence the velocities are interchanged. (viii) Let V be the velocity' of the second sphere after impact. By principle of conservation of momentum we have 1 x 2 + 2 x 0 = 1 x 0 + 2v or, 2v = 2 or, V = 1 ft/sec. Again, from Newton's experimental law, V - 0 = -e(O - 2) or, 2e = V or, 2e = lor, e = 1/2. (ix) Let VI and V2 be the velocities of the two balls after impact then by the principle of conservation of momentum, 2VI + 3V2 = 2 x 6 + 3 x 3 or, 2VI + 3V2 = 21 (1)
CH.10:
DYNAMICS OF PARTICLES
10.15
From Newton's experimental law, 2
"3 (6 - 3) or,
V2 - VI =
V2 - VI =
2
(2)
Solving (1) and (2), we have VI = 3 ftlsec and V2 = 5 ft/sec. . . . 1 1 99 Imtlal KE. = 2' x 2 X 62 + 2' x 3 X 32 = 2 ft. poundals. . 1 Fmal KE. = 2' x 2
loss of KE.
X
99
32
1
+ 2'
x 3
X
. 93 52 = 2 ft. poundals
93
= 2 - 2 = 3 ft. poundals.
Exercise 1 (i) From what height must a heavy elastic ball be dropped on a floor so that after rebounding once it will reach a height of 16 metres? (e = 0.5). (ii) A heavy elastic sphere falls from a height h on a smooth ground and after rebounding three times on the floor it comes to a height 49 cm. If e = 0.5 determine the height h. (iii) An elastic ball falls from a height of 27ft on a fixed horizontal floor and rises 3ft after two rebounds. Find the coefficient of elasticity. (iv) A. ball is dropped vertically on a fixed horizontal plane from a height of 100 ft. If the coefficient of restitution be 0.4, find the height of the second rebound. [Ans. (i) 64 m, (ii) 3136 cm, (iii)
1/-13,
(iv) 2.56 ft.]
Ex. 2 A nonelastic particle of mass M kg moving with a velocity of V metres per sec impinges, on another particle of mass m kg at rest. Find the subsequent velocity. Since the bodies are inelastic, so, after impact the bodies adhere together. This single body of mass (M + m) moves with a velocity V (say) after impact. As there is no loss of momentum, •
SOLUTION:
(M + m)V = Mv, or, V = bodies after impact.
10.5
M
M+m
V
which gives the common velocity of the
Motion in a Straight Line
Ex. 1 (i) Write done the equations of motion of a particle moving under gravity in a ver~ical plane. Hence show that horizontal velocity is constant.
V.G.
10.16
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(ii) A stone is dropped from a certain height and is observed to fall the last h cm in t sec. Show that the total time of fall is
(!2 + ~) gt
secs.
(iii) A motor car driven with a constant force at all speeds meets with air resistance proportional to the square of the velocity. If u be the maximum speed of the car, show that its acceleration when running at a speed v varies as (u 2 - v 2 ) . •
SOLUTION:
(i) Let m be the mass of the particle. The equations of motion in vertical and horizontal directions are
my = mg and x = O. Integrating x = 0 we get x = constant. Hence the horizontal velocity is constant. (ii) Let A be the starting position and B be the point at height h from the ground (0). Let AB = Hand BO = h. The velocity v at B is v 2 = 2gH or v = J2gH. Let T be the time taken by the stone to reach at B .
~gT2 or ' T'=
. H = .. 2
J2H. 9
Again during the motion from B to 0, h = vt + !gt 2 1 1 or, h = J2gHt + 29t2 or, J2gH = (h - 29t2)/t. total time
= t + T = t + J2H - = t + h _lgt 2 9
h 1 t -t=:; gt 2 2 (iii) The equation of motion is = t
+- -
2
gt
h
+ -gt
sec.
dv = F _ kv2 dt 1 where F is the constant force.
For maximum speed, Thus (1) becomes
~~
(1)
~: = 0 and v = u. 2
2
= k(u - v ).
Hence the acceleration varies as u 2
-
v2 .
Then F
= ku 2 •
CH.10: DYNAMICS OF PARTICLES
10.6
10.17
Simple Harmonic Motion
Note 1 The general equation of simple harmonic motion (S.H.M.) is x +w2 x = O. Its general solution is x = acos(wt + c) or x = asin(wt + c). 271" Where amplitude = a, periodic time = - , epoch=c, frequency of oscillation w
(the number of complete oscillations per unit time) = .!:!.... 271" Ex. 1 (i) The displacement of a particle at any time t of its motion in a straight line is given by x = a cos kt + b sin kt. Show that the motion is simple harmonic and find its time period. (ii) The speed v of a point moving along the x-axis is given by v 2 = b2 - x 2.
Prove that the motion is simple harmonic and find its amplitude. (iii) The speed v of a particle moving along the axis of x is given by v 2 = 64(8mx - x 2 - 12m 2). Show that the motion is simple harmonic with centre at x,= 4m and that the period of oscillation is ~. (iv) A particle describe a simple harmonic motion in a line with 2cms as its amplitude. Its velocity when passing through the centre of oscillation is 12 ems/sec. Find its period . •
SOLUTION:
(i) Given x = acoskt + bsinkt. Differentiating w.r.t t, x = -ak sin kt + bkcoskt Again differentiating w.r.t t, x = -ak 2 cos kt - bk2 sin kt = -k2x. or x + k 2 x = 0 This shows that the motion is simple harmonic with periodic time 271"/k. (ii) Given v 2 = b2 - x 2. .. dv = - 2x ·er Dl11erentlatmg w.r.t x, we get 2v dx or, x + x = O. This shows that the motion is simple harmonic. When v = 0 then x == ±b. Therefore the amplitude is b. (iii) We have v 2 = 64(8xm - x 2 - 12m2) = 64{4m 2 - (x - 4m)2}.
Differentiating w.r.t x, 2v ~~ = -128(x - 4m) or, x = -64(x - 4m). This shows that the motion is simple harmonic about the line x = 4m. .' . d· 271" 71" The tlme peno IS 8 = '4.
10.18
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iv) If a be the amplitude then v 2 = w 2(a 2 - x 2). Here a = 2.. '. v 2 = w 2 (4 - x 2 ). It is given that, when the particle passes through the centre of oscillation (x = 0) then its velocity is 12 cm/sec. 2 . . . 144 = 4w or , 'U'2 = 36.
. . . . 27r 27r 7r Hence the perIod of oscIllatIOn IS - = - = -. w 6 3
Ex. 2 Write down the equation of motion for damped simple harmonic motion where the damping force is proportional to the velocity.
• SOLUTION: The component of the damped force in the positive direction of the x-axis may be written as -2mj1x, where j1 is a positive constant. Then equation of motion is mx = -mw 2 x - 2mf.Lx, w is proportional constant, or, x + 2f.Lx + w 2 x = O. Ex. 3 A particle moving in a straight line is executing S.H.M. what is the effect of supper-imposition of another S.H.M. of equal time period on it?
• SOLUTION: Let the displacement of the two motions be given by Xl = al cos( wt + cd and X2 = a2 cos( wt + c2). Then the resultant displacement is x = Xl + X2 = al cos(wt + cl) + a2 cos(wt + c2) = cos wt(al cos Cl + a2 cos c2) - sin wt(al sin Cl + a2 sin c2) = A{coswtcosE - sinwtsinE) = Acos(wt + E), where A cos E = al cos C1 + a2 cos C2 and A sin E = al sin Cl + a2 sin C2. Thus the resultant of two S.H.M of the same period is a similar motion of same period whose amplitude and epoch are given by A and E respectively. Exercise 1 (i) The speed v of a point moving along the x-axis is given by v = J 4 - x 2 . Prove that the motion is simple harmonic and find its amplitude.
(ii) The displacement of a particle at any time t of its motion in a straight line is given by, x = cos kt + sin kt. Show that the motion is simple harmonic and find its amplitude. (iii) The distance x of a particle moving in a straight line from a point in time t is given by x = 3 sin t + 4 cos t. Show that the motion of the particle is simple harmonic. (iv) The speed v of a point moving along the x-axis is given by v 2 Prove that the motion is simple harmonic.
= 16 -- x 2 .
CILLO: DYNAM"
10.19
OF PARTICLES
j
(v) If a particle moves in a straight line and its velocity v at a distance x from the origin is given 1)2 = 2 - x - x 2 , show that the motion is an S.H.M. [Ans. (i) 2, (ii) v'2.]
10.7
Motion in a Plane
Motion in Cartesian Coordinates Note 1
. 1 acce1eratIOn . IS . dv (1') The tangentla dt' 2
(ii) The normal acceleration is ~, where p is the radius of curvature. p
Ex. 1 (i) If the tangential and normal acceleration of a particle moving in a plane curve are equal, find the expression for velocity. (ii) A particle describes a circle of radius a with unifor' speed v. Show that at any instant the tangential acceleration is zero and the normal , . . v2 acceleratIOn IS - . a
(iii) The position of a moving particle at time t is given by x = a cos nt, y = a sin nt. Find the path of the particle, its velocity and acceleration. (iv) The velocities of a point parallel to the axes of x and yare (u + wy) and (v + w' x) respectively where u, v, wand w' are constant. Show that the path is a conic section. (v) A particle is moving with a constant velocity parallel to Y-axis and a velocity proportional to y parallel to X -axis. Prove that it will describe a parabola. (vi) A particle starts from the origin and the components of its velocity parallel to the x and y axes at time tare 2t + 3 and 4t respectively. Fi:p.d the path . •
SOLUTION:
2
ds (1') A ccord'mg to the problem v -dv = -v = v 2d'IjJ -d [as p=-] ds
dv or, - =d'IjJ
p
s
d'IjJ
v Integrating we get, log v = logA + 'IjJ or, log(v/A) = 'IjJ Of, V = Ae-rP, where A is arbitrary constant.
V.G.
10.20
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(ii) Since the velocity is constant,
i;
= O. The tangential acceleration is
dv = 0 dt .
We know that the radius of curvature (p) at any point on the circle is equal to its radius (a). v2 v2 the normal acceleration is - = - . p a (iii) Here x = a cos nt, y = a sin nt (1)
x = -an sin nt, iJ = an cos nt x = -an2 cos nt, ii = -an2 sin nt
(2) (3)
Eliminating t in (1) we get x 2 + y2 = a 2 , which is the path of the particle. Velocity = J x2 + y2 = Va 2n 2 = an.
= "/;;2 + i? = Va 2n 4 = an 2. We have x = u + wyand iJ = v + w'x Acceleration
(iv)
•
Dividing we get dy v+w'x = dx u + wy
or, (v
+ w'x)dx =
(u
+ wy)dy
Integrating we get ,x2 y2 e vx + w '2 = uy + w'2 + 2 or, x 2w' - y 2w + 2vx - 2uy = e which is the required path and it represents a conic. (v) Let iJ
= e and x = ky, where e and k are constants.
Dividing we get dy e a - = - = - or, ydy = adx, where a dx ky y Integrating we get y2 A - = ax + - or, y2 = 2ax + A,
2
= elk.
2
which is the path of the particle and it represents a parabola. (vi) Here x = 2t + 3 and iJ = 4t Integrating we get x=t 2 +3t+a
(1)
CH.10: and y
= 2t2 + b or,
From (1), 3t
t2 = 2
=x- t -
Squaring we get 9t 2 = ( or,
DYNAMICf r'F PARTICLES
10.21
y-b 2
--. a
=x-
y-b a - -2-
2x - 2a - y
2
+ b) 2
9(Y - b) _ (2x - y - 2a + b)2 2
-
4
or, 18(y - b) = (2x - Y - 2a + b)2 which is the required path. Ex. 2 (i) Show that there are two different angles of projections for the same range of a projection when the velocity of projection is constant.
(ii) The maximum height attained by a particle is equal to the range. Find the direction of the projection . •
SOLUTION:
(i) The range of a projectile is R =
u 2 sin 20:
, where
.
u and 0: are the veloclty
9
and angle of projection respectively. If u is constant, the value of R remains unchanged if we write 0: for 0:. From this it follows that the same range can be acquired with two angles of projection which are complementary to each other.
i-
(ii) If u and 0: are respectively the velocity and angle of projection then maximum height attained by the particle is
u 2 sin2 0:
u 2 sin 20: and range is R = - - 29 9 2 2 2 u sin 0: u sin 20: If they are equal - - 29 9 2 or, sin 0: = 2.2sinacoso: or, sina = 4cosa H =
or, tana = 4 or, a = tan- 1 4 which is the angle of projection. Exercise 1
(i) The coordinates of a moving point on a plane are given by
x = a(t + sin 2t), Y = a(l + cos 2t), prove that its acceleration is constant in magnitude. (ii) The velocities of a particle at a point (x, y) parallel to the axes of x and yare (2 - 3y) and (3 + 4x) respectively. Find the path of the particle. (iii) The velocities of a particle at a point (x, y) parallel to the axes x and y are - ILY and AX respectively. Find the path of the particle. [Ans. (ii) 4x 2 + 3y2
+ 6x -
4y"= c, (ii) Ax2
+ ILy2
= c.]
10.22
V.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Motion in Polar Coordinates Note 2
' = dr (1') R a d'1a1 ve lOC1ty dt = r..
(ii) Transverse (cross radial) velocity
r-
(iii) Radial acceleration =
re
2
= r ~~ = reo
•
(iv) Transverse (cross radial) acceleration . .. 1 d 2' = 2r(J + r(J = - -d (r (J). r
t
Ex. 3 (i) If the angular velocity of a moving point about a fixed origin be constant, show that the transverse acceleration varies as its radial velocity.
(ii) A particle describes a curve r = ail with constant angular velocity. Show that its transverse acceleration varies as the distance from the pole. (iii) A particle describes a curve r = ae 9 with constant angular velocity. Show that the radial acceleration is zero. (iv) If the radial velocity is proportional to transverse velocity, find the path of the particle in polar coordinates. (v) If the path of a particle be a circle find its radial and cross radial acceleration. (vi) If the radial velocity is n times the transverse velocity find the equation of the path of the particle in plane coordinates. (vii) A particle describe a parabola r = a sec 2 ~ such that cross-radial velocity
d2 r
is constant. Show that dt 2 is constant . •
SOLUTION:
e= constant = c (say).
(i) If the angular velocity is constant then . 1d . Now transverse acceleratIOn = --d (r2(J) r
t
1d
= --d (r 2c) r t
= c!2r ddr = 2cx radial velocity.
r t transverse acceleration varies as its radial velocity.
(ii) Here r
= ae9 .
Given that angular velocity
e= constant = c (say).
1 d 2' Now transverse acceleration --d (r (J) r
= !dd (r 2c)
r t
dr
t
= 2c dt = 2c(ae 9 .e) [as r = ae9 ]
CR.I0: DYNAMICS OF PARTICLES 2
10.23
.
= 2c r [as {} = c].
Hence the transverse acceleration varies as distance from the pole. (iii) Given that the angular velocity iJ is constant. Let iJ = c. We have r = aeo .
. r = aeo.iJ = rc. Again differentiating we get
r = rc = (rc)c = re 2 . Thus radial acceleration =
r-
riJ 2 = re2
-
re 2 = o.
(iv) Given that radial velocity is proportional to transverse velocity.
... rex riJ or, r = kriJ, where k is the proportional constant. dr d{) or, -d = kr- or, t dt Integrating log r = or, log( r / c) =
k{}
dr - = kd{}. r
k{}
+ log e
or, r = ee kO , which is the path of the particle.
(v) Let the equation of the circle be r = a.
... r = O,r = o. Hence radial acceleration =
r - riJ 2 =
=
2riJ + rO = rO = aO.
-riJ 2 = -aiJ 2 and cross radial acceleration
(vi) The radial velocity = r, transverse velocity = riJ. According to the problem r = nriJ dr d{} dr or, -d = nr- or, - = nd{} t dt r Integrating we get logr = n{} constant.
+ loge
or, log(r/e) = n{} or, r = ce nO , where c is arbitrary
(vii) Here r = asec 2 ({}/2). As cross radial velocity constant, i.e., riJ stant = e (say). 1 . Now, r = 2a sec( {} /2).2" sec( {} /2) tan( {} /2).{} = asec 2 ({}/2)tan({}/2).9 = riJtan({}/2) = etan({}/2).
Again differentiating we get e 2 . er· 2 r = 2" sec ({}/2).{) = 2"-;;{} [r = asec ({}/2)]
= con-
V.G.
10.24
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
c2 2a
.r
is constant.
Exercise 2 (i) If the radial velocity is four times the transverse velocity find the equation of the path of the particle in plane coordinates. (ii) Find in polar coordinates the path of the particle whose radial velocity is equal to its transverse velocity. [Ans. (i) r = ce40 , (ii) r = ceo.J
10.8
Central Orbit
Def. 1 Central Force: A force F is called a central force if (i) it is always directed towards or away from a fixed point. (ii) the magnitude of the force F is a function of the distance r of the particle from the fixed point. Q. 1 Define central force.
Def. 2 Areal Velocity: When a particle moves along a plane curve, the rate of change of areas traced out by the radius vector joining the particle to a fixed point is called the areal velocity. The magnitude of areal velocity is 20 =
!r
!vp.
Q. 2 Define areal velocity. Def. 3 Apse: An apse is a point on a central orbit at which the radius vector drawn from the centre of force is a maximum or a minimum. Apsidal Distance: The length of the radius vector corresponding to apse is known as the apsidal distance. Q. 3 Define ..apse and apsidal distance.
Note 1 (i) The differential equation of the path of a particle moving in a plane under a central force is
where F is the force per unit mass and u
=
~.
CR.10:
DYNAMICS OF PARTICLES
10.25
(ii) The law of force F of a particle moving under central force is given by
(in polar curve)
or F Ex. 1 a
r
=
h 2 dp
3-d p
r
(in pedal curve)
(i) A particle describes a curve whose equation is . = () 2 + b, under a force to the pole. Fmd the law of force.
(ii) Find the law of force of the particle which describes the curve r()2 under a force to the pole.
=2
(iii) A particle describes the parabola p2 = ar under a force which is always directed towards its focus. Find the law of force. (iv) Find the law of force towards the pole under which a particle describes the curve r = ae li cot 0:, a being a constant. (v) The central orbit described by a particle moving under a central force is the conic
~r = 1 + e cos ().
Find the law of force.
(vi) A particle describes an equiangular spiral whose pedal equation is p = r sin a under a force F to the pole, find the law of force . •
SOLUTION:
a ()2 + b (i) Here - = ()2 + b or, u = - r a 2 . du _ 2() d d u _ ~ . . d() - a an d()2 - a'
Then law of force 2 F -- h 2u 2 [d - U + u ] -_ h2 u 2 d()2 a (ii) Here r()2 = 2 or u = ()2/2.
[2 + u] -_ -hr22 [2-a + -r1] -_ h22r + a. -ar3
Differentiating w.r.t () we get du d()
= () and
d2 u d()2
= 1.
V.G.
10.26
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iii) We have p2 = ar. 1 1 .. p2 ar Differentiating w.r. t r we get
2 dp
1
- p3 dr
ar2
h2 dp h2 1 F=--=-p3 dr 2a' r2
1 or, F ex: 2' r
(iv) Here we have r = aeOcoto: or, - log u = log a + 8 cot a
Differentiating w.r.t 8 we get 1 du du - - - = cot a or - = -u cot a u d8
' d8
Again differentiating w.r.t 8, we get d 2u d8 2
-
=
du d8
2
--cota = ucot a
3
2
= h 2 u cosec a
h2
2
cosec a = --::--r3
1 F ex: 3' r
(v) Here the equation of the path is or, lu
= 1 + e cos 8.
~r = 1 + e cos 8
Differentiating w.r.t 8 we get
du . I d8 = -esm8.
Again differentiating we get
d2 u
I d8 2 = -e cos () 2
u] +u
2 2 d .'. F = h u [d(}2 2 2
= h
u
. 1 .. F ex: 2' r
[lu - e cos ()]
1
2 2
= h u
[e - 1 cos 8 + u ] 2 2
h u = -1- [as I u - e cos 8 =
1]
10.27
CH.I0: DYNAMICS OF PARTICLES
(vi) H ere p
. 1 1 = rsma or, - = -.p rsma
Differentiating w.r. t r we get 1 dp 1 - p2 dr = - r2 sin2 a 1 dp 1 or, - - = p2 dr r2 sin a F _ h 2 dp _ h 2 1 _ h2 - p3 dr - p r2 sin a - r3 sin2 a' . F
1 oc 3' r
Ex. 2 Prove that pv = h in a central orbit where symbols have their usual significance. •
SOLUTION:
2. 2dO 2 dO ds We know h = r 0 = r - = r - dt ds dt 2dO . ",.. dO . "'] = vr ds = vrsm,/-, lsmce r ds = sm,/-, = vp [since p = r sin cPl.
Ex. 3 Prove that for a particle moving in a central orbit the angular momentum about the centre of force remains constant. SOLUTION: If m be the mass of the particl~ moving with a velocity v at any instant of its motion then the magnitude of the moment of momentum of the particle about a point is mvp, where p is the length of the perpendicular from the fixed point upon the tangent. Hence the angular momentum mvp = mh [since h = vp] which is constant (since h is constant). •
. 1 1 Ex. 4 Assummg the formula p2 = r2
+ r14 (dr)2 dO ,prcve
that at an apse of
the central orbit of a particle p = r. • SOLUTION: Let p be the length of the perpendicular from the centre of force on the tangent.
Then
1
p2 =
1 r2
1 (dr)2 dO =
+ r4
At an apse, where
U
+
(dU)2 dO
is a maximum or a minimum,
du
dO = O. . 1 2 . . -p2 = u
2
U
1 = -r2' or p = r .
10.28
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Exercise 1 (i) A particle describe a curve whose equation is ~ = emB under r a force to the pole. Find the law of force.
(ii) The central orbit is defined by 2a = 1 + cos 0, find the law of force. r
(iii) The central orbit is rectangular hyperbola defined by the law of force.
r2
= a 2 sec 20, find
[Ans. (i) F oc r- 3 , (ii) F oc r- 2 , (iii) F oc r.] Ex. 5 Show that the central orbit is a plane curve.
• SOLUTION: In central orbit, the force vector and hence the acceleration, are parallel to the radius vector and so velocity, acceleration and radius vector lie in a plane. The particle describing the motion can never leave this plane since there is no component of the velocity out of this plane.
10.9
Planetary Motion
Theorem 1 Kepler's laws of planetary motion:
(i) Each planet describes an elliptic orbit with the sun situated at one of its foci. (ii) The radius vector drawn from the sun to a planet describe equal areas in equal times in the same orbit. (iii) The squares of the periodic times of various planets are proportional to the cubes of the respective semi-major axes of their orbits.
Q. 1 State Kepler's laws of planetary motion. Ex. 1 (i) Find the velocity of an observer on the equator, taking the radius of the earth to be 4000 miles.
(ii) Prove that a planet has only radial acceleration towards the sun (use Kepler's laws). (iii) If VI and V2 be the respective velocities of a planet when it is nearest· and farthest from the sun, prove that (1 - e)VI = (1 + e)V2' where e is the eccentricity of the orbit.
CH.10: DYNAMICS OF PARTICLES •
10.29
SOLUTION:
(i) If the observer at the equator then, the distance from the centre of the earth to the orbit of observer = a, a is the radius of the earth. If v 2
be the velocity of the observer in its orbit, we have mv a 2
= ga , so that v = J9ii = v'32
GM
or,
2
mv
ga 2 .m --2-
= GM~ but a
-- = or, v = ga a a x 4000 x 1760 x 3 ft/sec = 2.5997 x'10 4 ft/sec. 2
(ii) From the second law of Kepler's, we see that the rate of description of sectorial area by the planet about the sun is constant. Hence r 2 0 = h (say), is constant. The cross-radial acceleration of the central orbit which IS
!!£(r 20) = ! dh =
O. r dt r dt Thus the- planet has only radial acceleration towards the sun. (iii) Let S be the position of the sun and C be the centre of the elliptic path. Also let A and A' be the nearest and fartheHt points from the sun . ... SA = AC - CS = a - ae = a(l - e) and SA' = A'C+C'S = a+ae = a(l+e). Now from the relation h = vp, for the nearest planet we have h = VI.SA = VIa(l - e) and for the farthest planet h = V2.SA' = V2a(1 + e). Thus V I a(l - e) = V2a(1 + e) or, (1 - e)VI
10.10
Figure: 10.9.1
= (1 + e)V2.
Artificial Satellites
Def. 1 Artificial Satellite: An artificial satellite is an object which is placed in an orbit round the earth outside the earth's atmosphere. Q. 1 Define Artificial satellite.
Def. 2 Parking Orbit: If the period of the satellite on an orbit be equal to the period of the earth's rotation, the satellite will be seen to be stationary over the same place on the earth while the earth rotates. This is known as the parking orbit. Q. 2 Define parking orbit.
10.30
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Ex. 1
(i) Define escape velocity. What is escape velocity for the earth?
(ii) Show that the velocity of escape from earth's gravitational field is y'2'ia, a being the radius of the earth and 9 is the acceleration due to gravity
on earth's surface. or What is 'escape velocity for earth' ? or Write a short note on 'escape velocity from earth' . •
SOLUTION:
(i) Escape velocity: The velocity required to allow escape of a body from the gravitational attraction of the earth is called the escape velocity. The escape
velocit~,
for earth v = y'2g(i.
If 9 = 9.8 m/sec'? and a = radius of the earth v = J2 x 9.8 x 6.4 x 106 m/sec = 11.2 x 10 3 m/sec .
6.4 x 106 m then
. '. escape velocity = 11.2 k.m/sec. Thus when a body projected with a velocity of 11.2 k.m/sec then it will completely escape from the gravitational attraction of the earth. (ii) Let us assume that a body of mass m be thrown with a velocity v from the earth surface so that it just escapes from the gravitational attraction of the earth. Then work done by the force = m x potential difference between infinity and . f GM t h e pomt 0 escape = m x - - . a
The K.E. of the body is
~mv2.
Since the K.E. is equal to work done by the force, !mv 2 = m x GM or v = 2
a
.j2GM. a
. GM Agam, -2- = g. a Hence v = J2ga, which is the escape velocity. Ex. 2 (i) The radius of the earth is ).2g, where 9 is the acceleration due to gravity at the earth's surface. Show that the velocity of a satellite of the earth moving in a circular orbit whose radius is approximately equal to that of the earth is ).g.
CI-l.10: DYNAMICS OF PARTICLES
10.31
(ii) An artificial satellite is circling round the earth at a height 700 km. from its surface. Find the horizontal velocity of the satellite. Radius of the earth is 6.4 x 103 km. and the acceleration due to gravity on the earth's surface = 980 cm. / sec 2. (iii) Assuming the moon to describe a circular orbit of radius 4 x 10 5 km. round the earth in 27.3 days, cQ.lculate the periodic time of an artificial satellite of the earth near the earth's surface. [radius of the earth = 6400 km.] (iv) Prove that a body projected from the earth's surface with speed exceeding 7 miles per second will not in general return to the earth (variation of gravity being taken into account) . •
SOLUTION:
(i) If v be the velocity of the satellite then v 2 = ga, where a is the radius of the earth.
... v
= vga = vf>.292 = >.g.
(ii) Here a = radius of the earth = 6.4 x 103 k.m, the distance of the satellite from the earth surface (h) = 700 k.m and 9 = 980 c.m/sec2. If v be the velocity of the satellite then 2 ga_ v2 = _ a+h
or,
v= aJ a! h = 6400
X
10
5
980 (6400 + 700) x 105
980 7100 x 105 = 751906.5 c.m/sec = 7.52 k.m/sec. (iii) LetTI and T2 be the periodic time of the moon and the satellite respectively. Let rl and r2 be the radii of the orbit of the moon and the satellite so that rl = 4 x 105 k.m and r2 = 6400 k.m. Tl = 27.3 days. T2 r3 From Kepler's third law, we have T,~ =
-i r2
2
(27.3)2
T1
or, or,
(4 x 105 )a = (6400)3
T,2 _ (27.3)2 X (6400)3 = (273)2 2 -
. '. T2
64
X
1015
.
= 27.3 x 64 x 10-5 x JIO =
X
642
X
0.055 days
(iv) The escape velocity on the earth surface is
J"2gO, = .../2 x 32 x 4000 x 1760 x 3 ft/sec
10-9
.
= 1.32 hours.
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
10.32
=
8 x 200 x 4v'33
1760 x 3
. mlles/sec
.
= 6.96 mIles/sec.
Hence if the velocity of projection exceeds 7 miles/sec the particle will never return to the earth again. Ex. 3 Define a Geo-synchronous satellite. Mention one advantage of such a satellite. • SOLUTION: Geo-Synchronous Satellite: A satellite with circular orbit round the earth in the equatorial plane with the same period time of the earth about its axis is called a Geo-Synchronous satellite. These satellites are used, as communication satellites, to transmit radio, television etc. signals.
Ch.IIII Probability and Statistics 11.1
Measure of Central Tendency
A sample (univariate) can be classified into three types: (i) Simple distribution: A sample of size n is represented as Xb X2, . .. ,X n . (ii) Ungrouped frequency distribution: The sample of n observations can be represented as: value (x)
frequency (I)
h h
Xl X2
In (iii) Grouped frequency distribution: The sample of n observations can be represented as: lower limit / boundary (l)
upper limit / boundary (u)
frequency (I)
h
Ul
l2
U2
h h
In n
The sum of frequency
Arithmetic mean Generally denoted by x.
L: Ii is denoted by N. i=l
(i) Simple distribution:
(ii) Ungrouped frequency distribution:
11.2
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iii) Grouped frequency distribution: 1 n X= -
N
If Yi =
Ex. 1
x·-A 'd then
x=
Ii
L,xdi'
+ Ui
Xi= - - .
2
,
i=1
A + dY·
(i) Find the mean of the following numbers 4, 5, 6, 7, 10, 21.
(ii) Calculate the arithmetic mean of family size from the following frequency distribution:
123 4 8 12
Family size No. of families
(iii) Find the mean of the following distribution: Class boundary
Frequency 3 5 6
10-20 20-30 30-40
(iv) Calculate the mean of first n natural numbers . •
SOLUTION:
.
(1) Mean
x=
1 1 53 ;; L, Xi = 6(4 + 5 + 6 + 7 + 10 + 21) = 6'
(ii) Sum of frequencies N = 4 + 8 + 12 = 24. 1 1 '56 Mean x = N L, xdi = 24 (1 x 4 + 2 x 8 + 3 x 12) = 24
7
= 3'
(iii) Calculation is shown below: Class boundary
Middle value
10-20 20-30 30-40
(Xi)
Frequency (Ii)
xdi
3 5
45 125 210 380
15 25 35
6
14
Total Therefore, mean
x=
1
N
380
L, xdi = 14
=
190
7'
(iv) The mean of first n natural numbers is 1...... '1 1 n(n+1) x = ;; LXi = ;;(1 + 2 + ... + n) = ;;. 2
n+l
= -2-'
CH.ll: PROBABILITY AND STATISTICS
11.3
Ex. 2 (i) Prove that the arithmetic mean of any sampling distribution is dependent on change of origin and change of scale. (ii) A sample of size nl has the mean Xl and a sample of size X2. Find the arithmetic mean of their combined sample . •
n2
has mean
SOLUTION:
(i) Let the sample values Xi and Yi are connected by Yi = aXi 1 1 1 or, - L:Yi = a- L:Xi + - L:b n
n
or, y =
+ b.
n
ax + b
This relation shows that the mean (y) of Yi is depend on a and b, i.e., depend on scale (a) and origin (b). (ii) The sample size of combined group is nl + n2. The sum of sample values of combined group is .
.
nixi
+ n2X2.
+ n2 x 2
nlXI
IS -=--=--~..:;.
Hence the reqUlred mean
nl +n2
Median Simple distribution: Let the given sample be Xl, X2, ... , Xn and the corresponding sorted list be xi, x~, ... , x~. Then if
x((n+l)/2) ,
Median
=
I
{
Ex. 3
n
is odd
I
+2X(n/2+l) , if n is even --'--'-"_:-'--'---'x(n/2)
(i) Find the median of 10, 12, 32, 11, 2, 45, 50, 11, 10.
(ii) Find the median of 3.1, 2.6, 5.0, 4.7, 2.4, 3.9, 5.1 and 3.6. (iii) Find the median of 88, 72, 33, 29, 70, 54, 86, 91, 57, 61. •
SOLUTION:
(i) The sorted form of the given sample is 2, 10, 10, 11, 11, 12, 32, 45, 50. Sample size is 9. Therefore the median is the
9;
1 = 5th value, i.e., 11.
(ii) The sorted list is 2.4,2.6,3.1,3.6, 3.9,4.7, 5.0, 5.1 and the sample size is 8. Therefore the median is the average of ~ = 4th and = 3.6
+ 3.9 2
= 3.75.
(~ + 1) =
5th values
U.G.
11.4
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iii) The sorted list is 29, 33, 54, 57, 61, 70, 72, 86, 88, 91 and the sample size is 10. Therefore the median is the average of 5th and 6th values = 61 ; 70
= 65.5
Quartiles Grouped frequency distribution The kth quartile Qk is given by
where h = lower boundary of the kth quartile class, F = cumulative frequency preceding of kth quartile class, f m = frequency of kth quartile class, c = width of the kth quartile class. k = 1,2,3 gives the first quartile Qb second quartile Q2 (median) and third quartile Q3 respectively.
Mode The mode of a sample is it sample value whose frequency is maximum among the frequency of other values. (i) Simple distribution Working principle: Sort the sample valUt'i-i and count the frequency of each of them. If the frequency of a1l items an- equal then there is no mode. If Ii is the maximum frequency of Xi then Xi is the mode. (ii) Ungrouped frequency distribution Find the maximum frequency. If Ii is the maximum then Ex. 4
Xi
is the mode.
(i) Find the mode of the distribution 10, 10, 20, 11, 12, 10.
(ii) Find the mode of the distribution 1, 1, 2, 11, 2. (iii) Find the mode of the distribution .3, 4, 5, 3, 5, 4. (iv) Find the mode of the distribution 10 4
20 5
30
7
40 6
50 3
(v) Find the mode of the following ungrouped frequency distribution
CH.ll: PROBABILITY AND STATISTiCS
11.5
1 2 3 4 5 Ji45663
Xi
•
SOLUTION:
(i) Here the frequency of 10 is three, which is maximum. So mode is 10. (ii) Here the frequency of 1 is 2, 2 is 2 and 11 is 1. The maximum frequency is 2, occur at 1 and 2. So the modes are 1 and 2. (The dis~ribution is bimodal) (iii) Here the frequency of all sample values are equal, i.e., 2. Hence the distribution has no mode. (iv) Here the frequency of 30 is 7 (maximum). So the mode is 30. (v) Here the frequency of 3 and 4 is 6. So modes are 3 and 4.
Relation between mean, median and mode Mean - Mode = 3 (Meau - Median). Ex. 5 (i) Find the mean of a distribution where mode is 64.2 and median is 70.5.
(ii) Find the mode of a. distribution whose mean is 64.1 and median is 65.5. (iii) Give one simple example in each case to show that (a) mode of a distribution does not always exist, (b) mode of a distribution is not always unique . •
SOLUTION:
(i) We know mean - mode= 3 (mean - median) or, mean =
3 x mt'dian - mode 3 x 70.5 - 64.2 7365 2 = 2 =..
(ii) We know mean - mode= 3 (mean - median) or, mode = 3 x median - 2 x mean = 3 x 65.5 - 2 x 64.1 = 68.3. (iii) (a) Let the distribution be 2, 2, 3, 3, 4, 4. All the sample values have equal frequency, so mode of this distribution does not exist. (b) Let the distribution be 2, 2, 3, 4, 4, 5. The frequency of 2 and 4 are 2 (maximum). So, 2 and 4 both the mode, i.e., the mode is not unique.
11.2
Measure of Dispersion
Moments Moments about origin
11.6
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(i) Simple distribution The kth order moment about origin is 1 n
ak=-Lxf, ao=1,a1=x. n i=1 (ii) Vngrouped frequency distribution The kth order moment about origin.is 1 n
ak
=N
L xf Ii,
n
N
i=1
=L k i=1
Central moments (i) Simple distribution The kth order central moment is 1 n IJk = - L(Xi - x)k .
.n
i=1
(ii) Vngrouped frequency distribution The kth order central moment is 1 n IJk = N L(Xi - x)k Ii,
n
N=Lk
i=1
i=1
Note that lJo = 1, 1J1 = O. Q. 1 Define moments of a distribution. Ex. 1 (i) Find the first four central moments from the following data: 4, 8, 11, 13, 9. (ii) Find the first two central moments of the observations 4, 5, 6, 7, 8, 9,
10. (iii) Distinguish between moment about an arbitrary constant A and central moment . •
SOLUTION:
(i) Sample mean
x=
g(4 + 8 + 11 + 13 + 9) = 9.
Xi
Xi -X
4 8
-5 -1 2 4 0 0
11
13 9
Total
x?
(Xi - X)3
(Xi - X)4
25 1 4 16
-125 -1
625 1 16 256
0
0
0
46
-54
898
(Xi -
8
64
CH.11: PROBABILITY AND STATISTICS
The four central moments are 1" 2 46 J.Ll = 0, J.L2 = ;; L)Xi - x) ="5 J.L3
" = ;;1L,..,(Xi -
x)
J.L4
= ;;1 "L,..,(Xi -
_4 x)
(ii) Mean x =
3
11.7
= 9.2
-54 = -5= -10.8
898 =5 = 179.6.
~(4 + 5 + 6 + 7 + 8 + 9 + 10) = 7. Xi 4 5 6 7 8 9 10 Total
Xi - x -3 -2
(Xi - X)2 9 4
-1 0 1
2 3
1 0 1 4 9
0
28
The first two central moments are 1 J.Ll = - L(Xi - x) = O. n J.L2
= -n1 L (Xi
- x) 2
= -28 = 4. 7
(iii) The kth order moment about A is Ok =
central moment is J.Lk =
.!.. 2)Xi -
.!.. L(Xi - A)k and the kth order n
x)k.
n The e two moments are equal if x = A otherwise they are different.
Exercise 1 Find the first four central moments from the following data : 2, 3, 11,6,9.
[Ans. J.Ll
= 0, J.L2 = 11.76, J.L3 = 5.1368, J.L4 = 201.6672.]
Quartile deviation . t'lOn= Q3 -2 Ql . Q uartI'1e deVla Mean deviation (i) Simple distribution
V.G.
11.8
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
about mean Mean deviation = 1
n
-L
IXi -
medianl, about median
n i=1 (ii) V ngrouped frequency distribution 1 n N
L
IXi - xiii,
about mean
i=l
Mean deviation = 1 n N
L IXi - medianlli,
about median
i=l
n
where N =
Lk i=1
Ex. 2 (i) If 45.5 and 40.3 be the third and first quartiles of a distribution then find quartile deviation of this distribution.
(ii) Find the mean deviation about mean of the following distribution 4, 5, 6, 9. (iii) Find the mean deviation of 7, 9, 14, 24, 26 measured from their Arithmetic Mean. (iv) Find the mean deviation about mean of the following distribution
(v) Find the mean deviation about median of the following distribution 4, 5, 6, 9, 10, 14, 16 . •
SOLUTION:
.. Q3 - Ql 45.5 - 40.3 .) . (1 Quartlle devlatlOn = 2 = 2 = 2.6.
(ii)
Meanx=~(4+5+6+9)=6. Mean deviation 1 1 = ;;; IXi - xl = 4(1 4 - 61 + 15 - 61 + 16 - 61 + 19 - 61)
L
1 3 = 4(2+ 1 +0+3) = 2'
11.9
CH.U: PROBABILITY AND STATISTICS
(iii) Arithmetic mean
=~(7 +
9
+14 + 24 + 26) = 16.
Mean deviation about mean 1 = 5(17 - 161 + 19 - 161 + 114 - 161 + 124 - 161 + 126 - 161) 1
= 5(9 + 7 + 2 + 8 + 10) =
(iv) Mean x =
~L
x,fi =
36
5
~(6 + 12 + 10) =
= 7.2.
4.
Mean deviation about mean 1 =
N L IXi - xlii 1
= 7(13 - 41
=
x 2 + 14 - 41 x 3 + 15 - 41 x 2)
1 4 7(2+0+2) = 7'
(v) The median of this distribution is the 4th value, i.e., 9. Therefore, mean deviation about median =
.!.n L
=
1 7(14 - 91
IXi -
median I
+ 15 -
91 + 16 - 91 + 19 - 91
+110 - ~I + 114 - 91 + 116 - 91) 1 25 = 7(5+4+3+0+1+5+7) = 7'
Exercise 2 (i) Find the mean deviation from the mean numbers 31, 35, 29. 63, 55, 72, 37.
or the
following
(ii) Find the mean deviation from the median of the following numbers: 8, 15, 53, 49, 19, 62, 7, 15, 95, 77. (iii) Find the Mean deviation of 7, 9, 14, 26 measured from t.heir A.M. [Ans. (i) 104/7, (ii) 27.2. (iii) 6.] Standard deviation (s.d) (i) Simple distribution The standard deviation is generally denoted by
(J"
and is defined as
11.10
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(ii) Vngrouped frequency distribution
1 n " x~f' NL..-J , ,-
q=
x2
i=1
Variance Variance=q2. Ex. 3
(i) Find the standard deviation of the six numbers 4, 4, 5, 5, 6, 6.
(ii) Find the standard deviation of the seven numbers 5,7,9, 11, 13, 15, 17. (iii) Find the standard deviation of first •
f!,
natural numbers .
SOLUTION:
(i) Mean (x)
1
= "6(4 + 4 + 5 + 5 + 6 + 6) = 5.
Standard deviation =
J~
L(Xi - x)2
= ~[(4 - 5)2 + (4 - 5)2 + (5 - 5)2 + (5 - 5)2 + (6 - 5)2 +(6 - 5)2]1/2 =
Ii
=
1[.
1 (ii) Mean (x) = 7(5 + 7 + 9 + 11 + 13 + 15 + 17) = 11.
Standard deviation =
J~ L xr - x
2
=
J~(52 + 72 + 92 + 112 + 132 + 152 + 172) -
=
J9~9
-
121 = 4.
(iii) The mean of first n natural numbers
(x)=.!.(1+2+"'+n)=.!..n(n+l) = n+l. n
Standard deviation =
n
2
J~ L xr - x
2
2
112
11.11
CH.11: PROBABILITY AND STATISTICS (n
+ 1)(2n + 1)
(n
+ 1)2
6
4
/n+l n-l_
=
Y-2-'-6- -
x-a Note 1 If y = - d - then O':r; = dard deviation of x and y. Ex. 4
IdlO'y,
~
V:12'
where O':r;, O'y are respectively the stan-
(i) When does the standard deviation vanish?
(ii) Two variables x and yare related by y = 10 - 3x. If the s.d. of x is 4, what will be the s.d. of y ? (iii) The s.d. of y is 6, find s.d. of x, if the two variables x and yare related by y - 5x = 7 . •
SOLUTION:
(i) If the standard deviation vanish then or, 2:(Xi - x)2
= 0,
J~
I)Xi - x)2 = 0
or, Xi - X = 0 for all i.
or, Xi = X for all i. That is, if all the values are equal then their standard deviation vanish. (ii) Given relation is y = 10 - 3x. Therefore,
O'y
= 1-
310':r;
= 3 x 4 = 12,
i.e., the standard deviation of y is 12.
= 7 or, y = 5x + 7. 0' 6 Therefore, O'y = 1510':r;, or, O':r; = ; = 5'
(iii) Given relation is y - 5x
which is the standard deviation of x. Exercise 3 (i) Determine the standard deviation of the six numbers 5, 5, 5,7,7,7. (ii) Find the standard deviation of the five numbers 1, 2, 3, 4, 5. (iii) Two variables x and yare given by y = 11- 5x. If the standard deviation of x is 4, calculate the standard deviation of y. [Ans. (i) 1, (ii) y'2, (iii) 20.] Def. 1 Standard error: If Xl, X2, lution with mean JL and variance
0'2
. .• , Xn
are the samples drawn from the pol-
then the standard error is
E{ (:/Jnf}.
U.G.
11.12
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Q. 2 Define standard error. What is the expression for standard error of sample mean ?
Ex. 5 (i) The first two moments of a distribution about the value 5 of the variables are 2 and 28. Find their mean and variance. (ii) In a certain distribution with 50 observations mean and standard deviation are found to be 40 and 3 respectively. Find the sum of squares of the observations . •
SOLUTION:
(i) The first and second moments about 5 are 1 - ~)Xi - 5) = 2 n
.!:. ~)Xi n
5)2 = 28
(2)
1 ' From (1), Xi - 5 = 2 or, x = 7. n 1 1 From (2), - L(x~ 5)2 = 28, or, n n 1 or, {(Xi - 7)2 + 4 + 4(Xi - 7)} = 28, n 1 or, - L(Xi - 7)2 + 4 + 0 = 28
L
-
L(Xi - 7 + 2)2 = 28
L
n
or, variance = 28 - 4 = 24.
(ii) Here n = 50, x = 40, s.d.=3. Therefore,
J~ L x~
or , .!:. n '" L...t x~~ or,
(1)
- x2 = 3
x2 = 9
Lx; = n(9 + x
2
)
= 50(9 + 40 2) = 80450.
Therefore, the sum of squares is 80450. Coefficient of variation . .. standa.rd deviation x 100 CoeffiCIent of vanatlOn= mean Coefficient of quartile deviation . ... quartile deviation d' x 100 CoeffiCIent of quartIle devlatIon= me Ian Coefficient of mean deviation Coefficient of mean deviation
CH.l1: PROBABILITY AND STATISTICS
=
11.13
mean deviation about mean (or median) . x 100 mean (or medlan)
Skewness (i) Moment measure
~~2'
Skewness (1'1) =
11-2
(ii) Pearson's first measure mean- mode Skewness (1'1) = - - - - - - standard deviation (iii) Pearson's second measure Skewness (1'1) =
3(mean - median) . standard devlation
(iv) Bowley's measure Skewness (1'1) =
Q3 - 2Q2 - Ql QQ' 3 -
1
Kurtosis Kurtosis (')'2) = 11-~ - 3. 11-2
Ex. 6
(i) Find the coefficient of variation when variance =4 and mean=40.
(ii) Find the coefficient of mean deviation when mean deviation about mean is 5 and mean is 2.
(iii) Find the skewness of the distribution 4, 4, 5, 5. (iv) Find the skewness when mean=4, mode=3 and standard deviation = 2. (v) Find the skewness when mean=4, median=5 and standard deviation = 3. (vi) Find the skewness when mean=4, median=4 and standard deviation = 3. Also explain its significance. (vii) Find the kurtosis of the distribution 2, 2, 4, 4 . •
SOLUTION:
. . s.d. 2 (i) CoefficIent of vanation = - - x 100 = 40 x 100 = 5. mean (ii) Coefficient of mean deviation = mean deviation about mean x 100 = ~ = 250. mean 2
U.G.
11.14
(iii) Mean
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
=~(4 + 4 +
5 + 5) = 4.5.
Median = 4.5 Therefore, skewness = 3(mean - median) = 3(4.5 - 4.5) = 0 standard deviation standard deviation . mean - mode 4- 3 1 (iv) Skewness = .. - -- - standard devIatIOn 2 - 2· 3(mean - median) 3(4 - 5) (v) Skewness = = = -l. standard deviation 3 (vi) Skewness = 3(mean - median) = 4 - 4 = o. standard deviation 3 This value indicates that the distribution is symmetric about mean. 2+2+4+4 (vii) Mean of given sample is (x) = 4 = 3. Then /-l2 =
.!.n L(Xi -
X)2
1 2+2 = "4[2 x (2 - 3)2 + 2 x (4 - 3)2] = -4- = l. /-l4 =
.!.n~ "(Xi -
x)4 =
~[2 x 4
(2 - 3)4 + 2 x (4 - 3)4J
=2+2=l. 4 Therefore, kurtosis = /-l~ - 3 = 1 - 3 = -2. /-l2
Ex. 7 (i) Find the mode of a distribution whose mean, s.d. and skewness are 29.6, 6.5 and 0.32 respectively. (ii) If coefficient of skewness =-0.375, mean=62 and median=65, find the value of standard deviation.
(iii) The first four central moments of distribution are 0,6,12,120. Find the coefficient of skewness and kurtosis. (iv) For a distribution, mean=25.45, standard deviation=9.035 and mode=26.72. Calculate the value of skewness. What is the physical significance of negative value of skewness ? •
SOLUTION:
(i) We know skewness =
mean - mode d dd . . stan ar eVIatIOn
CR.ll: PROBABILITY AND STATISTICS
or, O.32
11.15
= 29.6 - mode
6.5 or, mode = 29.6 - 6.5 x 0.32 = 27.52. k k 3(mean - median) (1'1') "'IXT vve now s ewness = d dd . . . stan ar eVIat!on .. 3(mean - median) or, standard devIatIOn = ---'---=-k-------'s ewness (iii) Here f-tl = 0, f-t2 = 6, f-t3 = 12, f-t4 = 120. . f-t3 12 2 skewness = "2" = 3/2 = . Ili and f-t2 6 v6 . f-t4 120 1 kurtosIs = f-t~ - 3 = 36 - 3 = 3'
- 65) = 3(62 = 24. -0.375
mean- mode (iv) We know skewness = - - - - - - - standard deviation _ 25.45 - 26.72 _ _ 4 9.035 - 0.1. The negative value indicates the distribution is left sided about mean.
11.3
Theory of Probability
Def. 1 Random Experiment: Random experiments are those experiments for which we know a priori the set of all different possible results or outcomes and which are such that it is impossible to predict which one of the set will occur at any particular performance of the experiment. In rolling of a die, the number of outcomes is known to be six, but one cannot predict which face will occur at any particular rolling of the die. Def. 2 Events: The results or outcomes of an experiment are called events. Def. 3 Simple or elementary event: An event which cannot be decomposed is called a simple event. In rolling of a die face 'one', 'two', ... , 'six' are all simple events. Def. 4 Compound event: An event which can be decomposed into some other events is called a compound event. , In rolling of a die, 'even face', 'multiple of three' are compound events. The event 'even face' can be decomposed into simple events 'two', 'four' and 'six'. Def. 5 Certain event: An event which is sure to occur at every performance of the experiment is called a certain event. In rolling of a die the event 'one' or 'two' or 'three' or 'four' or 'five' or 'six' is a certain event.
11.16
V.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Def. 6 Impossible event: An event which cannot occur at any particular performance of the experiment is called an impossible event. In rolling of a die the event 'seven' never occur and is then an impossible event. Def. 7 Mutually exclusive events: If two events are such that they cannot occur simultaneously for any random experiment are said to be mutually exclusive events. In rolling of a die events 'even face' and 'odd face' are mutually exclusive events. Def. 8 Complementary event: An event which consists with the negation of another event is called the complementary event to the latter event. The complement event of any event A is generally denoted by A or A C or
A'. In rolling of a die the complementary event of 'multiple of three' is 'no multiple of three'. Def. 9 Event points: The simple events connected with an experiment E are called event points or simply points. The event points in rolling of a die are 1, 2, 3, 4, 5, 6. Def. 10 Event space: The set of all possible simple events connected with a random experiment E is called the event space S connected with E. The event space in experiment 'rolling of a die' consists of six points 1, 2, 3,4,5,6. Ex. 1 Define the following terms with example. (i) random experiment, (ii) event, (iii) simple event, (iv) compound event, (v) certain event, (vi) impossible event, (vii) mutually exclusive events, (viii) complementary event, (ix) event space, (x) event points. Def. 11 Classical Definition of Probability: Let E pe a random experiment contains n (finite) event points, all of which are known to be equally likely or mutually symmetrical and exhaustive. If any event A connected with E contains m of these event points then the probability of A is denoted by P(A) and is defined as m P(A) =-.
n
Def. 12 Frequency definition of probability: The frequency ratio f(A) = N(A)
N
CH.ll: PROBABILITY AND STATISTICS
11.17
of an event A connected with an experiment E tends to a definite limit as N -+ 00 and this limit is called probability of the event A to be denot'ed by P(A) that is P(A) = lim f(A). N-'fOO
Q. 1
(i) Give the classical definition of probability.
(ii) Give the frequency definition of probability. Def. 13 Axioms of probability: Let E be a random experiment described by the event space S and A be any event connected with E. The probability of A denoted by P(A) such that the following axioms are satisfied: I. P(A) 2:: O. II. The probability of certain event S is 1, i.e., P(S) = 1. III. If A!, A 2 , .. ... , be a set of pairwise mutually exclusive events, that is, Ai Aj = cP (i i= j) then P(A 1
+ A2 + ... ) =
P(A 1 )
+ P(A 2) + ....
The axiom III is called the axiom of complete additivity. Q. 2 State axioms of probability. Def. 14 Stochastically impossible event: If an event A is such that P(A) = 0 then the event A is called stochastically impossible event. If P(A) = 0, we cannot conclude A = cP or A is impossible, in this case, we say A is stochastically impossible event. Def. 15 Stochastically certain event: If an event A is such that P(A) = 1 then the event A is called stochastically certain event. Q. 3
(i) Define stochastically certain event.
(ii) Define stochastically impossible event.
Ex. 2
(i) Show that for any event A, P(.A)
= 1-
P(A).
(ii) Prove that probability of any event lies between
a and 1.
(iii) Assuming that the probability of a certain event is 1, find the probability of an impossible event using the axioms of additivity.
(iv) If A
~
B then show that P(A) :::; P(B).
+ B) = P(A) + P(B) - P(AB). Using the relation P(A + B) = P(A) + P(B) - P(AB) prove that P(A + B + C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(CA) + P(ABC).
(v) Prove that P(A (vi)
(vii) If A and B are two mutually exclusive events then show that P(A+B) = P(A)
+ P(B).
11.18
U.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(viii) Prove that for any two events A and B, P(A + B) :::; P{A)
+ P(B).
(ix) For any two events A and B. Prove that peA) :::; P(A + B). (x) For any two events A and B. Prove that P(AB) :::; P{A + B). (xi) Prove that the relation p(A + fJ) = 1 - P{AB). (xii) Prove that P(A.B) = P(B) - P(AB). (xiii) Prove that the relation p(A.fJ) = 1 - P(A) - P{B) (xiv) Prove that p(A + B) = 1- P{A)
+ P(AB).
+ P{AB).
(xv) If A, Band C are three events such that P{A) = P{B) = P{C) = 1/4; p(AnB) = p{CnB) = 0, p(AnC) = 1/8 and p{AnBnC) = 1/8. Evaluate P(A U B U C). (xvi) Al and A2 are two events related to an experiment E. Given P(Ad = 1/3,P(A 2) = 1/3 and P(A I n A 2) = 1/4, determine P(A~ n A~) and P{A~ U A~).
(xvii) Write down the sample space when a coin is tossed three times. (xviii) What is the probability that a card drawn at random from the pack of playing cards may be either a queen or an acc ? (xix) Two unbiased dice are thrown simultaneously. What is the probability that the sum of the numbers on the faces is 8 ?
(xx) Two dice are thrown. Find the probability of obtaining a total of more than 10 . •
SOLUTION:
(i) Let A be any event and S be the certain event . . A+A=s. or, P(A + A) = P(S) or, P{A) + p(A) = 1 (by axioms II and III) (the events A and A are mutually exclusive) or, p{A) = 1 - P(A). (ii) Let A be any event and S be the certain event . . A+A=s. or, P(A + A) = P(S) Since A and A are mutually exclusive events, P(A) + P(A) = 1 or, p{A) = 1 - P(A). Since p{A) :2: 0 (by Axiom I), 1 - P{A) :2: 0 Of, P{A) ::; 1 Again by Axiom I, P{A) :2: 0 .'. 0:::; P(A) :::; 1.
CH.U: PROBABILITY AND STATISTICS
11.19
(iii) Let Sand ¢ be certain and impossible events respectively. Since S = ¢, P{¢) = P{S) = 1 - P(S) = 0 (as P(S) = 1). or, P(¢) = 0, i.e., probability of an impossible event is zero. (iv) Since A ~ B, B = (B - A) + A. Since B - A and A are mutually exclusive, P(B) = P(B - A) + P(A) or, P(B) - P{A) = P(B - A) 2: 0 or, P(B) 2: P{A). (v) The events A - AB, AB and B - AB are pairwise mutually exclusive and we have A = (A - AB) + AB, B = AB + (B - AB) and A + B = (A - AB) + AB + (B - AB). Then P{A) = P{A - AB) + P{AB) or, P(A - AB) = P{A) - P{AB). Similarly, P(B - AB) = P{B) - P{AB). Also, P{A + B) = P{A - AB) + P{AB) + P{B - AB) = P(A) - P{AB) + P{AB) + P{B) - P{AB) = P{A) + P{B) - P{AB). (vi) P{A + B + C) = P{A + B) + P(C) - P({A + B)C) = P(A + B) + P(C) - P{AC + BC) = P{A) + P{B) - P{AB) - {P(AC) + P(BC) - P{ABC)} + P(C) [since AB.AC = ABC] = P{A) + P(B) + P{C) - P(AB) - P(BC) - P{CA) + P{ABC) (vii) Let E be a random experiment and N(A) be the number of frequenl:Y of the event A in E. Since A and B are mutually exclusive, then N{A + B) = N(A) + N{B) N(A + B) N(A) N(B) or, N =~+~. Taking limit N -+ 00, we get P{A + B) = P(A) (viii) We know for any events A and B, P(A + B) = P{A) + P(B) - P{AB) Since P{AB) 2: 0, P(A + B) ~ P(A) + P(B).
+ P{B).
(ix) Since A + B = A + (B - AB) .'. P{A + B) = P(A) + P{B - AB) [As A and B - AB are mutually exclusive.] or, P{A + B) - P{A) = P{B - AB) 2: 0 or, P{A + B) 2: P(A). (x) We have A + B = (A - AB) + AB + (B - AB). Since A - AB, AB, B - AB are mutually exclusive. P{A + B) = P{A - AB) + P(AB) + P(B - AB) or, P{A + B) - P(AB) = P(A - AB) + P{B - AB) 2: 0 or, P{AB) ~ P{A + B).
11.20
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(xi) P(A
+ B) = P(AB)
(by De Morgan's law)
= 1- P(AB).
(xii) We know B = AB + AB Since AB and AB are mutually exclusive, P(B) = P(AB) + P(AB) or, P(AB) = P(A) - P(AB). (xiii) P(A.B) = P(A + B) (by De Morgan's law) = 1 - P(A + B) = 1 - P(A) - P(B)
+ P(AB).
(xiv) Since (A + B) + AB = S .'. P(A + B) + P(AB) = P{S) (As A + Band AB are mutually exclusive) or, P(A + B) = 1 - P(AB) Again A = AB + AB .'. P(A) = P(AB) + P{AB) or, P{AB) = P{A) - P{AB) Hence P{A + B) = 1 - P(A) + P(AB). (xv) For three A, B, C, P(AUBUC) = P{A) + P(B) + P(C) - P{AB) - P(BC) - P(CA) 1 1 1 1 1 3 = 4 + 4 + 4 - 0 - 8' - 0 + 8' = 4' (xvi) P{A~
n A~) =
1 - P(A 1 U A 2)
= 1- P(A 1 )
1
= 1P{A~ U A~) =
+ P(ABC)
'3 - '3 -
P{(AI
1
= 1-
1
-
P(A2) + P{AI 1 7 4 = 12'
n A2)C) =
1 - P{AI
n A2)
n A 2)
3
4 = 4'
(xvii) The sample space is {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}. (xviii) Total number of cards is 52. Total number of queens is 4 + 4 = 8. . d pro b a b'l' . 52 8 = 13' 2 t h e requlIe Ilty IS (xix) Total number of possible cases when two coins are tossed is 62 = 36. The cases when the sum of the faces 8 are {(2,6),{3,5),{4,4),{5,3),{6,2)}. Total number of favarable cases is 5, Hence the required probability is 5 36'
CR.l1: PROBABJLITY AND STATISTICS
11.21
(xx) The total possible cases is 6 2 = 36. The cases where the sum of total more than 10 are (5,6), (6, 5), (6, 6)}, i.e., total three cases. the required probability = 33 = 11 , 6 2 Def. 16 Conditional probability: The conditional probability of an event B on the hypothesis that another event A has occured is denoted by P{BjA) and defined by, P{AB) . P(B j A) = P(A) , provldedP(A)
=I o.
Q. 4 Define conditional probability for two events.
Ex. 3 Give the frequency interpretation of conditional probability. • SOLUTION: For a long sequence of repetitions of the random experiment under the uniform conditions, the conditional frequency ratio, f{B j A) is taken to be an approximate value of the conditional probability P{BjA). Def. 17 Stochastically independent or independent events: Two events A and B are defined to be stochatically independent or simply independent if P(AB) = P{A).P{B) even if P{A) = 0 or P{B) = O. Q. 5 Define independent events. Q. 6 Explain the significance of "stochastic independence" of tW()olevents.
Def. 18 Mutually Independence: A set of n events AI, A 2 , •.. ,An connected with an experiment E are said to be mutually independent if the following relations will hold P{AiAj) = P{AdP{Aj), for all i,j = 1,2,3, ... , nand i =I j P{AiAjAk) = P{Ai)P{Aj )P{A k ) for all i, j, k = 1,2,3, ... ,n and i =I j =I k
Theorem 1 Bayes' theorem: If AI, A 2 , .•• , An be a set of pairwise mutually exclusive events, one of which certainly occurs, then for any arbitrary event X, such that P(AdX) =
nP{Ai)P{XjAi )
2: P(Ai)P{XjAi) i=l
provided P{X) =I
o.
U.G.
11.22
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Q. 7 State Bayes' theorem. Ex. 4 (i) Using the definition of conditional probability prove that P(ABC) = P(A)P(B/A)P(C/AB).
(ii) If for two events A and B, P(A) then find P{A/ B).
= ~,P{B) = land
(iii) If A and B are two independent events then show that independent.
P{A U B)
A and iJ
= ~,
are also
(iv) If A and B are independent events prove that A and B are also independent. (v) If A and B are independent events, with P{A) find P{A U B). (vi) Given P{A) = ~,P{B) = l, P{AB) = the events A and B are independent ? (vii) For two events A and B, P{B) find P{A). (viii) If P(A) = 0.'37, P{B) between A and B . •
= 0.48
l,
= 0.5
and P{B)
= 0.3,
find P{A+B) and state whether
= 1/4,P{A/B) = 1/2,
and P{A U B)
= 0.85,
P{B/A)
= 1/4,
find the relation
SOLUTION:
...
(i) From the definition of conditional probability we have for two events A and B, P{AB) = P{B)P{A/ B) . ... P{ABC) = P{AB)P{C/AB) = P{A)P{A/B)P(C/AB). (ii) We have P{A U B) = P{A) + P{B) - P{A or, P{A n B) = P(A) + P{B) - P{A U B) 1 1 2 1
Now,
n B)
= 2" + 3" - 3" = 6· P(A n B) 1/6 1 P(A/ B) = P{B) = 1/3 = 2".
(iii) Since A and B are independent, P{AB) = P{A)P{B). P(A.iJ) = P(A + B) = 1 - P(A + B) = 1 - P(A) - P(B) + P{AB) = 1 - P{A) - P(B) + P{A)P(B) = 1 - P(A) - P(B)(1 - P(A)) = (1 - P(A))(1 - P(B)) =P(A)P{B) A and B are independent events.
11.23
CH.ll: PROBABILITY AND STATISTICS
(iv) We know A +.A = S or, .AB + AB = SB = B or, P(.AB) + P(AB) = P(B) (since .AB and AB are mutually exclusive) or, P(.AB) = P(B) - P(AB) = P(B) - P(A)P(B) (A and B are independent)
= P(B)(l- P(A)) = P{.A)P{B) . ... .A and B are independent.
+ P{B) - P{A n B) = P(A) + P(B) - P(A)P{B) = 0.5 + 0.3 - 0.5 x 0.3 = 0.65. 1 1 1 P{A + B) = P{A) + P{B) - P{AB) = 2" + 3" - 4 =
(v) P(A U B) = P{A)
. (VI)
Again P{AB) =
7 12·
~ 1= P{A)P{B).
Thus A and B are not independent. (vii) From the definition of conditional probability 1 1 1 P{AB) = P{B)P{A/ B) = 4 x 2 =
s·
Again P{AB) = P{A)P{B/A) P{AB) 1/8 1 or, P{A) = P{B/A) = 1/4 = 2· (viii) P{A U B) = P{A) + P{B) - P(AB) or, 0.85 = 0.37 + 0.48 - P(AB) or, P(AB) = 0 . .. A and B are mutually exclusive events.
Exercise 1 (i) If for two events A and B, P(A) = 0.5, P(B) = 0.6, P{A U B) = 0.7, find P(A/ B). (ii) If A and B are two independent events, calculate P(A+B) when P(A) = 0.37 and P{B} = 0.48. (iii) If for two events A and B, P(A) then find P(A/ B).
=
~,P(B)
(iv) If for two events A and B, P(A) = ~,P(B) = find P(A/ B).
=
! and P(A + B) =
i and P(AUB) =~, then
(v) If A and B are two independent events then show that A and independent. [Ans. (i) 2/3, (ii) 0.6724, (iii) 3/4, (iv) 1/3.]
7 1 2'
B are also
11.24
-11.4
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Distribution Function
Def. 1 Random variable: By a random variable (r.v.) we mean a real number X connected with the outcome of a random experiment. Random variables are of two types: discrete random variable and continuous random variable.
Q. 1 Define random variable. Def. 2 Discrete random variable: If a random variable takes at most a countable number of values, it is called a discrete random variable.
Q. 2 Define discrete random variable. Def. 3 Continuous random variable: A random variable X is said to be continuous if it can take all possible real values between certain limits.
Q. 3 Define continuous random variable. Def. 4 Probability distribution function: The distribution function of a random variable X is a function of a real variable x, to be denoted by Fx(x) or F(x) defined in (-00,00) by
F{x) = P{ -00 < X :s; x). Q. 4 Define probability distribution function. Important properties of distribution function: (i) The distribution function is monotonic non-decreasing everywhere. i.e., F(x) :::; F(y) whenever x < y. (ii) F(-oo)
= 0 and F(oo) = 1.
(iii) The distribution function F(x) has a jump discontinuity on the left at X = x, provided P(X = x) '# o. i.e.,
P(X = x) = F(x) -
lim F(y).
y-tx-O
(iv) The distribution function is continuous on the right at all points. i.e.,
F(x) = lim F(y). y-tx+O
Q. 5 Write down two important properties of distribution function.
CH.l1: PROBABILITY AND STATISTICS
11.25
Def. 5 Probability density function: Consider the small interval (x dx/2, x + dx/2) of length dx round the point x. Let f(x) be any continuous function of x so that f (x) dx represents the probability that X falls in the interval (x - dx/2, x + dx/2). That is, dX) dx P ( x-2~X~x+2 =f(x)dx. The function f(x) is known as probability density function of the random variable X and is usually abbreviated as p.d.f. The probability density function f(x) must satisfy the following conditions: (i) f(x) ~ 0, for all x, and
i:
(ii)
f(x) dx = 1.
Q. 6 Define probability density function. Def. 6 Probability mass function: Suppose X is a random variable taking at most a countable infinite number of values Xl, X2, .... With each possible outcome Xi, we associate a number P(X = Xi) = Ii, called the probability of Xi. The function f is called the probability mass function of the random variable X and is generally abbreviated as p.m.f. The probability mass function Ii must satisfy the following conditions: (i) Ii ~ 0 for all i and (ii) fi = 1.
L i
Q. 7 Define probability mass function. Note 1 The probability of the event a P(a < X
~ b) =
<X
~
b may be computed as
lab f(x) dx = F(b) -
F(a),
where f(x) and F(x) are respectively the probability density function and the distribution function of the random variable X. Ex. 1
(i) A random variable X has the following probability distribution: X:
f: Find k.
0 0
1 k
2 2k
3 2k
4 3k
5 k2
6 2k2
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
11.26
(ii) Find the value of the constant k such that
°
f(x) = { kx(l- x), <x 0) being two parameters. The random variable of the normal distribution is some times referred as N(m, 0'). Def. 11 Standard Normal distribution: A particular case of normal distribution is standard normal distribution. The probability density function of this distribution is
f(x) =
1
1
2
~e-2x, y21l'
-00
< x < 00
The standard normal distribution or the random variable of this distribution is referred as N(O, 1). Def. 12 Gamma distribution: The probability density function of this distribution is
O<x 0) being the only parameter. The random variable is often referred as ,(I) variate.
U.G.
11.30
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Def. 13 Beta distribution of first kind: The probability density function of this distribution is Xl-1(1 _ x)m-1 B(l, m) ,0
f(x) = {
<x 0), m(> 0) and the random variable is called a /31(l, m) variate. Def. 14 Beta distribution of second kind: The probability density function of this distribution is xl-1 f(x)
=
B(l, m)(l
+ x)l-m'
{ 0,
O<x 0), m(> 0) and the random variable is called a /32(l, m) variate.
Q. 8
(i) Define Binomial distribution.
(ii) Define Poisson distribution.
(iii) Define uniform (or rectangular) distribution. (iv) Define Normal distribution. (v) Write down the probability density function of a standard normal distribution. (vi) Define gamma distribution. (vii) Define beta distribution of first kind. (viii) Define beta distribution of second kind. Ex. 3
(i) If X is a Poisson /-l-variate and P(X
= 0) = P(X = 1), then find
/-l.
(ii) If X is a Poisson variate with parameter /-l and P(X = 0) = P(X = 1) = k, prove that /-l = 1 and k = e- 1 .
(iii) If the random variate X has a Poisson's distribution such that P(X = 2) = P(X = 3), find P(X = 4). [Given e- 3 = 0.0498] (iv) For a binomial (6,p) variate P(X
= 2) = 9P(X = 4), find p.
CR.II: PROBABILITY AND STATISTICS •
11.31
SOLUTION:
-/A
i
= i) =~. 2. Given P(X = 0) = P(X = 1).
(i) We know P(X
or, e-/A = J.L.e-/A, or, J.L = 1. (ii) Since X is a Poisson variate with parameter J.L, e-/A l P(X = i) =-+. 2.
Given P(X = 0) = P(X = 1) = k or, e-/A = J.L.e-/A = k, or, e-/A(I - J.L) = 0 or, J.L = 1. Also k = e-/A.J.L
= l.e- 1 =
e- 1 .
-/A
(iii) For Poisson's distribution P(X
i
= i) = ~. 2.
Since P(X
or,
1
= 2) = P(X = 3),
J.L
2 = '6' or, J.L =
Now P(X ,
= 4)=
3. e
-I-'
4
-334
.J.L = _e_._ 4! 24
= 0.0498 x 81 = 0.168. 24 (iv) For binomial (n,p) variate, P(X = i) = nCi pi(l_ p)(n-i).
Here n = 6. We have P(X = 2) = 9P(X = 4). or,6C2 p2(1 - p)6-2 = 9. 6C4 p4(1 _ p)6-4 or, 6C2 p2(1- p)4 = 9. 6C 2 p4(1 - p)2 [since nCr = ncn_ r] or, (1- p)2 = 9p2, or, 9p2 = p2 - 2p + 1, 2p - 1 = 0, or, 4p(2p + 1) - 1(2p 7-1) 1 1 or, (4p - 1)(2p + 1) = 0, or, p = 4' -2' or, 8p2
+ 4p -
. Since p > 0, so the required value of p is
=0
l.
Def. 15 Distribution function in two dimensions: Let X and Y be two random variables. The joint distribution function Fx,y(x, y) or simple F(x, y) of X and Y is defined by
F(x,y) = P(-oo < X
~
x, -00 < Y
~
y),
U.G.
11.32
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
where the events (-00 < X :::; x, -00 < Y :::; y) means the joint occurence of the events -00 < X :::; x and -00 < Y :::; y.
Q. 9 Define distribution function in two dimensions. Def. 16 Density function in two dimensions: The probability that the point (x, y) will lie in the infinitesimal rectangular region, of area dx dy is given by P(x - dx/2 :::; X :::; x + dx/2, y - dy/2 :::; Y :::; dy/2) = dFx,y(x, y) and is denoted by f x ,y (x, y) dx dy, where the function f x ,y (x, y) is called the joint probability density function of X and Y. Q. 10 Define density function in two dimensions. Note 2 If a function f(:c, y) is a p.d.f then (i) f(x, y) (ii)
~
i: i:
0 for all .r and y and
f(x, y) &1' dy
= 1.
Independent random variables If the events -00 < X :::; x and -00 < Y :::; yare independent for all x, y, then P(-oo < X:::; x,-oo
0), O"y(> 0) and < p < 1) are the parameters of the distribution.
(i) Define rectangular or uniform distribution in two dimensions.
(ii) Define Normal distribution in two dimensions.
Ex. 4
(i) Determine the value of the constant k which makes f(x, y) = kx 2y, 0 < X < 1,0 < y < 1 a joint probability density function.
(ii) Determine the value of the constant k which makes f(x, y) = kxy2, 0 < x < 1,0 < y < 1 a joint probability density function. (iii) Determine the value of the constant k which makes
f(x, y) = kxy, (0
< x < 1, 0 < y < x)
a joint probability density function . •
SOLUTION:
(i) Here f(x,y) = kx 2y, 0 < X Since f(x, y) is a p.d.f then
i: i: 1
or,
10 10
or, k 10
1
< 1,0 < y < 1.
f(x, y) dx dy = 1.
1
2 kx y dx dy = 1.
y[~3]~ dy =
1
or,
k r "310 ydy =
or,
k 1 "3k [y2] "2 01 = 1 or, "3' '2 = 1 or, k = 6.
1
1
(ii) Here f(x,y) = kxy2, 0 < X Since f(x, y) is a p.d.f then
< 1,0 < y < 1.
11.34
i: i:
or,
U.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
f(x, y) dx dy
= 1.
fo1 fo1 kxy2 dx dy = 1.
fo1 [~2]~y2 dy = 1 or, ~ r y2 dy = 1 2 10 or, k
1
or,
k [y3] 1 k 1 2 "3 0 = 1 or, 2·"3 = 1 or, k = 6.
(iii) Here f(x,y) = kxy, 0 < x < 1,0 Since f(x, y) is a p.d.f then
i: i:
f(x, y) dx dy
< y < x.
= 1.
f(x)
From figure, y ranging from 0 to x and x ranging from 0 to 1. or, foX kxydydx = 1.
fo1 r1 [y2] or, k lox 2
kln
x 0
dx = 1
Figure: 11.4.2
1
or x 3 dx = 1 , 2 0
or,
11.5
4 k [X Jl k 1 2 4 0 = 1 or, 2·4 = 1 or, k = 8. Mathematical Expectation
Def. 1 Mathematical expectation: If g(x) is a continuous function, then the mathematical expectation or mean value of the function g(X) of the random variable X, to be denoted by E{g(X)} and is defined as
E{g(X)} =
f=
1i:
g(Xi) fi,
.
for a discrete distribution
i=-oo
g(x) f(x) dx, for a continuous distribution
provided the series or the infinite integral converges absolutely.
Q. 1 Define mathematical expectation of a random variable X.
CH.II: PROBABILITY AND STATISTICS
11.35
Some properties of Expectation: (i) Show that E(a) = a, where a is a constant. (ii) Show that E(aX) ::: aE(X), where a is a constant.
(iii) Prove that E(91(X) •
+ 92(X))
= E(91(X))
+ E(92(X)) .
SOLUTION:
(i) Let X be a discrete random variable. 00
00
L
.'. E(a) =
a fi = a
i=-oo
L
fi
= a.I = a.
i=-oo
(ii) Let X be a discrete random variable. 00
.'. E(aX)
=
L
00
aXi fi
L
=a
i=-oo
Xi fi
= aE(X).
i=-oo
(iii) Let X be a discrete random variable. 00
E(91(X)
+ 92(X))
=
L
(91(xd
+ g2(Xi)) Ii
i=-oo 00
L
=
i=-oo
00
91 (Xi)
fi +
2:
92 (Xi) Ii = E(91(X))
+ E(92(X)).
i=-oo
Def. 2 Mean: The mean of X or that of the corresponding distribution is denoted by E(X) and is denoted by mx or simply m, i.e., m = E(X). The mean represents the centre of mass of the probability mass distribution. It gives a rough position of the bulk of the distribution and is called a measure of location. Q. 2
(i) Define mean of a random variable. or Define mathematical expectation E(X) of a random variable X.
(ii) What is the physical significance of the mean of a random variable X? Mean of some standard distributions: Distribution Binomial Poisson Normal Ex. 1 (ii) If y
Parameters
Mean
n,p
np
JL
JL
m,(J
m
(i) Calculate the mean of Poisson distribution.
= ax + b, where a, b are constants, prove that E(y) = aE(x) + b.
U.G.
11.36
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(iii) x is a discrete random variable which takes values 1, 2, 3, ... ,n. If the probability mass function of x be ax, find the value of a and the mean value of x. (iv) A random variable X has density function
f(x) = { 30x4(1 - x). 0:::; x ~ 1 otherwIse .
0,
•
SOLUTION:
(i) The p.m.f of Poisson distribution is
Ii
e-I-'.J.l i
.
= -.-,-, z = 0, 1,2, ..... .
z.
The mean (m)
E(X)
=
00.
L
00
Zli
=L
i=-oo
e-J1. .J.li
•
Z' - " -
i=O
Z.
+ 1'1!!:. + 2. J.l2 + 3. J.l3 + ... ) 2! 3!
=
e-J1. (0
=
J.le-I-' ( 1 + J.l
+ -J.l2 + -J.l3 + ... ) 2!
3!
(ii) Let X be the discrete random variable. 00
L
E(Y) =
00
yi/i =
i=-oo
a
Ix
= ax. Since
+ b)1i
00
00
i=-oo
i=-oo
Ii
+ b.l = aE(X) + b
is a p.m.f then
n
00
L
Ix
(axi
i=-oo
LXiii + b L
aE(X)
(iii) Let
L
Ix
= 1, or,
x=-oo
or, a(1
Lax = 1 ;1'=1
+ 2 + 3 + . ,. + n) = 1, or, a.
2 or,a=n(n+l)'
n(n + 1) 2
= 1.
(as
L Ii = 1).
11.37
CH.ll: PROBABILITY AND STATISTICS
n
= a
L
00
n
x=-oo
x=l
= Lxix = L x.ax
Now, mean (m) = E(X)
x 2 = a(12
+ 22 + 32 + ... + n 2)
x=l
=
n(n+1)(2n+1) 2n+1 =-n(n + 1)' 6 3 2
Def.3 Moments: (i) The kth moments of a random variable X about a fixed point 'a' is defined to be the mean value E{(X - a)k}. (ii) The kth moments about origin of the distribution ofthe random variable X denoted by ak(X) or ak is defined by ak = E(Xk). Obviously ao = 1, al = m. (iii) The kth order central moment J.Lk(X) or J.Lk which is the kth moment of X about the mean is given by
J.Lk = E{(x - m)k}. We have J.Lo
Q. 3
= 1, J.Ll = 0 for all distribution.
(i) Define moment about origin.
(ii) Define central moment. Def. 4 Variance: The variance of a random variable X denoted by Var(X) is the second central moment J.L2 of the distribution, i.e.,
Var{X)
= J.L2 = E{(X -
m)2} where m
= E{X) =
mean of X.
Def. 5 Standard Deviation: The positive square root of the variance is called the standard deviation of X to be denoted by O'(X) or O'x or 0', i.e., 0'
Q. 4
(i) Define variance.
(ii) Define standard deviation.
= +VVar(X).
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
11.38
Standard deviation of some standard distributions:
Distribution
Ex. 2
Parameters
Standard Deviation
Binomial
n,p
Jnp(l- p)
Poisson Normal
J.L
m,u
VJi
(i) Show that Var(aX
U
+ b) =
a2Var(X).
(ii) Show, for a probability distribution, that Var(X) = E(X2) - m 2. or
Prove that the variance of a random variable can be put in the form Var(X) = E(X2) - {E(X)}2. (iii) The mean and variance of binomial (n,p) distribution are 20 and 16 respectively. Find the values of nand p. (iv) Find the mean and variance of rectangular distribution. (v) A random variable X has density function
f(x) =
3+2x -18 { 0,
otherwise.
Calculate mean . •
SOLUTION:
+ b) = aE(X) + b = am x + b where mx = E(X). Var(aX + b) = E(aX + b - am x - b)2 = E{a(X - m x )}2
(i) E(aX
= a2E(X - m x )2 = a2Var(X).
(ii) Var(X) = E(X - m)2
+ m 2) = E(X2) - 2mE(X) + m 2 2m.m + m 2 = E(X2) - m 2.
= E(X2 - 2mX = E(X2) -
or, Var(X)
= E(X2) -
{E(X)}2 (as m
= E(X)),
(iii) The mean and variance of binomial (n, p) variate are respectively np and np(l - p), .'. np = 20 and np(1- p) = 16
or, 20(1 - p)
16
4
1
= 16 01',1- P = 20' or, p = 1- S = S'
CR.11: PROBABILITY AND STATISTICS
. Agam, np = 20 or, n
11.39
= -20 = 20 x 5 = 100. P
Therefore the value of nand p are respectively 100 and ~. , 5 (iv) The p.d,f of uniform distribution is 1 -a
f (x) Mean (m)
= E(X) =
< x < b.
= -b- , a
b
b
r x f(x) dx = Jra xb -1-a dx Ja
1 X2]b = b-a[T a 1 b2 - a 2 b+ a = b- a 2 2
=
E(X2) =
r x 2f(x)dx = Jar x 2_b -1 Ja b
b
[X 3]b
1
_1_~.(b _
a)(b2
1
3
dx a
b -a =b-a3a=b-a' 3
=
b- a3
3
+ ab + a2) =
,', Var{X) = E(X2) _ m2 = a
2
a
2
2
+ ab + b
.
3
2
+ ~b +b
_ (a; b)2
4(a 2 + ab + b2) - 3{a2 + b2 + 2ab) = 12 2 2 a +b -2ab (b-a)2 = = 12 12 4 3 + 2x (v) The mean (m) = J x~ dx 2 2 4 = 2.. (3x + 2x2) dx = 2.. x 18 J2 18 2
r
[3
r
= 2.. [~(16 _ 18 2
4)
+ ~(64 _ 3
+ 2X
3
3
8)] = 1 + 56 27
] 4
2
= 83. 27
Def. 6 Standardised Variate: The variable X* = X - m has mean zero (T and standard deviation unity is called the standardised or normalised random variable corresponding to the random variable X. Q. 5 Define standardised variate. Ex. 3
(i) Show that mean of a standardised variate is O.
V.G.
11.40
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(ii) Show that standard deviation of a standardised variate is 1. •
SOLUTION:
(i) The standardised variate X*
=
X - m where m and
(J'
are respectively
(J'
mean and standard deviation of the variate X. Now, mean of X* is E(X*) = E(X - 1n) (J'
1 1 1 = -E(X - m) = -(E(X) - E(m)) = -(m - m) = O. (J'
(J'
(J'
(ii) The mean of standardised variate X* = X - m is (J'
E(X*) = E(X - 111)
= .!.E(X -
(J'
m)
(J'
1 1 , = -(E(X) - E(m)) = -(m - m) = O. (J'
Var(X*)
(J'
=E(
X - m
)2 =
- 0
1 1 "2E(X - m)2 = "2(J'2 (J'
(T
(J'
= 1.
(i) If the random variable X has mean m and variance X-m if X* = , calculate E(X*) and E(X*2).
Exercise 1
(J'
and
.
(ii) If the random variable X has mean
and E(X
(J'2
mand variance
(J'2
find E (X
~ m)
~ m)2.
[Ans. (i) 0, 1, (ii) 0, 1.] Def. 7 Expectation for a bivariate distribution: Let X and Y be two random variables. The expectation or mean value of a continuous function g(X, Y) of X, Y is defined by
E{g(X, Y)} =
I
f: f:
g(Xi' Yj)!ij,
for discrete case
i=-oo i=-oo
i : i~(x, y)f(x, y) dx dy, for continuous case
provided the series or integral are absolutely convergent.
Q. 6 Define expectation for a bivariate distribution.
CH.U: PROBABILITY AND STATISTICS
11.41
Ex. 4
(i) If X and Y are random variables, then show that E(X + Y) = E(X) + E(Y).
(ii) If X and Yare independent variate then show that E(XY) = E(X)E(Y). (iii) If X and Y be two independent variates then prove that Var(aX +bY) = a 2Var(X) + b2Var(Y), where a and b are constants. (iv) If independent random variates . Xl + X2 + X3 variate z = 3 . •
Xl,
X2, X3 have mean m, find mean of the .
SOLUTION:
(i) Let X and Y be two continuous random variables. Now, E(X + Y) i : i : (x
+ y) f (x, y) dx dy
i : i'!f(x,y)dXd y + i : i'if(x,Y)dXdY
=
= i : x fz(x) dx + i : y fy(y) dy
+ E(Y) = i:f(x, y)dy
E(X)
=
where fz(x)
and fy(y)
= i:f(x, y)dx.
(ii) Let X and Y be two independent continuous variates. Then f(x,y) = fz(x).fy(y). Now, E(XY)
= = =
i:i: i: i: i:
xyf(x,y)dxdy xyfz(x)fy(y) dxdy
xfz(x) dx i : yfy(y) dy = E(X)E(Y).
(iii) Let Z = aX + bY. Then m z = am z + bm y. Now, Var(aX + bY) = Var(Z) = E(aX
+ bY -
= E(Z -
m z )2
am z - bmy)2
+ b(Y - my)F 2 = a E(X - m x )2 + b2E(Y - m y )2 + 2abE{(X - mz)(Y - my)} = a2Var(X) + b2Var(Y) + 2abE{(X - mx)(Y - my)} = E{a(X - m z )
V.G. MATHEMATICS (SHORT (,JUESTIONS AND ANSWERS)
11.42
= a2 Var(X)
+ b2 Var(Y).
(since X and Yare independent, so
E{(X - mx)(Y - my)} = E(X - mx)E(Y - my) = 0). (iv) Here given that E(Xl) = E(X2) = E(X3) = m. Mean of Z is 1 1 E(Z) = "3E(X1 + X 2 + Xa) = "3{E(Xl) + E(X2)
+ E(Xa)}
1
= "3(m+m+m) = m.
11.6
Correlation and Regression
Def. 1 Covariance: The covariance betweeJ:l_ two random variables X and Y is denoted by Cov(X, Y) and is defined by Cov(X, Y) = E{(X - mx)(Y - my)}.
Q. 1 Define covariance. Der. 2 Correlation Coefficient: The correlation coefficient between two random variables X and Y is generally denoted by r or by p and defined by p=
Cov(X, Y) , (1x·(1y
where (1x and (1y are respectively the standard deviations of X and Y.
Q. 2 Define correlation coefficient of a bivariate statistical data. Ex. 1 (i) Show that for two independent variables, their correlation coefficient is zero.
(ii) Interpret the cases when p = 0 and p = 1. •
SOLUTION:
(i) Let X and Y be two independent random variables. Then
Cov(X, Y) = E{(X - mx)(Y - my)} = E(X - mx).E(Y - my) = O. . '. p
= Cov(X, Y) = O. (1x·(1y
(ii) When p = 0 then the random variables X and Yare uncorrected. When p = 1 then the random variables X and Yare connected by a linear relation.
11.43
CH.ll: PROBABILITY AND STATISTICS
Ex. 2
(i) Define regression coefficients on y on x and x on y.
(ii) Write down the regression equations of y on x and x on y. (iii) Write down the regression equations of x on y and y on x and prove that correlation coefficient is the geometric mean of the two regression coefficients. or
Show that p = ±Jbxy.byx , where bxy and byx are regression coefficients . •
SOLUTION:
(i) byx = p a
y
~
is the regression coefficient of y on x and bxy =:P ax is the ~
regression coefficient of x on y, where p, ax, a y are the correlation coefficient, standard deviation of x and standard deviation of y respectively. (ii) The regression equation of yon x is y - fi = byx (x - x). The regression equation of x on y is x - x = bxy(y - fi). bxy and byx are called the regression coefficients of x on y and y on x respectively.
(iii) The regression equation of yon x is y - fi = byx (x - x). The regression equation of x on y is x - x = bxy(y - fi), y where byx = p a and bxy = Pax, called the regression coefficients. ax ay Now, bxy.byx = p2 or, p = ±Jbxy.byx ,
i.e., the correlation coefficient is the geometric mean of regression coefficients. Note 1 P = +Jbxy.byx when both bxy and byx are positive and p = -Jbxy.byx when both bxy and byx are negative. Ex. 3
(i) Find the correlation coefficient of the following bivariate sample
(ii) Find the regression coefficient of y on x of the following distribution
V.G.
11.44
•
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
SOLUTION:
(i) The correlation coefficient
x
1 2 3 6
Total Here n
E XiYi
6
Y 4 5 6 15
xy
x:.G
y:.G
4 10 18 32
1 4 9 14
16 25 36 77
15
= 3. x = 3 = 2, Y = "3 = 5. = 32, E X[ = 14, E Yl = 77. 32 _
. r=
2x 5
3
.. J(14/3 - 4).J{77/3 - 25) (ii) Regression coefficient of Y on x is
~ 3
=
J2/3.J2/3
=1
.
_ Sy _ Cov(.r, y) Sy _ Cov{x, y) _ ~ E XiYi - x.y byx - r - 1 Sx Sx.Sy Sx S; n Ex; - x2 x
Total Here n = 3, x = 0, y = ..·byx --
-1 0 1 0
Y 2 1 2 5
X2
xy
1 0 1 2
-2 0 2 0
5
3.
O-Ox£ ~
_ 0 3-0 - .
3
Ex. 4
(i) Find the regression equation of Y on x from the following data x = 10, i} = 15 and byx = 2.50.
(ii) If 4u = 2x + 7 and 6v = 2y - 15 and the regression coefficient of Y on x is 3 then find the regression coefficient of v on u.
CH.ll: PROBABILITY AND STATISTICS
(iii) If two random variables are linearly related, then show that p
11.45
= ±1.
(iv) If the correlation coefficient p between two variables be 1, find the relation between the variables by which tht-y are related. (v) If u = 2x and v = 3y then show that correlation coefficient between u, v is equal to the correlation coefficient between x and y. (vi) If the regression equation of y on x be y = 0.57x + 6.93 and x on y be x = 1.12y - 2.46, find the correlatio'1 coefficient between x and y. (vii) Two regression lines are given by 3X the variance of X and Y.
= 4Y +9, 1.5Y = 6X +7. Determine
(viii) Out of two lines ofregression given by X+2Y-5 = Oand2X+3Y-8 = 0, which one is the regression line of X on Y ? •
SOLUTION:
(i) The regression equation of yon x is y - y = byx (x - x) or, y - 15 = 2.5(x - 10) or, y = 2.5x - 10. (ii) bvu
Sv
2/6.Sy
(if y = ax + b then Sy = lalSx )
= rvu Su = ryx 2/4.Sx
_ Sy 2 _ 2 _ 2 _ or, bvu - ryx Sx· - 3·byx - 3. 3 - 2. 3 (iii) Let Y = aX + e be the relation between the variates X and Y.
= E(Y) = aE(X) + e = amx + e ... S; = E(Y - m y )2 = E(aX + e - amx - e)2
Now, my
= E{a(X - m x )}2 = a 2E(X - m x )2 = a 2S;.
Again, Cov(X, Y)= E{(X - mx)(Y - my)}
= E{(X
- mx)(aX
+e-
amx - e)}
= E{(X - mx)a(X - m x )}
= aE(X - m x )2 = as; . . p = Cov(x, y) = a.s; .. Sx.sy Sx·lalSx
(iv) Let X
..
= X -Sxmx
an
d Y'"
= ~ =.±1.
lal
= Y -Symy .
Suppose (X* - y*)2 = 0
S;
1 E (X - mx )2 = S2 = 1 an d Now, E (X *)2 = S2· x
x
U.G.
11.46
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
E(X*Y*) = E{(X
~2mX)(X ~2mX)} x
=
x
E(X - mx)(Y - mU) 52 82 y
x'
= p.
Also E(X'" - y*)2 = E(X*)2 - 2E(X*Y"')
+ E(y*)2
= 1 - 2p + 1 = 2 - 2p .
. '. E(X· - y*)2
= 0 or,
Thus when p = 1 then (X'" or,
X-m x Sx
=
= 0 or, p = 1. Y"')2 = 0 or, X* = Y·
2 - 2p
Y-my 5 • y
Which is the equation of a straight line passing through the point (mx, my). (v) mu
= E(U) = E(2X) = 2m x.
Similarly, mv S~
= E(U -
Similarly,
= 3m y.
mu)2
= E(2X -
2mx)2
= 4E(X -
m x )2
= 48;.
S; = 9S~.
Cov(U, V) = E{(U - mu)(V - mv)} = E{(2X - 2mx)(3Y - 3my)} =
6E{(X - mx)(Y - my)}
= 6Cov(X, Y).
.
. . Puv
=
Cov(U, V) Su.Sv
=
6 Cov(X, Y)
2Sx .3Sy
=
Cov(X, Y) Sx.Sy
= Pxy'
i.e., the correlation coefficient between u, v is equal to the correlation coefficient between x, y. (vi) The regression equation of y on x and x on yare respectively y = 0.57x
. '. byx
+ 6.93
and x = 1.12y - 2.46 .
= 0.57 and bxy = 1.12.
The correlation coefficient is
p = +Jbxy.byx
(vii)
= )0.57 x 1.12 = 0.80. We know, if aX = bY + c then a2Var(X)
=
b2Var(Y).
Therefore from the given relations we have 9Var(X) = 16Var(Y) and 2.25 Var(Y) = 36 Var(X). Solving we get Va7'(X)
= 0 and Var(Y)
= O.
CH.U: PROBABILITY AND STATISTICS
(viii) Let Y
= _!X + ~ and X = -~Y + 4.
Here by:z; =
2
2
.l
-"2
2
and bxy =
3
-"2'
Since bxy.byx =
-~. - ~ = ~ < 1,
the line X = -
~ Y + 4 is the regression line of X
11.7
11.47
on Y.
Sampling Distribution
Def. 1 Population: The collection or the aggregate of objects or the set of results of an operation is called population. In theory of sampling the population means the larger group from which samples are drawn. Def. 2 Sample: A part of the population selected from it with the object of investigating its properties is called a sample. The number of individuals taken in the sample is called sample size.
Def. 3 Population Parameters: The constants with a probability distribution such that for different values of those constants one gets different distributions of same kind are called parameters of characteristic of the probability distribution. For example, nand p are the parameters of the binomial (n, p) distribution of X given by P(X = i) = nCi pi(1 - p)n-i, i = 0,1, ... , n.
Def. 4 Sample Characteristic: Let X be a random variable associated with an experiment E and X be the random variable for the mean of sample objects corresponding to the distribution of the sample values Xl, X2,'" ,Xn . The characteristic such as mean, variance, moments etc. of the empirical distribution of X are called sample characteristics.
Def. 5 Statistic: Any function of the sample values Xl, X2, ••. , Xn of a ample 'of size n drawn from a population of a random variable X, is called a statistic. For exampIe, sampIe mean
x=
1 ~. . . - L.J Xi, IS a statlstlc.
n
i=l
Def. 6 Estimate or Estimator: A statistic a = a(xl' X2, .•. ,xn ) will be called an estimator of the population parameter a of the probability mass in the sampling distributioll if the statistic a is concentrated near the point a.
11.48
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Defo 7 Consistent Estimate: A statistic a = a(xl, X2, . .. , xn) is said to be a consistent estimate of a population parameter 0: if A ~ a as n ---+ 00, where A is the random variable corresponding to the statistic a. For example, sample mean is a consistent estimate of the population mean. Defo 8 Unbiased Estimate: A statistic a = a(xl, X2, ... , xn) is said to be an unbiased estimate of a population parameter 0: if E(A) = 0:, where A is the random variable corresponding to a. For example, sample mean is an unbiased estimate of the population mean. Qo 1 Define the following terms: (i) Population, (ii) Sample, (iii) Parameters, (iv) Sample characteristics, (v) Statistic, (vi) Estimate or estimator, (vii) Consistent estimate, (viii) Unbiased estimate. Exo 1 Show that the sample mean is unbiased estimate of population mean . • SOLUTION: Let whose mean m.
Xl, X2,""
be a sample of size n drawn from a population
Xn
.!.
t
Xi. We have E(X i ) = m for all i, where Xi n i=l is the random variable corresponding to the sample value Xi. Now,
The sample mean is i' =
1
E(X)
=-
n
n
2:E(Xd
1
=-
i=l
n
n
2:m = m. i=l
Hence the sample mean is unbiased estimate of population mean.
11.8
Testing of Hypothesis
Defo 1 Statistical Hypothesis: Any statement or assertion about a statistical population or the values of its parameters is called a statistical hypothesis. Hypothesis are of two types: simple and composite. Defo 2 Simple Hypothesis: A statistical hypothesis which specifies the probability distribution and all parameters of the population is called a simple hypothesis. Defo 3 Composite Hypothesis: A statistical hypothesis which does not specify the population completely, i.e., either the probability distribution or some parameters are unknown is called composite hypothesis.
CH.ll: PROBABILITY AND STATISTICS
11049
Defo 4 Null Hypothesis: A statistical hypothesis which is set-up depending of the given hypothesis and whose validity is tested for possible rejection 011 the basis of sample observations is called a null-hypothesis. It is denoted by Ho and tested against alternative hypothesis. Tests of hypothesis actually deal with rejection or acceptance of null-hypothesis (not with the given hypothesis) only. Defo 5 Alternative Hypothesis: A statistical hypothesis which differs from the null-hypothesis is called an alternative hypothesis and is denoted by HI. The alternative hypothesis is not tested, but its acceptance or rejection depends on the rejection or acceptance of the null-hypothesis. Defo 6 Test Statistic: A function of sample observations whose computed value determine the final decision regarding acceptance or rejection of the null-hypothesis is called a test statistic. Defo 7 Critical Region: The set of values of the test statistic which lead to rejection of the null-hypothesis is called critical region of the test. Defo 8 Level of Significance: The maximum probability with which a true null-hypothesis is rejected is known as level of significance of the test and is denoted by a. Defo 9 Type-I ~rror: This is the error committed in rejecting a nullhypothesis by the test when it is really true. The critical region is so determined that the probability of Type-I error does not exceed the level of significance of the test. Def. 10 Type-II Error: This is the error committed in accepting a nullhypothesis by the test when it is really false. The probability of Type-II error depends on the specified value of the alternative hypothesis and is used in evaluating the efficiency of a test.
Q. 1 Define the following terms: (i) Statistical hypothesis, (ii) Simple hypothesis, (iii) Composite hypothesis, (iv) Null hypothesis, (v) Alternative hypothesis, (vi) Test statistic, (vii) Critical region, (viii) Level of significance, (ix) Type-I error, (x) Type-II error.
Ch.12 II Numerical Methods 12.1
Errors
12.1.1
Sources of error
In numerical computation, error can be classified as two types: (i) Truncation error: The error which is inherent in a numerical method itself is called truncation error. The truncation error arises due to the replacement of an infinite process by a finite process. (ii) Computational error: This error arises during arithmetic calculation due to the finite representation of numerical numbers. For example, t is represented by 0.333333, which is not equal to it is correct up to six decimal places.
i, but
The computational error again can be classified into two types - round off error and significant error. Significant figures The figures (digits) which are used in representing a number are called significant figure or significant digits. The digits 1, 2, 3, ... , 9, 0 are significant figures or digits. The numbers 1.235, 2.0235, 378 contain respectively 4, 5 and 3 significant figures. But, in the number 0.000587 has 3 significant figures, the leading zeros after decimal point are not significant, they are used to fix the position of the decimal point. Rule for rounding-off a number to n significant digits Discard all the digits to t.he right of the nth digit and if the discarded number is greater than half a unit in the nth place, add 1 to the nth digit, if the discarded number is less than half a unit in the nth place, leave the nth digit unchanged, if the discarded number is exactly half a unit in the nth place, leave the nth digit unaltered if the digit is even, but add 1 with nth digit if it is odd. If a number is rounded using the above rule then the number is called correct up to n significant figures. The following numbers are rounded-off correctly to five significant figures:
U.G.
12.2
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Exact number 25.367~35
28.353215 3.7853Q3 5.R354Q3 6.7354Q 4.8327Q 0.005834518 38567Q4 2.37 8.99991 9.9999~
Round-off number 25.368 28.253 3.7854 3.8354 6.7354 4.8328 0.0058346 38568x 102 2.3700 9.0000 10.000
Round-off error
The error committing due to the rounding-off a number is called round-off error. Significant error
The error committing during arithmetic computation due to the loss of significant digits. This type of error generally occurs (i) when two almost equal numbers are subtracted, and (ii) when a number is divided by a very small number compared to the dividend.
12.1.2 Suppose tively.
Absolute, relative and percentage error XT
and
XA
denotes the true value and approximate value of x respec-
Absolute error:
The absolute error (Ea) of a number is the difference between true value and approximate value, i.e., Ea = XT - XA or XA - XT· Relative error:
The relative error (Er) of a number is equal to absolute error divided true or approximate value, i.e., Er
_ Ea
-
XT
or
Ea XA
12.3
CH.12: NUMERICAL METHODS
In other words, relative error is the absolute error in unit measurement.
Percentage error: The percentage error (Ep) of a number is the relative error multiplied by 100,
i.e., Ep = Er x 100 =
XT-XA XT
x 100.
Percentage error is thE' absolute error committed to measure 100 unit. When Er is very small then Ep is used. .
Theorem 1 If a number is correct up to n significant figures and k is the first significant digit then the maximum relative error is k
12.1.3
1 X
lO n -
1.
Generation of round-off errors
Ex. 1 (i) Show that the absolute error in summing two numbers is the sum of individual absolute errors.
(ii) Show that the relative error in product of two numbers is equal to the sum if individual relative errors. (iii) Show that the relative. error in division between two numbers is equal to the difference of individual relative errors . •
SOLUTION:
(i) Let XT and YT be two exact numbers whose approximate representations be XA and YA respectively. Let
f1
and
f2
be the absolute errors in
XT
and
YT
respectively.
Then XT = XA + f1 and YT = YA + f2· The sum XT + YT = (XA + YA) + (f1 + f2) or, (XT + YT) - (XA + YA) = (f1 + f2). Here, (XT + YT) which is equal to
+ YA) is the absolute error of added numbers and + f2), i.e., sum of two individual absolute errors.
(XA (101
(ii) Let XT and YT be the true value and XA, YA be the corresponding approximate values. Also, let f1 and f2 be the absolute errors in XT and YT respectively. Then
XTYT - XAYA = (XA
(neglecting
f1f2
+ f1)(YA + f2) -
as small number)
. XTYT -XAYA = ~ +~. 1.e., XAYA
XA
YA
XAYA
=
f1YA
+ f2 X A
12.4
U.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Thus, the relative error in product of two numbers is equal to the sum of individual relative errors. (iii) Let XT and YT be the true value and XA, YA be the corresponding approximate values. Also, let Cl and C2 be the absolute errors in XT and YT respectively. 'rhen
XT
= XA + C1
XA
and
YT
= YA + C2·
YA
(as relative error ~ is much less th.m 1). Thus the relative error in division is equal to the difference of their individual relative errors. Ex. 2 Write down the approximate representation of ~ correct up to four significant figures and find (i) absolute error, (ii) relative error, (iii) percentage error . •
SOLUTION:
~=
0.3333 (up to foul' significant figures).
(i) absolute error Ea
=~-
(ii) relative error Er
= ~~ = 0.000099 ::= 0.00010.
0.3333
= 0.000033.
(iii) percentage error Ep = Er x 100 = 0.01 %. Ex. 3 Find the sum of the following approximate numbers, if the numbers are correct to the last digit. 3.58, 4.835673, 1.5673, 1.05534, 8.36. SOLUTION: Here the least correct numbers are 2.58 and 8.36. Roundingoff the other numbers up to 4 decimal places (more than two decimal places relative to 2.58, 8.36) and adding them, we get, 3.58 + 4.8357 + 1.5673 + 1.0553 + 8.36 = 19.3983 ::= 19.40. •
Ex. 4 Find the relative error in computing Y = 3x 7 - Sa: at a: in x is 0.005.
= 1, if the error
CH.12: NUMERICAL METHODS
12.5
• SOLUTION: Let f:l.y and f:l.x be the absolute error in y and x respectively.
Then, f:l.y ~ ~~ f:l.x = {21x 6 - 5)f:l.x. At x = 1, f:l.y = (21 - 5) x 0.005 = 0.08. Again, at x = 1,y = 3 X 17 - 5 x 1 = -2 (true value). The relative error in y is
I~y I = I0.~81 = 0.04.
Ex. 5 Find the number of significant figures in x A with respect to XT where XA = 3.56737, XT = 3.56738. • SOLUTION: Here Ea = XT - XA = 3.56738 - 3.56736 = 0.2 x 10- 4 < 0.5 X 10- 4 . . '. XA is correct to four decimal places and XA is correct up to five significant
figures. Ex. 6 Find the number of significant figures in 13.3651 given its absolute error 0.25 x 10- 2 . • SOLUTION: Here Ea ::::: 0.25 X 10- 2 < 0.5 x 10- 2 . . '. XA = 13.3651 i:' correct to two decimal places. Hence x A is corree': t \. four significant figures.
Ex. 7 Find the.· relatiw error in computation of x + y for x = 15.23 and y = 8.58 having absoluh' error f:l.x = 0.002 and f:l.y = 0.004 respectively. • SOLUTION: Let:; = J: + y. Then, f:l.z x. . f:l.z = 0.002 + 0.004 = 0.006. Z = x + Y == 15.23 + 8.58 = 23.81.
The relative error in
J'
+ Y is ~~~~~
12.2
Interpolation
12.2.1
Lagrange Interpolation
= f:l.x + f:l.y,
where f:l.x is the error in
= 0.00025.
Let Yi = !(Xi), i = 0,1,2, ' .. ,n. Then the Lagrange interpolation polynomial IS
¢(X)
=
+
(x - xt}(x - X2) ... (x - xn) (xo - Xl )(XO - X2) , .. (XO - Xn) Yo {x - Xo)(X - X2).·· . (x - Xn) Yl (Xl - XO)(XI - X2)'" (Xl - Xn)
+ ...
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
12.6
+
(x - xo)(x - Xl) ... (X - Xn-l) (xn - XO)(xn - Xl) ... (xn - Xn-l) Yn
=
LLi(x)Yi,
n i=O
where
(X - xo)(x - Xl) ... (X - Xi-I)(X - Xi+!) ... (X - Xn) (Xi - XO)(Xi - Xl) ... (Xi - Xi-I)(Xi - Xi+d ... (Xi - Xn )· Lagrange interpolation formula for equal points Substituting X = Xo Then Li(X) =
+ uh
and Xi
= Xo + ih.
u(u - 1)(u - 2) .. · (u - n) ( - 1) n _.~z."( n - z')'( . u - z.)
n (-I)n- i u(u - 1)(u - 2)· .. (u - n) In thIS case, ¢(x) = Yi· i=O u Z z. n z. .
L
( _.) "( _ ')'
Ex. 1 Calculate f(1.2) using the following table X
Y
1 0.0175
2 0.0349
3 0.0523
• SOLUTION: From Lagrange interpolation formula we have ¢(x) = Lo(x)yo + LI(x)Yl + L 2(x)Y2. Here Yo = 0.0175, YI = 0.0349, Y2 = 0.0523 and X = 1.2.
(X (xo (x (Xl (x (X2
Lo(x) =
-
xd(x - X2) = (1.2 - 2)(1.2 - 3) = 0.72, XI)(XO - X2) (1 - 2)(1 - 3) xo)(x - X2) = (1.2 - 1)(1.2 - 3) = 0.36, XO)(XI - X2) (2 - 1)(2 - 3) xo)(x - Xl) = (1.2 - 1)(1.2 - 2) = -008 XI)(X2 - xt) (3 - 1)(3 - 2) ..
f(1.2) :: ¢(1.2) = 0.72 x 0.0175 +0.36 x 0.0349 + (-0.08) x 0.0523 = 0.0210. Ex. 2 For the following table find a polynomial by Lagrange interpolation formula and hence find the value of Y when X = 1.5. X
Y
:
0 3
1 5
2 9
3 15
12.7
CH.12: NUMERICAL METHODS
• SOLUTION: Here Xo and Y3 = 15.
= O,Xl = 1,X2 = 2,X3 = 3 and Yo = 3,Yl = 5,Y2 = 9 n
The Lagrange interpolating polynomial is y(x)
= I: Li(X )Yi. i=O
Now, Lo(x)
LI(X)
(x - Xl)(X - X2)(X - X3) (x - 1)(x - 2)(x - 3) (xo - ;VI)(XO - X2)(XO - X3) (0 - 1)(0 - 2)(0 - 3) x3 - 6x2 + llx - 6 -6 (x - O)(x - 2)(x - 3) x3 - 5x2 + 6x
~--~~---=~--~~=
=
..:.,---~-~--:'-=
(1 -0)(1 - 2)(1 - 3)
2
(x - 0) (x - 1) (x - 3) x3 - 4x 2 + 3x -':-(2---0-'-)(:-2--~1)'-:'-(2---3-:-:'-) = - 2 (x - 0) (x - 1) (x - 2) x 3 - 3x 2 + 2x -':---'--'---~--7- = - - - - - (3 -0)(3 - 1)(3 -2) 6
L 2(x) L3(X)
Therefore,
Vex)
+
x 3 - 6x 2 + llx - 6 x 3 - 5x 2 + 6x x3+ 2 x5 -6 x 3 - 4x 2 + 3x x 3 - 3x 2 + 2x ----- x 9 + x 15 = x 2 + -2
6
X
which is the required polynomial. y(1.5) = (1.5)2 + 1.5 + 3 = 6.75.
12.2.2
Finite differences
Forward differences The forward difference operator is denoted by. ~ and defined as ~f(x) =
f(x
+ h) -
f(x),
h is the spacing. Substituting, x = Xo and letting f(xo) = Yo, f(xo + h) = f(XI) = YI· Then ~Yo = YI - Yo· Similarly, ~YI = Y2 - YI, ~Y2 = Y3 - Y2 and so on. These differences are called first order differences. The second order difff'rences are ~2yO = ~(~Yo) = ~(YI - Yo) = ~YI - ~Yo = (Y2 - YI) - (YI - Yo)
+ 3,
12.8
U.G. MATHEMATICS (SHORT QUESTIONS = Y2 - 2YI
Similarly, 6. 2 YI
AN~
ANSWERS)
+ YO·
= 4-(6.YI) = 6.(Y2 -
= Y3 - 2Y2
+ YI
YI) = (Y3 - Y2) - (Y2 - YI) and so on.
In general,
Forward difference table (Diagonal difference table) 6.'ly fj. ?'Y 6.y 6. 4y x Y Xo Yo 6. Yo 6. 2yO Xl YI 6.YI 6. 3 Yo 6. 4yO X2 Y2 6. 2 YI 3 6. YI 6.Y2 2 X3 Y3 6. Y2 6.Y3 X4 Y4 Ex. 3 Find a forward difference table for the following data.
2
X Y
3 18
10
4 22
5 35
6
48
• SOLUTION: The forward difference table is X Y 6.y 6.'ly !:l. ?'Y !:l. 4y 2 10
8 3
18
-4 13
4 4
9
22
-22
13 5
35
-9 0
13 6
48
Backward differences The backward difference operator is denoted by \l and is defined as \1 f(x) = f(x) - f(x - h),
h is the spacing.
CH.12: NUMERICAL METHODS
12.9
Denoting f(XI) = Yb Xl - h = XO, f(XI - h) = f(Xn) = Yo. Then \/ f(XI) = f(xt) - f(XI - h) or, \/YI = YI - Yo· .Similarly, \/Y2 = Y2 - YI, \/Y3 = Y3 - Y2 and so on. These differences are the first order backward differences.' The second order backward differences are \/2Y2 = \/(\/Y2) = \/(Y2 - yd = \/Y2 - \/YI = (Y2 - YI) - (YI - Yo) = Y2 - 2YI + Yo = t::.. 2yo. Similarly, \/2Y3 = Y3 - 2Y2 + YI = t::.. 2YI, and so on. In general, \/2Yk = Yk - 2Yk-1 + Yk-2 = t::.. 2Yk_2' For any k, \/kYi = Yi - kCIYi_1 + kC2Yi_2 - ... + (-I)k Yi _k = t::..kYi_k. Thus,
Backward difference table (Horizontal difference table) X Xo Xl X2 X3 X4
Y Yo YI Y2 Y3 Y4
\/y
'\i2y
\/3 y
\/4 y
\/YI \/Y2 \/Y3 \/Y4
\/2Y2 \/2Y3 \/2Y4
\/3 Y3 \/3 Y4
\/4Y4
Ex. 4 For the following data write a backward difference table:
0 20
X
Y
•
2 15
4 12
6 7
8 3
SOLUTION: The backward difference table is X 0 2
4 6
8
12.2.3
Y 20 15 12 7 3
\/y -5 -3 -5
-4
\/'ly
\/3 y
2 -2 1
-4
Properties of
3 ~
\/4 y
7
and \/
(i) t::..e = 0 and \/e = 0, e is a constant. Let f(x) = e. Then t::..e = t::..f(x) = f(x + h) - f(x) Similarly, \/ e = O.
= c-
e
= O.
12.10
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
(ii) Ll{kf(x)} = kLlf(x) and V{kf(x)} = kVf(x). Let g(x) = kf(x). Ll{kf(x)} = Llg(x) = g(x + h) - g(x)
= kf(x + h) - kf(x)
= k{f(x
+ h) -
f(x)}
= kLlf(x).
Similarly, V{kf(x)}
= kVf(x).
+ c2h(x) + ... + ckfk(X)} + c2Llh(x) + ... + ckLlfk(X) and V{Clit(X) + c2h(x) + ... + ckfk(X)} = Cl V it (x) + C2 V h (x) + ... + Ck V fk (x). Ll{J(x)g(x)} = f(x + h)g(x + h) - f(x)g(x) = f(x + h)g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x)g(x) = f(x + h){g(x + h) - g(x)} + g(x){J(x + h) = f(x + h)b.g(x) + g(x)Llf(x).
(iii) Ll{cdl(X)
= ctLlit(x)
(iv)
(v) b..V = Ll- V. Ll.Vf(x) = Ll{J(x) - f(x - h)} = Llf(x) - b.f(x - h) = .{f(x + h) - f(x)} - {J(x) - f(x - h)} = Llf(x) - V f(x) = (Ll- V)f(x)
or, Ll.V = Ll- V. Ex. 5 If f(x) = x 3 then show that Ll3f(x) = 3! h3. •
Llf(x) = f(x + h) - f(x) = (x + h)3 - x 3 = x 3 + 3x2h + 3xh2 + h 3 - x 3 = 3x 2h + 3xh 2 + h 3 = g(x) (say). 2 Ll f(x) = b.{Llf(x)} = Llg(x) = g(x + h) - g(x) = 3h(x + h)2 + 3(x + h)h2 + h 3 - 3x 2 h - 3xh2 - h3 = 3h(2hx + h 2) + 3h 3 = h(x). Ll3 f(x) = Ll{b. 2f(x)} = Llh(x) = h(x + h) - h(x) = 3h2{2(x + h) + h} + 3h3 - 3h2(2x + h) - 3h3 = 9h 3 - 3h 3 = 3!h3. SOLUTION:
12.2.4
Shift operator E
.The shift operator E is defined as
Ef(x) = f(x
+ h)
f(x)}
CH.12: NUMERICAL METHODS
tl.f(x)
= f(x + h) -
f(x)
= Ef(x) -
f(x)
= (E -
12.11
1)f(x)
== E - 1 tl. + 1 == E.
or, tl. or,
E2 f(x) = E{f(x + h)} = f(x E3 f(x) = f(x + 3h).
+ 2h).
In general, Ek f{x) = f(x + kh). Similarly, E- k f(x) = f(x - kh). Properties of operator E (i) Ec = c.
(ii) E{cf(x)} = cEf(x). (iii) EP Eq{f(x)} = Ep+q f(x)
= EqEP{f(x)}. (iv) E{cdl(X) + c'4h(x) + ... + ckfk(X)} = c1E{h(x)} + C2 E {h(x)} + ... + CkE{fk(X)}.' (v) Etl. == tl.E Proof. Etl.f(x) = E{f(x + h) - f(x)} = f(x = tl.f(x
+ h) =
+ 2h) - f(x + h)
tl.Ef(x).
. Etl. == tl.E. (vi) 1 + tl. == e
hD
,.0 ==
tx'
Proof. By Taylor's series,
f(x+h) or Ef(x)
= = = =
Thus E
-
f(x) + hJ'(x) +
:
0 0
1 3
2 8
h3
h2 h3 f(x) + hDf(x) + 2fD2f(x) + aTD3f(x) + ... h2 h3 (1 + hD + _D2 + _D3 + .. ·)f(x) 2! 3! hD e f(x) ehD .
Ex. 6 Find the missing term in the table:
x f(x)
h2
2f r(x) + aT flll(X) + ...
3 15
4 -
12.12 •
U .G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
f(x) = ax3 + bx2 + ex + d. f(O) = d = 0 f(l) = a + b + e + d = 3 f (2) = 8a + 4b + 2e + d = 8 f(3) = 27a + 9b + 3e + d = 15
SOLUTION: Let
(i) (ii) (iii) (iv)
Using (i), (ii) to (iv) become,
(v) (vi) (vii) (viii)
a+b+e=3 4a+ 2b+e = 4 9a + 3b + e = 5
Subtracting (v) from (vi) we get 3a + b = 1 and from (vii) and (vi), 5a + b = 1 Solving, we get a = 0, b = 1. From (v), e = 2, d = O. . '. f(x) = x 2 + 2x. Hence, f(4) = 16 + 8 = 24.
12.2.5
(ix)
Newton interpolation formulae
Newton forward interpolation formula The Newton forward interpolation formula for equispace points xo, Xl,
... ,X n
IS
¢(X)
u
=
X - h Xo ,Xi
u(u - 1)
Yo
+
u(u - 1)(u - 3) 11 3 3! Yo
+ h were
+ ullyo +
=
=
2
2!
Il Yo
+ ...
u(u-l)(u-2)···(u-n-l)
, n.
+ z'h' ,z =
Xo
n
Il Yo,
0 , 1, 2, ... , n.
Newton backward interpolation formula The Newton backward interpolation formula for equispace points xo, Xl, ... IS
¢(X) + +
h were v =
X -
v(v+l)
2
+ vVYn + 2! V Yn v(v + l)(v + 2)V 3 Yn + ...
Yn
3! v(v + 1)(v + 2)··· (v , n.
h Xn ,Xi = Xo
+ z'h',z =
+ n=1)~-m
0 , 1, 2, ... , n.
v
Yn,
, Xn
CH.12: NUMERICAL METHODS
12.13
Ex. 7 Given x : Y:
-1 0 1 3
1 10
2 18
find (i) y( -0.5) and (ii) y(1.8)1 • SOLUTION: The difference table is
2
035 7 1 10 1 8 2 18 ·)H (1 ereu= Therefore,
y( -0.5)
X-Xo
h
-0.5-(-1) 05 1 = ..
=
f)..
+u 1 + 0.5
u(u - 1) f)..2
+ 2! x 2 + 0.5(0.~ -
_
1) x 5 + 0.5(0.5 - 1/(0.5 - 2) (-4) 2. 3. 1 + 1 - 0.625 - 0.25 = 1.125.
x -
y( 1.8) =
_
h
Xn
=
1.8 - 2
1
Yo
u(u - l)(u - 2) f)..3
Yo
Here v = Therefore,
Yo
+
=
= .. ) ( 11
-4
\1 + v(v + 1) \12 v(v + l)(v + 2) r73 + v Y3 .2! Y3 + 3! v Y3 18 + (-0.2) x 8 + -0.2(-~.2 + 1) x 1
Y3
+
-0.2( -0.2 + 1)( -0.2 + 2) (-4) 3!
18 - 1.6 - 0.08
+ 0.192 = 16.512.
Ex. 8 Calculate f(1.1) from the following table:
x
f{x)
:
Yo
= -0.2.
2.
=
3!
0.5 0.22245
1.0 0.25031
1.5 0.27799
2.0 0.30546
• SOLUTION: The forward difference table is
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
12.14
x 0.5
f(x) 0.22245
b.. 2 f(x)
b..f (;r) 0.02786
0.25031
1.0
-0.00018 0.02768
1.5
0.27799
2.0
0.30546
-0.00021 0.02747
Since x = 1.1 which is near to 1.0. We take Xo = 1.0. - 1.0 = 0 2 T h en u = x -h Xo = 1.1 0.5 .. The Newton forward interpolation formula is
+ ub..f(xo) +
f(x)
=
f(xo)
or, f(1.1)
=
0.25031
u(u- 1) 2 2! b.. f(xo)
+ ...
- 1) ( ) + 0.2 x 0.02768 + 0.2(0.2 2! x -0.00021 0.25031 + 0.005536 + 0.0000168 = 0.2558628.
=
Ex. 9 Find the value of y(2.5) using Newton forward interpolation formula. Given that
0 5
x y
1 15
2 30
3 35
• SOLUTION: The given table can be written as X
:
y
:
3 35
2 30
1 15
0 5
The forward difference table is
-5 2
30
-10 -15
·1
15
15
5 -10
o
5
Here u
x-xo
= -h- =
2.5-3 1
= -0.5.
12.15
CH.12: NUMERICAL METHODS
Then from Newton forward formula we have
+u
A
u(u - 1) A2 2! Yo
=
Yo
or, y(3.5)
=
35 + (-0.5) x (-5)
+
Yo
+
y(x)
u(u - 1)(u - 2) A 3 3! Yo
+ -0.5( -2~·5 -
+ ., .
1) x (-10)
-0.5( -0.5 - 1)( -0.5 - 2) 3! x 15
35 + 2.5 - 3.75 - 4.6875
=
+
= 29.0625.
12.3
Numerical Differentiation
12.3.1
Differentiation based on Newton forward interpolation
f '(x)
:
1 [A
h +
f
"(X)
Yo
4u 3
-
+
2u - 1 A2 2 Yo
+
3u
2
-
18u 2 + 22u - 64 A 4 24 Yo 6u - 6 A3 6 Yo
6u + 2 A3 6 Yo
+ ...
]
2
12u - 36u + 22 A4 24 Yo
=
1 [A2 h2 Yo
+
=
1 [ 2 h2 A Yo
+ (u -1)A 3Yo + 6u
+
and so on. . by u = x -h Xo ,Xi = Xo u and Xi are gIven At x = Xo, u = O. Then.
2
-
18u + 11 4 12 A Yo
+ z'h',z =
+ ... ]
+ ... ]
0 , 1, 2, ... , n.
f'(xo) !"(xo) When the point x (at which derivative is to be computed) at the beginning of the table then the above formulae are used.
12.3.2
Differentiation based on Newton backward interpolation
L et v = x -h Xn ,Xi = Xo
f
'() X
-
+ z'h',z =
0, 1, ... , n.
2 !h [t"7v Yn + 2v 2+ 1 t"72 3v + 6v + 2 t"73 v Yn + 6 v Yn
D.G.
12.16
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
3 2 ] + 4v + 18v 24+ 22v + 64 n4 v Yn + ...
1 [V2 6v + 6 V 3 12v h2 Yn+-6Yn+ 2 1 [ 2 3 6v h 2 V Yn + (V + l)V Yn +
J"{x)
and so on. At x = X n , Then
V
2
+ 36v + 22 V4 24
Yn+'"
]
+ 18v + 11 V 4Yn + ... ] 12
= O.
These formulae are used when the point at which the derivatives to be determined at the end of the table.
~~ and ~:;
Ex. 1 Find the values of in the table:
•
x
1
2
3
4
Y
3 19
25
32
SOLUTION:
x
Y
1
3
2
19
3
25
at x
= 1 for the function Y = f{x) given
The forward difference table is
Ay
A'ly
A 3y
16
-10 6
11
1 7
4
32
Here h = 1,x = Xo = 1,u = Then dy
/'(1) =
dx =
X,il!Q
= O.
~ [AYO - ~A2yO + ~A3yO h
2
3
i1[1 16 - 2( -10) + 1] "3 x 11 =
!A4yO 4
24.6667.
+ ... ]
12.17
CH.12: l';UMERICAL ~1T8THODS
Ex. 2 Find
•
~ and ~ at x = 8 for the function Y = J(x) given in the table:
x
2
4
6
8
f(x)
5
20
30
50
The backward difference table is
SOLUTION:
x Y Vy 2 5 4 20 15 6 30 10 8 50 20 Here h = 2,x Then dy dx
V 2y
-5 10
V 3y
= Xn
15 = 8,v =
=
JI (8)
=
=
2"1 [ 20 1 + 2" J"(8)
= O.
h1['VYn + 2"1 'V 2 Yn + 3"1 V3 Yn + ... ] ] x 10 + 1 3" x 15
= 15.
:2 [V2Yn + V Yn + ... ] = ;2
2
d y dx 2
z-hZn
3
=
[10 + 15]
= 6.25.
Ex. 3 A rod is rotated about a fixed point. The angles of rotation dian) at different time t (in sec.) are given below: t
:
f)
0 1
1 5
2 10
3 17
Find the angular velocity of the rod at t = 3 sec. •
SOLUTION:
t
f)
0 1 2 3
1 5 10 17
The backward difference table is
Vf)
V2f)
V 3f)
4 5
1
1.
2.
The angular velocity
1
f)
(in ra-
U.G.
12.18
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Ex. 4 A particle is moving along a straight line. The displacement x at some time instances t are given below:
t
0 5
x
1 8
2 12
3 17
4 26
Find the velocity and acceleration of the particle at t = 4. •
The backward difference table is
SOLUTION:
t 0 1 2
3 4
x 5 8 12 17 26
'\1 2 X
'\13 X
5
1 1
0
f!
i
a
'\1 X
'\1~x
3 4
a
The acceleration 2
d x dt 2
12.4
=
1 [ 2 h 2 '\1 Xn
+ '\1 3 Xn + 11 12'\14 Xn + ... ]
1 [
= 12 4 + 3
] +1 12 1 x 3
= 9.75.
Numerical Integration
Trapezoidal rule
(b
la
f{x) dx =
h
2[yo + 2{YI + Y2 + ... + Yn-l) + Yn],
where h = b~a, n is the number of subintervals, Yi = Xi = Xo + ih, i = 0,1,2, ... , nand Xo = a, Xn = b.
f{Xi),
Geometrical interpretation of Trapezoidal rule In this method, the curve Y = f{x) is replaced by the straight line joining the points A{xo, YO) and B{xI, yt} where a = Xo, b = Xl. Thus, the val ue of the integration f (x) dx is the area of the trapezium form by the lines x = Xo = a, x = Xl = b, x-axis and AB, shown in the Figure 12.4.1.
Y = f{x)
Y
B
.f:
Yo
o
YI X
Xo
Xl
Figure 12.4.1
CH.12:
l":-MERICAL
MF'':HODS
12.19
Simpson's rule
iar f(x) dx = '3h [Yo +4(Yl + Y3 +Y5 + ... +Yn-l) + 2(Y2 + Y4 + ... + Yn-2) +Yn], b
where h = b~a, n is the number of subintervals which is an even integer, Yi = f(Xi), Xi = Xo + ih, i = 0,1,2, ... , nand Xo = a, Xn = b. Geometrical interpretation of Simpson's 1/3 rule In Simpson's 1/3 rule, the curve Y = f (x) is replaced by the parabola passing through the points A(xo, Yo), B(XI, YI) and C(X2' Y2), where a = Xo, b = X2. Thus, the value of the integral f (x) dx is the area bounded by x = Xo = a, x = X2 = b, x-axis and the parabola ABC. The shaded region ofthe Figure 12.4.2 represents the integration obtained by Simpson's 1/3 rule.
Y
J:
o
Xo Xl x2 Figure 12.4.2
x
Weddle's rule
iar
b
3h f(x) dx = 10 [Yo + 5YI + Y2 + 6Y3 + Y4 + 5Y5 + Y6],
where Xo = a, X6 = b, h = Yi = f(Xi), Xi = Xo + ih, i Note 1
6
b a,
= 0,1,2, ... ,6.
(i) The error in Trapezoidal rule is -
of Simpson's 1/3 rule is -
~;t" (e), Xo < e< Xl and that
~~iV(e),xo < e< X2.
(ii) The error terms show that the Simpson's 1/3 rule gives exact result if the function f(x) is a polynomial of degree at most three, while the Trapezoidal rule gives exact result if f(x) is a linear function.
Ex. 1 Evaluate f~(2x - x 2) dx, taking 6 intervals, by (i) Trapezoidal rule, (ii) Simpson's rule and (iii) Weddle's rule . • SOLUTION: Here n = 6-, a = O,b b-a 3-0 So h = -n- = -6- = 0.5.
= 3,y = f(x) = 2x -
x 2.
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
12.20
Xi Yi
Xo 0.0 0.0 Yo
X2 1.0 1.0 Y2
Xl 0.5 0.75 YI
X3 1.5 0.75 . Y3
X4 2.0 0.0 Y4
XSX6 2.5 3.0 -1.25 -3.0 Ys Y6
(i) By Trapezoidal rule: 3 2 h fo (2x - X ) dx = 2"[YO + 2(YI + Y2 + Y3 + Y4
=
0~5 [0 + 2(0.75 +
+ YS) +
Y6]
1.0 + 0.75 + 0 - 1.25) - 3.0]
= -0.125. (ii) By Simpson's rule: 3 2 h fo (2x - X ) dx = 3[YO + 4(YI + Y3 + YS) + 2(Y2 + Y4) + Y6]
=
0~5 [0 +
=
0.5 [ "3 0+1+~ -
4(0.75 + 0.75 - 1.25) + 2(1.0 + 0.0) - 3.0] ] 3 =
o.
(iii) By Weddle's rule: 3 . 2 3h fo (2x - X ) dx = 10 [YO
=
+ 5YI + Y2 + 6Y3 + Y4 + 5ys + Y6]
3 x 0.5 [ 10 0+5xO.75+1.0+6xO.75+0
+5 x (-1.25) - 3.0] = O. l
Ex. 2 Evaluate son's rule .
r 10
- 11 dx, taking n = 4, by (i) Trapezoidal rule, (ii) Simp-
• SOLUTION: Here n
b-a h = -n Xi Yi
+X
= 4,a = O,b = l,y =
1-0
= -- =
Xo 0.0 1.0 Yo
4
Xl 0.25 0.8000 YI
(i) By Trapezoidal rule:
f(x)
= 1!:Z:'
0.25.
X2 0.50 0.6667 Y2
X3 0.75 0.5714 Y3
X4 1.00 0.5000 Y4
12.21
CH.12: NUMERICAL METHODS
r 10
1
1 h 1 + x dx = 2"[YO =
+ 2(Yl + Y2 + Y3) + Y4J
0.~5 [1 + 2(0.8000 + 0.6667 + 0.5714) + 0.5000J
= 0.6970.
(ii) By Simpson's rule:
r
1
10
1 h 1 + x dx = [Yo
a
+ 4(Yl + Y3) + 2(Y2) + Y4] + 4(0.8000 + 0.5714) + 2 x
= 0.:5 [1
+ 1.3334 + 0.5000J =
0.25 = -3- [1 + 5.4856
12.4.1
0.6667 + 0.5000J 0.6933.
Euler-Maclaurin sum formula
The sum of the series n
f(xo)
+ f(Xl) + f(X2) + ... + f(xn)
L f(Xi)
=
;=0 IS
~ -
1:
1
f(x) dx + ~ [f(xn) + f(xo)]
~30 [/,II(Xn) -
where Xi = Xo
+ 1~ [/'(xn) -
f'(xo)]
+ ... ,
f"'(xo)]
+ ih.
Ex. 3 Find the sum of the series using Euler-Maclaurin sum formula. (i) 13 + 23 + 33 + ... + k 3 ") 1 (11 12
1
1
1
+ 22 + 32 + ... + 102'
• SOLUTION: (i) Here f(x) = x 3 ,f'(x) = 3x 2,!,,(x) = 6x,f"'(x) = 6,xo = l,x n = k,h = 1.
=
tlk
x 3 dx
+ ~[f(k) + f(I)J + :2 [f'(k) - /,(I)J
--I-[f"'(k) - f"'(I)] 720
k4
1
1
3
1
4 - 4 + 2'[k + 1] + 12 [3k =
~2 [k 2 + 2k + 1] =
2
1
- 3] - 720 [6 - 6]
{k(k : 1) }
2 .
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
12.22
(ii) Here f(x)
=
1 , f'(x) x2
=-
2 , f"(x) x3
=
6
X4
,
24 f III ( x ) = -s'xo = l,x n = 10,h = 1. x n
1
r
lO
1 il
L.f(xd
1 x2 dx
1
1
+ 2[f(10) + f(I)] + 12 [f'(lO)
- f'(I)]
i=O
.:...2-[flll(10) - flll(I)] 720 1 ] 10
1 [1 102
+2
+ ...
1]
1 [
+ 12 + 12 __ 1 [_ ~ + 24] + . . . ' [ -:;
=
12.5
1
2 - 103
720 105 15 0.9 + 0.505 + 0.1665 - 0.033333
2]
+ 13
= 1.538167.
Solution of Algebraic and Transcedental Equations
Theorem 1 If f(x) is continuous in the interval (a, b) and if f(a) and f(b) are of opposite signs, then there is at least one real root of f(x) = 0 between a
~nd b.
12.5.1
Bisection method
< 0 then there exists a root between a and b of f(x) = O. Let Xl = a; b, the middle point of [a, b]. If f(Xl) = 0 then Xl is a root of f(x) = o. If f(Xl) i= 0 then either f(a).f(xl) < 0 or f(a).f(xl) > O. If f(a).f(xt) < 0, then a root li~s between (a, Xl) and if f(b).f(xt) < 0 then a We know, if f(a).f(b)
root lies between (Xl, b). Thus we reduce the interval from (a, b) to (a, Xl) or (Xl, b). The same method is repeated for the new reduced interval until the root is obtained" to the desired accuracy.
Ex. 1 Find a root of the equation x 2 + X - 7 : ;: : 0 by bisection method, correct up to 3 decimal places . •
Let f(x) = x 2 + X - 7. f(2) = -1 < 0 and f(3) = 5 > O. So, a root lies between 2 and 3. Now we calculate the root by the formula SOLUTION:
CH.12: NUMERICAL METHODS
n 0 1 2 3 4 5 6 7 8 9 10
an (- ve) 2 2 2 2.125 2.188 2.188 2.188 2.188 2.192 2.192 2.192 (
bn (+ve) 3 2.5 2.250 2.250 2.250 2.219 2.204 2.196 2.196 2.194 2.193
12'.23
f(x n+1}
Xn+l 2.5 2.250 2.125 2.188 2.219 2.204 2.196 2.192 2.194 2.193 2.193
1.750 0.313 -0.359 -0.027 0.143 0.062 0.018 -0.003 0.008 0.002 0.002
.'. the root is 2.193 correct up to three decimal places.
12.5.2
Regula-Falsi method or Method of false position
This method is closely related to the bisection method. It is the oldest method for finding a real root of an equation. We choose two points Xo and Xl such that f(xo) and f(XI} are of opposite signs. The equation of the chord joining the points (xo,f(xo}) and (xI,f(xd) is
'y - f(xo} f(xo} - f(xd
=
X - Xo Xo - Xl
This method replace thE' curve y = f(x} by the above chord. The point of intersection of the chord with the x-axis is the approximation to ·the root. To, . find the point of intersection, putting y = 0 to the above equation and let ...... (X2, O) be such point. Thus,
f(xo} X2 = Xo - f(xd _ f(xo} (a: l
-
xo).
Which is the second approximation of the root. If now f(X2} and f(xo} are of opposite signs then the root lies between Xo and X2 and we replace Xl by X2 to the above equation. The next approximation is obtain as
f(xo} X3 = Xo - f(X2} _ f(xo} (X2 - xo). If f (X2) and f (Xl) are of opposite signs then the root lies between Xl and X2 and the new approximation X3 is obtain as
f(X2} X3 = X2 - f(XI} _ f(X2} (Xl
-
X2).
The procedure is repeatE'd till the root is obtained to the desired accuracy.
U.G.
12.24
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Note 1 If the nth approximate root (Xn) lies between an and bn then
Ex. 2 Find a root of the equation x3 + 2x - 2 = 0 using Regula-Falsi method, correct up to three decimal places . Let f(x) = x 3+2x-2. f(O) one root lies between 0 and 1. •
SOLUTION:
n 0 1 2 3 4
an (- ve) 0.0 0.67 0.757 0.769 0.77066
bn (+ve) 1.0 1.0 1.0 1.0 1.0
f(a n ) -2.0 -0.36 -0.052 -0.00724 -0.00097
= -2 < 0 and f(l) = 1> O.
f(b n ) 1.0 1.0 1.0 1.0 1.0
Xn+l 0.67 0.757 0.769 0.77066 0.770882
Thus,
f{xn+l) -0.36 -0.052 -0.00724 -0.00097 -0.00013
a root of the equation is 0.771 correct up to three decimal places.
12.5.3
Iteration method
To find a root of the equation f(x) = 0, we rewrite this equation in the form
I x = ¢(x).
e.
Let Xo be an approximate value to the desired root Putting x = Xo to the above equation, we obtain the first approximation Xl = ¢(xo). Similarly, X = Xl gives X2 = ¢(XI). The successive approximations are given by
This successive iteration repeated till the root is obtained to the desired accuracy.
Note 2 There is no guarantee that this sequence Xo, Xl, X2, ... will converge. The function f{x) = 0 can be written as X = ¢(x) in different ways. Among them we choose a ¢(x) such that 1¢'{x)1 < 1 within the interval where the root lies. This is the sufficient condition. For example, the equation x 3 + x 2 - 1 = 0 has a root lies between 0 and 1. This equation can be rewritten in the following ways:
12.25
CH.12: NUMERICAL METHODS
_ 1 - x2
X---
x2
_ 2 1/3 _ -,x-(I-x) ,x.
H-
x2 _ ~-3 _ - - , x - V 1 - x ,xx
1 .Jl+x' l+x
etc. Ex. 3 Find a root of the equation x 3 + 2x2 - 1 = 0 by iteration method, correct up to four decimal places .
• SOLUTION: Let f(x) = x 3 + 2x2 - 1. f(O) = -1 < 0 and f(l) = 2 root lies between 0 and 1. The given equation can be written in the form x 2 (x + 2) = 1 1 . or, x = JX+2 = ¢(x) say. Now, 1¢'(x)1 =
1- ~ (.t:
+12)3/21
= 2(X: 2)3/2 < 1 for 0 < x < 1.
Here the iteration scheme is xn+1
= ¢(x n ) = vi
The iterations are shown below: n
Xn
¢(x n )
0 1 2 3 4 5
0 0.7071 0.6078 0.6192 0.6179 0.6180
0.7071 0.6078 0.6192 0.6179 0.6180 0.6180
> O. A
1
Xn
+2
, Xo
= O.
a root is 0.6180, correct up to four decimal places.
12.5,4
Newton-Raphson method
Let Xo be an approximate root and Xl be the exact root of the equation f(x) = O. Then Xl = Xo + h, where h is the error. Since Xl is an exact root, f(Xl) = O. Expanding f(Xl) or f(xo + h) by Taylor's series, we obtain 2
2iJ
f(xo) + hf '( xo) + h
"( xo)
+ ... =
o.
Neglecting the second and higher order derivatives, we have
f(xo)
+ hf'(xo) =
0 or h = _ f(xo) .
f'(xo)
12.26
V.G.
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
Then
f(xo} Xl
= Xo - f'(xo}'
To compute the value of h, we neglect second and higher powers of h, so f(xo}.IS no t correct, 1't IS . . t h e vaIue 0 f h = - !'(xo) an approXImate vaIue. S0, Xl obtained from the above equation is better approximation of x.
Note 3 If the derivative of the function f(x} is either zero or tends to zero any point in the neighbourhood of the root then the method fails to give the root. a~
Note 4 This method converges rapidly than other methods. The order of convergence of this method is quadratic. Theorem 2 The rate of convergence of Newton-Raphson method is quadratic. Proof. If e is an exact root then f(e} = O. or, f(e - Xn + xn) = 0 or, f(xn)
+ (e -
f(xn) or, - f'(x } = n
1
xn)f'(x n ) + 2(e
(e -
xn)
-
xn}2 f"(xn)
+ ... =
0
1 } 2 f"(xn}
+ 2(e -
Xn
f'(xn} '
(neglecting third and higher order derivatives).
1
en+!
2
f(xn)
= 2cn f'(x n ) .
So that the Newton-Raphson process has a second order or quadratic convergence. Ex. 4 V se Newton-Raphson method to find a -root of the equation x3 + X -1 =
O.
CH.12: NUMERICAL METHODS
12.27
• SOLUTION: Let f(x) = x3 + X - 1. Then f(O) = -1 < 0 and f(1) = 1 > O. So one root lies between 0 and 1. Let Xo = 0 be the initial root. The iteration scheme is
2x~ 3x;
f(xn) x~ +x n -1 Xn+1 = Xn - f'(x n ) = Xn 3x; + 1 n 0 1 2 3 4
Xn 0 1 0.7500 0.6861 0.6823
+1 + l'
Xn+l 1 0.7500 0.6861 0.6823 0.6823
a root of the equation is 0.6823 correct up to four decimal places.
Ex. 5 Find an iteration scheme to find the kth root of the number a. • SOLUTION: Let x be the kth root of a. That is, x Let f(x) = xk - a. The iteration scheme is
Xn+1
=
f(x n ) Xn - !'(x n ) k
or, Xn+l
=
= a 1/ k or xk - a = O.
Xn
k
k
!
- a _ k xn - xn + a _ [(k _ 1) k-l k-l - k Xn k Xn kXn
_ ·J:: n
.
+ _a_] k-l . Xn .
Ex. 6 Find the square root of the number 2. • SOLUTION: Let x = y'2 or, x2-2 = O. Let f(x) The Newton-Raphson iteration scheme is
Xn+l = Xn -
= x 2 -2. Then f'(x) = 2x.
x2n - 2 x2n + 2 2 = ~Xn 2x n
Let Xo = 1. n 0 1 2 3
Xn 1 1.5000 1.4167 1.4142
Xn+l 1.5000 1.4167 1.4142 1.4142
the value of y'2 is 1.4142, correct up to five significant figures.
Ex. 7 The following expression xn+1 = 3 X~8 + 2 is an iteration scheme to find a root of the equation f(x) = O. Find the function f(x).
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
12.28
• SOLUTION: Let a be the root obtain by performing the iteration scheme Xn+l
=
3x~ + 2 8
lim Xn = a. Then from the given iteration scheme we have n-+oo
lim
n-+oo
or, a
X n +1
=
~8 [3 n-+oo lim x; + 2]
1
= S[3a 2 + 2J
or, 3a 2
+ 2 = 8a or,
3a 2
8a + 2
-
= O.
. "" the required equation is 3x2 -8x+2 = 0 and hence f(x) = 3x2 -8x+2.
12.6
Solution of System of Linear Equations
12.6.1
Gauss-Elimination method
Suppose there are n linear equations with n unknowns , Xl, X2, ... ,Xn . In this method, remove Xl from second equation to last equation by elimination method. Then remove X2 from third equation to last equation. Remove X3 from forth equation to last equation and so on. The variable Xn will present only on tne last equation. That is, last equation contain only X n , (n - l)th equation contain Xn and Xn-l, (n - 2)th equation contain Xn,Xn-I,Xn-2 and so on. The first equation contain all the variables. Now, by back substitution we can determine the values of all the variables. ~
Ex. 1 Solve the following system of equations by Gauss-EliminatlOn method:
+ X2 + X3 Xl - X2 + 2X3 2XI + 2X2 - X3 2XI
=
4
=
2
3.
• SOLUTION: The augmented matrix
A= --+
[~
[~
~ [~
12 4] 2
1 -1
2 -1 1
3 1
-3/2 3/2 1. -2 1
1
-3/2 3/2 0 -1
~] ~ = R, - !R, -1
R3
~]
-1
= R3 -
R3
RI
= R3 + ~R2
CH.12: NUMERICAL METHODS
12.29
The equivalent equations are 2XI + X2 + X3 3 3 --X2 + -X3 2 2 -X3
From the last equation, we get . From the equation
3
X3
3
From first equation 2XI = 4 - X2 Hence the required solution is Xl
12.6.2
4
0 -1.
= 1.
+ '2X3 = 0 or,
-'2X2
= = = . X2
= X3 = 1.
= 4 - 1 - 1 = 2 or, Xl = 1. = 1, X2 = 1, x3 = 1.
- X3
Gauss-Seidal iteration method
For simplicity,
W{;
c0n o;ider the following system of equations:
+ al2 x 2 + al3 X 3 = bl a2l x I + a22 x 2 + a23 x 3 = ~ a3l X I + a32 x 2 + a33x3 = b 3 · an Xl
We assume that one of the following conditions is satisfied by the above system of equations: n
L
\aij\ < \ajj\, j = 1,2,3
I=lj;i:i or n
L
\aij\ < \aii\, i = 1,2,3.
J=lj;i:i
The above conditions are sufficient to converge the iteration process to the solution. The above equations can be written as
1
Xl = -(bl - al2 X 2 - a13 x 3)
au 1
X2
= - ( b2 - a2l X l - a23 x 3) a22
1
X3
= -(b3 a33
a31 X I - a32 x 2).
12.30
V.G. MATHEMATICS (pHORT QUESTIONS AND ANSWERS)
Let X~O), X~O), x~O) be the initial solution. The first approximate solution (x~I), x~I),: is obtained as (1)
xl
(1)
x2
(1)
x3
The (k
1 (
-
=
- ( b2 - a2I x i a22
=
-(b3 -
all
bi -
(0)
=
aI2 x 2
1
(1)
1 ass
(0)
- a13 x 3 )
(0)
- a23 x 3 )
(1)
(1)
aSlxI- -
aS2 x 2 ).
+ 1)th approximate solution is obtained by the following equations ~(bi
=
-
all
1
- ( b2 a22
aI2x~k)
-
(HI)
a2I x i
aIsx~k») -
1 (HI) -(bs - aSIxi -
(HI)
X3
(k)
a2S x s ) (HI)
aS2 x 2
).
aS3
The iteration process will terminate when the consecutive two values of Xl, X2, X3 will be very close.
Ex. 2 Find the Gauss-Seidal iteration scheme to solve the following equations.
Xl + 5X2 + Xs = Xl - X2 + 3xs -8XI + 3X2 + X3 =
8 2 5.
• SOLUTION: The coefficient of Xl. X2, Xs are maximum in third, first and second equation respectively. So we arrange the equation in the following order.
-8XI + 3X2 + X3 = Xl + 5X2 + Xs = Xl - X2 + 3xs =
5 8 2.
Here, 131 + 111 < 1- 81, 111 + 111 < 151, 111 + 1-11 < 131· Thus the sufficient condition for convergence is satisfied. Then
Xl = X2
=
Xs =
.'
1 --(5 - 3X2 - xs) 8 1 -(8-XI-XS) 5 1 '3(2 - Xl + X2).
CH.12: NUMERICAL METHODS Thus for (k
12.31
+ 1)th iteration scheme is -L(5 _
(k+l}
xl
8
3x(k) _ x(k») 2 3
(k+I)
__
51 (8 -
(k+I)
=
~(2 _ (k+l) 3 xl
X2
x3
(k+I) Xl -
(k»)
X3
+ X 2(k+1») •
Ex. 3 Solve the following equation by Gauss-Seidal iteration method. 3XI + X2 = 3, 2XI - 5X2 = 5. • SOLUTION: The given equation can be written as
X2
=
1 -(3 - X2) 3 1 -5(5 - 2XI).
(HI) xl
=
~(3 - x~k»)
=
-~(5 - 2x~k+1»).
Xl
The iteration scheme is
(HI)
X2
Let x~O)
k 0 1 2 3 4 5 6 7
= 0, x~O) = 0 be the initial solution.
Xl 0 1.0000 1.2000 1.1733 1.1769 1.1764 1.1765 1.1765
X2
0 -0.6000 -0.5200 -0.5307 -0.5292 -0.5294 -0.5294 -0.5294
the required solution is Xl decimal places.
= 1.1765,
X2
= -0.5294,
12.7
Solution of Differential Equation
12.7.1
Taylor's series method
We consider the differential equation
y'
dy
.
= -dx = J(x,y)
correct up to four
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
12.32
with initial condition y(xo) = Yo, i.e., when x = Xo then y = Yo. Let y(x) be the exact solution of the above equation. The Taylor's series for y(x) about x = Xo is given by
y ( x ) = Yo
' Xo )Yo
+ (x -
+ (x -2! x~) Yo" + ...
Therefore, determine y(x) we have to compute the value of y", y"', etc.
"
Y
=
y'"
= =
, af af dy , = ax + ay dx = 1:1: + fyy = 1:1: + fyf. (fxx + fxy}) + f(fyx + fyyf) + fy(fx + fyf) fxx + 2ffxy + f2fyy + fxfy + ff;. f
Similarly, we obtain the higher order derivatives of y. Ex. 1 Use Taylor's series method to solve the equation y' = x - y, y(O) = 1 at x and at 0.1.
• SOLUTION: The Taylor's series for y(x) is 2
Y( X) = Yo
3
4
x " x ", X iv + xYo, + 2TYo + 3fYo + 4! Yo + ...
We have
y'(x) = x - Y y"(x) = 1 - y' y'" (x) = -y"
Yo =-1 yg=1-Yo=2 yg' =-2 ybV = 2
yiv (x) = -y'" and so on. The Taylor's series becomes
x2
x3
x4
y(x) =
1- x
+ 2. 2 + 6·(-2) + 24. 2 + ...
=
1- x
+x
2
-
x3
x4
3 + 12
- ...
which is the Taylor's series solution of the given differential equation.
y(O.l)
2
(0.1)3
(0.1)4
+ (0.1) - -3- + 12 - ... 0.1 + 0.01 - 0.0003333 + 0.000008333
=
1 - (0.1)
=
10.909675,
correct up to six decimal places.
CH.l·/
12.7.2
12.33
NUMERICAL METHODS
Euler method
Let Y' = f(x, y) be the given differential equation with initial condition y(xo) = Yo, i.e., when x = Xo then y = Yo. The value of y at x = Xl is given by YI = Y(XI) = Yo + hf(xo, Yo), where Xl = Xo + h. Y2 = Y(X2) = YI + hf(XI, YI). In general, Yn+1 = Yn + hf(xn, Yn), where Xn = Xo + nh. Ex. 2 Find y(O.Ol), y(O.O;2) from the following equation Y' = x2 + y2, y(O) = 1. • SOLUTION: Here Xo = 0, Xl = 0.01, X2 = 0.02, h = 0.01, Yo - 1 and f(x, y) = x2 + y2. YI = y(O.Ol) = Yo + hf(xo, Yo) = 1 + 0.01(0 2 + 12) = 1.0100. Y2 = y(0.02) = YI + hf(XI, YI) = 1.0100 + 0.01(0.01 2 + 1.01 2 ) = 1.0202.
12.7.3
Modified Euler method
In this method, the solution of the differential equation of the form Y' = with y(xo) = Yo is given by (HI) _ h. (k)_ YI - Yo + 2"[f(xo, Yo) + f(XI, Yl )], k - 0,1,2, ...
yi
O
) = Yo where In general,
(HI)
Yi+1
+ hf(:co, yo), -
(0)
and Yi+1
yi
k
)
f (x, y)
is the kth approximation to Yl·
h
(k)
Yt
+ Z[f(Xi' Yi) + f(Xi+l, Yi+1)]' k
Yt
+ hf(Xi, Yi).
_
- 0,1,2, ...
Ex. 3 Determine the value of Y when X = 0.1 and 0.2 given that y(O) = 1 and Y' = x2 - y . •
SOLUTION:
Let h
= 0.1, Xo
= 0, Yo
x2 _ y. Then eO)
Y1
(1)
Yl
(2)
Yl
= 1, Xl = 0.1, X2 = 0.2 and
=
Yo
+ hf(xo, Yo)
=
Yo
h (0) + 2"[f(xo, yo) + f(xl, Yl )]
=
1+
=
Yo
f(x, y) =
= 1 + O.lf(O, 1) = 1 + 0.1 x (0 - 1) = 0.9000.
0~1 [(0 2 -1) + (0.1 2 -
0.9)]
+ ~[J(xo, Yo) + f(Xl, yP))]
= 0.9055.
D.G.
12.34
(3)
Y1
Y1
=
1+
=
Yo
_
1+
MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
0~1 [(02 _
1) + (0.1 2
-
0.9055)] = 0.9052.
h (2) + 2[f(xo, YO) + f(X1, Yl )]
0~1 [(02 -
1) + (0.1 2
-
0.9052)] = 0.9052.
= y(O.l) = 0.9052. (0) - Y1 + hf(xl, Y1) = 0.9052 + O.lf(O.l, 0.9052) Y2 2 = 0.9052 + 0.1 x (0.1 - 0.9052) = 0.8157. (1) Yl + ~[J(Xl, Y1) + f(X2, y~O))] Y2 (2)
Y2
(3)
Y2
=
0.9052 + 0;1 [(0.1 2
=
Y1
-
0.9052 + 0;1 [(0.1 2
=
Y1
=
0.9052 + 0;1 [(0.1 2
_
0.9052)
+ (0.2 2 -
.-.
0.8157)] = 0.8217.
h (1) + 2[f(Xl, yt} + f(X2, Y2 )] _
0.9052)
h
+ (0.22 -
0.8217)] = 0.8214.
(2)
+ 2[J(Xl, Yl) + f(X2, Y2 )] -
0.9052)
+ (0.2 2 -
0.8214)] = 0.8214.
Y2= y(0.2) = 0.8214.
12.7.4
Runge-Kutta methods
Second order Runge-Kutta method Let Y' = f (x, y) be the given differential .equation and initial condition be y(xo) = Yo· Then the value of y at Xl = Xo + h is given by
where kl
= hf(xo, YO) and k2
1
+ 2(k1 + k2), = hf(xo + h, Yo + k1 ).
Y1 = Yo
Ex. 4 Given y' = y2 - x 2, where yeO) = 2. Find y(O.l) and y(0.2) by second order Runge-Kutta method . 2 • SOLUTION: Here h = O.l,xo = 0, Yo = 2,f(x,y) = y2 - x . Then k1 = hf(xo, YO) = 0.1(2 2 - 02 ) = 0.4000. k2 = hf(xo + h, Yo + kd = 0.1 x f(O + 0.1,2 + 0.4000) = 0.1 x (2.42 - 0.1 2 ) = 0.5750.
CH.12:
N~MERICAL
METHODS·
12.35
Yl = Yo + ~(kl + k2) = 2 + ~(0.4000 + 0.5750) = 2.4875, Thus, y(O.l) = 2.4875.
Second part:·· To determine Y2 = y(0.2), we take
Xl
= 0.1 and Yl = 2.4875.
kl
=
hf(x},yt) =0.1 x f(0.1,2.4875) =0.1 x (2.4875 2 -0.1 2 ) =0.6178.
k2
=
hf(Xl + h, Yl + kl) = 0.1 x f(0.2, 2.4875 + 0.6178) 0.1 x f(0.2,3.1053) = 0.1 x (3.1053 2 - 0.2 2 ) = 0.9603.
=
... Y2 = YI + ~(kl + k2) = 2.4875 + !(0.6178 + 0.9603) = 3.2766. Hence, y(0.2) = 3.2766. Fourth order Runge-Kutta method Let y' = f(x, y) be the differential equation with initial condition y(xo) = Yo. Then the YI at X = Xl is given by 1 Yl = Yo + 6(kl + 2k2 + 2k g + k4), where kl = hf(xo, Yo), k2 = hf(xo + h/2, Yo + kl/2), kg = hf(xo + h/2, Yo + k2/2), k4 = hf(xo + h, Yo + kg).
Ex. 5 Given y' = x 2 + y2 with X = 0, y = 1. Find y(O.l) by fourth order Runge-Kutta method . • SOLUTION: Here h = O.l,xo = O,yo = 1,f(x,y) = x 2 +y2.
hf(xo, Yo) = 0.1 x (02 + 12) = 0.1000. = hf(xo + h/2, Yo + kl/2) = 0.1 x f(0.05, 1.05) = 0.1 x (0.05 2 + 1.052) = 0.1105. - hf(xo + h/2, Yo + k2/2) = 0.1 x f(0.05, 1.0553) = 0.1 x (0.052 + 1.05532 ) = 0.1116. =
kl
k2 kg
=
k4
=
hf(xo + h, Yo + kg) = 0.1 x f(O.l, 1.1116) 0.1 x (0.1 2 + 1.11162 ) = 0.1246.
Therefore, Yl
1
+ 6(k1 + 2k2 + 2k3 + k4)
=
Yo
=
1 1 + 6(0.1000 + 2 x 0.1105 + 2 x 0.1116 + 0.1246) = 1.1115.
12.36
U.G. MATHEMATICS (SHORT QUESTIONS AND ANSWERS)
'Exercise 1
(i) Use Lagrange interpolation to solve the following problems:
(a) Compute (1.015)-49 from the given table of u(x) = (1.015)-x. x 48 50 52 ~ u(x) 0.4893617 0.4750047 0.4610689. (b) Estimate the population in the year 1936 by. appropriate interpolation formula from the following data. Year 1921 1931 1941 1951 Population 20 27 39 52. (in crores) (c) Find the value of v'5.5, given that V5 = 2.236, V6 = 2.449,.j7 = 2.646. (d) Find from the following table the value of y when x = 1.05. x 1.0 1.1 1.2 1.3 y 0.24197 0.21785 0.19419 0.17137.
(ii) Use Newton's interpolation formula to solve the following problems. (a) Find from the following table find the value of y when x = 1.11 x 1.1 1.2 1.3 1.4 y 2.7183 3.3210 4.0552 4.9530. (8) The following table gives the values of x and corresponding f(x). Determine f(0.27) x 0.20 0.22 0.24 0.26 f(x) 1.6596 1.6698 1.6804 1.6912. (c) Find from the following table find the value of y when x = 4.05. x 4.0 4.2 4.4 4.6 y 1.3863 1.4351 1.4861 1.5261. (iii) Use Trapezoidal and Simpson's 1/3 rule to evaluate the following integrals. (a) f~ 1!~2' (b) fOI m;!~:)dx, (c)f~·2In(1 + x2)dx, (d) f~} ~,
1 (e) f01l'0cdx, (f) fOI (1+:;)3/2' (g) f0 1:jx, (h) liS A~:r:' (i) fl1+J;dx, (j) fol VI + x + x 2dx, (k) fOl(y'x + 1)3/2dx, (1) fl sinx 2dx, (m) fo1r/2 )1- 0.162sin2 rp drp, (n) f~ logg~:2)dx, rO,5 dx ( ) r 1r / 2 cosxd () r·5'. /1-0.25:r: 2 d ( o ) JO V(I-:r: 2 )(1-0,75:r: 2 )' p JO 1+:r: x, q JO V 1-:r:2 x, dx ( ) r1r /2 d:r: ( ) J2 d:r: (r ) J2r3 1+\Iiii"i' s JO :r:+v'cosx' t 1 3+2:r:-:r:2 ' (iv) Find a root of the following equations by (a) bisection, (b) iteration, (c) Newton-Rapson and (d) Regula-falsi methods. (a) sinx - x + 1 = 0, correct upto three decimal places. (b) x 3 - 5.2x 2 + 3 = 0, correct upto five significant digits.
CH.12: NUMERICAL METHODS
12.37
(c) X 3 - 5 = 0, correct upto five significant figures. (d) x 3 - 8x - 4 = 0, correct upto five significant figures. (e) x 3 + 2x - 1 = 0, correct upto five significant figures. (f) tan x + x = 0, correct upto five significant figures. (g) x 3 - 5x + 4 = 0, correct upto five significant figures. (h) x 2 = sinx, (between 1/2 and 1) correct upto five significant figures. (i) x = tan x, correct upto five significant figures. (j) xe x - 1 = 0, correct upto five significant figures. (k) x 2 - 5 = 0, correct upto five significant figures. (1) sin2 x = x 2, (m) x 3 + 2x - 1 = 0, (n) 2x -loglO x - 7 = 0, (0) cos x = xe x . (v) Find the solution of the following differential equation by (1) Euler's method and (2) Modified Euler's method (a) Find y(l) taking h = 0.2,.where ~ = xy,y(O) = l. (b) Find y at x = 0.01, when y satisfies ~ = x+y, with initial condition Yo = 1 at Xo = 1 (choose h = 0.01). (c) Find y(O.l), where y' = x 2 + y2, y(O) = O. (d) Evaluate y(O.04), given y' + y = 0, y(O) = l. (e) Compute y(O.02), y(0.04) and y(0.06), given y' ..:. y = 0, y(O) = l. (f) Evaluate y(4.4), given 5xy' = 2 - y2,y(4) = l. (g) Find the values of y(l.l), y(l.2) and y(l.3), given y' + ~ = ~,Y(l) =
O.
(h) Compute y(2.5), given y' = x 2 - y2, y(O) = 5.