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0 then the curve is increasing dx
< 0, then the curve is decreasing in that region.
. · ·· . by -dy 7. TIle po liltS 0f maxima an dminima are given
dx
=
0.
Curvature and Curve Tracing
261
y
~o+-------------------~X
dy dx
-
=
o.gIves
XI,X)
-
If in an interval for x 2
(i)
~
el
~-
> 0 then the curve is concave upwards
(X), X 4 ' .... in -
the figure) and has
a minima in that interval. 2
(ii)
el
~
llx-
< 0 the curve is concave downward (XI' xw .... in the figure) and has a
-
maxima in that interval. 2
(iii)
If at a point
X)
-
(say) in the interval d
~
dx-
=
0 then that point is called a point
of inflection and at that point the curve changes it concavity to the opposite direction.
4.5.1 Example Trace the curve y
X2
+2x
.... ( 1)
=
x+ 1 Solution: Substitution of
X
=0
in (1) gives y
F(x,-y):f:- F(x,-y):f:- F( -x, y) etc 1.
=
:. The curve is not symmetric about any line
can be written as y (x + 1) - x (ie) x
2
-
0:. origin lies on the curve.
2
-
2x = 0
.... (2)
xy - (y - 2x) = 0 Lowest degree term is y - 2x = 0
Equation of the tangent at (0,0) is Y Substitutingy
=
0 in (1) we have x(x+ 2)
= 2x .
= o,x = O,x =-2
Engineering Mathematics - I
262 :. The curve crosses x -axis at (0,0) and (-2,0)
Since x =0 in (I) gives y = 0 only, there is no intercept on y-axis
x :. The curve is defined in the region R -
The
curve
is
not
defined
-I.
at
Hence
it
IS
discontinuous.
{-I}
Coefficient of highest degree term in x «i.e.) of Xl) Hence there is no asymptote parallel to x -axis.
IS
I which is a constant.
Coefficient of highest degree term in y is x + I. :. x + 1=0 is the asymptote parallel to y- axis. From
(2)
dy dx
which
is >
0 always except
at
x
Hence the curve is increasing in the regions. 00
< x < -2, -2 < x < -1, -1 < x < 0 and 0 < x < 00 \ \
\~
\..-
\j.:, \ \ \ \ \ I \
/
\ /
1
2
Fig. 4.7
4.5.2 Example 2
Trace the curve
y-
x +1
- Xl
-1
Solution: ....... ( I)
= -1
Curvature and Curve Tracing
263
°
x= in (I) gives y y intercept is -I.
r (-x,y) = r (x,y) y=
°
gives x
2
=
-I :. Hence curve does not pass through (0,0), and
curve is symmetric about y-axis
°
+ 1= =>
= ± i curve does not cross
x
Coefficient of highest degree term in x is y-I
x -axis.
(from( I»
llence y = I is the asymptote parallel to x -axis Coefticient of highest degree term in y is to y-axis
dy =
Hence y
dx
I
=0
x2 -]
.
gives x
=0
x = ±] are asymptotes parallel II
and Yx=o =
-34 < 0 and =I:- 0
:. The maximum point on the curve is (0,-1) and there are no points of inflection.
y
~
~
l~Y.,
I'J
Ix
I I I
I I
--------1---_ I
----r--------I
I I I
I
I I
I
I I
0
I --------1---
X
I I __ ..JI ________ _ (0,-1) Y=-1
I I I
I
I I I I
I I I
I
Fig. 4.8 The curve is decreasing in the interval
.. an d ·· IS IIlcreaslllg 111
-00
°
a
l
is negative (ie) the curve does not exist for
Tangents at the points
ddx2y ) at x ( 2
=
°
gives
x
2
)
= 0 ~ y = ±x
(-(1,0)
From (I) it can be seen that
Ixl > a
= 0, ± .i2' points of maxima and III inima.
a x = ± J2 to the curve are parallel to x-axis .Further
a . . = J2 IS negative.
in 0< x y
= ±~.x
There are two real, distinct tangents to the curve at (0,0)
y
=0
in (1) gives
The curve intersects x-axis at
x2 ( a - x) = 0 => x = 0, a
(a, 0) 3
There is no asymptote parallel to x-axis as coefficient of x is a con y
I I I I
--~I~--------~~--~~~~----~x I I I l X=-b I
Fig. 4.18 Coefficient of highest degree term in y is b + x
:. x + b = 0 is the asymptotes parallel to y-axis Substituting y
= mx + c
x 3(1 + bm
in (I) and rearranging
2) + .... --0'I.e., V'3"'(m)-0' - gIves m -_+_i - Jb
Hence no oblique asymptote to the curve.
t!) can be written as y regions x > a or x < -b
= ±xJa - x b+x
which shows that y is imaginary in the
275
Curvature and Curve Tracing
Hence the curve completely lies in the region -b < x
dy
From (I) - = ±
dx
The tangent
( - 2X2 - 3bx + ax + 2ab) 2(b+x)
3/2
1/)
~
a
which shows that dy ---) 00 as x ---) a
dx
(a-x)-
x = a to the curve at (a, 0) is parallel to y-axis.
Thus the curve forms a loop between
(0,0)
and
(a, 0) .
4.5.13 Example: Trace the curve x 2(y2 - a 2) - b2/
....(\ )
=0
Solution:
x = 0 in (1) gives only y = O. Curve passes through origin.only and does not intersect either of the axes:
F(-X,y)= F (x,y) = F (x, -y) :. Curve is symmetric about both the axes. 2
Lowest degree terms equated to zero gives a 2x 2+ b y2 showing that there are no tangents to the curve at
=0
(0,0)
Coefficient of highest degree terms in y (i.e.) of y2 is x 2- b 2 :. x
= ±b are the asymptotes parallel to y-axis
Coefficient of highest degree terms in x (i.e) of x 2 is y2 _a 2
= ±a are the asymptotes parallel to x- axis. Substitution of y = mx + c in (1) gives :. y
x 4 .m 2 + x 3 (2mc) +X2 (c 2 _a 2 _b 2m2)_ 2b 2mcx-b 2c 2 = 0
¢3 ( m ) = 2mc does not provide value of c since m = 0
276
Engineering Mathematics - I
±a which are parallel to x-axis.
The asymptotes are y =
Hence there are no oblique asymptotes. From (I) x
~ ±~
by
Similarly for
IYI < a , x
For
-a2
y2
Ixl < b,
is imaginary
y is imaginary. Hence curve does not exist for
Ixl < b,
Iyl x =-a
2
278
Engineering Mathematics - I
Tangents to the curve at x For
= 1 + Fs a (A and B in fig) are parallel to 2
0< x < a
y is imaginary
a<x -=---2 r-+ a SIn dB dB r
Engineering Mathematics - I
294
dO r ff tan¢ = r - = r. 2 = -cot20 = tan(-+ 20) dr _a sin 20 2
tan¢ = co
Thus when O=O}
O=ff Hence the tangents at 0
=0
= ff
and 0
are
.lIar
to initial line
When (i.e.,) the radius vector 0
(': 0
=
=
%is tangential to the curve at the pole
%gives
r
= 0)
= 3ff
Similarly the radius vector 0
also is tangential to the curve at the pole.
4
4.2.6 Example: Trace the curve
r - a cos 30 = 0 - - - -(1)
Solution: F (r, -0) = r -acos30 == F (r,O)
Curve is symmetric about initial line Line 0
=0
intersects the curve at
(a, 0 )
Since Icos 301 ~ 1 , from (1), Irl ~ a .Curve completely lies within the region of the circle r
r=O when 30=0
=a
:.30=(2n+l)ff/2, n=0,1,2, .....
(i.e.) curve passes through the pole when 0
= (2n + 1)ff/6( n = O/oS)
These six lines are tangential to the curve at the pole The part of the curve from 0
= to 0 to
ff is AA, 0 .
6
Curvature and Curve Tracing
295
This has a reflection AA20 in the initial line. Thus AAIO A2A is a loop. F (r,4n/3 -B) = r -acos( 47l' - 3B) = F (r,B) The curve is symmetric about B = 27l'/3
OBIB is the part of the curve
from B = 27l'/2 to 27l'/3. This has a reflection OB2B in line 0 ThusOBI BB2 0 forms a loop.
= 27l'/3
F(r, 8; -B ) = r -acos(87l' - 3B) = F(r,B)
Curve is symmetric about B = 4~ 0= 1t/2 B
.........
A2
..... .........
(a,O) () :::
.... ..... 111t/6
....
(-a,1t/3)
Fig. 4.33 OCIC is the part of the curve from B = 7l'/6 to 7l'/3
and
OC2C is its reflection in B = 4~ :.OCpC20 forms a loop. It can be seen that
OCPC20 is a reflection of OBI B820 in the line B = 7l'
tan¢ =
¢ ~ 00
acos3B -1 = -cot3B -3asin3B 3 when B ~ 0
:. The tangent at (a,O) to the curve is
..LIar
to the initial line.
296
Engineering Mathematics - I
4.2.7 Example: Trace the curve r2 - a 2 sec 2 f)
=0
Solution: F(r, 0) = r2 cos 2 f) - a 2 = 0 - - - -(1)
The given curve can be written as f)
=0
gives r
F(-r,f)
= ±a .Curve docs not pass through the pole.
= F(r,f)
:. Curve is symmetric about the pole
F(r,-f)
=,.2 cos 2 (-f)-a 2 = F(r,f)
Curve is symmetric about the initial line f) F(r,TC -0) = r2
COS
2
=0
(TC -f)-a 2 = F(r,f):. Symmetric about f) = TC 2
2
Since cos 0 S; 1 (i.e.,) IcosOI S; 1 we get r2 2:: a Curve does not exist in the region l' --) 00
,.2 < a
TC as f) --) -
TC
±a
l'
dl'
r=-
df)
J
= a-
1
sec- f) tan f)
2
TC
0
2
:.0 = - is an asymptote to the curve.
2 f)
2
TC
TC
TC
-
-
-
-
6
4
3
2
+ 2a
±J2a
±2a
00
-J} df)
-
dr
l'
= -) cos 2 f)cotf) a-
tan¢=rdf) =cotf)=tan(TC -f)) :. ¢=TC/2 when f)=O showing that dr 2
the tangent ate a, 0) is ~/ar to the initial line .
297
Curvature and Curve Tracing
(-a,O)
-\---~-~I-
Fig. 4.34
4.2.8 Example Trace the curve
,.2 - a
2
cos 2(} = 0
Solution:
() = 0
gives
r = ±a :. curve intersects initial line at (±a,O)
,. = 0
gives cos 2fJ
= 0 ~ B = ± 7r/4
pole lies on the curve
and () = ± 7r /4 are two tangents to the curve at the pole
Icos201 S 1 ~
IrI sa. Curve completely lies within the region of the circle
I'
Curve is symmetric about the initial line. F
(-1',0) =(rr _a 2 cos20 = F (1',0)
F
(r,7r-(})=r 2 -a 2 cos(27r-20)= 0
0 ±a I'
±Jr/6
±7r/4
±a/J2
Curve does not exist in the regions
0
symmetric about the pole F
(1',0)
symmetric about
±7r/3
±Jr/2
imaginary
7r/4 < B < 3Jr/4
and -
0=7r/2
±37r/4 0
3Jr/4 < 0 < -Jr/4
=a
298
Engineering Mathematics - I
The curve forms a loop bctween (a ,0) and (0,0) and has reflection in B = 1(/2 ~
Diffcrentiating (1)2r-
dO
=
0= 0 gives ¢
= -2a 2 sin20
tan¢
dO =r- = dl'
~2 2
a sip 20
tan (1(/2 + 20) =
1(/2
Thus the tangents at 0
=
0 (ie) (a ,0 ) and ( - ( l ,0) are perpendicular to the initial
line. y
x
Fig. 4.35
4.2.9 Example: Trace the curve
r - {a + beosB} = 0
Solution: F (r, -0) = r -
= -eot20
{a + beos ( -B)} = F (r, 0)
Curve is symmetric about the initial line. Since leosBI ~ I we have
IrI ~ a + b
The curve completely lies within the region of the circle. r
= a+b
o =0 gives r=a+b :.curvemeets 0=0 in A(a+b,O) o= 1( gives r = b curve meets B = 1( in B ( a - b, 0) {l -
.. Curvature and Curve Tracing
Thus when When
299
a*- b curve meets the initial line at A and B
a = b we get r = 0 curve passes through the pole.
Curve meets the line B = 1£/2 at (a, n/ 2) . Curve passes through the pole when () = cos--!
(-{l/ b) (since r = 0)
Hence the tangent to the curve at the pole is the line B = cos"! ( (i)
does not exist when a > b
(ii)
is B
-a/ b)
which
cos-!(-l)=n when a=band
=
(iii) exists when a < b
0
0
n/3
1£/2
2n/3
r
a+b
a+b/2
a
a-h/2
dB
tan ¢ = r -
dr
:. ¢
= n /2
= a+hcosB -hsinB
.
showmg that tan¢
(ie) the tangent to the curve at
= 00
(a + b, 0)
when is
n
--
a-b
0
.l/llr
=
0
to the initiallinc.
Further ¢ *- 0 for any value of 0 and hence no tangent to the curve is parallel to the initial line When a =h tan ¢ When Wht1f\
0
=
=
l+cosO . . -cotB/2=tan(n/2+0/2). smO
n we get ¢
=
n (ie) the initial line is a tangent to the curve at (0, n)
a*- h we have tan ¢
~ 00
as () ~ n (i.e.) The tangent to the curve at
(a - b, n) is also perpendicular to the initial line.
300
____E_n--:9:::...ineerin9 Mathematics - I
I (a+b,O) (2a,0)
I I a=b
a>b
Fig. 4.36 (ii)
Fig. 4.36 (i)
a x ± yJii 2
Curvature and Curve Tracing
301
--------------------=-----~------------------------------------
x = 0 in (3) gives y = 0,0,2 a y-axis cuts the curve at (0,2 a) and (0,0) is an isolated point Equating coefficient of the highest dt'gree terms in y to zero gives y = () (i.e.) x-axis is an asymptotes to the curve. F (-x, y)
=
F
(x, y) :. curve is symmetric about y-axis
Differentiating (3) W.r.t 'x',
dy
2x ( a - y) . (dY ) dx = 3/ +X2 -4ay .. dx (O,la) = 0
The tangent at (0, 2a) is parallel to x - axis (0,2a)
y = 2a
o
x
Fig. 4.37 From (3)
J2ay y-a
x = ±y
when y < a or when y > 2a ,clearly x is imaginary (i.e.,) The curve exists only in the region a < y ~ 2a
Exercise 4.2 Trace the following curves 1. r 2 cos20=a
l
4. r=3+2cosO 7. r=asin40
2. r=l+J2cosO 5. r=asin30 l 8. r=a(coso+----) cosO a2 cos20
)0. r = a(I-sinO) 11. r2 = - -
3. r=asin 2 0secO 6. r=a(cosO+secO) 9. r2 =asin20
Engineering Mathematics - I
302
4.3.0 Tracing of curves when the equation is given in parametric form: Suppose the equation of the curve is in the tonn x
= ./; (t)
and y
= ./; (t)
..... (1)
where t is the parameter. The study of the following points are useful for tracing the curve I.
If for some value for I , say I) ,
.I; (t) = 0 and
.t; (I) = (l
then the curve passes through the origin. 2.
If.1; (I) is an odd function and
.t; (I)
is an even function
then the curve is symmetric about y-axis. 3.
If
J; (I)
is even and
J; (I)
is odd
then the curve is symmetric about x - axis 4.
Intercepts on X(Y)a.xis are obtained by solving
.1;(1)=0 (.t;(t) =0)
respectively. 5.
Greatest and least value of
.I; (I)
and
.t; (I)
give the region in which the curve
exists.
It (t)
6.
If
or
7.
(ddxY)I~II = 0
1; (I) tends to
00
as I -) t) then t = t) is an asymptote to the curve.
indicates that t = t) is a tangent to the curve parallel
to the
x-axis. ) -) 00 as I -) ') then I
= I)
is a tangent parallel to the y-axis.
8.
(:
9.
If It(t+a)=j;(1) and .t;(t+a)=.t;(t) then the curve is periodic of period a
10. Sometimes it may be convenient to transform the equation of the curve from parametric to cartesian form for tracing the curve
4.3.1 Example: Trace the curve x
= at 2 ,y = 2at
Curvature and Curve Tracing
303
Solution: Eliminating t from x and Y
l
= 4a 2t 2 = 4a.at 2 = 4ax y
~~+-
(0,0)
__
~
______-+x
l = 4ax ---...;..Fig. 4.38
This is a parabola with vertex at origin y aXIs is a tangent to the curve at the origin.
4.3.2 Example: Trace the curve x = acos/,Y = bsint
Solution: Eliminating t from both x and Y
(= J+( ; J~
1 wh ich is an ellipse (standard fonn) y (O,b)
---------+----~~----~--------~.x (-a,O) (a,O)
(O,-b)
Fig. 4.39
4.3.3 Example: 3
Trace the curve x
1
= (2 ,Y = t-3
304
Engineering Mathematics - I
--------------------------------------
Solution: Eliminating't'
2 y2 =t (I-t/{f
=x(l-73'f =i(3-X)2
9i = X(3-X)2 -----(1)
(i.e.,)
y
=0 gives x =O,x =3.
The curve passes through origin and again intersects x - axis
F(x,-y)
at (3,0).
= 9y2 -x(3 _X)2 = F(x,y)
The curve is symmetric about x - Q.'Cis and hence it forms a loop between (0,0) and (3,0) y
=±
(3-x)Fx 3
which shows that the curve does not exist for x < 0
(i.e.,) to the left of Y-axis Lowest degree terms equated to zero gives x curve at origin.
=0
(i.e.,) y-axis is a tangent to the
Coefficient of x 3 is I (a constant) and coefficient of
i
is a constant.
Hence, there are no asymptotes parallel to either x or y-axis Substituting, y
= mx + c 3
in (I) and rearranging 2
_x + x 2 (9m +6) + ........ = 0
fA (m) =-1
a constant,
fA (m) =0 gives
:. No oblique asymptotes exist
Fig. 4.40
m
=±J{i
Curvature and Curve Tracing
305
--------=-----------------------------------------dy =±![l-X] £Ix 2 ~
Y (d )
=
dx ,~I
elY) ( dx
0:. The curve is extreme at x = 1
1 J3
=+ ___ (3,0)
-
:. J3Y = ±(x - 3) are tangents to the curve at (3,0) 4.3.4 Example: Trace the curve x = a(t + sin I), Y = a(l + cos t)
Solution: For no value of f both x and Y simultaneously vanish :. The curvc does not pass through the origin. x is an odd function while y is even :. The curve is symmetric about y -llxis
x(t) = 0 gives t
=0
.Correspondingly x = 0 and y = 2a
dy =dYidt =_ llsint =-tany,;=tan(TC-t/) dx dx/dt a(i+cost) .2 12 When 1=0 slope tan¢ = tanTC ~ ¢ = TC :. The tangent at (0,2a) is parallel to x - axis
t
-TC
X
llTC
y
_5% -~ -a(~+ 5;) -a(; +1)
° a(l+ ~)
0
0
0
2a
5% a(I+;) a(~+ 5:) ~
a
(/(1+~)
TC
(lTC
°
Engineering Mathematics-I
306
y(t) = 0 gives cosl = -1 ~ I = ±7l",±37l", ..... . (i.e.,) 1= ±(2n + 1)7l"
corresponding x is
n=1,2,3, ....
±a7l",±3a7l", ....
Further x ~ 00 as I ~ 00 and
IYI ~ 2a
:. The curve completely lies within the region 0 ~ Y ~ 2a and is period of period 27l" y
Fig. 4.41
4.3.5
Example:
Trace the curve x
= a(t -sinl);y = a(l-cost)
Solution:
X(I)
=0
only when 1=0
y(t) = 0 gives 1= ±2n7l" x(O)
n=0,1,2, ....
= 0 = y(O):. curve passes through origin.
Since Icos'l ~ I we have y 2 0 The curve is completely above x - axis and lies within the region 0 ~ y ~ 2a
(y = 2a when' The curve meets x -
= ±(2n + 1)7l",n = 0,1,2, .. ) axis at x = ±2an7l", n = 0,1,2, ....
y is an even function and x is odd
307
Curvature and Curve Tracing
:. curve is symmetric about y-axis
LI x(t) t~oo
= 00
but LI y(t) is finite l~cIJ
:. There are no asymptotes to the curve
= cot ~ = tan ( ;
tan ¢ = :
Jr
-
~)
I
~¢=---
2 2
The tangents to the curve at t
= ±2nJr, n = 0,1,2, .......
are parallel to y-axis
The curve is periodic of period 21f y
~------~------~--~--~--------~----~x
-4a1t
-2a1t
2a1t
o
4a1t
Fig. 4.42
4.3.6 Example: Trace the curve x
= aeos 3 e,y = bsin 3 e
Solution: The functions cos 3 e and sin 3 e are periodic of period 2Jr . Hence it is sufficient to trace the curve for one period. For no value of
e both
x and
y vanish.
x( e)
=0
gives
e=±~ ,
yeO)
=0
gives
e = O,±Jr
(0,0) does not lie on the curve
Engineering Mathematics - I
308
:. The curve meets x - axis at x = ±a and y-axis at y = ±b Since
Icos
01 s:; 1 and Isin '01 s:; 1 we have Ixl s:; a, Iyl s:; b
3
e
0
Jr/6
x
a
3J3 --a
y
h
8 dy
tan ¢ = -
dx
tan¢
a/8
0
3J3 b
8
0
Jr/2
=0
2Jr/3
5Jr/6
Jr
-2J3a
8
1 -b
Jr/3
b
= - - tan
when
a
-a/8
8
~J3 b
-a
b
-
8
8
0
e
e = ±(2n+ l)Jr
(i.e.,) The tangent to the curve at these points is parallel to x - axis tan ¢ ~ 00 when
e
~±
2n+l) Jr
(-2-
:. The tangent to the curve at these points is parallel to y-axis It can be observed that for corresponding to x for some and corresponding to x for some
e there is
-
y
for - e .
:. The curve is symmetric about both the axes (O,b)
o =...nE-_ _-+ __~3!Jii-(a,o) (-a,O)
0=0
0= -n12 (O,-b)
Fig. 4.43
e there is -x
for Jr -
e
Curvature and Curve Tracing
309
4.3.7 Example: Trace the curve x = a[ cose +
~ log tan ~ J,:I = asin e 2
Solution:
yeO) = 0 but x(O):f:- 0 . (0,0) docs not lie on the curve yeO) = 0 gives
e= 0
=a
and lyllll'Lx
x is an even function and y is odd. So the curve is symmetric about the x - axis Corresponding to each x( e) there is -x for 1C -
e
:. Curve is also symmetric about y-axis Since Isin el -(I
S;
S; 1
we have
IYI S; a.:.
The
curve entirely lies in the region
Y S; a .
x( Similarly,
~) = a [ cos ;
+ log tan : ]
= a [0 + log 1] = 0
x[ -;] = 0
Corresponding y values are y(; )
=a
and y( -; )
=-£1
Thus the curve intersects the y -axis at (0, a) and (0, -(I)
tan~ =
2 Hence,
¢ =0
1
l
a -sine+ - .~ [ = sinti = cos' 0 acosO
J
= cot 0 = tan (1C _
sinOcosO
e
when
e = ~ and ¢ = 1C
when
e = -~
2
e)
Engineering Mathematics - I
310
(i .e.,) The tangents to the curve at Further .". X -
e~
00
or
-00
e =:= ± 7r/2
according as
are parallel to x - axis
e ~ 0 or 7r
axis is an asymptote to the curve.
(O,-a)
Fig. 4.44
4.3.8 Example:
3al 2 l+t'
3at l+r'
Trace the curve x = - - , ,y = - - , a > 0
Solution: X
3
3 3 = (3a)3 - , .I,y = (3a)3 -3 .1 6
l+r
.". x 3 + y3
1+1
3 3 3a )3 27a t 3 ) = = ( __ 1\1 +/ 3 3
?
1+1
(l+t
t
3at 3at 2 l+t l+t
=3a'--3 ' - - 3 = 3axy 3
Thus the Cartesian form of the curve is x + For tracing the curve see example 1.3
4.3.9
Example:
Trace the curve
x
=
a(l-t 2 ) y 1+ t 2 '
2bl
= - -2 1+ t
i
=
3axy
Curvature and Curve Tracing
311
Solution:
(1_12]2 i
2
x a2 Thus
= 1+t 2
'
2 2 ~ + L -_ I a
b
b2 ( see examp Ie 3.3 .L-'"')
2
Exercise 40
= aCt +sin t),y = a(l + cost) x = a sin 21(l + cos 2t), Y = a cos 2t(l- cos 2t)
I. x
2.
3. x = a[ cosO -log(l +cosO) ],y
= asin 0
4. x = aU -sint),y = a(l-cost) Exercise 3.1 I . An asymptote to the curve (i) x-a=O
i (a + x) = x (3a - x) is
(ii) x+a
2
=0
(iii) x-3a=O
(iv) none of these
I ADS: (ii») 2. The curve a 2y2 (i) x-axis
=x2 ( a2 -
Xl) is symmetric about
(iii) both the axes
(ii) y-axis
(iv) none of these
I ADS: (iii) I 3
3. The curve x +
i = 3axy
is symmetric about the line .............. .
[ADS: y = x) 4. If the tangents to the curve
i
= (x - a)( x - b)2 ,( b > a,)
at the point x = bare
inclined at angle 0, tan 0= .............. .
[ADS: 5. The two tangents to the curve
i
±.jb-a )
= x 3 at the origin are real and distinct. I ADS: false)
Engineering Mathematics - I
312
6. For the curve r
= a sec B tan B the origin is a cusp. [Ans: true)
7. The curve x
= a (t + sin t), Y = a (1 -
cos t) is symmetric about x-axis. [Ans: false)
8. x - 2a = 0 is an asymptote to the curve
l
(2a - x) = x 3 [Ans:true)
5 Application of Integration to Areas, Lengths, Volumes and Surface areas 5.1
LAGRANGE'S METHOD OF UNDETERMINED MULTIPLIERS The definite integrals are useful in formulating important physical applications like (i) lengths (ii) areas and (iii) volumes and surgace areas of sol ids otrevolution. 5.1.1 (a) The area bounded by a curve y = f(x} , the axis of x and the two ordinates x = a and x = b is given by the definite integral. S:ydx= J:f(x)dx y
= f(x)
. x'-~----!---
o
"'---=--x
y'
Fig.5.1.1(a)
Similarly the area bounded by a curve x = fry) the axis of y and the two abscissae y = c and y = d is given by the definite integral.
S:xdy = S:j(y)dy
Engineering Mathematics - I
314
x' ---:,-+----------------, o x y'
(b)
Fig.5.1.1(b) If the equation of the curve is in prametric form, then the area =
f
ll
I,
(c)
dx y - dt or dt
f
l
II
2
dy x - dt dt
where tl' t2 are the limits of integration depending on the boundary of the curve. The area enclosed by the polar form of the curve r = f(B) and two radii vectors (} = a, (} = f3,, is
= fPCd(}=~ a
2
2
fP[r«(})Jd(}
(area of sector OPQ
a
o~~----------------
Fig.5.1.1(c)
5.1.2 Example Find the total area within the curve c?y
"
=
c?x2 -
x4
,,
Y
/ , / 'y
Y = - x" " " "", ,,
,
=x
...........8
A
x ' ---~---~~~~~~~-~x
,,
,,
,
y'
Fig. 5.1.2 --
,,
,
,,
=
1 2
2
- r d(})
Application of Integration to Areas, Lengths, Volumes and Surface Areas
315
Solution: The curve has two equal loops between the lines x = a and x = -a and is symmetric about x axis. The total area within the curve = 4 (the area of the half of the loop i.e., 'OABO')
= 4 Laydx =4
ax
1~2--2
fo -va a
-x dx
=~x_l fa~2_x2j2(_2x}lx -2
a
0 3
( 2 _ -2\a -x 2 )-2 4( 2~=--\O-a J2 3 3 a 2 o
=
4
3a
2
Sq.units
5.1.3 Example: Find the complete area of the curve a 2y2
=
:x3(2a - x)
y x = 2a B
x'
x
0 (0, 0) A(2a,0)
y'
Fig. 5.1.3 Solution: The curve is symmetrical about x-axis it cuts the x-axis at the points 0(0,0), A(2a,0). The curve consists of a loop lying between x = 0, x = 2a The required area =20ABO 2
= 2 =
Substituting x
fo ;dx
2f =
2a X 3/ 2.J2a - x
o a 2a sin2 ()
dx
Engineering Mathematics - I
316
We have
dx = 4as in e cose x=O=>O=O x=2a=>O=Jr/2
Area
=~ a
=
Jll 0
32a 2
2
(2a)12 sin 3eEa cose.4asine cose de
J:
2 sin
4
e cos 2 e de
/2 2 = 32 a 23.1.1 --Jr = 1111 Sq. units
6.4.2 5.1.4 Example: Find the area of the segment cutoff from the parabolay y = 4x -1. Solution: The given curves are y = 2x Y = 4x - 1
=
x'-----Ib-~_1_-=--------
2x, by the straight line
..... (i) ..... (ii)
x
y'
Fig. 5.1.4 Solving (i) and (ii) we get the points of intersection as
A(.!2'I}'Lll8' J.! -.!} 2 The required area
=
area AOBA (shaded region)
= area CBEAOOC - area BOADOCB =
J ! {y + 1)dy- J 1 -124
1 -12
y2 dy (from equations (i) and (ii) respectively), 2
~ ~[~2 +yL -[Y:L 9
= 32 Sq. units
Application of Integration to Areas, Lengths, Volumes and Surface Areas
5.1.5 Example: Find the area between the curve.xlY
=
a2(y
y
- x 2)
and its asymptotes.
B
x' ---j===::;>+-:;::::=::::f-..!..--- x
'y
Fig. 5.1.5 Solution: The given equation can be written as
~2X2 2
y2{a 2 _X2)= a 2x 2 => y2 =
a- -x The asymptotes to the curve are given by
a 2 _x 2 = O=> x::;:±a The curve is symmetric about both the axes and passes through the origin. The shaded portion is the required area. The complete area = 4(areaOAB in the first quadrant)
= 4 Laydx =
4J" 12ax 0
= -
va -x
2a X
~2
2
dX=-2J" -2ax ,~dx oi2 2)1/~
2 - X )
1
12 a
10
-x
= -4a[:2 -J a -
x
2
Jl0I
2 =
--4a[O - a]
=
4a2 Sq.units.
5.1.6 Example: Find the area bounded by the curves
y
= 9x, x 2 = 9y,
Solution: Solving
y
=
9x and.xl
=
9y
The points of intersection of the two curves are 0(0,0) A(9,9)
317
318
Engineering Mathematics - I
x'----+----~=F=-.:=--~-----x
Fig. 5.1.6 The required area = area OBACD =
area ODACO - area OBACO
S:=oydX - S:=oydx (parabola y = 9x) (parabola x 2 = 9y)
=
X2 = f 93 J";dx- f9 dx = 27 sq. units 009
5.1.7 Example:
Find the area bound by the cissoid x
=
asin 2t, y
=
asin 3 t . and Its asymptote. cost
Solution: The equation of the curve is
x
=
asin 2t, y
asin 3 t
= ---
cost y
x=a x,-----l..~;;;;;;;~~~--x
o
y'
Fig. 5.1.7
y
=
a 2 sin 6 t cos 2 t
a-x
Application of Integration to Areas, Lengths, Volumes and Surface Areas
319
The equation of the given curve is transformed into cartesian form. The curve is symmetric about x-axis and x = a is the asymptote. The required area = 2(shaded area)
=2Jao ydx =2J: _x_ a-x 0 (
Writing
x
1/2 )
' dx
=
asin 2 B, dx=2asinBcosBdB
=
4a
2fl! 2Sin. 4ede = 4a 2.-.-.3 I n 042 2
4 5.1.8 Example: Find the area included between the cycloid x = a(e - sinO), y = a(1 - cosO) and its base. Solution: The equation ofthe curve is x = a(B- sinO) y = a(1 - cosO) y
x ' - - - - - - - - - : : ' I ' -.......~.........'-'-'~.........'---'-:::-...".----- x
e =0
:
A(2an ,0)
'y
Fig. 5.1.8 The required area = area OABO =
f02;, dx
=
J2l! Y dx .dO o
dO
=
f ;(1 -
=
4a
2 0
cose )a(I - cose)de
2f 2l! Sin. 4-de e = 8a- fl!·Sin 4-de e ?
o
2
e = 2n
0
2
Engineering Mathematics - I
320
f
",2
? 8a-.2
=
371a2 Sq.units
0
()
sin 4 tdt
=
(taking -=t) 2
5.1.9 Example: Find the area between the curve x
=
a(O + sinO), y
=
a(1 - cosO) and x-axis.
Solution: The required area =
2 area OAB
=
2f"ydx .dO =2f"aQ-cosO)aQ+cosO)dO o dO 0
=
2a .2
2
f
"/2
0
sin 2 0dO y .-------f------. A
x,---=--~~~~~~l--- x o =-1t 0 0 =0 8 'y
Fig. 5.1.9 =
2
I
1t
4a .-.-
2 2
= mJ Sq. units 5.1.10 Example: Find the complete area of the curve given by the equations.
x
=
acos3 fJ, y
=
bs in 3 ()
Application of Integration to Areas, Lengths, Volumes and Surface Areas Y 8(0, b)
x'---~?-+=~----x
A(a, 0)
Y' Fig. 5.1.10
Solution: The required area
= 4 area OABO
=4Jaydx =4Jo °
=
y dx de de
4 I:'2bsin3B~3Bcos2BsinB)
= 12ab =
1t 2
1t/2
4
2
J° sin e cos e 3 1 1
de
1f
12abx-.-.-.-
642 2
3
= 81fQb Sq. units. 5.1.11 Example: Find the area of loop of the curve
-? =
a 2 cos2f)
e =1[/4
e =0
x'----~~--------~~------~~~~
8
c Y'
Fig. 5.1.11
x
321
322
Engineering Mathematics - I
Solution:
The required area =
4 areaOABO
=
2f"4~9 = o
=
"!4
fo
f" 4r2d9
2
0
2
a cos 29 d9
a2
,,'4
= -~in29]o 2
a2 2
5.1.12 Example:
Find the area of the curve r
=
a(l + cosO)
Solution:
B
c
() = 1t
A
Fig. 5.1.12
Required area 2 . (area OABCO) r2 =2 -d9 =
f
"
2
o
= f: a Q+cos9 )2d9 2
=
J:
= 4a 2 = 4a 2 =
8a
4(cos %Jd~ f"o cos ~9 2 f cos t .2dt
a
2
2
2
4
"/2
0
3 I 1r 422
4
3Jlll 2
---=--
2
(where t = 0/2) •
Sq. umts
Application of Integration to Areas, Lengths, Volumes and Surface Areas
323
5.1.13 Example: Find the area of the portion included between the carbo ids , a(l - cosB').
=
a (l + cosB') and
, =
Solution: Required area =
4(OCBDO)
[I
1t 0
=4
2,2 d8
2
1
4-~-~r::---_~ r = aQ
A
+ cas8 ) 8 =0
Fig. 5.1.13
I
a-V -cas8 )2d8
=
2a 2
L1t/2a 2 Q- c~s8 )2d8
=
2a
= 2
1t / 2 0
1 {,
~!2 +~.~}= a2 (3n -8) Sq.units 2
2
{
[8 - 2sin8]
5.1.14 Example: Find the area afthe curve
r2 =
a2sin28.
8=~ 4
8
=n/4 x
Fig. 5.1.14
324
Engineering Mathematics - I
Solution: The total area of the curve = 2 area of one loop of the curve 1 = 2 . -1 fll .!2r 2de = f1l 2? a- sin 2ede
2
=
0
0
2
2
2
2
~ [- cos28] ~ 2 =~ n+ 1]=a2 Sq . units Exercise 5 (A)
i
2
1.
x Find the whole area of the ellipse ~ + [;2 = 1
2.
Find the whole area of the curve x 2(x 2 + .I)
3.
Find the area of the curve a2x 2
=
a 2(x 2 -
(Ans:
.I)
1t
ab]
[Ans: cl(Jr - 2)J =
y(2a - y)
5.
(Ans: Jra2 ] Find the area bounded by the curve xy = 4el(2a - x) and its asymptote. (Ans: 4Jra 2 ) Find the area included between .I = 4ax and y = mx
6.
8a 2 (Ans: - 33 ) m 4a(x + a) and
4.
Find the area included between the parabolas .I 4b (b - x)
=
.I =
(Ans:
7.
~(a + b}Fab )
3 Find the area common to the circle x 2 + y2 = 9 and parabola x2 = 8y
. [ADS: "31 {r;:; 2,,2 + 27 Sin 8.
Show that the area of the loop of the curve ely
9.
Find the area of the loop of the curve r
=
-1(2J2i} - 3- ) I 3JTa
=
2
r(2a - x)(x - a) is -8-
asin2B. JTa2
[Ans: 10.
Find the area common to the circles r
=
8
]
aJ2. - 2acosB [Ans: el(Jr - 1)]
Application of Integration to Areas, Lengths, Volumes and Surface Areas
325
5.2.1 Lengths of plane curves:
(a)
The length of the arc of the curve y = fix) included between the points whose abscissae are 'a' and 'b' is
(d)2 dx I+~
s=J h (b)
or
dx
<J
The length of the arc of the curve x = f(y) included between the points whose ordinates are 'c' and 'd' is 2
dx
d
s=J ('
1+dy ( dy )
(c)
The length of the arc of the curve x = fit), y = g(t) included between two points whose parametric values are 'a' and '/3' is
(d)
The length of the arc of the curve r = j(B) included between two points whose vectorial angles are 8/ and 82 is
s=
J-
9??
r-
9,
(e)
2
dr
+( r -
de
)
dt
The length of the arc of the curve 8 = j(r) from r = r/ to r = r2 is given by s=
J,,-,. ,
(de )2 dr
1+ r dr
5.2.2 Example:
Find the complete length of the curve x2(a2
-
x 2)
=
y
8a2
Engineering Mathematics - I
326
Solution: y
8
x'------4--------lIf-------+------ x A(a, 0)
A'(a,O)
'y
Fig. 5.2.2 The curve is symmetrical about both the axis, one loop is formed in between x and x = a and another loop is formed in between x = 0 and x = - a. The total length of the curve Total length = s = 4
f:
=
I+( :
=
0
4(1ength OABO)
J
.... ( I)
dx
The equation of the curve is
:x2(a2 - x 2 )
=
8a2y2
Differentiating with respect to 'x'
16a 2y dy =2xa2 _ 4x 3 dx dy _ x(a 2 _2X2) dx 8a 2 y
r
x2(a2 _2X2 dy )2 1+ ( dx =1+ 64a 4/
Substituting for y2 from the equation of the curve 8a2y2 = x 2(a 2 - x 2 )
Application of Integration to Area!, Lengths, Volutnes and Surface Areas
327
9a 4 -12a 2 x 2 +4X4 8a 2{a 2 -x2)
.... (2)
= s=
.J2 2a2 sin- I (I}=2ha. Jr a
.
2
.J27«1
5.2.3 Example:
Find the length of the loop of the curve
3ay2 = x(x - a)2 Solution: y
8 x'-------*J.U.jtJ.LL~~+_-- X
o
(0,0) 'y
Fit. S.l.3
328
Engineering Mathematics - I
The curve is symmetrical about x - axis, the loop is formed between lines x and x = a, differentiating 3ay = x(x - a)2 w.r., to x.
=
0
= x.2{x-a)+ {x-a)2
6ay dy dx
dy = (x-aX3x-a) dx 6ay dy )2 1+ (dx
=
(x-a)2(3x-a)2 1 + -'---------''------'::--::-----"-36a 2y2
... (1)
Substituting 3ay2 = x(x - a)2 in (I) we get
dy )2 {x-aY{3x-aY = I + -'---------''--'--------'-dx 12ax{x-a)2
1+ (-
12ax+9x 2 +a 2 -6ax 12ax
1+
(:r
=
(3~2+;)2
.... (2)
The length of the loop of the curve
=2
s=
r
(3x+a) dx o 2Fa-Fx
~Ja3-Fx+ax-12dx
2,,3a =
1 ,,3a
r:
r::;- L2x
0
32
+ 2ax
12] a 0
= ~ [2a 3/ 2 + 2aFa]= 4~ ,,3a
,,3
5.2.4 Example: Find the perimeter of the loop of the curve 3ay
=
r(a - x)
Application of Integration to Areas, Lengths, Volumes and Surface Areas
329
Solution:
x'-------;~t..J.J,.t..J.J,."'f__:__--
x
Fig. 5.2.4 The curve is symmetrical about x-axis and the loop of the curve lies between x and x
=
a.
Perimeter of the curve
s= 2
=
20ABO
(4a-3x) dx o 2J3a.Ja-x
f
a
s = _1-
afj
fa a+3{a -X~lx 0
.Ja-x
=
-1-[-2a{a-x) 2 -2{a-x)'2] ~
=
Ir:;- [0 + 2afa + 2a 3!2]= 4a/ fj
afj
a-v3
5.2.5 Example: Find the length of the arc of the parabola y2 = 4ax cut off by the line 3y = 8x.
=
0
330
Engineering Mathematics - I
Solution:
...------x
x'-----~!OII"I'!J
Y'
Fit. ,.2.6 Solving the equations of the J>araboLa anct the fine gives the points of intersections
0(0,0) and A
(~: ' 3; )
y=4ax 2ydy =4a dx
2a y
dy dx
-=-
1+
(dy)2 =1+L dx 40 2
The length of the arc OBA of the
s= .
=
_1
[y ~y2 +4a2+ 40 log~~~y24a2 }T2
2a 2 =
a
2
a[~~ + IOg2]
2..
Jo
Application of Integration to Areas, Lengths, Volumes and Surface Areas
5.2.6 Example:
Find the length of an arc of the cycloid x = a(t - sin t), y = a( I - cos t). Solution: y
------~o~--~a-n--~~a~n----A~------x
(2a1t,0 )
Fig. 5.2.6
dx dt
= a(l- cost), dy = asint dt
(~~J +(~J =~2Q-costY+a2sin2t] 1/2 =
2asint/2
The length of an arc of the cycloid
= 2
Io" 2asin 7i dt
=
4a[- 2cost/2] ~
=
8a
5.2.7 Example:
Find the total length of the curve X2/3
y2/3
a 2/ 3 + b 2/ 3 =I
331
332
Engineering Mathematics - I
Solution: y B(O,b) A(a,O) x'-----E:-~-~------x
B'(O,-b)
y'
Fig. 5.2.7 The parametric form of the curve is x = acos 3 (J, y = bsin3 0
The total length of the curve = 4. length OABO
s=4
f1t2[(-3asinecos2e)+~bsin2ecoseJ 2
0
:\2J12
de
Substituting and
clcos 2o + b 2sin 2 0 = z2 (b 2 - a 2) 2sinO cosO dO
s = 12
=
=
2z dz
z 12 Z3/ h z. ( 2 2) dz = 2 2 a ,b -a b -a 3 a
f
h
~ 4 ~ (b b--a-
3 _
a3
)= 4(a
2
2
+ b + ab) a+b
5.2.8 Example: Find the length of the arc of the curve given by x = asin21(J + cos21), y = acos2t(J-cos2t) measured from the origin to any point.
Application of Integration to Areas, Lengths, Volumes and Surface Areas
333
Solution: x
asin2t(1 +cos2t)
=
dx dt = 2acos2t(1 + cos2t) + asill2t( -2sin2t)
and
dx dt
= 2a(cos2t + cos4t)
dx dt
=
y
.... ( I)
4acos3tcost
acos2t(1 - cos2t)
=
dy dl
= -
2asin2t(1 - cos2t) + acos2t(2sin2t)
dy dl = 2a(sin4t - sin2t) = 4acos3tsint The length of the curve from origin to any point
=
L (~;
r r +(;
dt
Using (I) and (2)
f~ J(4a cos3/COS/)2 + (4acos 31 sin 1Ydl s = f'4a cos 3tdt = sin 3/11 s=
lIa
0 3 0
s=
4a . 3 3
-Sin I
5.2.9 Example: Find the perimeter of the cordioid r
=
a(1 + cosO).
Solution:
B
e = 1t
0=0
------~~------~-----------x
Fig. 5.2.9
.... (2)
Engineering Mathematics - I
334
r
=
a( I + cos 0)
dr
:=)
dB
=
-a sin 0
The perimeter of the cordiod =20ABO
+(~
J
=
2 fox
=
2f" Ja 2 Q+ COSO)2 + (- asinO )2dO
=
2
=
1'2
dO
o
L" J2a
2
Q+cosO )dO
f"o 4a COS2~2 dO 0 2 f 2a cos-dO 2
2
x
=
o
,..,
L
X
=
4a.2.sinB/
=
8a
20
5.2.10 Example: Find the perimeter of the curve r
=
2acosB.
Solution: The equation of the curve is r
=
2acosB.
It is a circle passing through pole whose centre is on the initial line at a distance 8/2 from pole.
0=0
Fig. 5.2.10
Application of Integration to Areas, Lengths, Volumes and Surface Areas
335
The perimeter of the curve
=20ABO =2J
n'2
r2+(dr)2dB dB
o
= 2
f: ~(211~-;;~OJ ~-(12
2asinB)1 dO
J(
=
4a2
=
2aJ(
5.2.11 Example Find the length of the arc of the parabola Ijr
Solution: LSL' = 21 is the latus rectum of the parabola.
Fig. 5.2.11 The length of the arc L' AL = 2AL
=2
I
n
o
;2
(dr)2 dB
1 r+ -
(IB
Equations of the curve is Ijr logr =
1 + cosOc logl + log(l + cosO) =
~ dr = 0 _ - sin 0 rdO l+cosO
=
1 -t cosOcut off by its latus rectum.
Engineering Mathematics - I
336
dr -=rtanB/2 dB The length of the arc = 2 =
f
1t/2 ~ 2
?
?
r + r- tan-9/2d9
0
1t/2 0
f rsec9j2d9 I 2 fo 1+ cos9 .sec9j2d9 2 fo 2cos 9 2 .sec9j2d9 9 I sec-.sec 9j2d9 fo 2 2
1t2
=
1t 2
=
/
2
2
1t!2
=
= / L1t/2 ~I + tan 2 9j2 sec 2 9j2d9 writing
tan Bj2
t
=
I
sec 2 B/2.-dB = dt 2
= 2{~F2 +~log(1 +F2)] =
l[ F2 + log(1 +
F2)]
Exercise - 5(8)
1.
Find the length of the arc of the parabola y
=
4ax cut off by its latus rectum. [ADS: 2a[F2+log(l+F2)]}
2.
Find the perimeter of the loop of the curve 9ay
=
(x - 2a}(x - 5ai
(ADS: 4.fja J
Application of Integration to Areas, Lengths, Volumes and Surface Areas
3.
337
Find the length of the arc of the parabola x2 = 4ay from vertex to one extremity of the latus rectum. [Ans: a[ 12 +Iog(l +
4.
Find the length of an arc of the parabola y = x2 measured from the vertex. [Ans:
5.
12)]J
Find that the length of the arc of the curve y
=
~Jl+4x2
+-.!.-sin- I {2x)] 4
2
log tanh (x/2) from x
=
1 to x 2
=
2
I)
e + [Ans: log ( -e- ] 6.
Find the length of the arc of the curve y
=
x(2 - x) as x varies from 0 to 2. [Ans: -.!.-log(2 + 2
7.
Find the length of the curve x
8.
Find the length of the arc of the curve x
B
9.
=
=
a(B + sinO) y =
=
J5)+ J5 ]
a(l- cosO)
eOsinB, y
=
[Ans: 8a) eOcosB from B = 0 to
1[/2
Prove that the loop of the curve (3
X = (2,
10. 11.
Y
= (- -
3
is of length 4fj
Find the perimeter of the cardioid r
=
a(l- cosO)
Show that the arc of the upper half of the curve r
e=
=
[Ans: 8a) a(l - cosO) is bisected by
21[/3
5.3.1 Volume and Surface of solids of revolution: (a)
Volume of the solid generated by the revolution of the area bounded by the curve y = f(x), the x-axis, the ordinates x == a and x == b about the x-axis is
(b)
Volume of the solid obtained by the revolution of the curve x abscissae y = c and y = d about the y-axis
Ld
1tX
2
dx.
=
J:
1t)12 dx.
fry), the y-axis the
338
(c)
Engineering Mathematics - I
Volume of the solid obtained by the revolution of the curve x x-axis is f
'2 ny2
dx dl dt
'] (d)
= f '2 n ~ Cr)r ']
=
}(t), y
=
$(t) about
=
j{t), y
=
¢i...t) about
dfCr ).dl dt
Volume of the solid obtained by the revolution of the curve x y-axis is f'21t\"2 dy dl = f'2n[f'Cr)Y d$Cr).dl
'] (e)
dt"
dl
Volume ofthe solid obtained by the revolution ofthe cure r = j{ fJ) about the initial line IS
=
02
f.
.
f\)2
S n\"sIno
d{rcose )d0 .
de
OJ
(f)
Volume of the solid obtained by the revolutuon of the curve r perpendicular to initial line =
f.
h?
81
=
f
82
f.
0)2
n \" COSo
0]
d{rsine),lO
de
.U'
The volume of the solid generated by the revolution about the initial line (x-axis) of the area bounded by the curve r = j{ fJ) with the radii vectors B = a, B = {3 is =
I
O=~
O=a
(h)
j{fJ) about the line
dv f rrx- dy = 1°2 n \" cose )2 -"-.de dO a
(g)
=
2nr 3 SIlIO '_0 d{\ 0 3
The volume of the solid generated by the revolution of the area about the line B = ni2 (y-axis) of the area bounded by the curve r = f(e) with the radii vectors
B=a, B={3is =
o=~ 2n -r3 cos OdO 8=a 3
I
5.3.2 Example: Find the volume of the solid formed by the revolution of the loop of the curve y(a + x) = x 2(a - x) about x-axis.
Application of Integration to Areas, Lengths, Volumes and Surface Areas
339
Solution: y
= 0 gives x = 0, a
:. The loop is formed between x = 0 and x = a x = - a is the asymptote to the curve. y
x=-a
Fig. 5.3.2 :. The volume formed by the revolution of the loop about x-axis
=
(Jr x2 (a-x)l 1(
a+x
o
ex
~ 1{- ~ + ax' -2a'x+2a'log(a+x)I = 21(a 3 [ log 2a - log a -1 ] 3
= 2Jra [IOg2-1] cubic units. 5.3.3 Example: Find the volume of the solid generated by revolving the curve xy y-axis.
=
4(2 - x) about
Engineering Mathematics - I
340
Solution: y
x' -------,o..-+----f.,-,,;-,~O)r- x
Fig. 5.3.3
The given curve can be written as
xcY+4)=8 The volume of the solid obtained by revolving the given curve about y-axis
Substitutingy
64
J y+4 )2 OC!
= 2n
o
(~
= tanO ~
= 8n
f
1t /
0
2
dy dO
dy
= 2sec 2 OdO
I +cos2e de
1
+2 sin28
=
8n [ e
=
4n2 cubic units.
]
1t
2
0
5.3.4 Example:
The part of the parabola y2 = 4ax cut off by the latus rectum revolves about the tangent at the vertex. Find the volume of the reel thus generated.
Application of Integration to Areas, Lengths, Volumes and Surface Areas
Solution: y
L(a,2a)
S
x'------------+-M+--+-------------x J?--+--4.I
L'
y'
Fig. 5.3.4 Volume of the reel generated
=
[ '}Y y-
211
=2fo
1[-
4a
Jf
Y-1
]2
.J2 cos(}dO =
cos/dt)
2.J27lZJ3 [4--I 4 --3 7C] -3
4 4-22
.J27C 2a3
- - - cubic units. 8
Exercise - 5(C) 1.
Find the volume of the solid generated by revolving the clips :: x (a)
About the major axis
(b)
About the minor axis
~: = I
[ADS: 2.
yea
The curve + x) = by the curve.
4
2
37lZJ b J
r(3a - x) revolves about the x-axis. Find the volume generated [ADS: ~(8Iog2 - 3)J
3.
Show that the volume of the solid generated by the revolution of the curve (a -x)y
7C 2a3
= a2x, a~out its asymptote is -2-. 4.
Find the volume of the solid generated by the revolution ofthe curve y(a2 + x2) about the asymptote
=
a3
Application of Integration to Areas, Lengths, Volumes and Surface Areas
5.
347
Find the volume of the solid obtained by revolving the loop of the curve a 2 == x 2(2a - x)(2a - x)(x - a) about x-axis.
1
(Ans:
6.
Find the vloume generated by the portion of the arc y ==
x
~
0 and x
=
JI + x"
60
lying between
4 as it revolves about the x-axis. (Ans: 767[/3)
7.
Show that the volume of the solid generated by the revolution of the cycloid
. . 3 , 1 87[' IS - 7[" £l - 2 3 Find the volume of the solid generated y revolving the cycloid x y = aU + cosO) about its base.
x
8.
==
. a( 8 + SinO), Y ==
aU -- cosO) about the y-axIs
=
a(8 + sinO), IAns: 5~£l31
9.
Find the volume of the solid generated by revolving
r
=
(Ans:
10.
Find the volume of the solid generated by revolving r
=
a2cos28about initial line. 7[(1'
Ii rl3log"\/2 ( h + 1)-"\/2h] I
12
acos38 between 8 == -7[/6
and e == 7[/6 about the initial line. (Ans:
197[(1' 968 )
5.4.1 Area of the Surface of revolution (a)
Surface area of the solid generated by the revolution about the x-axis of the area bounded by the curves y = fix), the x-axis the ordinates x = a, x = b is
h
I (b)
X=Q
21ty £Is ==
Ii> X=U
ds 21ty - £Ix
dx
Surface area of the solid generated by the revolution about y-axis of the area bounded by the curve y = fix), the y-axis and the abscissae y = c, y == £I is
I
Y=J 21tX cis == IJ
X=Q
Y={
cis (~V dy
21tX -
[ d~== dy
1+(dX)2] dy
Engineering Mathematics - I
348
(c)
The parametric form is
f
'=/2
/=/1
(d)
[ d~
ds .21ty-dt dt
dt
==
(dX)2 + ({~y)2l dt
dt
The polar form is
I - 21trsine (dS-de )de O?
91
5.4.2 Example: ? '0 Find the surface area of solid generated by revolving the curve x-i-' + y'? '3 == a-I?',
about x-axis. Solution:
The parametric form of the curve is x
=
acos3 e, y
=
asin 3 B.
Y B(O,a)
x
x'
A'(-a,O) 8'(0,-a)
Y' Fig. 5.4.2 The surface area of the solid due to revolution about x-axis == 2(surface area of the solid generated by revolving an arc in the first quadrant of the astroid is) =
ds " 21ty-.de f° de dx d 41t f Y - ) +(2) .de ° de' de 2
2
?
2
lt/2
=
=
(
41t
I
It/2
0
asin 3 e ~(\.3acos 20 sine )2 +( 3asin 2 e cose )2de
Application of Integration to Areas, Lengths, Volumes and Surface Areas
=
12na 2
"n
J
0' -
349
sin-le cosO dO ffn
sinSe - 12JlU 2 12JlU [ - 5 - j0 = - 5 - Sq. units. 2
=
5.4.3 Example:
Find the area of the surface formed by the revolution of the ellipse x2 + 4y about its: (a) major axis
(b) minor axis
Solution:
Equation of the ellipse is
ely dx
2
2
16
4
~ +L
=
I
x 4y
16y2
+X2
16/ (a)
Surface area formed by the revolution about major axis -l ds = 2 2ny-dx o dx
J
=
16
350
Engineering Mathematics - I
.Jl[~(~ r Jl] =
-x' +
=
8"[1 + 4"9
(b)
*
~: Sin-{ )I
Sq. units.
Surface area formed by the revolution about minor axis
=
41t
= 41t
=
L2 J
Xl
J:
+ 16yl dy 2
J16-4 y 2 +16y dy
8Jl. s: (1 r
+ y' dy
o
Application of Integration to Areas, Lengths, Volumes and Surface Areas
351
5.4.4 Example: Find the surface area formed by revolving cycloid x = a(B+ sinO),y = a = (/ - cosO) about the tangent at the vertex.
Solution:
e
y
=-1[
e
a1[
=-1[
a1[
0=0
x' - - -.........-.....;~"""'"--..I---- X y'
Fig. 5.4.4 Surface area required
= 2(surface area generated by revolving the arc =
2
I
x
o
= 4 IX II
in the first quadrant about OX)
ds 21ty- dB
de
y)2 de (dX)2 +(d de de
1t)'
. -cos e -e de =161t a-. IX sm?
ry
o
2
2
(%=t => dB = 2dt) = 321tary
32JlU 2
I
= --
3
X/2 0 sin 2
t cost dt
Sq. units.
Engineering Mathematics - I
352
5.4.5 Example: Find the area of the surface of reel thus generated by revolution of the part of the parabola = rax bounded by the latus rectum about the tangent at the vertex.
y
Solution: y
x'------+-~-+_------ x
Fig. 5.4.5
y
Surface area
=
4ax
dy
2a
dx
y
=
2
ds 21tX-dx o dx
f
a
d =2fa 21tX 1+.1:'. ( o dx
=
4n
)2 dx
fo" x.J x 2 + ax dx a
= 4n
J:[(x+~r -(~rl2dx
Application of Integration to Areas, Lengths, Volumes and Surface Areas
=
353
4lT
o = JZa
2
[3J2 -log(J2 + 1)] Sq.units.
5.4.6 Example: Find the area of the surface of revolution formed by revolving the curve r about the initialline~
Solution: Equation of the circle is
r
=
2acosB
dr= -2 ' B asm dB
y
....--x
x'-------~I----o
y'
Fig. 5.4.6
The surface area
=
L
lIi2
=
J" 21ty -dO dO 2
ds
o
dr 21trsinO r2 + ( dO
2 )
dO
J:/ 21trsinOJ4a 2cas 0 + 4a sin 0 dO = 81ta J sinO cosO dO =
2
2
=
2
"/2
0
4mr Sq. units
2
2
=
2acosB
Engineering Mathematics - I
354
5.4.7 Example:
Find the surface area of the solid formed by the revolution of the cardioid = a(J + cosO) about the initial line.
r
Solution:
Equation of the curve r
=
a(J + cosO)
dr . B -=-asm dB
0=0
The surface area =
Io"2ny-de de
Fig. 5.4.7
ds
= 2n
L" rsine
= 2n
I:
r2 +(*
J
2 rSineJa (i +cose
de
y +a 2sin 2e de
L" a(i + cose)sine .Ja 2(i + cose)2 + a = 16na I" COS3~ sin ~ cos ~ de o 2 2 2 e sm"2 . e d8 = 16na I" cos "2 =
2n
2
2
=
0
4
[25 2eJ"
16a 2n - - cos 5 -
0
32mi
= --
5
Sq. units
2
sin 2e de
Application of Integration to Areas, Lengths, Volumes and Surface Areas
355
5.4.8 Example:
Find the surface of the solid generated by revolving the lemniscate,.2 about the initial line. Solution:
Fig. 5.4.8
Equation of the curve is ,.2 Surface area
=
a 2cos2B
" ds fo 21ty-dO dO 4
=
2
41t
In
11/4
dr rsinO r2 + ( dO
1[/4 0 rsinO
f
=
41t
=
41ta 2 [_ cosO ] ~
2 )
dO
~ a4 • ~ r- +-~ sm-20 dO r-
4
=
a 2cos2B
356
Engineering Mathematics - I
Exercise - 5(0) I.
Find the total area of the surface obtained by revolving the ellipse
?
?
a-
b-
x~ + y~ = 1 about
its major axis.
2.
Find the surface ofa sphere of radius 'a'. (Ans: 4Jlu 2 ]
3.
Find the surface of the solid generated by revolving the arc of the parabolay bounded by its latus rectum about x-axis. (Ans:
4.
3 Find the area of the surface generated by the revolution of the cycloid x y = a(l - cost) about x-axis.
=
a(t - sint), 64
3
2
7m )
Find the area of the surface of the solid formed by the revolution of the cardiod r = a(l - cosO) about the initial line. (Ans:
6.
4ax
~7lU2(2J2 -I)J
[Ans: 5.
=
32
2
SJl"a )
The lemniscate? = a2cos2B revolves about a tangent at the pole. Show that the surface area generated is 47lU2 •
5.5.1 Double and triple integrals: Letj(x,y) be a continuous and single valued function of x and y within a region R bounded by a closed curve 'c' and upon the boundary c. Let the region R be subdi vi doo in rrty mrrtne- into n subregi ons of areas 8R\, 8R2 .... 8Rn •
Let (x" y,) be any point in the subregion of area 8Ri . Consider the sum
The limit of this sum as n ~ 00 (i == 1,2, .... ) is defined as the double integral ofj(x,y) over the region R and is written as
fff(x,y}iA
Application of Integration to Areas, Lengths, Volumes and Surface Areas
Hf(x,y}ixdy (a)
I f(x" yJ) R,
Lt
=
n~oo ,=\
II
Suppose the region R is described by the inequalities c ::::; y ::::; d and g(y) ::::; x ::::; heY) y y=d ...--_ _.,....::--....fX = h(y)
x = g(y)
y-c
x'---------+--------x o Y' Fig. 5.5.1 (a) Then t
="
ff f(x,y )dydx=Jff(x,y )dxdy = f fX=II(Y)f(r,y)dXdy x=g(y) II
(b)
Y~
II
Jfthe region R is described by inequalities a ::::; x ::::; band g, (x) ::::; y ::::; h,(x) y
x ' - - - - -....... oof----I----""----x y'
Fig.5.5.1(b) x=h
Then
II
(c)
y=lI\ (.)
ff r(x,y}ixdy = fff(x,y}iydx = f f y=g\ (.) f(x,y)dx.dy II
x=a
If the region R is bounded by the lines x = a, x = b, y = c, y = d (rectangle) Then
Hf(x,y}dxdy II
=
h
"
" h
a
c
( a
f ff(x,y}dydx= f ff(x, y}ix dy
Note: The order of integration is immaterial for constant limits
357
Engineering Mathematics - I
358 5.5.2 Example: Evaluate
I
2
o
I
2 I I{x + y2 }Ix dy
Solution:
I
=
8 I 2dy -+2y 2 ---y 03 3
I
il
o =
3
Y 7 I Y +- dy=-+-yl 3 3 3 0 7
2
J
1 7 8 333
-+-=-
5.5.3 Example: 4
Evaluate
K-~
I
Ixydydx o
Solution: 4
4
J4-x
I'=J4-x
I
fxydydx = f·
f [xydy] dx
I
0
0
4
x=l
y=K-~
f
fxydy.dx
x=l
=
0
2
4f~(4 -X}tx = 4x _~14
9
I
2
x=l
2
4
6
Application of Integration to Areas, Lengths, Volumes and Surface Are."'''
359
5.5.4 Example: 1 2- r
J Jx dXdY 2
Evaluate
o J~'
Solution:
v=1
J1(2- Y )' -l(J.i~)\~v
y~O
J8 - -
)'=1
= -1
3
3
V1
-12)1 + 6v2 --') V- l IV ~
\'~O
=
1[S-1-6+2-lJ 67 60
5.5.5 Example: Find the value of Hxy(x+ y}/xdy taken over the region enclosed by the curves
y
=
x and y
=
x2.
Solution:
y
y=x
x' _ _ _ _ _
~~:-:.....-----
y'
Fig. 5.5.5
x
360
Engineering Mathematics - I
R is the rogion bouuded by the curves y = x and J' =
1=
7
x~
ff'"y(x + y')dxcly II
=
x5 x5 x7 x8 I 10+]5-14- 24 10
=
I I I 1 -+-----
10
15
14
24
3
=-
56
5.5.6 Example: Evaluate
IfR 1- Y
dx{1y where A is the area in the positive quadrant of the circle
A
x2 + y2
=
1.
Solution: y
x'----------~r-~~~~~~~------x
8(1,0)
y'
Fig. 5.5.6
Application of Integration to Areas, Lengths, Volumes and Surface Areas
~ 1- (2
_I
±ofx[~ -/'2-~--] ()
= -
I
£Ix
I
fX{I[-Q-X 2)]2 -I}dx
= --
o I
f{x x}It 2
= -
-
o
-x 3 X-J ]1 - --+-
-
[
3
2 ()
-I
1
1
3
2
6
= -+-=-
5.5.7 Example: Evaluate ffe
tlHhY
dt dy over the triangle x
=
0, y = 0, ax + by = I
Solution:
x:: : Ol----t---" x'------------~----~~----~~----x
~
AU,a)
Fig. 5.5.7
361
362
Engineering Mathematics - I
1 = ffea + i o
0
+ Z1 }lwlydz
0
I [Ans: 20]
3.
, , J-XIY~ dydz
f II' ()
0
I
[Ans:
3"]
122
4.
2
f f fx yzdxdydz (J
U
I
[Ans: I]
III 5.
f f f(x + y + z)dxdydz
()
()
()
[Ans: I
6.
f o
3
2]
JI-x 2 ~~2_y2 f
fxyz dtdydz
0
0
I [Ans: 48]
Application of Integration to Areas, Lengths, Volumes and Surface Areas
7.
3
I
I
I x
Fv
JJ
Jxyz dxdydz 0
(Ans:] ( I
8.
13 1 1og3 ) 9-6 ]
I-x x+y
JJ
o
383
0
z
Je dxdydz 0
1 (Ans: -] 2
9.
4
x
!BY
o
0
0
JJ
Jz dxdydz (Ans: 16]
o
10.
J 02_x 2
J J
o _J02_x2
niX
2
Jz dxdydz o
"This page is Intentionally Left Blank"
6 Sequences of Series 6.0 Sequence A function fN ~ S, where S is any nonempty set is called a Sequence i.e., for each nE N, ::l a unique elementj{n) E S. The sequence is written asj{I),j{2), j(3), ..... j{n).... ,and is denoted by (j(n)}, or <j(n», or (/(n». If j{n) =an ,the sequence is written as ai' a 2 •••••an and denoted by , {an} or < all > or ( all ). Here j{ n) or all are the
d h terms of the Sequence. 6.1.1 Example: 1 ,4,9, 16, ......... n 2
, ••••• (or)
< n2 >
s
Engineering Mathematics - I
386
6.1.2 Example:
~3 ,~,-;-, ~ (or)(~) I 2- J n n •••••
••••
s
N
6.1.3 Example: 1, 1, 1. ..... I.. ... 0r
6.1.4 Example: 1 ,-1, 1, -1, ......... or (( - I
Note
1. 2.
r-
I )
If S c R then the sequence is called a real sequence. The range of a sequence is almost a countable set.
387
Sequences of Series
6.1.5 Kinds of Sequences 1. Finite Sequence :A sequence < all > in which an
= 0 Vn > mEN
is said to
be a finite Sequence. i.e., A finite Sequence has a finite number of terms.
2. Infinite Sequence: A sequence ,which is not finite is an infinite sequence. 6.1.6 Bounds of a Sequence and Bounded Sequence 1. If :3 a number 'M' ~ an:::; M, Vn E N, the Sequence < an > is said to be bounded above or bounded on the right.
Ex: 2. If
I,.!.), ,....... here all :::; 1Vn EN 2 3
:3 a number 'm' ~ a" ~ m, Vn E N, the sequence < a" >
IS
said to be
bounded below or bounded on the left. Ex : 1 , 2 , 3 ,..... here a" ~ 1 Vn EN
3. A sequence which is bounded above and below is said to be bounded. Ex:
Let
an = (-1
n
r (1 +;) 2
3
4
3/2
-4/3
5/4
2-1 __
o
2
3
4
5
n
-1 -
From the above figure (see also table) it can be seen that m :. The sequence is bounded.
=
-2 and M
=
l. 2
Engineering Mathematics - I
388
6.1.7 Limits of a Sequence A Sequence < (Ill > is said to tend to limit 'I' when, given any + ve number' E however small, we can always find an integer 'm' such that and we write Lt
=I
all
n->co
If an
Ex:
(an
or
=
n2 + 1 2
2n +3
"
Ian -II <E, Vn ;:::: m,
~ I)
then ~-.
2
n
6.1.8 Convergent, Divergent and Oscillatory Sequences ). Convergent Sequence: A sequence which tends to a finite limit, say' I' is called a Convergent Sequence. We say that the sequence converges to 'I' 2. Divergent Sequence: A sequence which tends to ±oo is said to be Divergent (or is said to diverge). 3. Oscillatory Sequence: A sequence which neither converges nor diverges ,is called an Oscillatory Sequence. Examples
3 -4 ,-,..... 5 here 1. Const'der the sequence 2, -,
234
1 n
an =1 +-
The sequence < all > is convergent and has the limit 1
1 1 1 1 an -1 = 1+ - -1 = - and - < E whenever n > n
nnE 1 Suppose we choose E= .001, we have - < .001 when n> 1000. n 2. If all
1 =3+(-lr-- converges to 3. 'n'
3. If an =n 4. If an 6.2
2
+(-lr .n, diverges.
= ..!. + 2 (-1 n
r '< an > oscillates between -2 and 2.
Infinite Series
6.2.1 If < UII > is a sequence, then the expression u1+ u2 + u3 + ........ + un co
infinite series. It is denoted by
I
Un
or simply
I
+ ..... is called an
Un
11=1
The sum of the first n terms of the series is denoted by sn i.e.,
sn = U, +u2 +u3 + ...... +un;sps2's3' ....sn are called partial sums.
Sequences of Series
389
6.2.2 Convergent, Divergent and Oscillatory Series Let be an infinite series. As ~ 00, there are three possibilities.
Iu"
n
COl1vergent series: As n
(a)
~
oo,s"
~
a finite limit, say's' in which case
the series is said to he convergent and's' is called its sum to infinity. Thus
Lt
8"
=S
=8
(or) simply LIs"
11----)00 00
This is also written as
+ 1I} + II] + ..... + 11" + .. .1000 = s. (or)
III
I
UII = S
11=1
(or) simply
IlIlI = s. 8 11 ~ 00
(b)
Divergent series: If
(c)
Oscillatory Series: If
or
-00,
the series said to be divergent.
does not tend to a unique limit either finite or
8 11
infinite it is said to be an Oscillatory Series. Note: Divergent or Oscillatory series are sometimes called non convergent series.
6.2.3 Geometric Series . 1 1 II-I TIle senes, +x+x- + ..... x + ... (i) Convergent when
•
IS
Ixl < 1, and its sum is
(ii) Divergent when x ~ 1. (iii) Oscillates finitely when x
=
_I_ I-x
-I and oscillates infinitely when x < -I.
Proof: The given series is a geometric series with common ratio 'x'
I-x"
sn
(i)
=- -
When
I-x
when x *1 [By actual division - verify]
Ixl < 1: LI s n;->oo n
=
1
( 1)
Lt - - - Lt (XII] - - =-1- x n->oo 1- x 1- x
n->oo
[ since xn ~ 0 as n ~ 00 ] . 1 :. T he senes converges to - -
I-x
(ii)
When x ~ 1: sn
x" -1
=- x-I
and
:. The series is divergent. (iii) When x = -1 : when n is even,
sn ~ 00
sn ~
:. The series oscillates finitely.
as n ~
00
0 and when n is odd,
sn ~
1
Engineering Mathematics - I
390
(iv) When x < -1, s"
~ 00
or
-00
according as n is odd or even.
:. The series oscillates infinitely.
6.2.4 Some Elementary Properties of Infinite Series 1. The convergence or divergence of an infinites series is unaltered by an addition or deletion of a finite number of terms from it. 2. If some or all the terms of a convergent series of positive terms change their signs, the series will still be convergent. 3.
Let
Iu"
converge to's'
Let 'k' be a non - zero fixed number. Then Ikll" converges to h.
I
Also, if 4.
Let (i) (ii)
I
II"
u" diverges or oscillates, so does converge to 'I' and
I
I
v" converge to
ku" '111 '.
Then
I
(u" + v,,) converges to ( 1+ 111 ) and I(u" -v,,) converges to (/-m)
6.2.5 Series of Positive Terms Consider the series in which all terms beginning from a particular term are +ve . Let the first term from which all terms are +ve be ul • Let
I
Un . be such a convergent series of +ve terms. Then, we observe that the
convergence is unaltered by any rearrangement of the terms of the series.
6.2.6 Theorem If Iu" is convergent, then It un n->oo
= o.
Proof:
Sn=u l +U2 +······+U" sn_1 = ul + u2 + ...... + U,,_I,' so that, un = sn - sn_1 Suppose
IU n =/ then It sn =/ and Lt sn_1 =/ n---+oo
It un = It (Sn-Sn_I); n~oo
n---+oo
n~oo
It s,,- It n~oo
S,,_I
=/-/=0
n~oo
Note: The converse of the above theorem need not be always true. This can be Observed from the following examples.
Sequences of Series
(i)
1 2
.
1 3
1 n
1+ - + - + ....... + - + .... ;
Consider the series,
But from p-series test ( (ii)
.
2.7) it is clear that
lin
1 n
= -, It
n~oo
lin
=0
I!n is divergent.
1 1 1 1 + -) + -) + ..... + -) + ..... . 1 2- 3n-
ConsIder the sertes,
-2
= ~, It
lin = 0, by p series test, clearly Ir n~OCJ Note: If Lt lin ::t:- 0 the series is divergent; lIl/
391
,
I
~ n-
converges,
IJ-too
Ex:
2n -1 1'" = - 2 ,here Lt n
lin
Il---Ht)
I
lin
=1
is divergent.
Tests for the Convergence of an Infinite Series In order to study the nature of any given infinite series of +ve terms regarding convergence or otherwise, a few tests are given below.
6.2.7 P-Series Test i ' sertes, . ~1 Th e '111f illite ~= -1+ 1 - +1 - + ...... , n~1
1"
n"
2"
IS
3"
(i) Convergent when p > I, and (ii) Divergent when p ~ 1. [JNTU Dec 2002, A 2003] Pro(~r:
Case (i) Let p> I . J) > 1 3" > 2'" => _1 < _1_
,
,
'3"
1
1
1
1
2
2"
3"
2"
2"
2"
2"
-+--+- =3 4 4 4 2 111111111 -+-+-+->-+-+-+-=567888882 11 111 11 -+-+ ....... - >-+-+ ..... - = - and so on 9 10 16 16 16 16 2 I_I = 1+(!+!)+(!+!+!+!)+ ..... n" 2 3 4 5 6 7 1 1 1 ~1+-+-+-+ ..... 2
The sum of RHS series is
00
2
2
1 (Since sn = 1 + n ~ = n; 1and /1~'" SII = 00 )
:. The slim of the given series is also
00;:.
f_l-n 11=1
P
(p
= I ) diverges.
Case(iii):Let p - - > - ...... and so on 21' 2' JP 3'
1 1 1 1 I->I+-+-+-+ ....... P n 2 3 4 From the Case (ii), it follows that the series on the RHS of above inequality is divergent.
I Jl is divergent, when n P
P< I
Note: This theorem is often helpful in discussing the nature of a given infinite series.
Sequences of Series
393
6.2.8 Comparison Tests 1. Let Un and
I
I
I
Then
be two series of +ve terms and let
Vn
I
VII
be convergent.
converges,
Un
1. If un ~ Vn' \:;f n EN
2. or
U
-.!!.. ~
k\:;fn EN where k is> 0 and finite.
vn
3. or
Proof: 1. Let
un vn
~ a finite limit> 0
I
Vn
Then, u l
=I
(finite)
+u2 + ....• +lln
+ ...... ~VI
Since I is finite it follows that
2.
un vn
ll
,
Ikvll is convergent:. IU
3. Since Lt
Then
and
UII
I
Un
VII
is convergent and k (>0)
IS
is convergent.
~ < k\:;fn EN vn
:. from (2), it follows that
I
n
I
is finite, we can find a +ve constant k,3
un n-'>oo vn
>0
is convergent
UII
~ k ~ UII ~ kv \:;fn EN, since
finite,
2. Let
I
+v2 + ..... VII + ..... ~l
I
IU
II
is convergent
be two series of +ve terms and let
Vn
I
VII
be divergent.
diverges,
* 1.
If
un ~ VII' \:;fn
or
* 2.
If
un vn
or
* 3.
If Lt ull is finite and non-zero.
EN
~ k, \:;fn EN where k is finite and"* 0
n-'>OO Vn
Proof: 1.
Let M be a +ve integer however large it may be . Science a number m can be found such that VI +V2 + ..... +vn > M, \:;fn > m
u +u2 + ...•. +un > M, \:;fn > m(un ~ vJ l
I
VII
is divergent,
Engineering Mathematics - I
394
I 2
11,
I
1I"
2:: kv,,\:;In
v" is divergent ~
I 3.
is divergent
I
kv" is divergent
u" is divergent
Since Lt
~ is finite, a + ve constant k can be found such that
un
> k, \:;In
(probably except for a finite number oftenns ) :. From (2), it follows that
I
is divergent.
Un
Note: I. In (I) and (2), it is sufficient that the conditions with * hold \:;In> mEN Alternate form of comparison tests : The above two types of comparison tests 2.8.( I) and 2.8.(2) can be culbed together and stated as follows: If
Iu"
I
and
Vn
are two series of + ve terms such that Lt "->00
where k is non- zero and finite, then
I
u" and
I
Vn
un =
vn
k,
both converge or both
diverge.
Note: I. The above fonn of comparison tests is mostly used in solving problems. 2. In order to apply the test in problems, we require a certain series Vn whose
I
nature is already known i.e., we must know whether divergent. For this reason, we call
I
Vn
I
Vn
is convergent are
as an 'auxiliary series'.
3. In problems, the geometric series (2.3.) and the p-series (2.7) can be conveniently used as 'auxiliary series'.
Solved Examples 6.2.9 Example Test the convergence of the following series:
3456 1 8 27 64
(a)
-+-+-+-+.....
(c)
00 [ ~ (n
4
+1) 1/4 -n ]
(b)
4567 1 4 9 16
-+-+-+-+ .....
Sequences of Series
395
Solution (a)
Step 1:
d"
To find "ul/" the
term of the given series. The numerators 3, 4, 5,
6 ...... ofthe terms, are in AP.
n'" term tl/
3 + (11- 1).1
=
=
n+2 n+2
3 2' 33 43 · Denomlllators are 1 ,-, , ..... n ,,, term = n \ ;
Step 2: To choose the auxiliary series
I
VI/ .
111/
= --3n
In 1I11 , the highest degree or 11 in the
numerator is I and that of denominator is 3. :. we take,
= ----:l=I = 2
VII
n
11
Step 3 :
~=
It 1/-->00
vl/
2 It n+2xn = Lt n+2 = It 3
I/~OO
n
I/~OO
11
I/~OO
(1+~}=1 11
'
which is non- zero and finite. Step 4: Conclusioll:
It ~=1 n-----}oo
and
:. IUI/
I
VI/
VII
I
,,-?1
= ~n-
v" both converge or diverge (by comparison test). But
is convergent by p-series test (p
=
2 > 1); :.
Ill"
is
convergent. (b)
4
5
6
7
1
4'-9
16
-+-+-+-+ ..... Step 1 : 4 , 5, 6, 7, ..... in AP , tn Step 2 : Let "~ v"
= -1
= 4 + (n -1) 1 = n + 3
. be tIle auxi'1'lary series
11
Step 3: Lt I/~'"
finite.
~ = Lt (11~3)xn = It (1 v" n,,~oo
n+3
I/~OO
+l) = 1, n
which is non-zero and
Engineering Mathematics - I
396
Step 4 : :. By comparison test, both
I
utI and
I
v" converge are diverge
together. But
Iv" =I-n1
is divergent, by p-series test (p
= 1); :.
Iu"
is
divergent.
=n
1+-'4n-.+ ~G-l)~+ . . _1 2! n
= 4~3 -
3;n 7 + .... =
~3 [~- 3;n
Here it will be convenient if we take v"
(! -
!, and I
4
=n[-'-.-~+ . . .J
+ ....
4n
32n
.J
=~ n
Lt utI = Lt -1-4 + ..... ) = which is non-zero and finite vn n~oo 4 32n 4
n~oo
:. By comparison test, by p-series test
IV
I
Un
1 n =-3
n
VII
both converge or both diverge. But
is convergent. (p
convergent.
6.2.10 Example
=
If u n
{j'--3n-2-+-1
~hn3 +3n+5
show that
I
Un-
is divergent
Solution As n increases, utI approximates to r:::-?
~ 3n 2 l
II
Jl3
2/
n73
II
3/ 3
1
~2n 3 = 2Y.; x nYt = 2}~ . nX2
=
3 > 1); :.
Iu"
is
Sequences of Series
397 Ii
= - 1-/
If we take vn
U
,
nl12
[(or) Hint: Take vn
Lt -.-!!.. n-->""VII
1
= ~ , where n' ,
)13
=-I
which is finite.
274
II and 12 are indices of 'n' of the
largest terms in denominator and nominator respectively of ull • Here
1
VII
1
=-3-2 =-1 ]
n4
3
nl2
I
By comparison test,
"L..,. VII
VII
and
I
U II
converge or diverge together. But
, 1 .IS d'Ivergent by p =, L..,.-I,-'
. test (smce . senes p
= -1 < 1)
Wl2
L un
12
is divergent.
6.2.11 Example Test for the convergence of the series.
Jf ~ J! ~ +
+
+
+ ......
Solution
=~
n:
Here,
un
Take
1 1 vn =-1-I =-0=1, Lt u. ----
n2
I
VII
1;
n
n-->"" V
~
Lt ) 11-->""
n
2
11
= 1 (finite)
1+n
is divergent by p - series test. (p = 0 < 1)
:. By comparison test,
I
Un
is divergent, (Students are advised to follow the
procedure given in ex. 1.2.9(a) and (b) to find" un" of the given series.}
6.2.12 Example 1 1 1 Show that 1 + IT + l2 + ....... + l!:! +..... is convergent. Solution
un
=
l!:!1
(neglecting 1sl term)
1 1 1 - - - - < -----==,--- = 1.2.3 ......n 1.2.2.2 .....n -ltimes (2n-l)
Engineering Mathematics - I
398
1
1
1
"1I" < 1+-+-? +-, + ..... . ~ 2 2- 2' which is an infinite geometric series with common ratio
" 2,,-1 1 IS . convergent. (1.2.3(a)). Hence :. ~
I "/I
~ 1 ); :. Iu"
n
is convergent.
6.2.14 Example If u" =
-J n4 + 1 - -Jn4 -1, show that I
Un
is convergent. [JNTU, 200s]
Solution 1
u"
= n2 ( 1+ n1)2 4
I
1)2
n 2 ( 1- n 4
~n2[(1+ 2~4 - 8:' + 16~12 - ....)+- 2:4 - 8:' -16~12 - .....)] =n Take
vn
2 [-;-
n
1
= -2 n
'
+
~ + .... J= ~ [1 + ~ + .... J 8n n 8n
hence Lt ,,~ao
u' -E...
v"
=1
Sequences of Series
399
I
:. By comparison test,
I
Vn
=~ n
and
lin
I
v" converge or diverge together. But
I
is convergent by p -series test (p = 2 > 1):.
Un
is convergent.
6. 2.15 Example
1 1 1 I-' · Test the senes - - + - - + - - + ..... lor convergence.
l+x
2+x
3+x
Solution
Take
1
un
=--;
VII
=-,
n+x
n
UII
then
_
n n+x
=
1+~ n
Lt
(_1_] 1; I =
n~""
VII
=
X
1+n
:. By comparison test,
I
I!n
is divergent by p-series test (p =1 )
is divergent.
Un
6.2.16 Example Show that
fSin(!)n
is divergent.
n=1
Solution Take
. (1) sint =
Yn
SIn U Lt --E... n~"" V n
n = n~"" Lt ( ) 1 _
Lt -
I~O
t
(where t
= 1n) = 1
n
Iu I ,
n
VII
.both converge or diverge. But
(p -series test, p
= 1);
:.
I
Un
is divergent.
I
VII
=
I
~
is divergent
Engineering Mathematics - I
400
6.2.17 Example Test the series
L sin
-I ( : )
for convergence.
Solution • _I 1 u =sm _. n n' 1 vn =n
Take
-{~)
u sin ( ()) ( . . _ 1 ) Lt 2..= Lt ( ) ;= Lt - . - =1 Takmgsm '-=() v tHoo 1 8---*0 sm () n
/1---*00
/I
_
n But
L v"
is divergent .Hence
L u"
is divergent.
6.2.18 Example
1
33
22
Show that the series 1 + - 2 + - 3 + - 3 + ..... is divergent.
234
Solution
1
22
33
Neglecting the first term, the series is 1 + - 2 + - 3 + - 4 + .....
n" U
-
-
234 nn n-
.Therefore nn
"~(n+l)"" ~ (n+l)(n+l) ~ n(l+~).n"(I+~J 1
1
Take vn =-
n
1 e
which is finite and
L vn = L!n is divergent by p -series test (p L un is divergent.
=
1)
Sequences of Series
401
6.2.19 Example . 1 3 5 . SIlOW t Ilat t Ile series - - + - - + - - + .......00 IS convergent.
1.2.3
2.3.4
1
3
5
1.2.3
2.3.4
3.4.5
3.4.5
Solution
--+--+--+ .......00
Iii
n
2n-1
term~ u" ~ n(n+l)(n+2) ~ n' ·(I+~)(I+~)
Take
1 n
Vn = - 2
it un
n~oo VII it HOO
Un Vn
= it HOO
=(
LVII =
L n1
-2
_I (2-~ ) (_1 ) n
2
(1 + ~'~)(1+ 2n)
2)(
0
1 +0 1+0
:. By comparison test But
1 ( 2- n )
)
Llln and
=2
n
2
which is finite and non-zero
LVII
is convergent. :.
+
converge or diverge together
LUll
is also convergent.
6.2.20 Example 1
LII~I J;;n + ,J;;+l n+ I 00
Test whether the series
Solution 1 L-==---== J;; +.j;;+1 00
The given series is
11=1
1
is convergent
Engineering Mathematics - I
402
Take
which is finite and non-zero
I
Using comparison test But
I I
Vn =
l
I
Un
and
I
Vn
converge or diverge together.
is divergent (since p
= Ii)
is also divergent.
Un
6.2.21 Example Test fo~ convergence
f[ ~n + 1 - nJ [JNTU 1996,2003, 2003s] 3
n=1
nih term
Un
= n[(1 =
f~L
Then
3:2- 9:5
I
Vn
+ ...... = :2
(1- 9:3 + ......) ;
X=n~Jlj - X
n 3 + ....
:. By comparison test, But
+~))~ -1) .~+ n -1] = n[1 +~+ 3n Jj U~ 1.2 n ..... -1]
I
Un
and
I
Vn
Let
1 n
Vn =-2
)= lj *0
both converge or diverge.
is convergent by p -series test (since p = 2 > 1)
convergent.
6.2.22 Example Show that the series, for p 5, 2
~ + ~P + ~P + ....... JP
2
3
is convergent for p > 2 and divergent
Sequences of Series
403
Solution III
••
n term of the given senes =
U
I
n +1
n ( 1+ In)
n"
n"
= --~ =
=
( 1+ 1~) 11
,,-1
Lt u = l:;t: O· I = _1_. n,,-I '11-"') /v ... L 1111 and L VII both converge or diverge by comparison test. BlIt LVII = L }~>-I converges when p -I> 1 ; i.e., p >2 and diverges when
Let
liS
take v
n/
'
lI
p -1 ~ 1 i.e p 6.2.23 Example
~
2; Hence the result.
+ L (?II 7-3J 3 +1 OCJ
Test for convergence
2 1
11=1
Solution
"" J 2"(1+ 3~,,)r' l3
11
1 (
+
'~3 1 ) J
" Jf. VII U
V
Take
n -
_.
3"'
I )1,3
1+ 3 2 - [ 1 1,' + n
--'!...-
Lt !!A = 1 :;t: 0; :. By comparison test, v
1l-7r:iJ
L un
and
L
VII
behave the same way.
ll
"
BlIt L... vn
=
~(2)"2 = ~- + -2+ (2)3 -
L... 11=1 3
common ratio
3
3
I. I. a geometflc···1 sefles Wit
2
+ ..... , W lIC 1
3
IS
1
)34, «I) :. LVII is convergent. Hence LUll
is convergent.
6.2.24 Example
. Test for convergence of the sefles,
1
4.7.10
+
4
7.10.13
+
9
10.13.16
+ ..... .
Solution
and
4,7, 10, .............. is an A. P; 7, 10, 13, ............ is an A. P; 10, 13, 16 , ............. is an A. P;
til
= 4 + (n -1) 3 = 3n + 1
'II = 7 +(n-l)3 = 3n+4 'II = 10 + ( n - 1) 3 = 311 + 7
Engineering Mathematics - I
404
n2
ll=
n
=---:-----:------:-----;----;----;-
(3n+1)(3n+4)(3n+7)
II
3n(1+
V
" 3n
).3n(1+4
3n
).3n(1+ 7/ ) ·3n
1
27n(1+
=!
Taking vn
Lt IHOO
n
,we get
~ = _1_:;t: 0; v
27
II
1, )(1+ 413n ;' )(1+ 7/ ) /3n
/3n
LUn and L vn
:. By comparison test, both
same manner. But by p -series test,
L
VII
is divergent, since p
=
behave in the I. :.
Lilli
is
L
is
divergent.
6.2.25 Example Test for convergence
",hn2 -Sn+1 ~ 3 2 4n -7n +2
Solution Iii
••
n term of the gIVen senes = un Let
vn
=-2
L
lin
=
.J2n2 - Sn + 1 3 2 4n -7n +2
n
L
l l
nloo ~ = IH~· L
nloo
~/ ~/?
n 2 - / n + / n- n 2 n3 /~ +~ I] :.
0
both converge or diverge. But
L un
is convergent.
6.2.26 Example
·" Test tI1e senes ~ un ,w hose n Iii term
• ( IS
I .) 4n 2 - I
VII
Sequences of Series
405
Solution
1 Ull
Let
VII
:. I
Ull
and
I
Vn
= (4n 2 -/.);
=-\, n-
Lt 11->00
_llil VII
=Lt lrn Il->ao
2 2
1
n /'n
(
4-
)1=-4 *0 " o
both converge or diverge by comparison test. But
convergent by p -series test (p
= 2 > 1) ; :.
I
UII
is convergent.
6.2.27 Example If un
= (; ).Sin (;), show that
IU
II
is convergent.
Solution Let
VII
=
1 -?
n-
,so that
'" VII is convergent by lJ -series test. ~
J
(1)n - It (-sin I ) (1 n ) - 1l~<X' I
ll Lt _" - LI sin
11->00 ( V
- n->ao n
(~J =1 * 0 :. By comparison test, I is convergent.
where t = lin, Thus Lt
Jl~
= II ~/; Lt [ulIl ] = 1 * 0
In
2
n--->oo
I VII
Hence by comparison test,
IIIIl converges as I
6.2.29 Example Show that fsin 11=1
(~s in above example)
2(~) is convergent. n
VII
converges.
I
VII
IS
Engineering Mathematics-I
406
Solution
. ?(I) ; Take vn =-?1
Let un =sm- n
Lt
I/~Cf)
( l n-
(~)= Lt
n~oo
V
n
sin 1/
. ')12 =Lt(smt) II . 12log2 72'
log 2 < 1 => 2log 2 < 2 => 1/
}jIOg3> X,.····Xlogn> Yn,nE N
L In 1/ Iogn > L In II ; But L II In
is divergent by p-series test.
By comparison test, given series is divergent. [If
u/I ~ vn"i/n then
IU
n
I
Vn
is divergent and
is divergent.]
(Note: This problem can also be done using Cauchy's integral Test. See ex 1.6.2) 6.2.31 Example 00
I
Test the convergence of the series
(c + n
rr (d + nr' ,
where c, d, r, s are all
n=1
+ve. Solution · Th e n Ih term 0 ft h e senes
Let
1
= un =
vII = -r+.\- Then
n
r
1
(c+n) (d+n)
U/' _
,
1
--------=------
Vn
n
r (
1+
~
J +~- J (1+~J (1+ ~J .n" (
1
Sequences of Series
Lf 11-)00
_
1111 -_
vJl
407
1 --," -;-- 0 :. L)III and " ~ v" both converge are diverge, by comparison
test.
L
But by p-series test,
:. L
II"
converges if (r + .\) > 1 and diverges if (r + s) ~
VII
converges if ( r + s ) > I and diverges if (/"1 s) ~ I.
6.2.32 Example Show that
~
~n
-(1+ 1 )
"is divergent.
1
Solution 11 "
Take
VII
=n
"=--
n.n 1n
1I 1 1 =- ; Lt -" = Lf -
n
11---)00
Il-)ct)
y
11-+00
"
n
In
= 17:- 0
II
VII
11-)00
«(:)
= eO = 1
using L Hospitals rule)
L
1I"
and
diverges (since p
=
I); Hence
By comparison test both
L
V
1 1 1 Lt -v = y say; log y = Lt - - .log n = - Lt -.-!1 = 0 n ,,-too n 1
For let
test,
1
-(1+ 1 )
LV" L
converge or diverge. But p-series lIlI
diverges.
6.2.33 Example
~
(n+ar II~I (n + b)" ( n+ c )"
Test for convergence the series ~
Solution
, a, h,
C
,p, q, r, being +ve.
Engineering Mathematics - I
408
v
Take
II
1 =- . It u = 1"* 0 .' I1P+lJ-r' II~OO V _II
II
Applying comparison test both Rut by p-series test,
I. Hence ~
I
lIlI
I
VII
I
and
lin
I
VII
converge or diverge.
converges if (p+ q -r) > I and diverges if (p + q -r)
converges if (p + q - r) > 1 and diverges if (p + q - r) ~ I.
6.2.34 Example Test the convergence of the following series whose nl" terms are:
(311+4)
(a)
( 211 + 1)( 211 + 3)( 211 + 5)
(c)
1
tan-;
(b)
11
(d)
11
(3 +5 (e)
11 )'
11. 311
Solution (a)
Hillt:Take
VII
=~;Iv" 11-
is convergent;
It
,,~oo
(u,,)=~"*o (Verify) v 8 "
Apply comparison test:
I
lIlI
is convergent [the student is advised to work out this problem fully]
(b)
Proceed as in 1. 2.16;
(c)
Hint: Take v"
1 11
=-2
Iu" ;
is convergent.
(u) = It ((l+J~lr = 3e = 1 "* 0
It -"
,,~oo
VII
II~OO
1 + 3/~
)11
-?
e
e-
v" = -\- is convergent (work out completely for yourself) 11
1
(d) U"
1
~ 3" +5" ~ 5"
II
1
+(~J]
1
;Take v" ~~ ; "If.
(Un)
~ ~l ,,0
Sequences of Series
LUll
409
L
and
VII
behave the same way. But
a geometric series with common ratio
:. LUll (e)
is convergent since it is
~ 1 and divergent if k < 1 (the test tilib
ll
'.I.
hen k
=
I).
I
Un
is convergent
Sequences of Series -~---~
~-----
----~----
413
-----
Solved Examples Tests for convergence of Series
6.3.3 Example X
X
X
\
-+--+---+ ................ .
(a)
1.2
2.3
3.4
Solution x" 111/
Xl/II
= n (11 + 1) ;111/+1 = (11 + 1)( n + 2)
~/!.!..'.l=
Xl/+I
(n+
1/"
Therefore
II
LI
_1/_,1
11----)00
1111
:. By ratio test
Ixl>
r
1
n(n+1)
1)(n + 2
,
~ (I + ~
x"
r.
=X
IUI/
is convergent When
Ixl
00
:. By comparison test Hence
I
IUI/
is convergent.
Ixl ~ 1
Un is convergent when
and divergent when
Ixl > 1
3
1+ 3x + 5x 2 + 7 x + ...... .
(b) Solution
Un
= (2n-l)xn-l;ul/+
1
=(2n+l)x"
ul/+ I L t (2n 1) x=x -+ L1-= 1/-->00
:. By ratio test
lin
I
1/-->00
2n-l
Un is convergent when
Ixl > 1 When x
= 1: ul/ = 211 -1; Lt un = 00; 11----)00
Hence IlIl/ is convergent when 00
(c)
1/
I-+-··········· n- + 1 1/;1
:.
I
UI/
Ixl < 1
and divergent when
is divergent.
Ixl < 1 and divergent when Ixl ~ 1
Engineering Mathematics - I
414
Solution U n
x" n +1' (
-;;: =
(n+l)2 +1
n+1
n +1 ) n 2+ 2n + 2 x 2
Lt un +1 = Lt
n~'"
n
(
1 +
>;:2)
II~'" n2(1+~+2) 2
Un
n
:. By ratio test,
When
I
=----
U
2
un +1
Hence
X"+
=2 -- .
L un
is convergent when
1 x=I'u = - - . Take • n n2 + 1 '
LU
:. By comparison test,
n
(x)=x
n
Ixl < 1
and divergent when
Ixl > 1
1
v =n n2 is convergent when
Ixl ~ 1
and divergent when
Ixl >1 6.3.4 Example Test the series
f (n:n -1}n, x> +1
n~'"
0 for convergence.
Solution
U" =(:::: )X";U"" = Lt un+1 n~'" un
[~:: :i: :: ]X'HI 2
2
= Lt [( 2n + 2n ) ( n2 + 1)] x n~'" n + 2n + 2 (n - 1) . 4 n (1+2/n)(1 +l/n) =nIf", [ n4(l+2/n+2/n2)(1-l/n)
:. By ratio test,
LU
n
is convergent when
n
n2 -1 n +1
= 2- -
=x
x< 1 and divergent when x> 1 when
x = 1, U
1
Take vn
1 n
=-0
Sequences of Series
415
Applying p-series and comparison test, it can be seen that
L un is divergent when
x= 1.
.'. L un
is convergent when x < 1 and divergent x ~ 1
6.3.5 Example
2P 3P 4P + l] +
Show that the series 1+ ~
l1 + ..... , is convergent for all values ofp.
Solution
Lt .un +1
n~oo
=
un
l.fl]=
Lt [(n+IY x In + 1 nP
n~oo
= Lt ( n~oo
L
lll/
1
Lt {
n~oo
(I
) x Lt 1+ - )P n + 1 n~oo n
1
(n + I)
(n+I)P} n
= 0 < 1;
is convergent for all 'p'.
6.3.6 Example Test the convergence ofthe following series
1 F
1 3
1 5
1 7
-+-+-+-+ P P P ........... . Solution
u = 1/
ul/+ 1
un
1
(2n-IY'
=
1
·u
=---
(2n+ly (2n-Iy _ 2P.n P(I-l/2nY. (2n+ly - 2PnP(I+l/2ny , 1/+1
Lt un+1 =1 HOO
un
.'. Ratio test fails.
1 = _. n nP'
Take v
U
-E... vn
=
nP
(2n -I)P
=
which is non - zero and finite 7 .'. By comparison test,
LU
n
ar.d
1
U . Lt -E... = -1P ( I)P , I/~OO V 2 , 2 P 1-n 2n
L
vn
both converge or both diverge.
EngineerinG M~th~ni1!!l'S~
·H6
I
But by p - series test,
-=
VII
I -I
'~onverges when p > I and di\ c,
11"
when p ~ I
:. I
iI"
I and divergent if p ~ I .
is convergent if /'
6.3.7 Example Test the com crgence of till'
" ' (n -1- l).\""
"cric~
L.. - 1/,
-,--.X
I
>0
11
Solution
( n + I) x" lin =
( n + 2) X"-l I
1
;1I1Itl - - - - , - , -
(11+1)
n
; I
By
i'".1i
j()
illi rl
LI
--
11->''0
111/
test,
I
=
II
..: ' ! ,,- _.. ,
L,
" ,.w
UII
~
;
1I
1
i
: , "" - - - ,
.x = X
+ 11 ( 1+ 1 )' n, \.. n
converges when x < 1 and diverges when x > 1 .
n+ I
When ..x= I ' IIII = n3Take v"
.'. I
= -~ ; By comparison test n
UII
I
UII
.is convergent ( give proof)
is convergent if x ~ 1 and divergent if x > I.
6.3.8 Example Test the convergence of the series
. ~(n2
1) (.. 1 2.5.8 2.5.8.11 ... ) 1 1.2 1.2.3 II) + - - + + ... (III - + - + - - + n 1.5.9 1.5.9.13 3 3.5 3.5.7
(I) L.. -+-2 II_I
2"
Solution
2 1 (n2 1) n (i) I ---;+-2 I-II + I-2 2 n 2 n 00
00
00
11=1
11=1
=
II-I
Let
:'t_ '
Sequences of Series
417
= (n+1)2
U
.1l11+1
2"rl'
11+1
2"tl·
U"
11
2
1(1 +-1)2 --< _1 1
_ L Lt1111+1 -t-. 2
I1~OO lin
convergent. :. The given series
2
n
11-+00
By ratio test
(ii)
= (n+1)2 ~
I
Un
(I
Un
is convergent .By p -series test,
+
I
v,,)
I
v"
IS
term
=
is convergent.
Neglecting the first term, the series can be taken as,
2.5.8 2.5.8.11 --+ + 1.5.9 1.5.9.13 Here, 1sl term has 3 fractions ,2 nd term has 4 fractions and so on . :. nih term contains (11 + 2) fractions 2.5. 8....... are in A. P.
t ( n + 2t (n + 2
term = 2 + ( 11 + 1 ) 3 = 3n + 5 ;
1. 5. 9, ....... are in A. P.
U
term = 1 + ( n + I ) 4 = 4n + 5
2.5.8 ..... (3n+5) 1.5.9 ..... ( 4n + 5)
=----'--~
n
un+1
2.5.8 ..... (3n+ 5)(3n+ 8) = 1.5.9 ..... (4n+5 )( 4n+9 )
un +1 _ (3n+8) . -;,:- - (4n + 9) ,
n(3+~)
Lt u +1 = Lt n HOO un HOO n ( 4 + By ratio test, (iii)
I
Un
00 2n + 2 )
Lt 1111+1 U//
IHOO
I
By ratio test,
x
= ),
u//
IHOO
)
2n .x=x 2n( 1+ 22n)
is converges when x < 1 and diverges when x > I when
the test fails.
Then
=
U
1.3.5 .... (2n -1)
2.4.6 ..... 2n
II
:. I
2n(1 + 1
Lt
=
U/I
is divergent. Hence
0) Test for the convergence of 1+ -2 x + -6 x 2 + ........ + ( 5
2/1+1
9
Solution
II,
Om itting 1" tcnn,
=( ~: :~ ) x'-' ,(n " 2)
'II,' arc all eve.
and
= (2"+1 - 2) x/l. Lt (~) = Lt (211+1 - 2) x ( 2// + 1) X
U IHI
(
2//+1 + 1)
'/1-->00
u
11-->00· 2"+1 + 1
ll
2n+1 (1- 12/1) 2// ( 1+ 12/1) =
//~~ [ 2//+1 (I +
Hence, by ratio test, When x Hence
I
Un
Un
1
l''2/1+1 )·2/1 (1- 2, 2/1 ).x = x ;
converges if x < 1 and diverges if x> 1.
= ), the test fails. Then
I
u/I
2/1-2 2 +1
= -n - ; Lt
un
= 1 ;f::. 0;
IHOO
is convergent when x < ) and divergent x > 1
6.3.12 Example
I
00
Using ratio test show that the series
/1=0
(3 -4i)" n!
converges
Solution
_ (3-4i)" un -
2/1 - 2 .
I .
_(3 -4i),,+1 /
In! ,un+1 -
. /(n+l)! '
:.
I
Un
diverges
Engmeerlng ---------------------------------=--
420
MC3tht::rnatl~~
1II1tl)_ 1 I (3-4i)_0 ---00
I
:. By ratio test,
(3n+4)(3n+5)
11->00
(1
f
1
)1 = 5 > 1
is divergent.
lin
6.3.15 Example 00
I
Test for convergence the series
n l - II
Solution
= n ; u +1 = (1)-11 n+ ; I-II
1I11
lI
':;,:' = = (n;I,!"
Lt un+1 = Lt
n~'" U
I
Un ,
n
1+ 1,'
,'n
is convergent
6.3.16 Example
I
00
Test the series
11=1
2
3
~ ,for convergence. ~
Solution
2n 3 Un
= ~ ;
U II
=~C:J
!.[_l_,]" =O}=O00
II
:. By ratio test
n(n:l)"
+I
2(n+ 1)3 = In + 1
e
Engineering Mathematics-I
422
lln+1
= 2(n+l)3 x
~
un
In + 1
2n
u
Lt
= (n+l)2 = (1+;~r
n3
3
n
= 0 < 1;
--.!!..±l
n-+oo Un
L u/l
:. By ratio test,
is convergent.
6.3.17 Example
2/1 ,
Test convergence of the series
L~ nn
Solution
n 1
+ (n + I)! /1 ( ~ )11 ~=2 (n + 1)"+1 . 211 n! n+ 1
un +1 = 2
1111
U
Lt
--.!!..±l
IHOO
ull
:. By ratio test,
= 2 Lt
LUll
(e)
3n + 7n ( 5n 9 +
2
11
Lt n-+oo
L un (b)
1.2.3 ....n )2 ( 4.7.10 .... .3n+3
(a)
< 1 (since 2 < e < 3)
e
where u/I is
x n- I
(c)
Solution
=-
is convergent.
n2 + 1 3/1 +1
3
2
Hoo(l+ )-~r
6.3.20 Example Test the convergence of the series (a)
1
I)
(d)
--,(a>O) (2n+lf ~ ~1+3n
)xn
u = Lt [(n+l)2 +1 x3/1-+1] (~ ull n-+oo 3n+1 + 1 n2 + 1
423
Seq uences of Series
1
-00
xn
un
n I -
n-->oo
aa
= Lt IHOO
By ratio test,
LUll
[ 2a na (l+ }Snf
convergence if x < 1 and diverges if x> 1.
When x = 1, the test fails; Then,
=
1I /I
Lt n-->OO
(Un) = Vn
(_n_)a =
Lt n-->oo
l
2 n (I+!' )a . 2n .x = X
2n + 1
un
If
; Taking v
1 = _1 a 2 + ;!~r 2
Lt n-->oo (
L
:. By comparison test,
1 (2n+
and
L
VII
= _1a
II
* 0 and finite
have same property
But p -series test, we have
and
(i)
LVII convergent when a> 1
(ii)
divergent when a
(ii)
LU x > 1, L
(iii)
x
:. To sum up, (i)
(iv)
x < 1,
n
un
~
1
is convergent Va. is divergent Va.
L x = 1, a ~ 1, L =
1, a> 1,
un
un
is convergent, and is divergent.
awe have,
n
(since a > 0).
Engineering Mathematics - I
424
Lt
U
I
~= IHoc> ltll
(c)
IHoc>
Lt [
=
IHoo
:. By ratio test,
[
Lt
I.2.3 ....n(n+I) 4.7.1O .... (3n+3)]2 x-----'-------- lIlI
LI
(e)
U
= Lt
--"..±.!..
n->oc>
LI
lr 3 ( n + 1)3 + 7 ( n + I
t
S( n + 1 + II
(I
{(I + .1/)3 +2_(1 + 1/)2} n 3n n
n->oc>
Sn :. By ratio test,
9
I
{(
Un
Sn 9 + II ] xx 3 3n + 7
x
~ )3 +7n2(1+ .1.1)2 Sn9(I+l!'s/ 9) ] .n n x n xX 3 9 Sn (I + V)9 + II 3n + 7;1 .n / 3n i
Lt
r
f3n3(1+
n->oc>
3
is convergent.
Hoc>
= Lt
U"
3n =
U"
I+ }~r + SI~_}
(1 I/VS
Sn 9 +
n
x
3n
3
(1 + .%n
U
= 3n3(1+ jjn)
" Taking
Vn =
Sn
9 (
1+ IYsn9)
=~ 6
Sn
~, we observe that n
3)
) xx
=X
3)
converges when x < 1 and diverges when x > 1.
When x = 1, the test fails, Then
9
1
(1+ l~n)
(I + IYsn
Lt
U" =
"->OC>~
9)
l;f: 0 S
Sequences of Series
425
:. By comparison test and p series test, we conclude that
:. L u"
L un
is convergent.
is convergent when x::::; 1 and divergent when x > 1.
Exercise - 1(b) 1. Test the convergency or divergency of the series whose general term is :
x"
(b)
nx n-I .......................... .
Ixl < legl, Ix ~Ildgt ] I Ans : Ixl < legt, Ix ~Ildgt I
(c)
(~:.
I Ans :
Ixl < legl, Ix ~Ildgt
I
(d)
(:::: )X" ......... .
[ Ans :
Ixl < legl, Ix ~Ildgt
I
(e)
lfl
[ Ans: cgt.]
(f)
4"·lfl
[ Ans: dgt.]
(a)
(g)
I Ans:
n
:n
X
"'
•••••
nil (
n 3 +1 )"
[ Ans: cgt.]
(3 + 1) n
2. Determine whether the following series are convergent or divergent :
(a) (b)
x3
x2
x
- + - + - + .............. 1.2 3.4 5.6 X x2 x3 1+-2+-2 +-2 + ............. . 234 1
x2
X
(Ans:
Ixl::::; legl,lxl > ldgt
]
I Ans :
Ixl : : ; legt, Ixl > Idgt
]
( Ans :
Ixl::::; legt, Ixl > Idgl
] J
(c)
- - + - - + - - + ..... . 1.2.3 4.5.6 7.8.9
(d)
1+-+-+-+ ..... - 2- + ....
(Ans:
Ixl::::; legt,lxl > ldgt
(e)
1.2 2.3 3.4 -+-2 +-3 + .............. .
[Ans:
Ixl > legt,lxl::::; Idgt]
X
X
2
x
2
5
x
X
3
10
x
x
n
n +1
Engineering Mathematics - I
426
6.4
Raabe's Test
Let
I u" be series of +ve terms and let E!. {n(:.:, - I)} ~ k
Then (i) If k > 1, ~>n is convergent. (ii) If k < 1,
L
un
is divergent. (The test fails if k = 1)
Proof: Consider the series
Case (i): In this case,
Lt n{~-l} =k
n-+oo
U + n 1
>1
We choose a number 'p' 3 k > p > 1 ; Comparing the series which is convergent, we get that
L
un
L
U"
with
L
will converge if after some fixed number
of terms un
Vn
U + n 1
i.e. If.,
> Vn+1
Un)
i.e., If
(n+ l)P
= -;;-
n ( - - 1 > p+ Un+1
p(p-l) 1 .-+ ......... from (I) L2 n
Lt n(~-I» P
n-+oo
U + n 1
Vn
Sequences of Series
427
Llln
i.e., If k > p, which is true. Hence
is convergent .The second case also
can be proved similarly.
Solved Examples 6.4.1 Example Test fe>r convergence the series
1
x 3
1.3
x
x 7
1.3.5
x+-.-+-.-+--.-+ ..... 2 3 2.4 5 2.4.6 7 Solution Neglecting the first tern ,the series can be taken as ,
1
x
3
1.3
x
5
1.3.5
x
7
-.-+-.-+--.-+ ..... 2 3
2.4 5
2.4.6 7 1.3.5 .... are in A.P. nih term = 1+(n-1)2=2n-1 2.4.6 ... are in A.p. n'h term
2 + (n
=
-1) 2 = 2n
3.5.7 ..... are in A.P nih term = 3+(n-1)2 = 2n+ I uti ( nih term of the series)
=
1.3.5 .... (2n-1)
X
21l 1
+
=----'----;------::--"--
2.4.6 .... (2n) 1l1l+1
=
1.3.5 .... (2n-1)(2n+ 1)
() 2.4.6 .... ( 2n)2n+2
X
2n
+
2n+1
3
. 2n+3
1.3.5 .... (2n+1) X 2n +3 2.4.6 ... .2n (2n+1) -;;:- 2.4.6 .... (2n+2)"(2n+3)"1.3.5 .... (2n-1)" X 2n + 1 lIn+1 _
(2n + 1)2 x 2 ( 2n + 2) ( 2n + 3)
1)2
4n-1 ( 1+-2n
II
LI ~ = LI IHOO
:. By ratio test, If
LU"
Ixl = 1 the test fails.
Un
n~oo 4n
converges if
2
(
1+ 22~ )(1+ ~~ )
x
2
1
= X-
lxl < 1 and diverges if Ixl > 1
Engineering Mathematics - I
428
Then
=1
x2
~ = (2n + 2)( 2n + 3 )
and
(2n+
1111+1
1r
~ -1 = (211 + 2)( 2n + 3) _ 1 = 6n + 5 (2n+1)2
111/+1
ul/ n -
LI tHOO
{ (
ul/+
1J}
(2n+l)2
1
2 6n + 5n I/~OO 4n 2 + 4n + 1
= l I (
1
n Lt
=
11-->00
By Raabe's test,
Ixl ~ 1
I
UI/
n
2
2 (
~)
6+
3 =- >
(4 + ~n + _n-\ ) 2
1
converges. Hence the given series is convergent when
an divergent when
Ixl > 1 .
6.4.2 Example Test for the convergence of the series
3 7
3.6 7.10
2
1+-x+--x +
3.6.9 3 x + ..... ;x> 0 7.10.13
Solution Neglecting the first term,
=
ul/
3.6.9 .... 3n 1/ .x 7.10.13 .... 3n+4 3.6.9 .... 3n(3n+3)
= 7.10.13 .... ( ) ( 3n+7 ).x 3n+4
lIl/+1
= 3n + 3.x
:. By ratio test,
; Lt
3n + 7
lIl/
When x
1/+1
ul/+ 1
I
UII
lIll+1
is convergent when x < I and divergent when x > I.
= I The ratio test fails. Then u 3n + 7. lin 1 _ ll
_
1I1/+1 -
=X
lIlI
1/-->00
3n+3' lIn+1
-
-
4 3n+3
LI {n(~-l)}= Lt (~)=i>l
11-'>00
lln+1
11-->00
3n+3
3
Sequences of Series
429
--~-----------------------------------------------------------
LUll
:. By Raabe's test,
is convergent .Hence the given series converges if x ~ 1
and diverges if x> 1.
6.4.3 Example Examine the convergence of the series
f 12.5 2.92.... (4n-3)2 42.g2.122 .... (4n)2
11=1
Solution
3)2 ~ ~ 4-.8-.12-.... 4n )2
12.5 2.92•••• ( 4n -
= ---------'-------'-
U II
7
(
12.5 2.9 2.... ( 4n -
litH
3)2 (4n + 1)2 1= -~--~---2--'---(-----')2'---(--'--------'-)24-.8-.12 .... 4n
Lt u
lI
+
1
=
un
n->oo
:. The ratio test fails. Hence by Raabe's test,
4n+4
( 4n + 1)2 ~ 11->00 ( 4n + 4 Lt
LU
t
n
= 1 (verify)
is convergent.
6.4.4 Example Find the nature of the series
L (l!:!)2 x
n
~
,
(x > 0)
Solution ull
=
ulI +1
(l!:!)2 ~
ull
n +1
:. By ratio test,
(l~±l)2 11+1
(n+l)2 X' (2n+l)(2n+2) '
_
Lt u 11->00 U
II
.x ; un + 1 = 12n+2 .x
n
L un
=
n2 ( 1+ 1/ ')2 ;' n
Lt n->004n2(1+}/ , 2n
)(1+2/) 2n
.x = ~
4
converges when:! < 1, i. e ; x < 4; and diverges when x >4;
When x = 4, the test fails.
Engineering Mathematics-I
430
x = 4 ~ = (2n + 1)( 2n + 2 ) '1l11+1 4(n+l)2
When
-2n-2
U
r
_" -1=
LI
Hence
I
Un
[n(~-11]= -12 if.>
-1
=
ntl
is divergent
lin
is convergent when x < 4 and divergent when x> 4
6.4.5 Example Test for convergence of the series
,,4.7 .... (3n+ 1)
~
1.2.3 .... n
n
x-
Solution
_ 4.7 .... (3n+1) lin -
n.
X
1.2.3 ....n
_
4.7 .... (3n+l)(3n+4)
'U n + 1 -
II Lt ~ = Lt [(3n+4) .x] n-->oo ( n + 1)
()
1.2.3 ....n n + 1 .
n+1
X
= 3x
n-->oo Un
:. By ratio test
I
Un
converges if 3x
x=
When
x ~:3' n u.:, -1
" lin ~
II
]
~n
[(n+ 1)3
3n + 1 -1
]
~
[-1 ]
n 3n + 4
n[~-I] = -!3 < 1
:. By Raabe's test, :.
3
3
1 [
Lt
3
!, the test fails
If
n--><Xl
II
11+1
IU
n
is divergent.
. convergent wl IS len I x I
II
Ifx= I, the test fails.
Lt
11
[
II
_tl -
1] -1 [ n(n+2) 1 =LtI1 =000
:. By Raabe's test
:. I
Zlil
X
2
11
IU
Then
1(1+ 2) 11 .x = (1 1 11 + )1
H~-'UJ
11 tl
:. By ratio test,
_11(11+2) (11+1)2 .x
lr
I
_H_'
H-N,
II'HI
'--;;,:-
III//
is divergent
is convergent when x < I and divergent when
x 21
6.4.7 Example
.
.
3 4
3.6 3.6.9 + + ......00 4.7 4.7.10
FlI1d the nature of the series - + -
Solution
3.6.9 ..... 311(311 + 3)
3.6.9 ..... 3n 1I
11
= 4.7.10 ..... ();1I1/+ = 4.7.10 ..... ( )( ) 3n+l 311+1 311+4
1111+1
ZlII
1
3 3 =~. LI llll+1 3n + 4' /1--'>00 llll
Ratio test fails.
LI 11 { - [ ,lll/+1 Zltl
tl--'>oo
I}]
) 3n = 1 11->00 3n( 1+ 4 ) 3n
= LI
3n(I+3
- LI [ n (3n+4 --/1--'>00 3n + 3
I)]
Engineering Mathematics - I
432
= Lt
11
,,->003(n+1)
:. By Raabe's test
L
U"
=
n Lt 11->003n(1+,]n)
1
=- () and the series
1 p 1.3.p(p+1) 1.3.5 p(p+1)(p+2) 1+"2 q + 2.4.q(q + 1) + 2.4.6 q(q + l)(q +2) + .... is convergent, find the relation to be satisfied by p and q. Solution
un
=
1.3.5 ..... (2n-1) p(p+l) ..... (p+n-1) . I [negle"tmg l' term 1 2.4.6 .... .2n q(q + 1) ..... (q + n-1)
lI"tl =
1.3.5 ..... (2n-1)(2n + 1) p(p+ l) ..... (p + n -l)(p+ n) 2.4.6 .....2n(2n+ 2) q( q + 1) ..... ( q + n-1)( q + n)
U,,+l _
--;z- Lt 11->00
(2n+1) (p+n). (2n + 2) (q + n) ,
~ = Lt [2n (1+ 12n ) n (1+ ~'~ ) J= 1 lI"
11->00
2n( 1+ 1"2n)" n( 1+ ~~)
:. ratio test fails. Let us apply Raabe's test
,~.H::':, ~1)]~ ,~.H ~::~)~~::~~ ~1}] 1 Lt [n {2q ( n + 1) - ~ (2n + ) + nIl n 2 ( 1+ ~/~ ) ( 2 + },~)
11->00
r
Lt 2q(1+ )~)-p(2+ II->ool 2
J~)+11=2q-2p+l 2
433
Sequences of Series
Since
~
.
L..,.llil IS
convergent, by Raabe's test,
Yi,
=> q - p >
2q-2p+l >1 2
is the required relation.
Exercise 1 (c) 00
1. Test whether the series
I
Un
is convergent or divergent where
I
II II
=
2 2.4 2.6 2..... ( 2n - 2 )2 3.4.5 ..... (2n-I)(2n)
2n
X
[ADS:
Ixl:::; legl, Ixl > Idgl ]
2. Test for the convergence the series
f 4 .7 . 10 ..... (3n+l) x" I
[ADS:
lfl
1 I Ixl < -egl,lxl ~ -dg/] 3 3
[ Ans : divergent] (ii)
3.4 5.6 3 + ..... ( x> 0) x +4.5 - x 2 +-x 1.2 2.3 3.4 [ ADS: cgt if x :::; 1 dgt if, x > 1 ]
(iii)
~ 1.3.5 ..... (2n -1) xn ( ) L.. . ( ) x>O 2.4.6 ..... 2n 2n + 2 [ ADS: cgt if x :::; 1 dgt if, x > 1
(iv)
X + (1l)2 x 1+ (U)2 II l1
2
+
(lJ)2 x
l2
3
+
(x > 0)
..... .
[ ADS: cgt if x < 4 and dgt if, x ~ 4
6.5
Cauchy's Root Test Let
L un
I'
be a series of +ve terms and let LI u/n = I . Then
when I < 1 and divergent when I> 1
n-->co
L un
is convergent
Engineering Mathematics - I
434
Proof: I,'
II
Lt u/" =13a +venumber 'A'(!rn
I
A" is a geometric series with common ratio < I and
therefore convergent. Hence
I
is convergent.
UII
II
(ii)
Lt 11-->00
UII
/11 = I> 1
:. By the definition of a limit we can find a number r 3
u/ > 1Vn > r 'II
i.e., ull > Vn > r i.e., after the 1sl 'r ' terms, each term is> 1.
Lt 11-->00
Note: When
Lt{ In) Un
=
2>11 =
00
LUll
:.
is divergent.
1, the root test can't decide the nature of
LUll' The
fact
n-->oo
of this statement can be observed by the following two examples.
1. Consider the series
I Yn3n :-LfU'/" =Lt 1/
11-->00
2. Consider the series
L Yn,
(
11-->00
in which
1 ) J~
-3
n
= Lt ( V1 11-->00
nl II
LfUlly" = Lt ~ = 1 11-->00
11-->00
n11l
1/
In both the examples 'given above,
LfU/" = 1. But series (1) is convergent
(p-series test) And series (2) is divergent. Hence when the limit= 1, the test fails.
Solved Examples 6.5.1 Example Test for convergence the infinite series whose (i)
1
n'"
(ii)
1
Oogn)"
)3 =1
nIh
(iii)
terms are: 1
[1 +~r
Sequences of Series
435
Solution
. Lt u ,
By root test
(ii)
u
=
n
I
Un
n~co
,
1n
II
1
= lI~oo Lt -n2 = 0 < 1· '
is convergent.
1 ~. 1 ·u n = - - . (log n)'" n log n '
By root test,
I
Un
,.
Lt u 'n
=
n
n->oo
1 n->oo log n
Lt --=0oo n/ n Un is convergent if Ixl < 1 and divergent if Ixl > 1 and when Ixi = 1 the test Lt
n-->oo
:. I fails. Then
Take
r
1 n
vn =-
r
Un = (n + 1 .n = (n + 1 = (1 + !)n ; vn n"+ 1 nn n :. By comparison test,
(I
Uti
11-->00 VII
is divergent.
diverges by p -series test)
Vn
Hence
I
U
I
Un
is convergent if
Ixl < 1 and divergent Ixl ~ 1
6.5.3 Example ,/
If un =
n
(n+l)
,,2'
show that
I
Ull
is convergent
Lt-"=e>l
437
Sequences of Series Ill
l
2
I n n II n /1 Lt u" " = Lt 2 ; = Lt = " = Lt (-)-" "~a) II~OO n + 1)" ,,~oo n + 1) ,,~oo n + 1
(
J
(
~
Lt [_1_]" = < 1 ;:. "u" converges by root test. ,,~OO l I e L. +n
=
r
6.5.4 Example
J
~ + ( % + ( % + .' ......
Establish the convergence of the series Solution U
=
11
(_n_)" . . . . . 2n+l
Lt u };,
,,~oo n
By root test,
I
UII
(verify);
= Lt (_n_)=! 1. When Ixl = 1 : un = J n ,taking vn = ~ and applying comparison test, it can n+I n :. By root test,
n
be seen that is divergent
I
U/I
is convergent if
Ixl < 1
and divergent if
Ixl ~ 1 .
Engineering Mathematics - I
438
6.5.6 Example
Show that
~ ( n~ _ I) n converges.
Solution
=( nYn -If
Un
Lt n~oo
:. L
Un
unYn
= Lt
n . . . . . ct)
(n~-I)=I-I=O 1
n
n+3
Lt
= Lt
n~<XJ
2)n (1+-
nn
( 1+-3 ) n
e
2
I
- 3 = - '" 0
:. L un
e
e
is divergent. Hence
and the terms are all +ve .
L un
is convergent if
Ixl < 1
and divergent if
Ixl~l. 6.5.8 Example Show that the series,
-HI +[~: -H' +[:: -H +. . . u" ~ [(n :":t -n: r~ ( n: T[( n: r r [~:
3
isconvergent
1
1
-1
Sequences of Series
Lt un
~ II
IHOC)
:. By root test,
439
1 1 1 e-l
1 e-l
=-.--=-- I, this limit is finite; ClIse (ii)
,,1 . . L..J- IS dIvergent. n"
If P < 1, the limit is in finite; Ca.\·e
(iii)
Ifp= 1, the limit Lt
logxl;' = Lt (logn)=oo;:. I_l-
11...--.)00
I_ln"
Hence
n"
11-)00
IS divergent.
is convergent if p > I and divergent if p :S
6.6.4 Example
n "-2 ~e" 00
Test the series
for convergence.
Solution 1111
¢ (n)
=
n = ¢(n){say); e"
-2
n increases. So let liS apply the integral test.
is +ve and decreases as
I e-,2
1¢(x){lx= x I
=! le- dt{t=x 1
IlIl1
-"21e
2
,dt=2xdx}
2I
I
=
By integral test,
dx 00
-I 1
I
1' W IlIC. h·IS fiillIte. . = -"21(0 --;;1) = 2e
is convergent.
6.6.5 Example
(7f)
1 Apply integral test to test the convergence of the series "-sin 00
~n2
11
Solution Let ¢ (n)
= ~2 sin ( :
); ¢ (n) decreases as
J-21
00
J¢(x)dx =
00
2
2 X
sin
(}x 7f X
11
increases and is +ve .
Engineering Mathematics - I
444
Letix =t
J
-! sintdt = !costl~/=! TC
~
TC
finite,
TC
/2
_TC/ / x
2dx = dt
Yx2dx=- ~dt :. By integral test,
I
converges x
Un
= 2 => t = ~ x = 00 => t = 0
6.6.6 Example Apply integral test and determine the convergence ofthe following series. (a)
(c)
(b)
Solution (a)
¢ (n) -
3n is +ve and decreases as n increases - 4n 2 + 1 2 j¢(x)dx= ~x dx (4X +1=t=>Xdx= Ysdt ) I 14x +1 x=l=>t=5,x=00=>t=00
j
= Lt [3-log t -log 5]= I¢ ( x ) dx = Lt [3-8 f-dt] t 8 I
00
n~oo
I
:. By integral test, (b)
¢ ( n) =
~n3
3n +2
j¢(x)dx =
I
Un
diverges.
decreases as n increases and is +ve
1~X3
dx 3x +2 1 OOfdt 1 = (; = (;[logt]5 = 00
I
00
n-->oo
5
1
t
00
4
[where t= 3x + 2]
5
By integral test, (c)
I
Un
is divergent.
¢ (n) = _1_ is +ve, and decreases as n increases. 3n+l
1 1 dt 1 f¢(x)dx= f-dx= f--[t=3x+l]=-logtl~=00 I I 3x+ 1 4 3 t 3
00
00
00
445
Sequences of Series
:. By integral test,
I
Un
is divergent.
6.7 6.7.1 Alternating Series A series, u l - u2 + u3 - u4 + - + ..... + ( -1 ),,-1 un + ..... , where un are all +ve, is an alternating series.
6.7.2 Leibneitz Test If in an alternating series
I (-1 ),,-1 .u
n '
where
un
are alI +ve,
> un+I ' Vn, and Lt ull = 0, then the series is convergent.
(i)
un
(ii)
n ........ oo
Proof: Let u l
-
u2 + u3 - u4 + ...... be an alternating series (' un 'are all +ve)
Let ul > u2 > u3 > u4 •••••• , Then the series may be written in each of the following two forms: ......... (A) (UI -U 2)+(U3-U4)+(U5-u6)+····· ul
(A) (B)
-(u 2 -U3 )-(U4 -u5 )-·····
............ (B)
Shows that the sum of any number of terms is +ve and Shows that the sum of any number of terms is < ul . Hence the sum of the series is finite. :. The series is convergent.
Note: If Lt ull n--.<XJ
-:t:-
0, then the series is oscillatory. Solved Examples
6.7.3 Example . 1--+--1 1 1 ..... Cons!·derthe series
234
In this series, each term is numerically less than its preceding term and ~O as n~oo. :. By Leibneitz's test, the series is convergent. (Note the sum of the above series is Loge2 )
6.7.4 Example Test for convergence
L (-I ),,-1 2n-1
nih
term
Engineering Mathematics - I
446
Solution The given series is an alternating series We observe that (i)
> 0, Vn (ii)
un
un
I (-1 ),,-1
un
,where
= _I_
U n
> Un+ I ' Vn (iii) Lt
Un
2n-I
=0
/1---700
:. By Leibneitz's test, the given series is convergent.
6.7.5 Example 111 Show that the series S - 1- - + - - -
3 9
27
+ ...... converges.
Solution The given series is series in which 1.
~ ~
un
( ),,-1 -1 1 3n-
= "'(_1)/1-1 u" L.J
u, = _1_1 is an alternating
where
I
> 0, Vn 2.
un
3/1-
I
> U/I+1' Vn and 3. Lt
un
= 0;
ll---7oo
Hence by Leibneitz's test, it is convergent.
6.7.6 Example X
x3
2
Test for convergence of the series, - - - ~ + - - 3 - +..... , 0 < x < 1 1+x 1+x- 1+x Solution n-1 n-1 ( -1 ) .xn The given series is of the form'" = "'(-1) u,
L.J
x"
where
UII=--II
. I+x'
Further,
U II
-u
L.J
n
Since O<x< 1, ulI>O,Vn;
x"
n+1
1+ XII
X"+1 I+XIl+1
=-n ----
l+x
x" _X"+1
x" (I-x)
=--:-----:--:------;-
( 1+ x" ) ( I + x"+ 1 ) 0< x < 1 +ve.
Again,
~
all tenns in numerator and denominator of the above expression are
x" ~ 0 as x" ~ 00 since 0 < x < 1; .'. Lt
:. By Leibneitz's test, the given series is convergent.
II~OO
o =0 1+0
U = --. II
447
Sequences of Series
6.7.7 Example
(-1 ),,-1
L ----r======== 00
Test for convergence
1I~2 ~11(11+1)(n+2)
Solution The given series is an alternating series "(-1)"-1 L..J ull where Again,
1
1111
~11 ( 11 + 1)( 11 + 2)
=
; u > 0, \1'11 ; ll
~(11+1)(n+2)(n+3) >~n(I1+1)(n+2) 1
1
~(11+1)(n+2)(11+3) < JI1(I1+1)(11+2) ,\1'11 I.e.,
Further,
Lt u IHOO
= Lt
II
By Leibnitz's test,
11-->00
1
J 11 ( 11 + 1)( 11 + 2 )
f( -1)"-1 un
=0
is convergent
2
6.7.8 Example Test for the convergence of the following series,
1
2
3
4
5
---+---+--+ 6 11 16 21 26 .... Solution
L (-1) 00
Given series,
n=1
n-I
11
- - = L (-1)
II-I
is an alternating series
Un
511 + 1
11 n 511 + 1 11 11+1
u =-->0\1'11 . -----
511+1
Again,
511+6
'
-1 (
511+1
11 IHOO 511 + 1
)(
511+6
)
~
Un
< Un+ I' \1'11
1 5
Lt u = Lt --=-:;t:0 n-->oo
n
Thus conditions (ii) or (iii) of Leibnitz's test are not satisfied. The given series is not convergent. It is oscillatory.
Engineering Mathematics - I
448
6.7.9 Example Test the nature of the following series.
I
00
(a)
1
(
r-
-1
I
I --'--::--'r--(
(b)
f;z +J;+i
-1
I
(c)
2
n +1
Solution (a)
u,,=
1
~>OVn
I
vn+vn+l
1
U" -
U,,+I
=
1
f;z + J;+i - J;+i + J n + 2
:. By Leibnitz's test the series converges. 1
1
1
u" =-2->0,Vn; - 2 - > 2 =>U" >un+I,Vn; n +1 n +1 (n+l) +1
(b)
Lt
UI/
n-->oo
(c)
un
=0 1
= I ~.l
1
~
:. By Leibnitz's test, given series converges.
> 0, Vn ;
1 < I ~.1 1 => UI/ > lln+P V n ~~
In + 2 > In + 1 => I ~
1 I
')
By Leibnitz's test, given series converges. 6.7.10 Example Test the convergence of the series
111 sJ2 - sJ3 + sJ4 - +..... .
Solution
I
00
The series can be written as
n=1
(_1)"-1 -'-;=====
sJ;+i
1
=---;==
U /I
SJn+l
(i)
(ii) (iii)
Lt un
n-->oo
=0
.
'
By Leibnitz's test, the given series converges.
Sequences of Series
6.7.11
449
Example Test for convergence the series, 1-
h + ~ - ~ + .....
Solution The given series can be written as
(-If (omitting \'1 term) I-2n
1 1 1 1 ->OVn; ->--=>u >lll/+I'Vn; Lt -=0 2n 2n 2n + 2 2n lI
IHOO
:. By Leibnitz's test,
6.7.12
(-If I-2n
is convergent.
Example
1 3!
1 5!
1 7!
Test for convergence the series, 1- - + - - - + ..... .
Solution
General tenn of the series is /-1)"-;
2n-I !
The series is an alternating series; (
1 ) > OV n
2n-I !
1 1 1 >( ) => Un> Un+ I' Vn EN; Lt ( ) =0 2n - 1 ! IHOO 2n - 1 ! ( 2n - I! ) By Leibnitz's test, given series is convergent.
6.8
Absolute convergence A series UII is said to be absolutely convergent if the series
I
convergent
6.8.1 Consider the series
1 1 1 IU =1--3 +3"--3 +-+ ...... 234 1 1 1 1 Ilun l=1+-3 +3"+-3 + ...... = I-3 2 3 4 I n By p - series test, IUn is convergent (p = 3 > 1 ) II
00
I
I
I
IU I IS II
Engineering Mathematics - I
450
I
Hence
Note: 1. If
I
Un
Un
is absolutely convergent. is a series of +ve terms, then
I
lIlI =
I
IUI/ I·
For such a series, there is no difference between convergence and ahsolute convergence. Thus a series of +ve terms is convergent as well as absolutely convergent. 2. An absolutely convergent series is convergent. But the converse need not he true.
I (_1)"-1.-!-n =1-1..+1.. -1..+ ..... . 2 3 4
Consider
1
This series is convergent (1.7.3)
I I(-1 ),,-1 )~I =1+ ~ + ~ + ~ + ....... is divergent (p-series test). Thus I is convergent need not imply that II I is convergent (i.e., I is not absolutely convergent). But
Un
UII
Uti
6.9 6.9.1
Conditional Convergence [fthe series is divergent and
Ilu,,1
I
Un
is convergent, then
I
Un
is said
to be conditionally convergent.
6.9.2
Consider the Series
1-1..+.!-1.. ....... 234 But
:. I
I
IU
II
is convergent by Leibnitz's test. (Ex. 1.7.3)
IUn I= 1+ ~ + ~ + ~ ..... Uti
is divergent by p - series test.
is conditionally convergent.
6.10 6.10.1
Power Series and Interval of Convergence A series, ao + a1x + a2 x 2 + ..... + anx n + ..... where 'an' are all constants is a power series in x. It may converge for some values of x .
. Sequences of Series
451
Lt n~<Xl
UII +I till
= Lt
a/HI •x (I st term is omitted.)
n~<Xl
all
Lt
=kx (say) where
n~'"
Then, by ratio test, the series converges when i.e., it converges '\Ix
E (
,)
(~1 ~
The interval
a ll +1 = all
k
Ikxl < 1 .
~1 , ~ )(k *- 0)
is known as the interval of convergence of the given
power series.
Solved Examples 6.10.2
Example
I \ '"
Find the interval of convergence of the series
11;1
II
n
Solution
"Lf.( U~:, )="Lf.(n: l)'·x =H.[ ~.~ 1
By ratio test, the given series converges when When x
= 1, L un = L
:. L un
in
3'
I
J:
Ixl < 1, i.e., x
= E
x
(-1,1)
which, is convergent by p series test.
is convergent when x
=1
Hence, the interval of convergence of the given series is (-1, I]
6.10.3
Example Test for the convergence of the following series. (a)
1-
YJ2+ YJ3- YJ4+··········
(b)
1+
X2- }i2- ~2+ Ys2+ ~2- (;2- h2+ . . . .
Engineering Mathematics - I
452
(c)
I-X~+X~-~+ ......... .
(d)
I(-I)"(n + l)x
n
,
with x
OV n (ii) Un > 1l,l+IVn and
(iii) Lt n~oo
Again
un = 0;:.
the
By Leibnitz test, the series is convergent. 1+
series
YJ2 YJ3 YJ4 +
+
+ ......
is
divergent,
p - series test. Hence the given series is conditionally convergent. 00
(b)
L lu,,1 = L;;:2
which is convergent by p - series test.
p=1
The given series is absolutely convergent. It is convergent.
(c) The given series is 2n-2
"(_I)n-l. x . ="(-1 ),,-I U · . Iu ~ (2n-2)! ~ n' •• n
-U n +1 = u"
By ratio test, the series
2n-2
I=_x __ (2n-2)!
1 1 21 ; Lt -lln+1 .x (2n -1 )(2n) tHoo ll"
Llu,,1 converges Vx;
=0 < 1
L u"
i.e.,
is absolutely
convergent Vx; (d) Here,
Iunl =(n + I)xn;1 un+1 1= (n + 2)x n+1 (neglect 1
sl
lu I
Lt ~ = Lt IHOO
Iunl
:. Llunl
IHOO
(n + 2)
(n + 1)
Ixl = Lt IHOO
term)
7n)~ Ixl =Ixl < 1 (1 + Yn)
(1 +
}i
(": x < 12 )
is convergent Vx, i.e., given series is absolutely convergent
and hence convergent.
by
453
Sequences of Series
6.10.4 Show that the series 1 + x + x/6 + x/6 + ..... converges absolutely \;Ix Solution
Lt '1->00
IU'HII =13 =0 < 1 when lUl I
n
Ilunl
:. By ratio test, When x
x*-O [since .
111" 1= (n-I)!' Ix"-II ·Iu 1= Ix"l ] n! IItl
is convergent \;Ix *- O.
= 0, the series is (1 + 0 + 0 + ...... ) and is convergent
:. Ilu,,1 converges ~ Ill" is absolutely convergent 6.10.5
\;Ix.
Example Show that the series, 1 -
~ + ~ - ~4 + ........ 3
3-
3
is absolutely convergent.
Solution , which is a geometric series with common ratio
~ 3-Fn+l >3,Jn,Vn.
1 1. :. 3.Jn+l n
Whenx= 1,
L ~n Hence I since
=
-(1+~+!+~+ ...... ) 234
which is divergent,
is divergent by p -series test (prove).
Un
is convergent when -1 < x ::; 1
Interval of convergence is ( -1 < x ::; 1 ).
Exercise -1 (e)
1. Use integral test and determine the convergence or divergence of the following series: 1
Ln
[ Ans : convergent]
2
1
00
2.
~ n{logn )2
2. Test for convergence of the following series: 1 1 1 l-~
'"
2.
+l:!: -l2 + ......................................
{-If-l
~(2n-I)(2n)
[ Ans : convergent]
[ Ans : convergent]
[ Ans : convergent] [ Ans : convergent]
I
3. Classify the following series into absolutely convergent and conditionally convergentseries: [ Ans : abs.cgt ] [ Ans : abs.cgt ]
3.
I
{-If n{logn)2
4. Find the interval of convergence of the following series 2n x n 1.
L
~
[ Ans : abs.cgt ]
[ Ans :
-00
< x < 00 ]
Engineering Mathematics - I
458
[ Ans : - 1 ::;; x ::;; 1 ]
r Ans : -1 < x ::;; 1 ] 111 - 2 + .... is absolutely convergent. 2 3- 4 111
(a) Show that 1- - 2 + - ? -
5.
(b) Show that 1 -
J2 + j j - J4 + ....
is conditionally convergent.
Tests of Convergence - A Summary 00
I. The geometric series
Lx
n I -
converges if 1x 1< 1 , diverges if x ~ 1, and oscilates
n=1
when x::;;-1 2. If
L u"
is convergent, Lt ull n->oo
f ~n
3. P - series test:-
=0
[The convergent need not be necessary ]
is convergent if p > 1 and divergent if p::;; 1
11=1
4.-C-omparison lest :- The series
LUll
and
L vn
are both convergent or both
~ is finite and non - zero.
divergent if Lt
n->oo VII
L un
5. D 'A/ember! 'sRatio test:-
Lt n->oo
(
U
2:':l.
u
< 1 or > 1
ll
Alternately, if Lt n->oo
6. Raabe's test:
converges or diverges according as
L un
~ >1 u ll
+
or < 1) . If the limit = 1, the test fails
1
converges or diverges according as
.~HU::1 -I}]> lor O) ........ l+x l+x l+x 2 3 3x 4x 6. 2x+-+-+ ..... (x > 0) .......................... X
8
27
7. 1+!+ 1.3 71.3.5 +....
................................. 2.4.6 32 32.5 2 32.5 2.7 2 8. 2" + -2-2 + 2 2 2 + ... ... ............................... 6 6 .8 6 .8 .10 2
2~4-
9. 3.4 + 4.5 + 5.6 +....
1.2
( dgt.]
2.3
3.4
................................
( cgt. if
x~l dgt. ifx> I]
( cgt. if x ~ 1 dgt. if x > 1) (dgt.]
[ cgt ] (dig.]
Engineering Mathematics - I
460
10.
(l!)2 (tJ)2 2 (lJr ~ tJ .X+ l1 x + 12 x + .... (X>O) .............
X x2 x3 II. 1+ 22 +j2+4T+ ..... (x>O)
(4
5
........................ .
12.
3x 5)2 x 2+(6)3 +x3+ ..... (x>o) 4+
13.
L(l+~r
............ .
[ege ifx I] [ egt if x < 1, dgt. if x 2 1 ]
[ dgt. ] [ egt. ]
15.
a" 2:--2,a or
1 ;
(If the Limit = I, The test fails).
8. Integral test: A series
L rj> ( n) of +ve terms where rj> ( n) decreases as n increases
frj> ( x ) dx is finite or infinite.
00
is convergent or divergent according as integral
1 00
9. Alternating series - Leibnitz's test: An alternating series
I
(-1 ),,-1 un (where
n~1
Un>
OVn ) is convergent if(i)
un
>
lin+1 '
Vn (ii) Lt
Un
=0
n~oo
10. Absolute/ conditional convergence: (a) LUn is absolutely convergent if II Un (b)
IU
n
is conditionally convergent if
I is convergent.
Iu"
is convergent and
II U" I
divergent. (c) An absolutely convergent series is convergent but converse need not be true. i.e., a convergent series need not be convergent.
is
465
Sequences of Series
Solved University Questions
1. Test the convergence of the series:
123 1.2.3 2.3.4 3.4.5
- - + - - + - - + ........ . Solution Let ull be the
d h term of the series;
Then,
U
II -
1 n(n+1)(n+2) - (n+1)(n+2)
n
1
Let
VII
=
-
U
then, Lt
--'!.. 11->00 V
-2 ;
n
Lt (
=
tHoo
II
11->00 (
LUll
But
L
VII
and
L
n +1 n +2
VII
)
1
Lt
=
Which is non-zero and finite. :. By comparison test, both
n2 )(
=1
1) (
2)
1+;; 1+;;
,
converge or diverge together.
is convergent by p-series test (p > 1):.
LUll
is convergent.
2. Show the every convergent sequence is bounded Solution Let
Lt
(all) all
be a sequence which converges to a limit '/ 'say.
= I =>
given any +ve number
E,
however small ,
1l~C()
we can always find an integer'
III ' , ) ,
Ian -II <E,
'ifn ~ m
jaIl -II < 1; I.e., (I - 1) < all < (i + 1), 'if n ~ m Let A = min {a"a 2 , •••••••• am _,,(1 -1)} ,and fl = max {a"a 2 , •••••••• am-,,(I + 1)} Then obviously, A ~ an ~ fl, 'ifn EN; Hence (an) is bounded.
Taking
E
=
1, we have,
3. Show that the series,
1 1 1 S =1--+---+ .......... converges. 3! 5! 7!
Engineering Mathematics - I
466
Solution
2)-1 ),,-1 u" '
The given series is an alternating series where u" (i) (ii)
ll"
Lt
=(
1
)
211-1 !
We observe that,
> 0 and u" > 1l,,+1' '1111 and II
,,--+00"
1 =0 IHoo ( 211 - 1) !
= Lt
:. By Leibneitz's test [7 .2 ] the given series converges. 00
4.
Show that the geometric series
I
qlll = 1+ q + q2 + ....... converges to the slim
111=0
1 - - when /q/ < 1 and diverges when /q/ ~ 1 l-q Solution See theorm 2.3 (replace' x ' by 'q') . 5. Define the convergence of a series. Explain the absolute convergence and conditional convergence of a series. Test the convergence of series
I[l+ J,;r' Solution For theory part, refer 2.1,2.2, 8.1 ,9.1, and 9.2
,
Problem:
Let
u" = (1 + _l_)-n ; Lt (u~{,) = Lt (1 + _1_)-n J;; J;; 11--+00
~ H-[l+ By Cauchy's root test,
6.
I
Un
1
J,;J
,,--+00
1
=7 O. Solution
1.3.5 ..... (211-1)
Omitting the first term of the series, we have, ------'--------'- x
2.4.6 .... .211
n
and
Sequences of Series
467
1.3.5.(2n-l) 1/
UI/ =
2.4.6 .... 2n
1.3.5 ..... (2n+l) ; UI/+ I =
X
(
2.4.6 ..... 2n + 2
n+1
).x;
L (2n + 1 )
L u n+ 1
1/100 ---;;: = 1/100 2n + 2 .x = x By ratio test,
LUI/
is convergent when x < 1, and divergent when x > 1
The ratio test fails when x = 1 Whenx= 1,
~-1 = 2n+2 -1 =_1_ 2n + 1
U n +1
2n + 1
Lt [n(~-I)]= Lt (_n 2n +
n~oo
:. By Raabe's test,
U Il +1
LUI/
IHOO
1
)=!
ll----too
1; when x = 1 , the test fails. When
:. LU
x
n
=
1, un
(1 + l/~r
=(
, )n;
1+ 3/~
Lt Un = ee = -e1* 0 -2
IHOO
is divergent.
:. is cgt. when x < 1 and dgt. when x 2 1 .
468
8.
Engineering Mathematics - I
nih
Test the series whose
term is
(3n -1) / 2n
for convergence.
Solution
= n
U
(3n-l)
U,,+I _
un :. By ratio test,
2"
. '
{3(n+l)-I}
U
=-'--------'-
(3n + 2) 2(3n-l)
L un
2"+1
IHI
- - -- 1 oo Un
is convergent.
L
1
<X)
9.
Show by Cauchy's integral test that the series
n;2
n(logn
r
converges if p > 1
and diverges if 0 < p ~ 1 Solution 1 Let ¢ ( x ) = ; x ~ 2 ; Then ¢ ( x) decreases as x increases in x(logxy <X)
00
ax
<X)
f¢(x)dx=f x(logx)
2
2
p
=
d
I --;;; U
[2,00]
I-p
II
=~ P
log2
. I [Takmg log x = u, - ax = du x = 2 => u = log 2 and x x Case (i) : p > 1 => 1- P < 0 => Integral is finite, and
1
00
log2
;
= 00 => U = 00 ]
Case (ii) : 0 < p ~ I => Integral is infinite. Hence, by integral test, the given series converges if p > I and diverges when O 0 and
> 1l1l+IVn EN.
1I"
= o. Hence, by Leibnitz test, the series is convergent.
11-»00
12.
. DISCLISS
. the convergence of the series,
1 J.
2'1'1
+
x2
x6
X4
r;; + r:; + t. + ......... . 3'1'2 4'1'3 5'1'4
Solution 2n
n
Iii
· term 0 f th e series
=
x r--:-:; u = (n+2)'I'n+l
"1I+1,J;;+2 rn+i
21 X llt
~= I/~OO lrVn.M. (+ 3.~) 2l=x LI
2
X
"II
ll
2
;
1
It"
= 1, u =
2
.x
LI
I converges if x > 1 , i.e., if Ixl > 1 ;
When x
(n + 3)
1=
By ratio test, x
1+2
(n + 3).J n + 2 ; ---;;: =
U
/HOO
2
( . . I sl term ) omlttmg
ll
2
< 1 , i.e., if
rn+i ;taking
1
(n+2) n+l
vn
Ixl < 1, and diverges if
= -3-'
,
n i2
]I
Lt
U _II
n~oo vn
=
n/ 2
Lt
n~oo n%
By comparison test,
I
Un
(1 + ,'n 2/) \fIT?n hl and
I
VII
=1
both converge or diverge together;
Engineering Mathematics - I
470
L vn is convergent by p-series test. :. L un is convergent if Ixl ~ 1 and divergent if Ixl > 1
But
211
00
13.
un+IVn and un
By Leibneitz's test
IU
~ 0 as n ~
converges when x =-1
II
.. Interval of convergence is [(-1, 1) i.e., -1 15.
00
~
x 1.
I
00
18.
=
I = LI -2n3-+X n- = 2"* 0
VII
and
VII
=
Test the series
II~I
(-I)" (log n ) 2
'
for absolute/conditional convergence
n
Solution
Un = (-1)" (;ogn) ; IUIII=(lo~n) n
n
Sequences of Series
473
"'flO~, x dx -- ' 'f le-'dl 2
x-
I/ [taking logx=l, x=e ' , /xdx=log/]
log2
~ -Ie -, + e-, I:.2 ~ 0 - [I-log 2].e -'"" ~ i(!Og 2 -I), which is finite.
I
I
:. By integral test ~]un is convergent => (Note that 19.
I
UII
1I11 converges absolutely.
is cgt. by Leibnitz's test).
Test the convergence of the series
I
I
(log log n )"
Solution
=
Given that u, I
LI u'1/n n-->oo n
= LI
n-->oo
[I] = log log n
I
By Cauchy's root test, 20.
I
(log log n)"
Un
0OO n 2( 4 + ~ + 6 n2
19
Ixl < 1 => -1 < x < 1
r
Engineering Mathematics - I
474
Then
~_1=(4112?+1011+6 -1)= ~11+5 u
ll
411- + 211 + 1
+1
Lt IHOO
411- + 211 + 1
[11(~-1)]= Lt n2(8+1~) lIn+1
By Raabe's test,
IHOO
LUll
n ( 4 + 7~ + /~~2 )
converges when x
:. Interval of convergence of
=2>1
2
L
lin
is
2
= 1, i.e., x = ±1.
(-1:::; X:::; 1)
7 Vector Differentiation
7.1.1
Vector Point function and vector field Let P be any point in a region 'D' of space. Let r be the position vector ofP. Ifthere exists a vector function F corresponding to each P, then such a function F is called a vector point function and the region D is called a vector field.
Note: In what follows i,j, k are unit vectors along X, Y, Z axes respectively For example, consider the vector function
F = (x - y) i + xyj + yzk
..... (1 )
Let P be a point whose position vector is r = 2i
+ j + 3k in the region D of space.
At P, the value of F is obtained by putting x = 2, Y = I, z = 3 in F.
I.e.
At P,
F = i + 2j + 3 k
Thus, to each point P of the region D, there corresponds a vector F given by the vector function (I). Hence F is a vecotr point function (of scalar variables x, y, z) and the region D
is a vector field.
Engineering Mathematics - I
476
Scalar point junctioll and scalar fielit Ifthere exists a scalar 'f' given by a scalar function 'f' corresponding to each point P (with position vector r) in a region 0 of space, of' is called a scalar point function and 0 is called a scalar field. As an example, let P be a point whose position vector is r = 2i + j + 3k. Consider f= xyz + xy + z Then the value of f at P is obtained by putting x = 2, y = I, z = 3
f= 2.1.3 + 2.1 + 3 = II
i.e., At P,
Hence the scalar' II ' is attached to the point P. The function 'f' is a scalar point function (of scalar variables x, y, z), and 0 is a scalar field.
Note : There can be vector and scalar function of one or more scalar variables.
7.1.2 Differentiation of a vector Ifr (u) = r,(u)i + r2(u}j + riu)k, (where rl' r2, r3 , are scalar functions of 'u') be a vector function of' u' , then,
Lt
dr du
Or
~Oou
ou
~ dr,.
dlj. du
Lt
r{ u + ou) - r{ u)
ou ~O
ou
dr2 . du
dr3 k du
= L.,-l=-l+-}+-
du
Example If r(u) = (3u 2 + 5u + 6) i + 3U7 - 4uk, Find dr , when u = 1 du
dr du
= {~(3U2 +5U+6)li+{~(3U2)}j+{~(-4U)}k du
1
du
du
Note: We can apply the above rule of derivative to the case of partial derivatives also ex : If A
=
a
(ryz)i + (xyz}j - (3;Jyz2)k
2 A find - - at the point (I, -I, 2)
axOv
Vector Differentiation
aA ay
-
477
)t.
=
{ - a (2 x yz ay
=
(x 2z)i + (2xyz)j - (6x1yz2) k
=
a {xy 2Z }. j -a- {3x·yz3 ') ')} k 1+Oy
Oy
2xz i + 2yzj - 18x2yz2 k
At the point (I, -1,2)
a2 A
ax~v
=
4i - 4j + 72k
7.1.3 Application to space curves Let ret) = x(t)i + y(t)j + z(t) k represent the position vector of a point (x. y. z) on a space curve whose equations are given by x time.
=
x(t), y = yet), z
= z(t), where 't' is (K + M, Y + ~Y, z + 6Z)
dr dx. dy. dz Then -=-I+-J+-k dt dt dt dt and x
(i) dr represents the velocity vector v (or tangent vector) of the point (x, y, z) dt (ii) d 2 r represents the acceleration vector a at the point (x, y, z) dt 2 Ex:
If a particle moves along a curve x
=
e-t, y
=
2 cos 2t, z = 2 sin 2t, where 't' is time.
I) find velocity and acceleration at time t = 0, and 2) find also their magnitudes Sol:
r = xi =
+ yj + zk
(e-t)i + (2 cos 2t)j + (2sin 2t)k
Engineering Mathematics - I
478
dr d v= - = dt dt =
(e-~i
d d + - (2cos 2t}j + - (2sin 2t)k dt dt
(-e-t)i -(4sin 2t}j + (4 cos 2t)k
d 2 r dv d d. d a= -=-=-(-e-t)i-(4sm2t}j+ - (4cos2t)k 2 dt dt dt dt dt = (e-t ) i-(8cos 2t)j-(8sin 2t) k
.... (2)
Putting t = 0 in (I), velocity as t = 0 is v = -i + 4k Magnitude =
10
putting t = 0 in (2) acceleration at t = 0 is a = i-8j Magnitude =
J65
7.2 Gradient of Scalar Function 7.2.1
The Vector differential operator 'DEL' or 'NABLA', denoted as 'V' ' is defined by V'
=i
a .a ax ay
k
a az
-+ }-+ -
(i,j, k are unit vectors in x, y, z directions)
This operator' V' ' is used in defining the gradient, divergence and curl. Properties of' V' ' are similar to those of vectors. The operator is appled to both vector and scalar functions.
7.2.2 Gradient If ~ (x, y, z) is a scalar function, defined at each point (x, y, z) in a certain region of space and is differentiable, the gradent of as,
.a .a ( ax ay
a) az
grad ~ = V' ~ = 1-+ }-+k-- ~
(which is a vector function)
~
(shortly written as grad
~)
is defined
Vector Differentiation
479
:. If ~ defines a scalar field, 'grad' ~ or V ~ defines a vector field.
7.2.3 Physical significance of 'grad
~'
:
If ~ (x, y, z) = c (c being a constant) represents a surface, then 'grad ~'represents the normal vector to the surface at the point (x, y, z) For, if r = xi + yj + zk, is the position vector of the point (x, y, z) on the surface, we have, dr = (dx) i + (dy) j + (dz) k which is in the tangent plane to the surface of (x, y, z)
(.,'
~
=c)
:. The vector' V ~' which is 1.- r to the tangent plane is the normal vector to ~ =
c at(x, y, z)
7.2.4 Directional Derivative
V~.a
If a be any vector,
lal
which represents the component of V
.'~
~
in the direction of
a is known as the directional derivative of' ~ , in the direction of a, (1) Physically the directional derivative is the rate of change of' ~ , in the direction
ofa. (2) The directional derivative will be maximum in the direction of V a=
V~) and the maximum value of the directional derivate = It~~
=
~
(i.e.,
IV~I·
7.2.5 Some basic properties of the gradient If
~
and \jJ are two scalar functions,
'J) grad (~+\jJ)
Proof: grad
=
(~+ \1')
grad ~ + grad \jJ(or) V(~+\jJ)= V~+V\I'
=
~
V( + \jJ)
=
{! (~+
\jJ)} i + { ;
(~ + \jJ)} j + {! (~+ \I')} k
Engineering Mathematics - I
480
8ljI. 8ljI )
o~. ~. ~ ) + (Oljl. = -I+-j+-k -I+-j+-k ox oy OZ ox ay OZ
(
= Y'~+Y'ljI
(2) grad (~ljI)= ~ (grad\fl) + ljI(grad~) (or) Y'(~ljI) = ~ (Y'ljI)+(Y'~)\jJ Proof: grad
=
{! (~ljI)}i
(~ljI)
Y'(~ljI)
=
+{;
(~ljI)} j + {! (~ljI)}k ar _~
ax
.. ar Slmtiarly 8y
ax
r
yarz =r'az r
=- .-
grad ~ = Vr 3
a 3. a 3 . a 3 ..1 ar 2 ar. 2 ar = -(r )l+-(r )}+-(r )k =3r-i+3r -}+3r - k ax 8y az ax 8y az
zk] =3rr
[x.
..1 y. =3r -l+-)+r r r
7.2.12
Evaluate grad r"
Sol:
Let ~= rn
= (xl + T + z2)nl2
~ =!!{x 2 + y2 + Z2 ax
2
r/
t
2 -
.2x =nx{x 2 + y2 z2
r/z-
t
Engineering Mathematics - I
484
..
sImilarly -o~ = ny ox grad rn = grad
~
(X 2 + y2 + Z2 )'?z-/,I and -o~ oz
(222)";-1
= nz x + y + z
o~. ~. ~k = - I + - j +ox 0' oz
= n, {x 2 + y2 + Z2 )'fi- I • (xi + yj + zk) = n rn-2.r Aliter:
g
o~ or. o~ or. o~ or 1+ (o~l. - j+ (~) - k=-.-I+-.-j+-.-k (-o~). ox oy OZ orox or0' oroz
~
rad
= nrn- 1 . !.. i + nxn- I . y j + nxn- I . !... k = nrn-2.r r
r
r
Ex.7.2.13
If A is a constant vector prove that grad (r . A) = A
Sol:
Let A = A I i + Aj + A3k r
= xi + yj + zk
r. A
=
A1x + A2Y + A3z = f(say)
of ox = AI'
81
Ex.7.2.14 Prove V Let f=
81
-
ox
=
~
",I
81
0'
:. grad (r.A)
Sol:
(A I' A 2 , A3 being constant functions)
= A2, OZ = A3
= gradf= 81 i+ 81 j+ 81 k ox 0' oz
=A i+Aj+A k=A I
3
~I(r).r
(~(r»=--
r
(r)
ar
(r) -
ox
'Y
=
~
or ox
I'X
(r).r
81 =~I(r)ar =~I(r).y 0' 0' r 81 = ~I(r) or = ~I(r).!... OZ
.. V
OZ
or
81. of. 81 0/ + 0' ) +
oz'''
<j>1(r)
=
W
r
or
y
0'
r
-az =-r (see Aliter of ex. 7.2.11)
r
(~ (r» = V f=
z
x
<j>1(r)r
(xi+yj+zk) =
W
485
Vector Differentiation
Ex.7.2.15 Find the equations for the tangent plane and normal line to the surface z = x 2 + at the point (2, I, 5).
.r
Sol:
Let r = xi + y) + zk be the position vector of any point P(x, y, z) on the surface. Let r l = xli + yJ + zl k be the position vector offixed point A(xI'YI' z,) on the surface. Then AP = (x - xI) i + (y - YI») + (z - z\) k
= r - r,
Let n be the normal to the surface at A. Then, since AP is perpendicular to n, we have, ..... (I)
(r-r\)n=O which is the equation to the tangent plane at A. Here, in the given problem r-r, =(x-2)i+(y-I»)+(z-5)k and 11 = V (xl + =
.r -
z) at A (2, I, 5)
(2xi + 2y) - k)I(2.).5)
= 4i + 2) - k
:. The tangent plane at (2, I, 5) is, (from (I», 4(x - 2) + 2(y - I) - I(z - 5)
=0
~4x
+ 2y - z = 5
.... (2)
From (2), the direction ratios of the normal Iine at A are 4, 2, -I :. Equation 10 tpt( pormal line at A are,
y- Y) z-z) he \ w re abc
x-x)
-- =
~,=--
(xl'YI' zl)
= (2,_1,5) and (a, b, c) = (4; 2, -I)
:. The equations of normal line are x - 2 4,
= Y -I = z 2
5
-I
From the above example. we have to remember the following : Let (I)
+
=
c be any given surface and (xl' YI' zl) be a point on it; then
Equation to the tangent plane to
+=cat(xI'Y"z,) is
486
Engineering Mathematics - I
Equations to Normal line at (xl,y\, zl) are
(2)
= Y- y, = z-z, 8/8x 8/ay 8/8z x-x,
Exercise 7(a) I.
If <jl
=
2xz3 - 3xlyz, find V and IVI at the point (2, 2, -I) [Ans: (i) 22i+ 12}-12k(ii)2M3]
2.
3.
If V = 2x i - 3y) + z 3k, and <jl = 2xyz - 3z2, find V. V and Vx V at the point (1,2,3) [Ans . (I) - 426 (2) 6i + 352} t 156k] Iff=2xyzandg=xly+z,fing V (f+g) and V (fg) at the point(I,-I,0) [Ans : -2i + ) - k; 2k ] 2
4.,Jr + 1 ~)
7
[AI:1S : (6 - 2~/2 - 2r -3) r ]
4.
Evaluate V (3r
5. 6.
If = r2 e- , show that grad = (2-r) e-r r Find a unit normal vector to the surface z = xl + Y at the point (I, -2, 5)
-
~r
r
I
[Ans: 7.
51 (2i -
4j - k)
Finq the equations to the tangent plane and the normal line to the surface xz2 + xly = z - I at the point (I, -3,2)
°
"
. .. x-I y+3 z-2 [Ans: (I) 2x- y- 3z + 1 = (n) -2 =-1- =-3-
8.
Find ~uations to the Tangent plane and normal line to the surface y at the point (1, 5, -2)
= xl + z2
x-I y-5 z+2 [Ans: (i) 2x-y-4z= 5 (ii) -2-=-=1=--=-4 9.
Find the directional derivative ofU = 4xz3 - 3x2yz at (2, 1, -2) in the direction of (3i - 2) + 6k). [Ans:
10.
Find the directional derivative of
384
7]
= 4e2x- y+z at the point (1, I, -1) in a direction
towards the point (2, 3, 1) [Ans : 8/3]
487
Vector Differentiation
II.
axl
Find the values of the constants a, b, c so that the directional derivative off= + byz + cz2x 3 at (1,2, -I) has a maximum magnitude 64 in a direction paraIlel to z-axis. [Hint:
/0:./ at (I, 2, -I) is IlleI to z-axis.
:. Equate coefficients of i and j to zero and IV ~ = 64. Thus get 3 equations in a, b, c and solve them] [Ans: a =6, b = 24, c = - 8] 12.
Find the acute angle between the surfaces xlz = 3x + z2 and 3x2 - l + 2z = I at thepoint(I,-2, I)
J3
[Ans : cos- I - - ]
7J2
13.
Find grad ~ if r = xi + yj + zk, r = Irl and
(i)
~
= Log r (ii)
~
I = r
(iii)
~
=r [Ans: (i) r/~ (ii) -r/~ (iii) r/r]
14.
Find the directional derivative of g = x 2l + lz2 + z 2x 2 at the point (I, I, -2) in the direction of the tangent to the curve x = e-t, y = 2 sint + I, z = t - cos t at t = o. [Hint: Tangent vector to the given curve is
dx. dy. dz k -l+-}+dt dt dt 2
[Ans: 15.
Find the acute angle between the normals to the surface xy = z2 at the points (1,9,3) and (3, 3, -3)
16.
J6]
If r
= xi + yj + zk and
4> = x 3 +
Y + z3 -
3xyz, show that r, grad 4> = 34>.
Engineering Mathematics - I
488
7.3
The Divergence of a Vector Function
7.3.1
If A = A)i + A2k is a vector function, defined and differentiable at each point (x, y, z) in a certain region of space [i.e., A defines a vector field], then the divergence of A (abbreviated as 'Div A') is defined as, DivA= V.A
. a }-+ . a k a) = (1-+ ax ay -az' (A )1'+A21'+A3k) =
(aAax i + aAay + aA3) az 2
I
(since i.i = j j = k.k = 1)
Note: (1) Div A is a scalar field (2) V.A
7:-
A.V
7.3.2 Physical significance of the divergence If A represents the velocity of fluid in a fluid flow, Div A represents the rate offluid flow through unit volume. (or) Div A gives the rate at which fluid is originating at a point per unit volume. Similarly if A represents the Electric flux or heat flux, Div A represents the amount of electric flux or heat flux that diverges per unit volume in unit time.
7.3.3 Some properties of Divergence If A, B , are vector functions and '/' is a scalar function, then, prove that (1)
Div (A + B) = Div A + Div B (i.e) V. (A + B) = -V . A + V .8
Proof Let A = A)i + Aj + A3k B = 8 1i + B~ + B3k A + B = (A) + 8) i + (A2 + 8 2 )j + (A3 + 8 3) k
a ax
a
a az
Div A = -(AI +B l )+-(A 2 + 8 2 )+-(A 3 + 8 3 )
ay
= (aA1 + aA2 + aA3)+(aBI + aB2 + aB3)
ax
ay
~
= Div A + Div B.
az
ax
ay
az
489
Vector Differentiation
(2)
Prove that, Div (fA) = (grad.f) .A + .f{Div A) i.e. V .(fA) = (V .f). A + .f{ V A) Proof: Let A = Ali + A'll + A3k then fA = fAli + fA'll + fA3k V .(fA)
a
a
a
= ax (fA.)+ ay (fA2)+ az (fA3) = fQ~+A. af + f aA2 +A2 af + f aA3 +A3 8j -Ox
ay
ax
ay
az
az
..... (1)
..... (2)
..... (3) (1), (2), (3) => V .(fA) = (v.f).A + f(v .A)
7.3.4 Solenoidal vectors: A vector A is said to be solenoidal if Div A = 0 Solved Examples
Ex. 7.3.5
If A = (.x2y) i + (xy2z)j + (xyz)k, find div A at the point (1, -1,2).
Sol:
A = (.x2y) i + (xyz)j + (xyz)k
. aA. aA 2 aA 3 DIVA=-+-+ax ay az =
a 2 a 2 a -(x y) + -(xy z) + -(xyz) ax ay az
=
2xy + 2xyz + xy
= xy (2z ,+ 3) :. At (1, -1, 2), Div A = (1) (-1) [4 + 3] =-7
Engineering Mathematics - I
490
Ex. 7.3.6
If V = 2xyi + 3x2yj - 3pyz k is solenoidal at (1, 1, 1), find 'p'.
Sol:
. Dlv V
=
a a 2 a -(2xy)+-(3x y)+-(-3pyz) ax oy oz
= 2y + 3xl - 3py At (1. 1, 1). Div V = 5 - 3p Since V is solenoidal, Div V = 0 :. p =
5
3"
Ex.7.3.7
Ifr = xi + yz + zk, and r = Irl, show that Div (r3 r) = 6 r3
Sol:
r= ~X2+y2+Z2
:. r3 r = (xl+y +z2)3/2 (xi+yj+zk)
= Ali + Aj + A3k (say) then, A I = x(xl + y + z2)312
A2
=
y(x2 + Y + z2)3/2
A3 = z(xl + y + z2)312
oA 3 2 .. l =3z r+r
oA2 .1. .. l similarly ~ =3y r+r~
OZ
. oA I oA2 oA3 DlvA= - + -+ax oy oz = 3r (xl + Y + z2) + 3r3 = 3r3 + 3r3 = 6r3 Aliter:
r3 r = r3 xi + r3 yi + r3zk.
a Or x -8 (r3 x) = r3.1 + x . 3r2 -8 = r3 + x . 3r2 . x'
=
r3 + 3xlr
x
r
Vector Differentiation
Similarly
491
o
ay
0
(~y) = ~ + 3yr and oz (r3z) = ~ + 3z2r
DivA=6~
Ex. 7.3.8
l
Evaluate Div [ r grad (r-3) ] or V. {rvC 3 )}
(x.
_ 4 or . ar . --4 -or k--3r _ --4 -i+-j+y. Z --3r -1-3r4 -j-3r ox c:y oz r r r
=
-3r-5 (xi + yj + zk)
r grad (~) = -3r--4 (xi + yj + zk)
= Ali + Aj + A3k
(say)
where AI = -3r--4x, A2 = -3r--4y , A3 = -3r--4z
aA I = ~ [-3r--4 x]
ax
ax
= -3r--4 . 1 + x. -3. -4 r- 5
. ::
= -3r--4 + 12 x r- 5
= -3r--4 + 12 x 2 r-6 Similarly,
oA c:y2
=-3r-4 + 12yr-6
:. Div (r grad r-3) =
oA ax
oA
oA
Oy
az
_I + __ 2 +_3
= -9r-4 + 12 r-6 (x 2 +
y + z2) = -9r-4 + 12 r--6 . r2 = 3r-4
(Alternate method is left to the reader).
.
~
k)
Engineering Mathematics - I
492
Ex.7.3.9
Show that V .(x" r)
Sol:
r" r
= (n + 3) r". Hence show that r/f3 is solenoidal.
= r" (xi + yj + zk)
V .(r" r)
a
= -a (xr")+ x
a
a az
+-
~, (yr") v)'
(zr")
z)
y + z.= 3r" + n ['1-1 ( x.-X + y.r
=
r
r
3r" + n r"-2 . r2 = (n + 3) r"
Ifn = -3, (r-3 r) = (-3 + 3) r-3
=0
r
:. 3 is solenoidal r
Ex.7.3.10 Prove that Div (CIA + C 2 B) = C I Div A + C 2 Div B, where CI' C 2 are constants.
Sol:
Let A
= A I i + Aj + A3k
B = Bli+ Bj+ Bl CIA + C2B = (CIA I + C 2B I) i + (C IA2 + C 2B2U + (C IA3 + C 2 B 3 )k Div (CIA + C2B) =
a ax
a
(CIA I + C2B I) + Oy (C IA2 + C 2B2)
=C (MI+M2+M3)+C(aBI+aB2+ I
ax
Oy
az
2ax
ay
aB
3)
az
Vector Differentiation
493
Ex.7.3.11 If A = 2xi + 3yj + 5zk and f= 2xyz, find div (fA) at (1,2,3). Sol:
2xyz (2xi + 3yj + 5zk) = (4x 2yz)i + (6xyz)j + (10xyz2)k
fA
=
a a a Div (fA) = ax (4x2yz) + ay (6xyz) + az (10xyz2)
= 40xyz :. At (\,2,3), div (fA) = 240 Aliter:
div (fA) = D.(fA) = (V f). A + f( V A) Vf
= 2yzi + 2xzj + 2xyk
V fA
= 4xyz + 6xyz + \ Oxyz = 20xyz
f(v .A) = 2xyz (2 + 3 + 5) = 20xyz Hence V .(fA) = 40xyz and at (\,2,3) V .(fA) = 240
Ex.7.3.12 Iff, g are scalar fields show that V fx V g is solenoidal
vfx V g
=
a.fjax agjax
j
k
aljay agjay
atjoz agjaz (Suffixes denote partial derivatives)
a
Div (V fx V g) = L ax Vygz - fzgy) =
L
(!y gxz + gzfxy - fz gxy -
=0 :. V fx V g is solenoidal
~fxz)
Engineering Mathematics - I
494
Exercise 6(b) I.
If V
=
(xlz)i - (2y3z1)j + (vz)k, find div A at the point (I, -I, I) [Ans. - 3]
2.
Ifr = xi + yj + zk, find div r
3.
If F = (3,xyz2)i + (2xy3)j - (xlyz)k, and ¢ = 3xl - yz,
[Ans.3]
find (i) Div F (ii) Div (¢F) and (iii) Div (grad¢); at the point (I, -I, I) [Ans. (i) 4 (ii) I (iii) 6 ] 4.
If V = (3xly - z)i + (xz 3 + y4)j - 2x3z 2k, find grad (Div V) at the point (2, -I, 0) [Ans. - 6i + 24j - 32k]
5.
Evaluate: (I) Div (r2r) (2) Div (r r) (3) grad Div (r/r) (4) div (r/r3) [ Ans. (1) 5r2 (2) 4r (3) -2r/r3 (4) 0 ] 3y4z2i + 4x3z7 + 6x2y3k is solenoidal
6.
Show that V
7.
Show that the vector F = (2xl + 8xyz)i + (3~y - 3xy)j - (4yz2 + 2x3z)k is not solenoidal, but G = xyz2 F is solenoidal.
8.
If a is a constant vector and V solenoidal vector.
9.
Determine the constant 'b' such that the vector, V = (2x + 3y)i + (by - 3z)j + (6x- 12z)kis solenoidal
10.
Ifr, and r2 are vectors joining fixed points A (x"YI' z,) and B(x2'Y2' z2) to a variable point P(x, y, z) prove that r, x r2 is solenoidal.
11.
If r
12.
If g = r-2n , find div (grad g) amd find 'n' such that 'g' is solenoidal.
=
= xi + yj + zk and r = Irl
=
a x r, where r
=
xi + yj + zk, show that V is a
show that, Div (grad rn)
= n( n+ I )rn- 2 .
[A ns.
7.4
2n(2n -I) I ·n-?n+2 ,- 2 ] r-
Curl of a vector function
7.4.1 If A is a differential vector function, then curl A is defined as, curl A = V' x A
If A
= Ali + Aj + A3 k , then
Curl A = a/ax AI
j
k
a/ay A2
a/az A3
495
Vector Differentiation
=
(aAJ _
0'
aA 2 )i + (aA I _ aA3)j + (aA 2 _ aA I )k az az ax ax 0'
Note: The curl of a vector is also a vector
7.4.2 Physical significance of curl Let r = xi + yj + zk be the position vector cf a point P(x, y, z) of a rigid body rotating about a fixed axis about the origin 0 with an angular velocity ffi = ffili + ffi.j + ffi)k. Then the velocity V of the particle P is given by,
V
= ffi
x r
=
j
k
ffil
ffi2
ffi3
X
Y
z
a/ax
Curl V=
=
i
i (ffi 1+ ffi I) + j
j
k
8/ay
a/8z
(ffi2 + ffi 2) +
k ( ffi 3 + ffi3) = 2ffi
Thus the curl of velocity vector is twice the angular velocity of rotation.
7.4.3 Irrotational Vector: A vector V whose curl is zero is said to be an irrotational vector. 7.4.4 Properties :(1) V x (A + B) = V x A + V x B (or) curl (A + B) = curl A + curl B Proof:
Let A = Ali + A.j + A)k and B
= Bli + B.j + B)k so that
Engineering Mathematics - I
496
i
Curl (A+B) =
j
k
ajax ajay AJ+BJ A2 +B2 j
ajaz A3 +B3 j
k
k
= ajax ajay ajaz + ajax ajay ajoz = curl A + B (2)
If ¢ is a scalar function and A is a vector function Curl (¢A) = ((curl A) + (grad¢)
x
A
(or) V x (¢A) = ((V x A) + (V {P) x A Proof: If A = Ali + Aj + A3k, then cj>A = cj>Ali + cj>Ai + cj> A3k
j L\ x (cj>A)
k
= ajax ajay ajaz cj>AI
cj>A2
cj> A 3
= i[~(q,A3)-~(cj>A2)]+ j[~(cj>AJ)-~(cj>A3)J+ k[~(cj>A2)-~(cj>AJ)] ay az az ax ax ay =
"i[.I."'ayaA3 ~
+ A aq, _.I. OA2 - A aq,] 3ay "'az 2az
= .I. L( aA 3 _ aA2)i + L(A aq, _ A acj»i '" ay az 3ay 2az j k cj>ajax ajay ajaz i
=
AI
A2
j k aq,jax aq,jay acj>jaz i
-?>
A3
=cj>( V x A) + (V cj» x A
7.4.5 Conservative vector field: A vector field F, which can be derived from a scalar field q, such that F = Vcj>, is called a conservative vector field and q, is called the scalar potential ofF.
497
Vector Differentiation
Solved Examples Ex.7.4.6 If A = (xy)i + (yz}j + (zx) Ii; find (a) curl A and (b) curl curl A at (1,2, -3)
Sol :
A
= xyi + yzj + zxk j
(a)
curl A = V x A =
k
a/ax a/ay a/az xy
yz
zx
a a].[ a a ] +k[-(yz)--(xy) a a] = I.[-(zx)--(yz) + } -(xy)--(zx) ay
= i(O =
&
&
&
&
ay
y) + j(O - z) + k (0 - x)
-yi - zj - xk
:. curl A at (1, 2, -3) = -2i + 3j - k (b)
curl A= V x (V xA)
=V
(-yi-zj-xk)
x
j
i
=
k
a/ax a/ay a/az -y
-z
-x
a].[ a a ] +k[-(-z)--(-y) a a ] = i -(-x)--(-z) + } -(-y)--(-x) [aya & fu fu fu ay = i(O - (-I) + j(O - (-1» + k(0 - (-I» =i+j+k
:. curIAat(I,2,-3)=i+j+k Ex.7.4.7 Show that V = xi + Ij + z3k is irrotational -
Sol:
Curl V = V x V
a 3)---(y a 2]) + }.[-(x)--(z a a 3]) +k[-(y a 2)--(x) a ] = I.[-(z ay
=0
az
:. V is irrotational
oz
&
fu
ay
Engineering Mathematics - I
498
Ex.7.4.8 If F = (4x + 3y + az)i + (bx - y + z)j + (2x + cy + z)k is irrotational, find the constants a, b, c i
j
k
ajax
ajay
ajaz
(4x+3y+az)
(bx- y+z)
(2x+cy+z)
Curl F =
Sol:
=
i[~(2X+CY+Z)-~(bXy+Z)]+ j[~(4x+3y+aZ)-~(2X+CY+Z)J ay az az ax k[~(bXy+Z)-~(4x+3y+az)] ax ay
+
= (c -
l)i + (a - 2)j + (b - 3)k
Since F is irrotational, curl F = 0 :. c - I = 0, a - 2 = 0, b - 3 = 0
i.e., a = 2, b = 3, c = 1 Ex.7.4.9 If r = xi + yj + zk, and r = Irl, find curl (rn r) Sol :
r = xi + yj + zk, r = Irl = ~ x 2 + y2 + Z2 rn r
=
(.xl + Y + :?)n/2 (xi + yj + zk) k
j
curl (rn r) =
ajay
ajax III
x(x 2 + y2 + Z2Y2
Y(X2 +
l
ajaz 1/ 1
+ Z2Y2
(? + y-?+z-?hJLY- Y."2n (2 x + Y 2 +z 2) 2z J
n x= I.[ z."2
n 2 + y 2 +z 2 ).2z-z."2(x n 2 + y 2 +z 2 ).2x ] + ).[ x."2(X
= i(O) + j(O) + k(O) =0
Z(X2 +
l
+ Z2 )"2
499
Vector Differentiation Aliter:
rn r
=
rn (xi + yj + zk)
curl (rn r) =
= "'.[ L..,l - a
ay _
-
'" .[
L..,l
j
k
ajax
ajay
ajaz
xrn
yr"
zr"
( zr n) - a ( yr n)] = "'.[ ar - y.nr n-I -ar] L..,l, z.nr n-I az ay az
z]
Y - y.n.r II_I -; f/zr n-I -;
= i( 0) + j( 0) + k (0) = 0 Ex. 7.4.10 Prove that, ifF
Sol:
Curl F =
= (x + Y +
ajax
j
k
a/ay
ajaz -x-y
x+ y+1 = i
I)i + j - (x + y) k, F. curl F = 0
[-I -0] + j[O + I] + k[O -1]
= -;
+j - k
:. F curl F = -I (x + y + 1) + 1.1 + 1(x + y)
=0
Ex. 7.4.11 P(x, y, z) is a variable point and Q(xl' YJ, zJ)' R(x2, Y2' z2) are fixed points. If U = QP and V = RP; Prove that curl (U x V) is equal to 2(U - V).
Sol:
P(x, y, z), Q = (xl' Yl' zJ)' R = (x 2, Y2' z2) :. QP = (x -xJ); + (y - YJN + (z - zJ)k = U RP
= (x -
x 2); + (y - Y2 N + (z - z2)k j
k
UxV= (X-XI)
(Y-YI)
(Z-Zl)
(x-x 2)
(Y-Y2)
(Z-Z2)
= ~{(y =~
=V
YJ) (z - z2) - (z - z\) (y - Y2)} ;
{y(z\ - z2) - z(yJ - Y2) + (yJ z2 - Y2 z J)}i
Engineering Mathematics - I
500
Curl (U
x
V)
=
I[ ~{x(y, -!
- Y2)- Y{X, -X2)+{X'Y2 -x2 y,)}
{Z{X, -X2)-x{Z, -Z2)+{Z,X2 -Z2XJ}}
= L(-(XI - X2) - (xI - X2); = L -2(x l - X2 ); = 2L(X2 - xI); = 2(U - V) Ex. 7.4.12 If F is a conservative vector field show that curl F = 0 Sol.
F is a conservative vecor field.
:. There exists a scalar field '~' such that F = j
k
ajay
ajaz 8$jaz
; :. Curl F = ajax a~jax
a~j~
i( a ~ a ~ 2
== L
V'"
'I'
8$. 8$. 8$ k = ax 1 + ~ J + az
2
~az
_
az~
)
== 0
Ex. 7.4.13 Show that F == (6xy + iJ)i + (3r -z)j + (3xz 2 - y)k is irrotational. Find ~ such that F == V ~ i
Sol:
Curl F =
ajax 6XY+Z3
j k aj8y ajaz 3X2 -z 3xz2-
J
== i (-1 + 1) + j (3z2 - 3z2) + k(6x - 6x) ==0 :. F is irrotational L tF== e
:. :
V~==
8$. 8$. a~k
-I+-J+ax ay az
== 6xy + iJ
..... (I)
Vector Differentiation
501
a~ -
Oy a~
-
az
=
3x2 -z ?
= 3xz~
..... (2)
- y
..... (3)
Integrating equation (1) with respect to x we get ~ = 3.x2y + xz 3 + fly, z)
... :. (4)
Differentiating (4) partially w.r.t 'y' a~
af(y,z)
ay = 3.x2 +
ay
..... (5)
From (2) and (5) we have
af(y,z)
ay
..... (6)
=-z
Integrating (6) w.r.t 'y' we get fey, z) = -yz + h(z) ~ = 3x2 + zx 3 - yz + h(z)
Hence
..... (7)
I
difffrentiating (7) w.r.t 'z' we get ..... (8)
Comparing (3) and (8) we have h l(z) = 0, :. h(z) = constant, 'c' say Hence ~ Aliter:
=
3.x2y + xz 3 - yz + c -~ =6xy+z3
ax
a~
ay
=
3.x2 - z
(from 7) .... (I)
.... (2)
..... (3)
Engineering Mathematics - I
502
Integrating the above equations respectively w.r.t x, y, and z, we get
= 3x2y + xz3 + f(y, z) ~ = 3x2y - yz + g(z, x)
.... (4) .... (5) ..... (6)
~
~= ~
xz3 - yz + h(x. y)
should satisfy all the above three equations simultaneously
(i.e.) (4), (5) & (6)
3x2y + xz3 - yz + c, where c is a numerical constant. [Herej= -yz, g = xz3, h = 3x2y will satisfy (4), (5) & (6) :. ~ =
Note:
Ex.6.4.14 Iff(r) is differentiable, show that f(r)r is irrotational Sol:
fir) r = fir) (xi + yj + zk)
Curl (f(r) r)
j o/Oy
%x
=
k
%z
j
xj{r) y j{r) zj{r =
..
I{; {zj{r)}- ! {y j{r)})
=
I{z./{r)~ - y./{r):)
=
I{z./{r)~ -y./{r);) =0
f{r)r is irrotational
Ex. 7.4.1SlfU and V are irrotational, prove that U Sol:
Let U = Uti + U2i +
x
V is solenoidal
Ui
V=V t i+V2i+ Vi U is irrotational
=> Li[OU 3
oy
..
sImilarly,
_
:. curl U
OU 2 ]
oz
=
=
0
'" .[OV 3 oV 2] = 0 Oy oz L-l -
- -
0
..... (1)
..... (2)
Vector Differentiation
503
}
k
VxV= VI
V2
V3 =1:;(U 2V 3 -U 3 V 2)
VI
V2
V3
..... (3) au 3 _ au 2 • au I _ au 3 . au 2 _ au I From (I) ,we have, - - - - - , - - - , - - - 0' az az ax ax 0' av3 _ av2 • aV _ av3 . av2
aV
an d from (2) we have - - - -I- - - - -I , '0' az' az ax' ax 0' _
..... (4)
..... (5)
Substituting the six equations of(4) & (5) in (3), we observe that all the 12 terms of (3) will get cancelled. Hence Div (U x V) = 0 ~
U x V is solenoidal Exercise - 7(c)
I. If V = (2xz2);_ (yz)j + (3xz3)k, andf= xlyz, find the following at the point (I, I, I) (a) curl V (b) curl (jV) (c) curl (curl V) [Ans. (a) ; + } (b) 5; - 3} - 4k (c) 5; + 3k] 2. If 'g' is a scalar function, show that curl (g grad g) = 0 3. Find the value of the constant 'p' for which the vector V = (pxy - z3) + (p - 2) xl} + (I - p) xz2k is irrotational. [Ans: 4] 4. Find the constants a, b, c so that the curl of the vector A = (x + 2y + az); + (bx - 3y - z)j + (4x + cy + 2z)k is identically equal to zero [Ans. 4, 2, -1] 5. If r =
xi + y) + zk, r = Irl, show that curl (;Z ) = 0 and find a scalar function of' such r
that 2"= -v J,.f(a) = 0 where a> r
o. [Ans :f= log
(~)]
Engineering Mathematics - I
504
6. If r = xi + yj + zk, and p, q are constant vectors, show that (I) curl [(r x p) x q] = ( p x q) and (2) curl [(p.q)r)] = o.
7.5
Laplacian Operator: V2
7.5.1
V2 =
=
V.V
=
(i~+ j~+k~) ax j~+k~).(i~+ ry az ax ry az
a2 + -a2' + -a22 ) .IS called the Laplacian . Operator ( ax 2 ay2 az -
'V2' can be applied to both scalar and vector functions as shown below. a2~
2
a·2~
a2~
V~=-+-+-
ax 2 ry2 az2
where '~' is scalar function If A = Ali + Aj + A3k, is a vector function, then V2 A = =
a2 a2 a2 ) ( ax2+ ay2 + az2 (Ali + Aj + A3k)
(V2AJ+(V2A2)j+(V2A3~
7.5.2 Vector Identities: we shall give below some vector identities with proofs. \. If '¢' is a scalar function curl grad
Proof: Curl grad cjI= V x
~
= 0, (or) V x V ~ = 0
acjl acjl. acjl ) (-i+-}+-k ax ry az k
j
ajax ajry ajaz acjljax acjljay acjljaz = L>[~(acjl)-~(a~)l ay az az ry
= 0, assuming that
'~'
partial derivatives.
possesses continuous second order
505
Vector Differentiation
(2)
IfY is a vector function, Div (Curl A) = 0 (or) V.(Vx
A) = 0
Proof: Let A = Ali + A-j + A3k CurIA=
3 -aA-, ) " .(aAay az ~I
"~a {aA3 aA)} ax -ay- - -aza~A3 a2 A2 a2A4 a~A3 a~A~ a~AI = -----+-----+----axay axaz ayaz ayax azax azay
Div (curl A) =
=0 (3)
Proof:
If A is a vector function, curl (curl A)
curlA=
grad (Div A) -
3 "~l.(aAaA, ) ay -az-
j
Curl (curIA)=
=
k
a/ax a/ay a/az aA3_ aA2) aA I _ aA3) aA2_ aAI) ( ay az ( az ax ( ax ay
I[~(aA2 _ aA I)_~(aAI _ aA 3)]i ay ax ay az az ax
V2 A (or)
Engineering Mathematics - I
506
=
(4)
-'1 2A + '1('1.A)
'1.(AxB)=B.('1xA)-A.('1xB), (A, B are vector functions) (or) Div(AxB)=B. curl A-A. curl B. Proof: Let A = A) i + A?J + A3k and B = B) i + B?J + B3k
j
k
B = AI
A2
A)
BI
B2
B)
i
A
x
=
I
i(A2B) - A)B2)
a ox
Div (A x B) = "-(A2B) -A3B 2) LJ = "
L:.
'1xA
=
(A OB3 + B OA2 + A OB2 + B OA3) 2 AX 3 AX 3 AX 2
ax
..... (1)
a/ax AI
..... (2)
507
Vector Differentiation
Similarly, A. (V'
x
" (BB3 0' - BB?) Bz-
B) = L.J AI
..... (3)
Expanding the summations of (I), (2) & (3), we observe that, Div (A x B) = B. (V' x A) - A. (V' x B)
7.5.3 Operation of V' on product of two functions Suppose ~ and 'P, are two scalar or vector point functions. When V' is operated on the product of ~ and \j1, the following rule is useful. V' (~'P) = V' (~o \{J) + V' (~'P 0) wherein the suffix '0' indicates that the function is not to be varied, that is, V' is not to be operated on that function. After the completion of operation of V' the suffixes are dropped. While proving the identities the following are useful. (a)
a.b = b.a
(b)
a x b = -b x a
(c)
axa=O
(d)
a.(bxc)=(axb).c
(e)
a x (b x c) = (a. c)b - (a. b) c
(f)
(a. c) b = a x (b x c) + (a . .b}C
(g)
a. a = a2
(h)
x operation is always between vectors only
vector identities using V' operation: (i) V' .(FG) = V' . (FG o) + V' . (FoG) = V' F.G o + Fo V' .G :. div (fG) = G. grad F + F div G (ii) V' x (FG) = V' x (FG o) + V' x (FoG) V' x (FG o) = V' F x Go V' x (FoG) = (V' x G)Fo=Fo(V' xG) :. V' x (FG) = V' F x G + F( V' x G) (i.e.) curl (FG) = (grad F) x G + F curl G
due to (h)
Engineering Mathematics - I
508
(iii) V'. (F x G) = V' .(F x Go) + V'. (Fo x G) V' . (F x Go) = (V' x F) .G o
due to (d)
Go .( V' x F)
due to (a)
=
V' . (Fox G) = - V' . (G x F0)
due to (b)
(G. F0)
due to (d)
= -
V'
x
=-(V' x G). Fo
due to (a)
=-Fo(V' x G) :. V'. (F x G) = G. (V' x F) -F. (V' x G)
Thus div (F x G) = G. curl F - F. curl G (iv) V' x (F x G) = V' x (F x Go) + V' x (Fo x G) V' x (F
x Go) = (V' . Go) F -
due to (e)
(V' . F) Go
due to (a)
= (Go· V') F - Go (V' . F)
due to (e)
V' x(FoxG)=(V'.G)Fo-(V'·Fo)G
due to (a)
= Fo (V' . 0) - (F o · V') G :. V' x (F x G) = (G . V') F - G( V' .F) + F (V' .G) - (F. V' ) G
Thus curl (F x G) = (G. V' ) F - (F. V' ) G + F (V' .G) - G (V' .F) "I-
(v) V' (F . G) = V' (Fo . G) + V' (F .G o) { V'(Fo.G)=Fox(V' xG)+(Fo. V')G
due to (t)
V' (F . Go)
due to (t)
=
V' (Go. F)
=
Go x (V' x F) + (Go. V') F
:. V' (F . G) = F x (V' x G) + G x (V' x F) + (F . V') G + (G . V') F
Thus grad (F . G) = F x curl G + G x curl F + (F . V') G + (G . V') F (vi) curl (grad F) = V' x (V' F) = (V' x V') F = 0 (vii) div (curl F) = V' .( V' x F) = V' x (V' . F) = ( V' x V' ). F = 0
due to (c) due to (d) due to (c)
(viii) curl (curl F) = V' x (V' x F) = (V' . F) V' - (V' . V' ) F =
V' ( V' . F) - V' 2 F
=
grad (div F) - V' 2 F
due to (e) (since V'. must not appear at the end)
509
Vector Differentiation
Solved examples
Ex.7.5.4 If/= x 21z2, find V 2/ at (1,2, 1) Sol:
Y
V
=
=
0
2
2
/
ox- 0/
0
2
/
oz-
21z2 + 6x2yz2 + 2x21
:. at (1, 2, 1),
Ex.7.5.5 Show that, if r Sol:
0
/
--J +--+-J-
v2j= 16+ 12+ 16=44. =
xi + yj + zk, r =
\rI, then
~ = (x2 + Y + z2)n12, (·:r = ~x2 + y2
V 2r ll = n(n + 1) rll
2
+Z2 )
... (1) 2
Similarly -02 (r n) = n ( n- 2)y 2r n-4 + nr n-2
... (2)
, oy
2
( n) J n-4 and -a2 I! =n ( n- 2)z-r +nr n- 2
... (3)
oz
=
3n,.n-2 + n(n-2) ~-4 (x 2 +y + z2)
= 3n~-2 + l1(n - 2) r"-2
= n(n +
1) ~-2
adding (I), (2) & (3)
(": x 2 +
y + z2 = r2)
Engineering Mathematics - I
510
ar 11-1 .x -a (r") =nr 11-1 .-=nr ax ax r
Aliter:
=
n,.rr2 . x
~~II)= n(r 2 az
ll
-
2.1 + x.{n _ 2)r"-3. ar) ax
~ n(,"-' + (n-2}r"-'. x:) =
nr"-2 + n(n-2) r"--4. xl
... (I)
2
Similarly -a2 (r ") =nr 11-2 +n (n- 2)r n-4 .y 2
... (2)
, ay
and
~2 ~n ) = nr n- 2 + n{n - 2 )r n-4.Z2
... (3)
az
Adding (I), (2) & (3) we get,
=> V2rn = 3nr"- 2 + n{n - 2)r n- 4.r2 = ,.rr2[3n + n2 - 2n] = n(n + I) r"-2 Note: Ifn =-1, we have
Ex.7.S.6 If V .U =0,
vt~ )= 0, which means that (.; ) satisfies the Laplace's equation
av
the wave equation V 2U Sol:
au
v .v =0, v xU=-at' v xV = at' show that Uand V satisfy a2u =
at 2
vx(vxU)= v x (_av) = -a{vxv)
at
at
2 -a(au)_ -a u
= at
at - ----ai2
... (1 )
Vector Differentiation
But
511
V x (V x U) = V (V .U) - V 2U
= - V 2U from ()) & (2),
(": V U
= 0)
... (2)
a,2 = V2U , which shows that U satisties the wave equation.
i'iu
Similarly; Vx (Vx and
au a at at
-a~v
v)= Vx- =-(Vx u)=-2
... (3)
at
vx(vxY)=V(V.Y)-V 2 Y=-V 1 Y(": V.Y=O) ... (4)
(3) and (4)
=> Y satisfies the wave equation.
Ex.7.5.7 If V.Y = 0, show that V x[V x {V x (V x V)}] = V 4y (or) curl [curl {curl (curl Y)} ] = V 4y
Sol:
We know that [from 7.5.2 (3)]
V x (V x Y) = V (V .Y) - V 2y = - V 2y (": V.Y = 0) = -
U (say)
Then the given expression = - V x( V x U) = V 2U - V CV .U) = V 2( V 2y) - V (V . U)
[from 7.5.2 (3)]
(": V 2y = U)
= V 4y - V (V . U) V . U = V .( V 2y) =
{~) ~}{V2V}
=
I {i ~ }.(V2(~i + V2i + V k))
=
V2(a~ + aV2 + aV3)
=
V2(DivV)= V2(0) =
3
ax
Hence the problem
0'
az
0
(": Div Y
= V . Y = 0)
Engineering Mathematics - 1
512
Exercise 7(d) (I)
(2)
ry( v. r"r) =-;-:t 2
(3)
Show that V-
(4)
Show that V-<j> r
ry () = ~ry d"<j> 2 df +-dr-
r dr
~'ry
1
show also that, if V-<j> = 0, then <j> = C] + -- where CI' C 2 are constants
r
xi + )1 + zk,
7.6
VECTOR INTEGRATION
7.6.1
Ordinary integration of vectors
(I)
If A(u) = A](u)i + A2(u}j + Aiu)k be a vector function of a scalar variable 'u' (A,(u), Aiu), A 3(u) assumed to be continuous in any given interval), the indefinite integral of A(u) is given by.
r
(k.
I grad -) = 0
If r =
prove that, curl
(k x grad
I -) + grad
(5)
r
fA(u)du = i fAl(ll)du+} fA" (u)du +k fA3(1l)du (2)
If there exists a vector B(u) such that A(u) =
d
du
(B(u)), we can write
fA(U )du = f:U (B(ll ))dll = B(u) + c (3)
where c is a constant of integration independent of u. The definite integral of A(u) between the limits 'a' and 'b' in the above, is written as, h
b
d
b
fA(u)du = f- (B(u))du = B(u)+cl du a a a
= B(b) - B(a)
7.6.2 Line Integrals: Let A(x,y,z) be a continuous vector function defined in the entire region of space. Let c be any curve in the region. Divide c into n intervals by taking points A = Bo, BI' B2 ... Bn(= B). Let Pi be any point in the interval B i_, Bi 0"
' •.;.;.
Vector Differentiation
513
A= 8 0
Let ro' rJ ...... rn be the position vectors of points Bo' B)' B2 ...... Bn respectively. Let us consider the sum,
The limit of this sum as n ~ 00 and
1&,:1 ~ 0
is defined as the line integral of A
along the curve c and is denoted symbolically by
f A.dr or
dr
f A.-dt; dt
If c is a closed curve, the integral is written as
which is a scalar.
fA.dr
Cartesian form of line integral: If A=AJi+A-j+Ai dr = (dx)i + ({Mi + (dz)k,
f A .dr = fA1dx + A2 dy + A3 dz c
Note:
fdr, and fAX dr are also examples of line integrals.
7.6.3
Physical appliations:
(1)
Work done by a force (I) If A represents a force and dr is an element of the path of the particle along a curve c, then the line integral {J
fA .dr
(P, Q are 2 points on c)
J'
represents the work done by force A in moving the particle from P to Q.
514
(2)
Engineering Mathematics - I
flow or circulation: (2) If A is the electric field strength, the line integral given above i.e., JA.dr, is called the flow of A along c. If c is a closed curve it is often referred to as circulation of A around c. {J
In general, the line integral JA.dr, will depend on the path from P to Q /'
7.6.4 T"eorem: Prove that the necessary and sufficient condition for the integral fA.dr, c
to be independent of the path c joining any two points is that A is a conservative vector field, (or) there exists a scalar field I/> such that A = VI/> (or) curl A = 0, [i.e., the work done by the force A in moving the particle from one point to another is independent of the path if A = VI/>]
Proof Let P = (x I ,y I ,Z I)' Q = (x2'Y2,z2) be any two given points on the curve c. Let A = VI/> where I/> is single-valued and has continuous derivatives. (J
(1)
Work done
=
JA.dr
I'
Q
fVI/>.dr=
p
G~ i-+j-+k01/> 01/> 01/» p ox ry oz
(dxi+dyj+dzk)
Qol/> 01/> 01/> (J J-dx+-dy+-dz= fdl/> = I/>(Q)-I/>(p) /' ox ry oz I' = ~(x2'Y2,z2) -1/>(xI'Yl'zl) i.e., the integral depends upon the two points only but not on the path joining them. (This is true only if I/> is single-valued at all points P and Q). (2)
The integral is independent of the path. Then, (x,y,z)
I/>(x,y,z)
=
f A.dr = (Xl ,YI ,Zl)
(x,y,z)
f
dr A. ds ds
(Xl ,)'1 ,zl )
Differentiation with respect to's' gives
dl/> =A. dr ds ds
...... (I)
Vector Differentiation
515
d<j> =V<j>dr ds ds
But
...... (2)
dr s
dr ds
(I) - (2) => (A - V<j». -d = 0, which is true irrespective of-
:. A= V<j>
Solved Examples 4
Ex. 7.6.5:
F(t) = (3t 2-t)j + (2-6t}j -
4tk, find
(a) JF{t}dt
(b) JF{t}dt 2
Sol:
(a) JF{t}dI=; J{3/
2
-/~t+ j J{2-61}dI-k J41dl
~ (I' - I~} +(21-31'); -2t'k+c (b) fF{t}dt
=(/3 -~lj + (21 -3t 2)j - 2t 2k + cj = 50; - 32j - 24k 2
2
2
x/2
Ex.7.6.6 Evaluate: n{3sin e)i + {2cose)j}ie
Sol:
o Given integral xl2
xl2
x/2
ri/2
=; J(3sine}de+ j J{2cose}de=-3cosel;
+2sinejJ o
0 0 0
Ex.7.6.7 The acceleration a ofa particle at any time 't' a
~
=3;+ 2j
0 is given by,
= e-tj - 6(t + 1}j + (3sint)k
Find the velocity V and displacement r at any time 't' given that V = 0 when t = 0 and r = 0 when t = O.
Sol:
a = e-t ;
:. V
-
6(t + l}j + (3sint)k =
d 2r -2
dt
~ : ~ Jadl ~ - e-t i - 6(1~ +I
lj -
(3cost)k + C,
(C 1 is constant of integration) But V
= 0 when t = 0
:.-i-3k+C 1 =0=> C 1 =;+3k Substituting in (I), we get Velocity V
= (I - e-t)i - (3t2 + 6t}j + (3 - 3cost)k
..... (1 )
516
Engineering Mathematics - I
Integrating, r
=
(I + e-t)i - (t3 + 3t2)j + (3t - 3sint)k + C 2
(e 2 being constant of integration)
But r
= 0 when t = O.
:. (2) => + i + C 2
=
0 :. C 2 =
-
i
From (2), r = (t - 1 + e-t)i - (t3 + 3t2)j + (3t - 3sint)k
Ex.7.6.8 Show that d2F dF Fx--=Fx-+c 2 dt dt We know that
f
Sol:
~(FxdF)=FX~(dF)+dFxdF dt
dt
dt
dt
dl
dl
Hence the result.
Ex.7.6.9 If A = 2ti + 3t7 - (4t + 1)k, and B = ti + 2j + t 2k, find 2
2
f(A x B}dt . o (i) A.B = (2t)t + (3t2) (2) - t2 (4t + 1) = 2t2 + 6t2 - 4t3 - t2 = 7t2 - 4t3
(i)
f(A.B}dt
(ii)
o
Sol:
i (ii) A x B = 21 t
j 2
3/ 2
=
(3t4 + 8t +2)i + (--4t2 - t - 2t3)j + (4t - 3t3 )k
=
-1--j-4k 5 3
196.
62.
..... (2)
Vector Differentiation
517
Ex.7.6.10 If F(2) = ; + 2j - 2k and F(3) = 6; - 2j + 3k,
dF
3
evaluate IF.-du
du
2
Sol:
From vector differentiation,
d dF dF (dF) We get, -(F.F)=F.-+-.F=2 F.- so that du
F. dF 'du
du
du
du '
=~~(F.F)=~~IFI2 2 du
2 du
But F(2) = ; + 2j - 2k ~
IFI2 = I + 4 + 4 = 9, when u = 2
and F(3) = 6; +2j - 3k ~ IFI2
= 36 + 4 + 9 = 49, when u = 3
3 dF I . IF.-du=-[49-9] =20 ., 2 du 2
Ex. 7.6.11 If F = (.xl
- 2y); - 6yzj + 8xz2k, evaluate IF.dr
from the point (0,0,0) to the
c
point (I, I, I) along the following paths (1) x
= t, Y = t2 , z = t3 .
(2) the straight line from (0,0,0) to (I ,0,0), then to (1, I ,0) and then to (1,1, I) and (3) the straight line joining (0,0,0) to (1, I, I).
Sol.
(1) x = t;
dx = dt;
Y = t 2; dy = 2t dt;
z = t 3,.
dz = 3t2 dt·,
when t = 0, the point is (0,0,0), when t = 1, the point is (1,1, I)
IF.dr= I(x 2 -2y}tx-6yzdy+8xz 2 dz c
c /;1
J~2 -2t 2 }it -6.t 2 .t 3(2t}dt + 8t~3 J{3t 2 }it
/;0
1
=
2
6
9
J-t dt -12t dt + 24t dt o
Engineering Mathematics - I
518
=
Irf
J~24t
9
o
6 2 \, 24t lO 12t 7 (3 I -12t -t p t = - - - - - - ! 10 7 30
12 12 1 252-180-35 37 = 5 7 3 105 105 Along C, F = (t2 - 2t2 )j - 6. t 2 t 3j + 8.t.t6 k = - t 2 j - 6tJ + 8t7k dr = (dx)i + (dy}j + (dz)k = dt i + (2t dt}j + (3t2 dt)k -----=
Aliter:
I
JF.dr c
=
9 6 2 J(- t -12t + 24t ~t
(Taking dot product of F and dr)
1=0
37 (2)
105 Let 0 = (0,0,0), P = (1,0,0), Q = (I, 1,0), R = (I, 1,1 ). Then along OP, y = 0, z = 0, dy = 0, dz = 0 and x varies from 0 to 1. I
I
2 2 :. JF.dr= J(x -2.0}tx-6(OXOXO)+8x(0)2(0)= Jx dx=± ~
x~
.... (I)
0
Along PQ, x = I, z = 0, dx = 0, dz = 0 and y varies from 0 to 1. I
:.
JF.dr=
J(12-2y~-6y(0)dy+8.1.(0)2(0)= Jo=O
.... (2)
y=O
I'Q
Along QR, x = I, Y = 1, dx = 0, dy = 0 and z varies from 0 to I. I
I
2 :. JF.dr= J(12-2.IXO)-6.1.z(0)+8.1.z2.dz= J8z dz=% QR
z=O
Adding (I), (2), (3), fF.dr
(3)
.... (3)
0
=~ + 0 +! = 3 3
3
The equation of the straight line joining (0,0,0) to (1,1, I) is
f
=
f
~ = = t (say),
so that x = t,y = t, z = t, dx = dy = dz = dt 't' takes values from 0 to I. I
I
.. JF.dr = n~2 2t)- 611 + 8tJ }it = J(8t c o o 2
3
-
2 5t - 21
'Vt
519
Vector Differentiation
Ex. 7.6.12 Find the total work done by a force F = 2xyi - 4zj + 5xk along the curve x = t2, Y = 2t + 1, z = t3, from the points t = 1 to t = 2. Sol: x = t2 , dx = 2t dt; y = 2t + 1, dy = 2dt; z = t\ dz = 3t2 dt Total work done = fF.dr c
= f(2xyi-4zj + 5xk}.{(dx}i + (dy}j + (dz}k} = f2xydx-4zdy+ 5xdz c 2
2
'~I
1
= fl2J 2(21 + I}J2Idl _{4./ 3 )2dt + {5/2 )3/2 dl = f{8/ 4 + 4/ 3 -81 3 + 15/ 4)dl
{ 713 638 23/ 4 -4/ }11= [1 23--/4 ]2 =23 - (32-1)-(16-1)= - - 1 5 = f 5 5 5 5 5
2
3
1
1
Ex.7.6.13 If c is the curve y = 3x2 in the xy - plane and F = (x + 2y)i - xyj, evaluate fF.dr, from the point (0,0) to (I ,3). S'ol:
Since c is a curve in xy - plane, we take r = xi + yj, so that F.dr= {(x+2y)i-xyj}. {(dx)i+(dy}j} =(x+2y)dx-xydy. 1st Method: By taking the curve in parametric coordinates as, x = t, Y = 3t2, dx = dt, dy = 6t dt, so that t varies from
°
to 1 to get the points
(0,0) & (1,3): We have (1,3)
fF.dr = (0,0)
1
1
'~O
0
fv + 6t 2) dt -/{3t 2) (6tdt) = fv + 6t 2 -181 4)dt
t2 2
6/ 3 3
18t 5 1 5 0
1
18
5
18
11
2
5
2
5
10
= -+---1 = -+2--=---=-2nd Method:
>: =3x2, dy = 6x dx and x varies from
1
.. fF.dr c
=
f(x + 6x 2}Ix - x.3x 2.6x.dx
0
1
~( 2 , 11 = Jx+6x -18x 4 \px =-o 10
°
to 1.
Engineering Mathematics - I
520
Ex. 7.S.14 Find the work done by the force F in moving a particle once around the circle 'C' in xy - plane, ifthe centre of the circle is origin and radius is '2' and F = (x +y + z)i + (2x + y}j + (2x - y +z)k. Sol:
xy - plane is z = 0, :. F = (x + y)i + (2x +y}j + (2x - y)k and r = xi +yj F. dr = (x
+ y)dx + (2x + y)dy.
~
dr = (dr)i + (dy}j y
The equation of the circle is x = 2cosS, Y = 2sinS :. dx =
-
2sinS dS, dy = 2cosS dS
S varies from 0 to 2n Work done = fF.dr
=
2lt J(2COSS + 2sin sX- 2sinS)de + (2.2cosS + 2sin SX2cosS)dS o
-~
2lt 2lt 2 2 2 2 J{- 2sin2S -4sin S + Scos S + 2sin 2S}ie = J{scos S - 4sin S ~S o 0
2lt 2lt = n4(1 + cos2S)- 2(1- cos2S)}iS = J(2 + cos2S)dS = [2S + 3sin 2S~lt o 0 = 4n Ex.7.S.1S Show that the necessary and sufficient condition for a vector field V to be conservative is curl V = 0 Sol: a) Necessary condition: If V is conservative, :3 a 'cp' 3 V = V cpo curl V·= curl (V cp) = 0 (see 7.5.2 (1)) b) Sufficient condition: Let V = V I i + V'li + V 3k i
Curl V = V x V
~
0
=
~
j
k
8/ Ox 8/ By 8/ 8z = 0
I(8VBy 8V8z )i = 0 3 _
2
.... (1)
521
Vector Differentiation
The work done by the force field V in moving a particle from (x\,y\,z\) to (x,y,z) is IV.dr c
= Iv; (x,y,z)dx + V2 (x,y,z)ay + V3 (x,y,z)dz c
where V is a path joining (x\,y\,z\) to (x,y,z) Let us choose a particular path consisting straight line segments from (x\,y\,z\) to (x,y\,z\) to (x,y,z\) to (x,y,z) and denote the work done along this path by a scalar function ~(x,y,z); x
:.
z
y
~(x,y,z) = IV;(X'Yl,ZI)dx+ IV2 (x,y,zJly+ IV3 (x,y,z)dz XI
..... (2)
Yl
From (2), it can be seen that,
8cI>
.... (3)
8z = Vix,y,z)
.z
= V2 (X,y,ZI)+ V2(X,y,Z~
= V2(X,y,ZI)- V2 (x,y,z)- V2(X,y,ZI) .... (4)
from y
= V\(x,y\,z\) + V;(X,y,ZI~
Yl
= V \(x,y\,z\) + V \(x,y,z\) = V \(x,y,z)
(1)
z
+ V;(x,y,z~
Zl
V \(x,y\,z\) + V \(x,y,z) - V \(x,y\,z\)
.... (5)
(3),(4),(5) => :i+ : j + : k=V;(x,y,z)i+V2 (x,y,Z)j+V3 (x,y,z)k=V
=> V =
V~
Hence the proof.
Engineering Mathematics - I
522
Ex. 7.6.16a) Show that F = yi + (2xy +z2)j +2yzk is a conservative force field. b) Find its scalar potential. c) Find the work done in moving an object in this field from (1,2,1) to (3, 1,4)
Sol:
a) V x F= Max
i
j
k
a/fJy 2xy+Z2
a/az
2yz
= i(2z - 2z) + j(O - 0) + k(2y -2y) = 0 :. F is a conservative force field. b) Let 'cjl' be the scalar potential of F. 1st method:
2
:. acjl ax = y2
acjl_2 z az - y .... (3)
.... (1) :=2Xy +z .... (2)
Integrating (1) w.r.t x, (2) w.r.t. y, & (3) w.r.t. z, respectively, we get,
cjl =
xY + j(y,z),
cjl =
xY +yz2 + g(z,x), and
cjl
=
yz2 + h(x,y).
These equations will be consistent iff, g, h are taken as
j(y,z) Hence cjl
=
yz2,
g(z,x) = 0, h(x,y)
=
xy.
= xy + yz2 + constant
2nd Method: Since F is conservative, JF.dr is independent of path joining c
(Xl,yl,zl) to (x,y,z)
using method of problem 7.6.15(b),
cjl =
x
Y
Z
XI
YI
zl
J&12}ix + J(2Xy + z/ }ry + J2yzdz = xy~ ( +(xy 2 + Z12 y) r+yz2 f Xl
.Y]
= xYI - XlYI + xy +z 12y - ry I - ZI 2y I + yz2 - yz I = xy2 + yr - x1yl2 - Zl2Yl = xy2 + yz2 + constant.
2
3rt/ Method: Since F. dr = Vcjl.dr = acjl dx + acjl dy + acjl dz =dcjl ,
ax
fJy
az
zl
523
Vector Differentiation d~ == cY)dx + (2xy + z2)dy + (2yz)dz == cYdx + 2xydy) + (z 2dy +2yz)dz
== d(xy) +d(yz2) == d(xy2 + yz2)
=>
~ == xy + yz2 +constant
(0,1)8
1'2=(3,1,4)
fd~
f F.dr ==
c) work done ==
IHI,2,1)
c
(3,1,4)
~(P2) - ~(PI) = xy + xz2
==
'=:-:::----i----"'I\( 1,0)
I
(1,2,1)
=(3.1 2 +3.4 2)-(1j2+ 1.1 2)=51-5=46.
Ex. 7.6.17 Evaluate f(x 3dy + y 2dx) where c is the boundary of the triangle whose vertices c
are (0,0), (1,0), (0, I). Sol:
f(x 3dy + y2 dx)
Let I ==
c
I
== II +12 +1 3, where II ==
f, 12 == f, 1 f 3=
OA
(i)
Along OA, y == 0, => dy == . ••
(ii)
AH
Jx 3 •0 + 02 •dx
11 ==
AlongAB, o
HO
°
== 0
x+y=1 ,y=l-x=> dy=-dx,xvaries from
I toO.
:.1 2 = fx 3 {-dx)+{I-x)2 dx I
= J(x 2_x 3 +1_2X)dx=_x
3
o
(iii)
Along BO,
:. 1
3
x
_~o =0_(_~_~)= __1 4
I
3
4
=°=> dx =° :. 13 =fo.dy+ i.o =° 1
= II + 12 + 13 = - 12
Ex.7.6.18: If 1= xy2z2 ,evaluate fldr where the curve 'c' is given by
x = (, Y = (2,
Z
=(3
from t
=
°
to I .
12
524
Engineering Mathematics - I
f=xy2 Z2 =/(/2y .(/3
Sol:
Y=1
11
dr=dx i+dyj+dz k 2 2 dr =(dt)i + (2tdl )j + (3/ dl )k =(i + 2tj + 3/ k )dl 1
Jfdr
=
c
Jill
(i + 2tj + 3/ 2k )dl
1
=i
1=0
1
Jill + dl+ j J2/
0
12
1
dl+ k J3/ 13 dl
0
0
./12 I • 2/ 1 3/ 14 I 1. 2. 3 =1- J+l-J+k- J=-I+-l+- k 12 0 13 0 14 0 12 13 14 Ex. 7.6.18: If A = 3zi - 2xj + yk , and c is the curve given by 13
x = cos I, Y = sin 1 , Z = 2cos I, evaluate
JA x dr from I =0to I = 2
1t .
c
i j k A x dr = 3z - 2x y dx dy dz
Sol:
= (-2x dz- y dY)i+(y dx-3z dZ)j +(3z dy+2x dx)k x
= cos I, Y =sin I, z = 2 cos I ~ dx = (-sin/)dl,dy = (cos I) dl, dz = (-2sin/)dl
:. (1)
~
Ax dr = i[(-2cos/)(-2sin/) - sinl . cost]dt +j[(sin/)(-sint) - 3(2cost)
(-2sin/)]dl + k[3(2cos/)( cost) + (2cos/)(-sin/)]d/.
=i(3sint cost) dt + j[(12 sinl COS/) -
sin2/)] dl + k(6cos2/- 2sint cost) d/.
J
J
1C/23 ft/2[ 1 Axdr=i J-sin2Idt+ j 6sin2t--(1-cos2/) dl .. c o 2 0 2
!
1C12
+ k ~3(1 + cos2/)-sin 2/}it o
.3 (_COS2/)1C12.[ 1 sin 21 Jft/2 [ 3. COS2/]ft/2 = 1+ J -3cos2/--+-+k 3/+-sI02/+--
24
220
= {
0
3 -4 )(-1-1)+ j[-3(-1-1)- : + 0]+k[3;
= %;+(6-
:}+e;
-1)k
2
+O+~(-l- ])]
20
525
Vector Differentiation
Exercise - 7(e) 4
I)
IfU(t)=(2t2 -I)i+3tj+(2-t)k. Find (a)
JU(t)dt,(b) Ju(t)dt 2
l
12
3 2 t2 2/ \06 [Ans:(a) ( ~-I i+ 3t j + ( 2t2 k,(b) 3i+18j-2k] 11/2
2) Evaluate: J(6sillu)i -(3cosu)j +uk o 1t
2
[Ans: 6i-3j+-k] 8 2
3) If A(s) = si - s2j + (s + l)k, B(s)
=
2
2si + 6sk find (a) JA.Bds, (b) JA x Bds . o
[Ans: (a) 4) If A = ti - j + 2tk, B = t2 ; 1
-
0
100
3' (b) -
48i - 12j + 16k]
tj +2k, C = ; - 2j + 2k, evaluate 1
(a) J{A.(BxC)}dt,(b) J{Ax(BxC)}dt o 0
I 3'
5. 37. 8 k +-} +- ] . 2 6 3 5) The acceleration of a particle a at any time t ~ 0 is given by a = eli + (2cos2t)j + (2sin2t)k If the velocity V and displacement r are both zero at t = 0, find V and r at [Ans' (a) -- (b)
--I
any time t. [Ans: V = (el
r
-
I)i + (sin2t)j + (1 - cos2t)k,
=(i -t-I)+~(l-COS2t)j +(t -~Sin2t}]
3
6) Evaluate
fA. dA
2
du, given that, A(2) = 4i - 2j + 3k and A(3) = 2i + j + 2k
du
[Ans: - 10]
Engineering Mathematics - I
526
Exercise 7(f) I. If ~
= xyz, valuate I~dr, where c is the curve x = t3, y = t2, z = t, from t = 0 to I c
[Ans: 2. If F
= xi - yzj + z2k,
and c is the curve given by x
(i) IF.dl" and (ii) fF.dr, from t = 0 to t e
I. I k JI i +4J+ 7 1
= t, Y = t3, z = t 2, evaluate
= I.
C
.) [A ns: ( I
5. 7. 1\ k .. -71-I5J +1"2 ,(II)
23] 24
3. If A = (2x + 3)i + xyj + (zx - y)k, evaluate fF.dr along c where c is (a) the curve x
= t3,y = 2t2, z = t from t = 0 to I.
(b) The straight lines from (0,0,0) to (1,0,0), then to (1,0, I) and then to (1,2, I) (c) The straight Iinejoining (0,0,0) and (I ,2, I). 491 13 14 [Ans: (a) 105' (b) (c)
2'
4. If A = (3xy - 2r)i + (x - y)j, evaluate
"3]
fA.dr along the curve c in xy - plane given c
by y
= x 3 from the point (0,0) to (2,8). 1308
[Ans:
3.5]
5. IfF = (2x- y)i + (x- 2y)j, evaluate fA.dl" where c is the closed curve shown in the c
figure below. y
x
527
Vector Differentiation
6. If A = (3x + 2y)i + (x + y)j, and c is the boundary of the tringle whose vertices are (0,0), (1,0), (0, I), evaluate fA.dr. c
3
[Ans:
2" J
7. If A = (2x + y)i + (3x - 2y)j, compute the circulation of A about the circle C: xl + 4, traversed in the positive direction. [Ans: 81t] 8. Find the work done in moving a particle in the force field. F = 2x2i + (2yz - x)j + yk, along (a) the straight line from (0,0,0) to (3, I ,2) (b) the space curve x = 3t2, Y = t, z = 3t2 - t from t = 0 to t = I
1=
113
[Ans: (a)
58
6' (b) 3]
9. (a) Prove that V = (2x siny - 3)i + (x 2 cosy + z2)j + 2(yz + I)k is a conservative force field. (b) Find the scalar potential ofY. (c) Find the work done in moving an object in this field from (1,0,-1) to (2,
2"1t ,I).
[Ans: (b) x 2 siny + yz2 - 3x + 2z + constant, (c) 10. If A
= (9xly -
1t
2"
+ 5]
2xz3 )i + 3x3j - 3x2z2k, (a) prove that fA.dr is independent of the c
curve 'c' joining two given points. (b) show that there is a differentiable function <jl such that A = \l<jl and find it.
7.7
SURFACE INTEGRALS
7.7.1
Let S be a two-sided surface. Let one side be taken as the positive side. If S is a closed surface, the outer side is considered as the positive side. Let A be a vector function. Consider an element of area 'ds' in the surface. Let n be the unit normal vector to ds in the positive direction. It can be seen that A.n = A cose. (where '8' is the angle between A alld n and A = IAI) is the normal component of A. Let ds be a vector whose magnitude is ds and whose direction is that ofn. :. ds
=
n ds.
Engineering Mathematics - I
528
z
Then the integral,
ffA.ds = ffA.Il .ds
..... (1)
s
is an example of a surface integral which is also called as flux of A over s. If'f' is a scalar function,
ff4>ds
..... (2)
..... (3)
ffAx ds = ffA x nds
Note
..... (4)
are some other examples of surface integrals. (1) The surface integrals can also be defined in terms oflimits of sums. (see 7.7.2) (2) The notation
If is also used to denote a surface integral over the closed
surface s. (3) Sometimes the notation
f may also be used for surface integrals.
(4) Surface integrals can be conveniently evaluated by expressing them as double integrals over the projected area of s on one of the coordinate planes {see 7.7.3)
529
Vector Differentiation /
7.7.2 Definition of surface integral as the limit of a sum:
.,J------...... y
The area S is divided into 'L' elements of area Mill' m = 1,2, .... L, let
P'II =
(xm'YIII,zm) be any point in
&\)111.
Let A(xm,Ym,zl/.)= AHI
Let nlll be the positive unit normal to I'1S111 at Pm. Then «(Am.nm) is the normal component of Am at Pm. Consider the sum, /,
I
Am.nIllMm
..... (I)
1/1=1
The limit of the sum (l)as L ~oo such thatthe largestdimens~on of each I'1S", ~ 0 (if the limit exists) is known as the surface integral of the normal component of A
IIA.n.d'}
over S an d denote d by ,
7.7.3 Evaluation of a surface integral To evaluate surface integrals, it is convenient to express them as double integrals taken over the projected area of the surface S on one of the coordinate planes (xy,yz, or zx planes). If R is the projection of S on the xy - plane, it can be shown that
ffA.nds ,
= ffA.n d\XdY, /I
n.k
From 7.7.2, the surface integral is the limit ofdle sum /,
I
111=1
Am .nmMIII
..... (1)
Engineering Mathematics - I
530
z
;---~----~~~--.Y
The projection of ASmon thexy- plane is
InmA.S'II/.kl (or) Illm.kIMi'm
which is equal
:. The sum (I) becomes
By the fundamental theorem of integral calculus the limit of this sum as L
~ 00
in
such a manner that the largest Llx m and AYm approach zero is
dxdy
IfA.n In.kl
which is the required result.
R
Note: Similarly ifR is the projection ofS on yz and zx planes respectively, it can be seen as
ff A .nds = ffA.n dl'Y~ZI
S
and
R
nJ
ffA.nds = ffA.n dz~xl S
R
ln.]
7.7.4 Physical interprettion of surface integrals: Let A denote the velocity of a moving fluid. Let S be a fixed surface in the fluid. Let ds be an element of surface. Then A.n ds = A.dS represents the amount offluid that passes normally through dS in unit time at any point. If the direction ofn is outward or positive, the amount offluid flow is positive. Similarly if dS' is another element for which n is in the negative direction, the fluid flow is negative at that point.
531
Vector Differentiation
to fA.nds and it is known as the total flux of A through the entire surface S.
s A can be a vector denoting physical quantities such as electric force, magnetic A
n....---...---~C"
force, flux of heat or gravitational force etc. In all these cases,
fA.nds
denotes
s total flux of A through S.
Sioved Examples Ex.7.7.5 Evaluate
fJA.nd\" where A = {x + y2} - 2xj + 2yzk and S is the surface of the
plane 2x + y + 2z = 6 in the first octant.
Sol:
A = (x + y)i - 2.>.] + 2yzk Let ~ = 2x + Y + 2z - 6
z
V ~ = 8q>i + aqy + 8q>k = 2i + j + 2k
ax
~
az
V~
2i+j+2k Unit normal n to S = IV~I = ~22 + 12 + 22 2i+ j + 2k 3
x
Engineering Mathematics - I
532
= .![2x+ 2y2 - 2x + 2y(6 - 2x - y)] 3
[Substituting 2z = 6 - 2x - y ( ... 2x + y + 2z = 6)]
= .!(12y-4xy) 3
If R be the projection of S on the xy - plane.
dxdy
2
3
In.kl = -3 => ds = -l/l.kl- = -dxdy 2 .. IfA.nds = If(A.n) dXdY /I /I ln.k I If .!(12y - 4);y )~dxdy /I 3 2 =
If(6 y - 2Xy}dXlry
..... (I)
/I
To evaluate this double integral over R, (i) Keep x fixed and integrate w.r.t. y from y w.r.t. x from x = 0 to x = 3. :. Given integral
=
0 to (6 - 2x), ii) and then integrate
3 6-2x
J J(6y-2xy)1ydx
x=O y=O
3 3 3
=
TI3i _xy2 r:~x dx =
J(3 -xX6- 2xY dx =
2 3 J(J08-108x+36x -4x }Ix
= [\08x-54x2+12x3-x4]~ =324-486+324-81 =81 Ex.7.7.6 If F = 4xzi -
yJ + yzk, evaluate
IfF.nds where S is the surface of the cube s
bounded by x =
O~
x = I, Y = 0, Y = 1, z = 0 and z = 1.
533
Vector Differentiation
Sol.
The surface S can be divided into 6 faces (see figu --,
z
o
(i) S\: Face EPFA (ii) S2: Face OBOC
E f---+----(
(iii) S3: Face PFBO y
(iv) S4: Face OCEA (v) Ss: Face POCE
x (vi) S6: Face OBFA
IfF.nd5 = IfF.nds + IfF.nds + IfF.nds + IfF.nds + IfF.nds + IfF.nd.,. S ~ ~ ~ ~ ~ % On S \: n = i x = 1 1
1
IfF.nds =
JJ(4Zi - y2 j + yzk }idydz JJ4zdydz
SI
0
0
On S2: n =
-
1
0
0
1
1
o
0
=
i, x
1
JJ(- y2 j + yzk X- i)dydz JJ(O}lydz =
o
=
1
1
0
0
J2Z21~
0
1
1
1
1
1
J J(-I)1xdz= J-zi
.'13
0
0
0
0
0
0
OnS 4 :n=-j,y=O
:. HF.n.ds
1
1
o
0
=
1
1
o
0
J J(4xzi).(- j)dxdz J J(O)1ydz =
= 0
On Ss: n = k, z = 1 1
1
:. HF.n.ds=J J(4Xi-/j+yk}kdxdy= 0
8S
On S6: n = - k,
Z 1
=
0
=0 1
J J{- y2 j }(- k )dxdy =
0
0
0
. JJF.n.ds =2 + 0 - 1 + 0 + ~ + 0 = ~ .. 2 2
s
J2dY = 2 0
=I 1
.'16
=
= 0
HF.n.ds=J J(4xzi-j+ zk).jdxdz =
HF.n.ds
1
0
=
1
HF.n.d.,. = On S3: n = j, y
1
1
1
1
21
0
0
0
0
J Jydxdy= J~ I
1
dx= J-Idx=-l 0
Engineering Mathematics - I
534
Ex.7.7.7 If A = z 2i + xlj - yZk, and S is the surface ofthe cylinder xl + Y = 16 included
Sol:
in the first octant between z = 0 and z = - 5, evaluate jfA.n.ds s z
f
d~- ~~'
n
,. 5 R Q
r=-----'JL..-- y
,....
' 8 is indeterminate. These points on z-axis are known as singular points of the transformation. Ex. 7.9.11 Represent the vector A = 2xi - yj +z2k in cylindrical polar coordinates. Sol: The base vectors in cylindrical coordinates are
Note:
ep =(cos8)i+(sin8)j eo = (-sin8)i+(cos8)j
..... (2)
ez = k
..... (3)
Solving (\) and (2),. we get i=(cos8)ep -(sin8h j
Then
A = 2[(cosS)ep
= (sin8)ep + (cosS)eo
-(sin8)eo~-[(sinS)ep +(cos8~o}y+z2e~.
A = (2xcosS - ysinS)ep -[2xsin8+ ycosSh + z2e~
= (2pcos 2S - psin 28 ~p - [2psin8cos8 + psin8cos8to + Z2 ez (substitutingx= pcos8, =
...... (\)
y= psin8)
p(2cos2S-sin28~p -(3psinScos8)eo +z 2e:
Engineering Mathematics - I
566
Aliter: Since cylindrical coordinates form an orthogonal coordinate system, we can write A as A = a,ep + a2ee + a3ez From (1) we find,
A.ep =a"
A.eo =a2 ;
.•..•
(I)
whereat'~, a 3 are to
be determined.
A.ez =a3
..... (2)
For cylindrical coordinates, we have, x=pcos8, y=psin8, Z=Z given A becomes, A= (2pcos8)i-(psin8)j+z 2 k
..... (3)
ep = (cos8)i+(sin8)j } Also, eo = (-sin8)i+ (cos8)j
..... (4)
ez =k using (3) and (4) in (2), we get
a, = 2pcos 2 8 - psin 2 8; a 2 =-2psin8cos8-psin8cos8=-3psin8cos8; a 3
=Z2
A = a,ep + a2 ee + a3 ez = (2 cos 28 - sin 2 8) pep - (3p sin8 cos8) eo + z2 ez
Ex. 7.9.12 Sol:
Represent the vector A = xyi - zj + xzk in the spherical coordinate system. Since the base vectors er, eo' A
=
ecj>
are mutually orthogonal, we can write
a,er + a2 eO + a3eq,
..... (J)
where aI' a 2, a 3 are to be found. From (I) we get, A.er =a,. A.ee =a2 • A.eq, =a3 For spherical coordinates, we have, x = r sin8 cos<j>, y = r sin8 sin<j>, 2
2
z
•••••
= r cos8
2
A = {r sin 8sin <j> cos <j»i - {rcos8)j + {r sin 8cos8cos<j»k Also we have,
(2)
..... (3) ..... (4)
er = {sin 8cos<j»i + {sin 8sin <j»j + {cos 8 )k ee
={cos8cos<j»i + (cos 8cos 4> )j - {sin 8)k
eq,
= (-sin4»i+ {cosiJj
Using (4) and (5) in (2), we get,
a, = 1'2 sin 3 8sin 4>cos 2 4> - rsin 8cos8sin<j> + 1'2 sin 8cos 2 8cos4> . a2 = r2 sin 2 8cos8sin <j>cos 2 <j> - rcos 2 8sin cp - r2 sin 2 8cos8coscp a3
= _1'2 sin 2 8sin 2 cpcos<j> - r cos 8 cos cp
..... (5)
Vector Differentiation
567
Hence A = aler + a 2ee + 01 eq, =
{r
2
sin} 8sin $cos 1 $ - rsin 8cos8sin $ + r2 sin Ocos 2 8cosc/>k + {r 2 sin 2 8cos8sin $cos 2 $ - rcos 2 8sin $ - r2 sin 2 8cos8cos$ ~o 1
2
2
+ (-r sin 8sin $cos$-rcos8cos$h
Ex. 7.9.13 Prove the following:
Sol:
(i)
e = Seo+ {sin 8 )~eq,
(ii)
eo = -Ser + {cos8)~eq,
r
(iii) eq, = (- sin 8 )~er - (cos8 )~eo where' • ' denotes differentiation w.r.t. time 't'. (i) We know that er = (sin 8cos$)i + (sin 8sin $)j + (cosS)k " cos8cos$8"} i + { " sinOcosc/>.$"} j - (sin8)8k " er ={sin 8(- sin$)$+ cos8.sin$.8+ =
e{(cos 8 cos $)i + {cos8sin$)j - (sin 8 )k} + {{- sin8sinc/»i + {sin8cos$)j }
" " = 8e e + sin 8.$eq,
(ii) ee
={cos8cos$}i + (cos8sin C/>}j - {sin 8}k
ee =( - sin8.8cos$ -cos8sin$~ } + ( - sin8.8s;". + coseco.~++ )j -(cOS8.e)k = -
e{(sin 8cos$)i + (sin 8sin$}j + (cos 8 )k} + cos8.+{(- sinc/»i + {cos$)j}
= -
" " 8e r + cos8$eq,
(iii) eo!> =(-sin$)i+{cos$}j
eo!>
=-cos$~i-sin$~j (- sin 8 }~er - cos8~ee
= ~[- sin 8{sin 8cos$i + sin 8sin $ j + cos8k} - cos 8 {cos 8cos c/> i + cos8sin c/> j - sin 8 k}]
..... (i)
Engineering Mathematics - I
568
o
0
= - coscj).cj)i - sin cj).cj) j
= e~ Ex. 7.9.14
from (i)
If ep , eo are base vectors in cylindrical coordinates, show that
(i)(ep)=ee o and
(ii)(eo)=-eep,
where'.' denotes differentiation w.r.t. 't' Sol:
ep = (cosS)i+ (sinS)j
= e{(-sinS)i + (cosS)j}
eo = -sin Si + cosS j, eo = - S(cosS)i -e(sin S)j = -Sep
Ex. 7.9.15 Express the velocity V and acceleration a ofa particle in cylindrical coordinates. Sol:
In cartesian coordiantes, the position .vector r = xi + yj + zk .
. dr dx. dy. dz VelOCIty V = -=-I+-j+-k dt dt dl dt Acceleration a=
d 2r
d 2 x.
d 2y
d 2z
dt
dt
dr
dt
-=-1+-, j+-k 2 2 2
In cylindrical coordinates, x = pcosS, y = psin S , z = z and ep=(cosS)i+(sinS)j, eo =(-sinS)i+(cosS)j, ez
=k
~'. i = cosSep - sin Seo ; j :::: (sin S)ep + (cos e)eo (Solving above equations for i andj) :. r = (pcosS XcosSe p - sin See)+ psin e(sin Sep + cosSee)+ zez = pep + ze z
dr dp de dz . V = -=-e +P-P +-e .. dt dt p dt dt Z
(0) 0
o = pep +p Sea +zez
[
.: dez dt
=~(k)=O] dt
(from Ex. 7.9.14)
569 .
Vector Differentiation
dV
a = dt
d.
.
.
= dt [pep + pOeo + zeJ (since
....
..
..
....
= pep + pep + pOeo + pOeo + pOeo + ze z
ez = 0)
pBen + pep + pBeo + pBeo + pB(-Bep ) + ze z ' (using results of (ex 7.9.14»
=
[p- pe 2 ]e p +[pO + 2pB]eo + zez Ex. 7.9.16 If t
= xyz,
find 'gradf' in (a) Cylindrical coordinates (b) Spherical
coordinates Solution: In cylindrical coordinates (p,O,z), we have, x = pcosO,y = psinO,z =
Z
..... ( I)
and 2
f = xyz = p2 Z sin 0 cos 0 = p
2
Z
sin 20
af = pz sin 20· af = p2 Z cos 20. at = p2 sin 20 ap 'ao 'az 2 2
:. (I) gives, gradf = Vf = (pz sin 20)e p + (pz cos 20)eo + (~ sin 20)ez (b)
In spherical coordinates (r,O,tjJ), we have,
x = rsinOcostjJ,y = rsinOsintjJ,z = rcosO, and
Vf= af e +! af e +_1_ af e ar r r ao 0 r sin 0 atjJ ; Here
f = xyz = 3r 2sin 2 OcosOcostjJsintjJ :. af ='csin2tjJ{-sin 3 0+cos02sinOcosO} ao 2
..... (2)
Engineering Mathematics - I
579
aj
o~
=rJsin20cosO.cos2~
:. from (2) )
~l=(3r2sin20cosOsin+cos~~,. + '; sin2~{-sin30+2sinOcos20}ee + (r2 sinOcosOcos2~~~. E.7.9.17 Itf= pzcosO (in cylindrical coordinates) find Vf. Sol: f= pz cosO.
-aj = zcoso of
op
Hence Vf
,
=
aj
00
=
-pzsin 0 aj = pcosO ' oz
I aj
of
op ep +p ae eo + oz ez
= (zcosO)ep -(zsinO)eo + (pcosO}ez Ex.7.9.18 Iff=
Sol:
,.'2
sin2Gsin+ in spherical coordinates, find Vf.
aj = 2rsin20sin4>
or
-8f = 2r2 cos 20sin 4>
ae
aj =r2 sin20cos~ o~
.. ~l
8j = =
Jo/
1
Of
or e,. +-; ae ce + rsinO o~ e.p
(2rsin20sinq,)e,. +(2rcos20sin~)eij + (2rcosOcos4»e.p
Ex. 7.9.19 Show that the vectorfieldA =
z{(sin O)ep + (cosoh }-(pcosO}ez, in cylindrical
polar coord inates is solenoidal.
Sol:
IfA= Aiel' +A2 co +AJe z then, AI =zsinO; A2 =zcosO; A3 =-pcosO
571
Vector Differentiation
=
~[~(pz sinO)+~(zcOsO)+~(- p2 coso)l
=
~[zsinO-zsinO+O] = 0
as
p ap
aZ
J
p
:. A is solenoidal
I
1
I
)
Ex.7.9.20 IfA= ( r-cosOer +-eo +-.-Oe+ ,in spherical coordinates, find Div A. r
Sol:
rS1l1
I I Here A =r 2 cosO, A2 =-, A3 = - . -
r
I
rS1l10
1
3
=
=
[4r -sin20+cosO+O ,.- S1l1 0 2 I
1.
cotO
4rcosO+-2r
Ex.7.9.21 Ifj= p2Z2cos20, show that V2j=2p2cos20 Sol:
In cylindrical coordinates, 2
af ap2
1 af p ap
1 p2
a2 f
a2 f
ao
az
V j = - + - - + - -2- + 2
j
= p2 Z 2 cos20 2
Qf = 2 pz 2 cos2 0 . a j = 2 z 2 cos 20 . ap , ap2 ,
-
..... (I)
at ') as = -2p- Z 2 si1120·'
_'J
Engineering Mathematics - I
572
:. From (1) we get, yo2 f = 2Z2 cos 26 + 2Z2 cos 29 - 4Z2 cos 29 + 2p2 cos 29 = 2p2 cos 29 .
Ex.7.9.22 Using spherical coordinates, show that yo2r" = n{n + l)r,,-2 when n is a Sol:
constant and r = 0 if nO, and x-axis. [Ans: 3J[a 2 ] 6.
Evaluate f{2x 2 - y2 )dx + (x 2 +
i)
by Green' theorem where C is the boundary of
c
the surface in the xy plane enclosed by x axis and the semi-circle y
= ..Jl- x 2
[Ans: 4/3] 7.
Evaluate f(cos x sin y - xy)dx + sin x cos y,using Green's theorem where c is the c
[Ans: 0] 8.
VerifY Green's Theorem in the plane for f{x 2_xy 3 )dx+ &2 -2xy)where C is the c
square with vertices at ( 0,0) , ( 2,0) and (0 ,2) [Ans: 8] (2,1)
9.
f (12 x3- 2x/ ) dx - 3x2y2 dy along the path x3- y3 + Y - 4xy
Evaluate
(0,0)
[ Hint: Proceed as in aliter of 7.1 0.9] 4
2 31(2,1) (0,0)
[Ans: 3x - x y
= 44 ]
= 0
591
Vector Differentiation
7.11
Gauss Divergence Theorem
7.11.1 Gauss' Divergence Theorem Let (I) V be the volume bounded by a closed surface S (2) A be a vector function of position with continuous derivatives. Then,
fffV .A .dv ffA .nds 1A .ds =
=
where n is positive unit normal (outside drawn normal) to S.
Proof: z
t------y ..... R
,
x S is a closed surface Let any line II lel to the coordinate axes cut S in at most two points. Let z = gl(x, y) and z = gix, y) be the equations of the upper portion SI and lower portion S2 respectively. Let R be the projection of S on the xy-plane.
If A = Al i + A::J + A3 k
Engineering Mathematics - I
592
Then,
oA OA fff OZ 3 dv = fff OZ 3 dzdydx v
v
oA 0/ dz] dydt=
Iq(x,y)
ff[ S
z~g2( x,y)
II
ff
AJX,y,Z)1
z~g,
dydx
z~g2
II
f f[A 3 (x,y,g,)- A 3 (x,y,g2)]dydx
=
R
For the upper portion SI' dy dx = (cosyl)dS I = k. n l dS I since the normal n l to SI makes an acute angle YI with k. For the lower portion S2' dydx
= (-cOSY2) dS 2 = --k.n 2 dS 2,
since the normal n 2 to S2 makes an obtuse angle Y2 with k.
:. f f A3(X,y,g, )£lydx =:f f A 3k .n, dS, and R
"
So that, f f[A 3 (x,y,g,) -A3 (x, y, g2) ]dydx
= f f A 3k .n,dS, + f fA3k.n2dS2 8,
II
f
= fA3k.ndS s Similarly, by projecting S on the other coordinate planes, we get,
f f fO~, dv= f fAl nds v
S2
..... (I)
..... (2)
•
and f f f v
oA2
cry
dv = f f A 2j.n ds
..... (3)
s
(I) + (2) + (3) =>
fff(o~, + O~2 + o~3)dV=
ff(A ,i+A 2j+A3k).ndS s
v
i.e. f f fV.A dv = f fA.n dS which proves the theorem. S
593
Vector Differentiation
7.11.2 (a) Express Gauss' Theorem in words and (b) obtain its Cartesian form. (a)
Gauss' Theorem states that "The surface integral of the normal component of vector A taken over a closed surface is equal to the integral of the divergence of A taken over the volume enclosed by the surface. Let A=A 1i+A 2 j+A 3k,
(b)
.
Then dlv A = V.A =
aA aA? aA 3 + -"'- + --ax ay az
~-I
Let the unit normal n to S make angles
a,P,r
with the +ve coordinate axes so
that, cosa=n. i, cosp =n. j, cosr=n.kand cosa, cosp, cosr are the direction cosines of n
n = ( cos a )i +( cos P)+( cos r
)
[ n. i =
"I = cos a
etc.]
:. A.II =Alcosa+A 2 cosP+A 3 cosr Hence the divergence theorem can be written in Cartesian form as,
JJr{ aA + aA + aA )dXdYdZ = v
Jl ax
ay
az
=
J J(A I cosa + A2 cos p + A3 cosr)dr;
s
JIA,dydz + A 2dzdx + A dzdy 3
s
Solved Examples Ex. 7.11.3:
Evaluate
using
the
divergence
theorem = J J F .n
d~
where
s
F = 2xyi + yz2 j + xzk and S is the surface of the parallelepiped bounded by x = 0, y = 0, z = 0, x = 2, Y = 1 and z =3 . Sol: By the divergence theorem; HF.n ds = JHV.Fdv s Here, V.F
=~(2Xy)+~(yz2)+~(XZ) ax ay ay =(2Y+Z2 +x)
I IF.n ds = HI(2Y+Z2 +x)dv s
Engineering Mathematics - I
594
2
I
3
2
I
3
I I Ie 2 y + Z2 + x)dxdydz = I I [ I(2y + Z2 + x)dz] dxdy X;O y;O z;O 2 I
X;O y;O =;0 3
II2Yz+~+xz
=
3
00
2
3
2 I
dxdy== Ifc 6 y+9+3x)dxdy 0
00
I
2
I[ I (6y+9+3x)try]d\" = I 3i
=
X;O y;O
3 22 +9y+3xyl dx== I(12+3x)d\"=12x+ ; I
2
0
0
0
0
=24+6= 30
Ex. 7.11.4 Verify the divergence theorem for F = (4xy)i - (j2)j + (xz )k, over the cub( bounded by x = 0, x = I, Y = 0, Y = 1, z = 0 and z = 1. Sol:
By the divergence theorem II F.n dS == I I IV.F dv \
V
Here F
= (4xy)i - (j2)j + (xz)k
a a 2 a V.F==-(4xy)--(y )+-(xz) =4y-2y+x=x+2y ax ay az I
I
I
I
I
I
I I IV.Fdv == I I Ie x + 2y)dzdydx == I I [ Ie x + 2y)dz ]a:vdx x;Oy;Oz;O
v
x;Oy;O =;0 I
== I
I
fi xz + 2 y z1 dydx
x;Oy;O I
==
I
If I(x+2y)dy]dx=
X;O y;O
To evaluate
I
Ixy+y x;O
I
2 1
0
I
dx=
2
3
IeX+I)d~==~ +xl~==2 x;o
IIF.ndS
z
The surface S contains 6 faces (see figure) S) - Face EPFA, S2 - Face OBDC
S3 - Face PFBD, S4 - Face OCEA
J---+--+---y
Ss - Face PDCE, S6 - Face OBFA x
595
Vector Differentiation
The surface integral
IIF.ndS is equal to the sum of the surface integrals on the
above 6 faces.
x = I, F.n = 4y; dS = dy dz
on S" n = i, J
J
J
J
J
:. JJF.ndS= J J 4ydydz= J[ J4ydz]dy= J y;O :;0
on S2' n = -i, x = 0, F.n = :.
°
F-O
:~O
J
4yzl~dy= J 4ydY=2/1~ =2 0
y~O
IIF.n~S=O
on S3' n = i, y = I, F.n = -I, dS = dx dz. I
I
I
J
I[ I-Idz]dx= I-z dx= I-dx = --xl~ =-1
:. I IF.ndS=
on S4' n = -i, y = 0, F.n =
:. I IF.ndS=o
°
84
on S5'
11
= k, z = I, F.n = x ; dS = dx ely J
J
J
I
J
J
1
2 1
:. I I F.lldS = I I xdxdy = I [ Ix dy 1 dx = I xy dx = I xdx = ~ = 85 X~O y=o X;O y;O x;(} 0 x=O 2 0 on S6' n = k, z = 0, F.n =
:. I IF.ndS =
°
Hence :. II F.ndS
~
°
= 2 - 1 + ~ =%
8
The divergence theorem is verified
Ex.S.11.5 :
If r = xi + Yi + zk, evaluate :.
IIr.n dS where S is any closed surface. s
Sol:
By the divergence theorem
:. I Ir.n dS = I I Iv." dv ,where V is the volume enclosing S. s v
Engineering Mathematics - I
596
V.r
000
= -(x) + -(y) + -(z) = 3
ox
0'
oz
:. I Ir.n ds = I I I3 dv = 3v
Ex.7.11.6 :
If '\jI' is a scalar function, prove that I I IV\jIdv = I I'P n ds
Sol: By divergence theorem I I IV.Adv = I IA.n ds
..... (I)
Take A = \jIc, C = cli + c-j + c3k is any constant vector.
:.(l)=> JIIV.(lPc)dv= fI('Pc).nds
..... (2)
s
v
But,
= (V \jf).C
o )+-(c~)+-(c 0 0 l [ ":V.c=-(c ox oy- OZ3 )=0 ,
= c. V \jf and
(\jIc).n = c.(\jJl1)
cl' c2 ' c3' being constant] [ ": a.b = b.a]
so that (2) becomes,
IJI(c.V'P) dv = JI{c.('Pn)}ds
=> c. JI IV'Pdv = c. JI'-IJn ds => JI IVlPdv = JI'Pn ds v
(.: c is arbitrary)
Ex.7.11.7 : Ifs is a closed surface enclosing a volume V and ifF = (Ix)i + (my)j + (llz)k,
I, m, n being constants, evaluate JIF.nds
Vector Differentiation
Sol:
597
By Gauss' theorem HF.nds= Hfv .Fdv
a ax
a
a az
V.F = -(Ix) + -(my) + -(nz) = I + m + n :. HF.nds
ry
=
Hfu + m + 11) dv
=(/+m+n)V
Ex.7.11.8: IfE=curlA,evaluate HE.nds wheresisanyclosedsurface Sol:
By Gauss' theorem,
HE.n ds = Hf(div E) dv where V is the volume enclosed by s .\'
V
But Div E = div (curl A) = 0
[ See 7.5.2 (2) ]
:. HE.nds=O
Ex. 7 .11.9 : If s any closed surface and n is unit +ve normal to s, show that Hn ds = 0 Sol:
Consider a constant vector a
=
ali + ~ + a3k
( i.e.
a2, a3 are constants)
al'
Then, a·[Hndv] = H(a.n)ds =
Hf(V.a)dv
(by divergence theorem)
aa aa ax ay az
a.
[ ':V.a=-) +_2 +_3 =0]
=0
Thus a. a.[ Hn dv] = 0 is true where a is any arbitrary vector.
:. Hnds=O
Ex.7.11.10 : Evaluate H(F.n)ds, where s is the region bownded byy2 = lx, x = 2, s Z = 3,
and F = 2xi + 3yj +
1
"3 z3k, using Gauss' theorelll
Z
= 0,
Engineering Mathematics - I
598
Sol: By the divergence theorem,
ff(F .n) ds
= fff(V.F)dv
s
V.F
=~(2X)+~(3y)~~(Z3) = 5+z ax ay ay 3
From (I), ff(l1~.Il)ds s
2
=
.... (1)
v
2.[i-;
f f
2
= fff(5+z 2 )dv,
3 3
5z + ~
x=O Y= 2 Ex
dydx 0
~--------r-----X
2
=
) 2
f 96J2xdx = 96J2 ~
= 192J2 0
x=o
Ex. 7.11.11
ff( 2xi - 31 j + Z2 k ).n ds over the surface bounded by
Evaluate
s 2
x + y2
= 1, z = 0,
Z
= 2, using Gauss' theorem.
Sol: By the divergence theorem,
ff(F .n) ds
= fff(V.F)dv ,
s
Here,F= 2xi-3y2j+Z2k DivF
=~(2x)-~(31)+~(z2) ax ay ay = 2-6y+2z +1
ff(F.n)ds=
M
f f
2
f(2-6y+2z)dzdydx
x=ly=~O
Vector Differentiation
=
599
16[~~I-X2 +~sin-1 XJ1 2
Ex.7.11.12:
2
=8n
.1
Evaluate using the divergence theorem H(F.Il)dS where S is the surface
of the sphere x 2 + I + z2
Sol:
=
h2 in the first octant and F = yi + zj + xk
By divergence theorem,
f
:. H F.1l d~' = HV'·F dv
..... (i)
v
F = yi + zj + xk
V'.F= 0 :. H(V'.F) dv = 0
..... (ii) z
~--+B--Y
x Let us evaluate the surface integrals over the faces OAB, OBC and OCA. b
Jj;2~)
f fF.n ds = - f fx dx(ry OA B
x=O
y=O
(,,' n = -k)
Engineering Mathematics - I
600
=
If F.ndv
Similarly
DAB
If F.nds + If F.nds + If F.nds + If F.nds OAB
ABC
ABC
DCA
3
= -Jrb + If F.nds
.... (iii)
AIlC
From (i), (ii), and (iii), we get 3
0= -Jrb + If F.nds ABC
If F.nds
Jrb 3
=
ABC
VerifY divergence theorem for F=4xi-2y2j+z2k taken over the region bounded by
Ex. 7.11.13: 2
x + / = 4,
Z
= 0 and z = 3.
Sol: By the divergence theorem, we have
Iff DivFdv (1) DivF
=
IfF.~ ds
=~(4x)-~(2/)+~(Z2) =4-4y+2z Ox
L.R.S. 0[(1)
~
.... ( I )
s
,Ll"
By
~ ,t~~l,JL (4-4y+2z)dz ]dydt
[4z-4yz+z'JC
dydx~
)11
(21-12y)dy
f ~4_X2dx=84Jr
x~-2
[Do the integration w.r.t.x yourself, taking x (2)
.,.
8z
2
=42
..
Evaluate of surface integral IfF.~ ds s
= 2sin ()]
Jdx
601
Vector Differentiation Z
Z=3
y
x
The given surface of the cylinder can be divided into 3parts, namely (a)
8 1 : the circular surface z = 0
(b)
S2: the surface z == 3 (circular) and I
(c)
S3: the cylindrical portion of 8: Xl + i
= 4,
z == 0, z
=3
we now find JJF.~ ds over SI ,82 , S3 .If we add them, we get R.H.S of (I). (a)
on SI:Z=.O;
~=-k; F.~=-(4xi-2ij).k=0;:. JJF.~ds'=O. s,
(b)
- - ( ) dxdy on S2:z==3; n=k ; F.n= 4xi-2y2j+9k .k=9; d\'=-I':' -I =dxdy
n.k
:. JJF.1i (is' = JJ9dxdy = 9A , where A is the area of the circle s, x 2 + y2
(c)
s,
= 4, = 9J'l' (22) = 36J'l'
on S3: Let fjJ =x2+i-4=0;
-
V fjJ
2 ( xi + yi)
!VfjJ!
2~X2 + i
n =- - =
F.n=
=
xi + yi ( .
2
SInce x + y
2
= 4)
2
4 x2 -2y 3 22 3 =x-y; 2
To evaluate
JJF.1i ds, take x = 2cosB,y = 2sinB, S3
Engineering Mathematics - I
602
= 2dO dz
and ds
; limits of z are 0 to 3 and those of 0 are 0 to 27r . 2lf
Hence
fJF.~ ds S3
=
3
f f (8cos 0-8sin 0)2d 0 dz 3
2
(}=02=0
2"
=
16
f [( cos 0-sin O)z ]
3
2"
3
2
(}=o
2
3
(}=O
2=0
[ Do the integration w.r.t.
f (cos 0-sin O)dB =487r
dO = 48
e yourself]
:. R.H.Sof(I)= 0+367r+487r=847r;:. L.H.S= R.H.S Hence the theorem is verified.
Exercise - 7K 1.
Verify Gauss's divergence theorem for A
= (x 2 -
yz)i +
(i - zx) j + (Z2 - xy)k
taken over the rectangular parallalopiped 0 ~ x ~ 2, 0 ~ y ~ 3, 0 ~ z ~ 1.[Ans:36] 2.
Use the divergence theorem to find ff F.ndS, where --
f
s
= (3x + 2Z2)i - (Z2 - 2 y) j
+ (/ - 2z)k and S is the surface of the sphere with
centre at (2, -I, 3) and radius 2 units. [Ans:32 TC]
3.
Verify
the
divergence
theorem
for
the
unit A = (4xz)i - (i)j + (yz)k , taken over the x = 0, x = 1, y = 0, y = 1, z = 0 and z = 1.[Ans:3/2] 4.
If r
= xi + yj + zk , and S
vector cube
function, bounded
by
is the surface of the rectangular parallelepiped bounded
by planes x = O,y = O,z = O,x = a,y = band z = c, find the value of ffr.ndS s using Gauss's theorem. Verity your answer by direct evaluation of the integral. [Ans:3abc] 5.
Use the divergence theorem to evaluate ffA.nds for A where s is the sphere given by (x _1)2
6.
If
V=(lx)i+(my)j+(nz)k
ff V.ds s
= 327r (l + m + n). where 3
(X-3)2 +(y-2i +(z-I)2
= 4.
+ i + Z2 = 1 I, m, n
and
S
= (2x)i-(2y)j + (3z)k
IS
being the
[Ans:4 7r] constants surface
show the
that
sphere
603
Vector Differentiation
7.12
Stoke's Theorem
7.12.1 Stoke's Theorem Let (1) S be an open, two-sided surface bounded by a simple closed curve C. (2) A be a vector function having continuous derivatives Then, fA.dr
=
H(Y' x A).nds = H(Y' x A).d\'
where C travels in the +ve direction and n is the unit +ve (outward drawn) normal to S.
Proof:
z
Let S be the surface. Let the projections of S on the coordinate planes be regions bounded by simple closed curves. Let 'R' the projection ofS on xy plane be bounded by C l . (see the figure above). Let the equation of S be z differentiable function.
1(x, y) where 1 is a single valued, continuous and
=
j
Then
Y'x(Ali)=alax alay AI
0
Engineering Mathematics - I
604
{V x (A]i}}.n ds =(8A] _ 8A] ).nds = {8A] (n.}) _ 8A3 (n.k)}ds
8z
8y
8z
8y
.....
(.) 1
The position vector r of any point on S can be taken as r
=
xi + y) + zk
= xi + y} + t/>/x, y)k 8r _. 0+] and - - } + - k ay 8y 8r
But
ay
8r
being the vector tangent to S, it is .i, to n.
.. ay' n = 0
~
n.} =
3] -
8y (n.k)
. 8AI 84>1 8AI :. (I) => { V x A\,)} . n ds = (---n.k--n.k)ds 8z ay 8y ..... (ii)
on S,
A\(x y, z)
= A\(x, y, 4>\(Y» = G(x, y) (say)
8AI 8AI 8z aa -+-.-=ay 8z 8y 8y
:. (ii) becomes, { V
x
(A\i) ] .n ds
H(VxA\i)}.nds= ,I
aa
aa
= - ay (n.k) ds = - 8y dxdy
r·:
n.k. ds
=
dxdy]
Jf- aaay dxdy=fCdx-, by Green's theorem in the plane. Now (' R
at each point (x, y) ofC, the value ofG is the same as the value of AI at each point (x, y, z) of C, and since dx is same for both curves, we have
I
605
Vector Differentiation
H{V x(Ali}}·n ds = fAldx
Hence,
s
..... (iii)
c
lilly by projecting S on yz and zx planes, it can be shown that,
H{V x (A2i)}·n ds = fA 2dY s
c
H{V x (A3 k )}.n ds = f A 3dz c
s
Adding (iii), (iv) and (v),
HVx A.n ds = s
..... (iv)
..... (v)
fc A. dr
(since A. dr = AI dx + A 2dy + A 3dz) Hence the theorem is proved. Ex.7.12.1 :(a) Express Stoke's theorem in words and (b) obtain its cartesian form. Sol :(a)
The I ine integral of the tangential component of a vector A taken around a simple closed curve C is equal to the surface integral of the normal component of curl A taken over a surface Shaving C as its boundary.
(b) As in 7.11.2(b) A=A l i+Aj+A3k n = (cosa)i + (cos~)j + (cosy)k
Then,
Vx A = ajax
) ajay
k ajaz
A2
A3
Al
=(aA 3 _ ay
aA2)i + (aA I az az
_
aA3)} + (aA 2 _ aA I )k ax ax ay
aA 3 aA 2 aA aA 3 aA 2 aA I (V x A)Jl =( - -)cosa + ( -I - -)cos~ +( - -)cos y ay az az ax ax ay
Engine~ring
606
Mathematics - I
Hence the cartesian form of the Stoke's theorem can be stated as,
=fA ,dx+A 2dy+A 3dz
c
Solved Examples Ex.7.12.3 : Sol:
VerifY Stoke's Theorem for A = (x - 2y)i + yilj + y2zk, where S is the upper half of the sphere xl + y2 + z2 = 1 and C is its boundary.
The boundary of the projection ofS in the xy-plane is a circle with centre at origin and unit radius. Its parametric equations are x = cosS, y = sinS, Z = 0, 0:::: S < 2n
dx = (-sinS)d9, dy = (cosS)dS. f Adr =f(x-2y)dx+ yz 2dy+ y2 zdz c c 21t
(.: z =0)
= f[cosS - 2sinS] (- sinS )dS e~()
-_ 2~[-sin2S JI + I -cos2S]dS- cos2S - - + S -sin2SI - - -_2 n o 2 4 2 0 21t
i
Y' x A =
ajax (x-2y)
k
j
ajry ajaz = 2k yz2
y2 z
H(Y' x A).nds = H2(n.k )ds = 2 Hdxdy R
s
(n.k ds
=
dxdy and R is the projection of S on the xy plane)
Hence Stoke's theorem is verified.
607
Vector Differentiation
Ex.7.12.4: Prove that a necessary and sufficient condition that fF.dr for every closed ("
curve C is that '\I x F
Proof: (a)
=0 .
The condition is necessary: Let '\IF = 0; Then by Stoke's theorem, fF.dr
= If('\I x F).n ds = 0
("
(b)
S
The condition is sufficient:
Suppose that fF.dr
=0
around every closed path c.
"* 0
at some point P. then, assuming that '\IF is continuous,
('
Assume that '\I x F
there exists a region with P as its interior point where '\IF = 0 .Let S be surface contained in this region and let the normal n to S at each point has the same direction as '\IF . Then '\IF = an, (a being a +ve constant);
Let C be the boundary of S.
Then by Stokes theorem, f F .dr = If ('\IF).n ds =a Ifn.n ds > 0 c s s Which is a contradiction to the hypothesis that f F.dr
=0 ;
:. '\IF
=0
Note: It follows that '\IF = 0 is also a necessary and sufficient condition for the line 1'2
integral f F.dr to be independent of path joining the points ~ and ~ .(see 7.6.4) I;
Ex.7.12.S
If r = xi + yj + zk , show that f r .dr = 0 c
By Stoke's theorem, Jr.dr c
= If(Curl r).n ds ('
..... (1)
Engineering Mathematics - I
608
j
i
k
But curl r= a/ax a/ay a/dz =0
x
y
z
:. (1) => Jr.dr = 0 ('
Ex.7.12.6:
If'f and 'g' are scalar functions, show that Jf( grad g). dr =- Jg(grad j).dr ('
('
Sol: By stoke's theorem, J{grad (fg)·dr
= Jficurl{grad(fg)}ln ds = 0, since curl grad (fg) = 0
('
S
..... (1)
.. J{grad (fg)}·dr =0 ('
But grad(fg)
=
f(grad g) + g(grad j)
Hence the result
..... (2) /
[from (1) and (2)]
If A is any vector function, prove by stoke's theorem that div curl A = O.
Ex.7.12.7:
Sol: Let V be any volume enclosed by a closed surface S. Then by Gauss' divergence theorem. We get, JJJV.(curl A)dv = JJ(curl A).nds
Divide the surface S into two portions SI and S2 by a closed curve C. Then
JJ(curl A ).n ds
= JJ(curl A).ndst + JJ(curi A).n dS 2 = JA.dr c
JA.dr c
..... (1)
609
Vector Differentiation
= 0, by Soke's theorem, since the +ve directions along the boundaries ofS, andS 2 are opposite. :. (I)
~ fffV.(clirl A)dv=O
Since this is true for all volume elements V, we have, V . curl A = 0
~
div(curl A) = 0
Ex.7.12.8: Use Stoke's theorem and prove that curl gradJ= 0, where 'f' is a scalar function.
Sol:
IfS is a surface enclosed by a simple closed curve C, we have, by Stoke's theorem,
H{curl(grad J)}.nd.. = f(grad J).dr
.... (I)
('
Now, grad! dr
=(8f i + aJ j ax 0'
+ 8f k ).( dx i + dy j + dz k) az
= aJ dx+ 8f dy+ 8f dz=dJ ax ay az A
:. f(grad J).dr = fdf ('
A
= f(A) - f(A) = 0, where A is any point on C :. (1) ~ ff{curl(grad f)}·nds
=0
Since this equation is true for all surface elements S, we have,
Ex.7.12.9:
curl (grad j) = 0
Veri1)r stoke's theorem for A = yj - 2xyj taken round the rectangle bounded by x = ±b, y = 0, y = a y F~
x=b
x=-b
--____
__~__~y=a ______~B
~----~--~----~~--L---~~X
D
A
y=O
Engineering Mathematics - I
610
i
j
Sol: curl A = ajax
k
ajay ajaz = - 4yk
y2
-2xy
0
For the given surface S, n = k :. (curl A). n
4y
= -
Hence ff(curIA).n ds a
a
h
h
U
= ff- 4 ydxdy= f[ f -4ydx]dy= f -4xyl dx= f- 8bydy y=o
s
J.=~h
0
~h
0
a
= -4by 2/ =-4a 2b o fA.dr
=
C
J
f+ f+ f+ DA AB BF
..... (I)
FD
fA.dr = y2 dx - 2xydy y
AlongDA,
0, dy = 0, => JA.dr=O
=
( .: A.dr = 0)
DA
x= b, dx
AlongAB,
=
0
a
2 f A.dr= f-2bydy=-by2/: =-a b
:.
y=O
AB
y=a,dy=O
AlongBF, -b
:. JA.dr= BF
Ja
2
dx
=
-2a2b
b
x
Along FD,
=
-b, dx
0
=
o 0 2 :. fA.dr= f 2bydy=-bi/ =-a b FD
a
a
:. fA.dr= 0 - a 2 b - 2a2 b - a 2 b
= -
4a2b
C
From (1) and (2), fA.dr = ffcurl(A).nds
c
s
Hence the theorem is verified
..... (2)
Vector Differentiation
Ex.7.12.10:
611
Use stoke's theorem to evaluate the integral fA.dr where A = 2y.i +3xY('
(2x +z)k, and C is the boundary of the triangle whose vertices are (0,0,0), (2,0,0), (2,2,0). )
Sol:
k
Curl A = ajax ajay ajaz 2 2/ 3x -2x-z = 2}
+ (6x - 4y)k
Since the z-coordinate of each vertex of the triangle is zero, the triangle lies in the xy-plane.
:. n
=
k.
:. (curl A). n = 6x - 4y consider the triangle in xy-plane. Equation of the straight line OB isy = x By Stoke's theorem, f A .dr c
=
fsf(CurlA)nds
I
2
2
o
3
= f6xy - 2/ dx = f(6x - 2X2 )dx = 4 ~ x=O
Ex. 7.12.11
A(2,O)
(0,0)
2
1= 332 .
0
Use Stoke's theorem to evaluate
ff(curlA)nds, where A
=
2yi + (x- 2zx)j
+ xyk, and S is the surface of the sphere .xl + Y. + z2 = b 2 above the xy-plane. Sol:
The boundary C of the surface S is the circle.xl + The parametric equations of C are x By Stoke's theorem, we have,
fsf(CurIA).nds =cf A.dr
=
bcose, y
=
y. + z2 = b2, Z = o.
bsine, z = 0, 0 ~ e < 27t
612
Engineering Mathematics - I
= f2ydx+(x-2zx}dy+xydz = f2ydx+xdy
(0: z=O,dz=OonC)
('
('
2"
[-: x = bcos8 => dx = - bsin8 d8
f(2bsin8)(-bsin8)d8+bcos8.bcos8.d8 o
and y = bsin8 => dy = bcos8 d8]
= h 2 ](cos 2 8 -2sin 2 8~8 = b 2 2][1 + cos 28 - (l-cos28)18 o 0 2
r
J"
b2 [ 3sin28]b2 = - r(-1+3cos28)=+- -8+ =--.2n=-nb 2 2 222 o h221t
t
Ex.7.12.12
Apply Stoke's theorem to evaluate f A.dr .where A = (x- y)i + (2y + z}j +
(y-z)k and C is the boundary ofthe triangle whose vertices are
and
Sol:
(O,O,~)
LetA=
(~, 0, 0) (O,~, 0) z
(~,O,O)
B=
(O,~,O) C= (O,O,~).
The equation of the plane ABC is (by intercept form), x y z 1/6 + 1/3 + 1/2
=
..... (I)
I =>6x+3y+2z= 1
The direction ratios of the normal to (1) are 6, 3, 2 .. . 6 3 2 :. D IrectlOn cosmes are -,-,777 Ifn is the unit normal to the plane, n =
§..i +l j
A = (x - y)i + (2y + z}j + (y - z)k i
VxA= a/ax x-y
.. (Vx A}.n =3. 7
j
k
aj8y a/az =k 2y+z y-z
7
y
7
+3. k 7
Vector Differentiation
613
:. By Stoke's theorem,
fA.dr = Ifs (VF).n ds = ~7sIfds = ~7
..... (2)
(Area of triangle ABC)
('
To find the area of triangle ABC:
AB=
AC=
(H+(H ~ ~ (H+(H ~~
Direction ratios of AB are -1, Direction ratios of AC are
!,
0 .
6 3 -1 2
15' 15'
0
. . ratios . 0 f AC a r _-1 0.~Direction e..
flO' , flO
cos CAB = (
sin CAB =
Ts )(~ )+
0 +0 =
)so
.:i.-EO
:. Area of triangle ABC = 1. AB.AC sin CAB = 1.. 15 2
2
6
.JlO ._7_ = 2. 6 J50 72
JA.dr=~x-2..=_1 [from(2)]
(' Ex.7.12.13Evaluate
7 72
If(curl A).n ds
36
taken over the portion s of the surface
s
X2+/+Z2_2/x+ft=0
above
the
xy
plane
z
0,
if
A = ~::CX2 + /-z2)i and verity Stoke's theorem. Let'S' denote the portion of the surface, x 2 + y2 + Z2 - 2ft + ft = 0 above the xy-plane z= O. The surface S meets the xy-plane in the circle 'C', whose equations are
Solution:
2
x + / - 2ft = 0, Z
= O.
Engineering Mathematics - I
614
:. The parametric equations of 'C' can be taken as
x == j + j cosS,
y== jsinS,
(0~Scoscj> - ~sin28sin2cj>coscj»e.] 11. If/= pzsin28, find grad/in cylindrical coordinates [Ans: (zsin28)e p + (2xcos20)e a + (psin28)e z] 12. If A = (rcosO}er -(-;,sin 0
)eo + re4>, find curl A in spherical coordinates [Ans: (cot8)e r + 2eo - (sin8)e.]
13. Verify Green's theorem in the plane for J(x 3 c boundary of the square bounded by 0 ~ x
~
-l )tx + (x 2 -
2xy}ty, where C is the
I, 0 ~ Y ~ I [Ans: 1]
14. Using Gauss' divergence theorem, prove that
J
JF.nds = 1tr2[2 , where s
F = (y2 Z} + (xz)j + (Z2 ~ , and S is the surface bounded by
x2 + y2 = r2 , z =0, and
z=l 15. Show that the Stoke's theorem, when restricted to the xy plane, is Green's theorem in the plane [Hint: In Stoke's theorem, take A = Pi + Qj; n = k; and ds = dx dy]
"This page is Intentionally Left Blank"
8 Laplace Transforms 8.1
Laplace Transforms Introd uction The theory of Laplace transform is an essential part ofthe mathematical background required by engineers, physists and mathematicians. It gives an easy and effective means for solving certain types of differential and integral equations. It is the foundation of the modern form of operational calculus, which was originated in an attempt to justify certain operational methods used by an electrical engineer Oliver Heavinide, in the latter part of the 19th century for solving equations in electromagnetic theory. The Laplace transform reduces the problem of solving a differential equation to an algebraic problem. It is particularly useful for solving problems where the mechanical or electrical driving force has discontinuties, acts for a short time only or is periodic but not merely a sine or cosine function.
8.1.1 A class of linear transformations of particular importance is that of the integral transforms. The Laplace transformation (L.T) is a special case of an integral transformation and is defined h~' cO
F(s) = L [f{t)1
=
Je-stf(t) dt o
624
Def.
Engineering Mathematics - I
Piecewise continuity ofa function: A function off(t) is sectionally continuous (Piecewise continuous) in an interval o~ t ~ b iff(t) is continuous over every finite interval 0 ~ t ~ b except possibly at a finite number of points, where there are jump discontinuties and at which the function approaches different limits from left and right.
f(t)
a
Sectionally continuous function Def.
Functions of exponential Order:A function f(t) is said to be of exponential order cr, as t
3/f(t)/ < Meat, or equivalently, le--crtf(t)1 ~ M,
\;f
\;f
~
00
if 3 M > 0
t~0
.... (1)
t ~ o.
Remark: eat and sinbt are of exponential order, while e
t2
is not of exponential order since
which can be made larger than any given constant by increasing 't' indefinitely.
8.1.2
Conditions for L. T to exist By using above remarks we can prove the conditions that are sufficient for the existence of Laplace transforms.
Theorem: Iff(t) is piecewise continuous in every finite integral [0, N] for N > 0, and of exponential order cr,that is If(t)1 ~ Meat, \;f t ~ 0 and M> o. (from I) then the Laplace transform ofF(t) exists for all s> 00
00
00
o
0
0
0'
Proof: LetI = /F(s)1 = Je-stf(t)dt ~ ~ f(t)/ e-stdt ~ JM eate-stdt
=-=M
s
cr
Here, s > 0' since I is non-negative everywhere. This shows that above conditions are sufficient for the existence of L [f(t)].
625
Laplace Transforms
Properties of Laplace Transforms
8.1.3
Linearity property: Let f,(t) and fit) be two functions on [0, co] such that the Laplace transforms L[f,(t)] and Lltit)] exist. If k, and k2 are two constants, then L[k/,(t) + klit)]
=
k,L[f,(t)J + k2 L[fit)J
..... (i)
This property is valid since 00
00
00
=
f[ktf; (t) + k 2 f2(t)]e-'1 dt fk\.!; (t)e-'I elt + fk2.f; 0 0 0
(t)e -,I lit
= k\L[j; (I)] + k 2 L[f2(t)] ~
This result can be extended to the linear combinations of more than two functions.
8.1.4
Iff(t) is piecewise continuous on [0, N] for each N > 0, and ttt) is of exponential order 0", that is by 8.10.1 (1), If(t)1 ::s Meat, V t ~ 0 and M > 0 and ifL [f(t)] = F(s) then we have (1) L [eatf(t)]
,
= F(s-a) V s > 0" + a va> 0
(2) L[f (tn.
=
e-aSF(s)
where f,(t)
=
{
and (3) L (ttat»
..... (i)
f(t -a) t > a} 0 t < a (a>O)
=
±F(~)
(for any a> 0)
_a_
Proof(l)
--~------~----~~s
From definition, we have 00
L[f(t)]
= fe-Mf(t) dt = F(s) o 00
so that
L[eatf(t)]
= fe-ot[eal f(t)dt o
00
= fe-(s-a)1 f(t)dt = F(s - a) 0
626
Engineering Mathematics - I
Th is property shows sh ifting on the s-axis and is called the first shifting property. This means that replacing s by s - a in the Laplace transform corresponds to multiplying the original function by eat a
00
(2)
00
L(fl(t) = fe-Ilf; (t)dt = fe-'If; (t)dt + fe-sl.t;(t)dt o 0 a a
00
fe-.II.Odt+ fe-lIf(/-a) dt
=
o
from definition offl(t)
a
00
00
= fe-·llf(t-a)dt== fe-I(p+a)f(p)dp o
(where t = p+a)
0 00
= e -a.1 fe -P' f(p )dp == e -a.1 F(s) o
(3)
This property indicates shifting on the t-axis and is called the second shifting property d L[f(at)] = fe-'I f(at)dt == fe-·I(p/a) f(p)-'£ [taking t = pia]
=
00
00
o
o
a
~ }e-IP/af(p)dP==~F(~)
a0 a a This property is known as change of scale property.
8.1.5
Laplace Transforms of standard functions
(a)
L (e-at )
Pro()f:
we have L(e-at ) =
I
= --
(s> a)
s+a
.
rJe o
-.1/
_ .e -aidt-
rJe
-(Ha)1
d _ - 1 [ -(Ha)I]OO t--- e 0
_
1 s
Ifa=OtheL(I)= -
1 s+a
---
s+a
0
...... ( I)
(s> 0)
..... (2)
Also changing the sign of , a' we get L(eat )
(b)
Pro()f:
I
= --
..... (3)
s-a
L(sinat) =
a 2
S
+a
2
and L(cosat) =
s J
s- +a
2
1 s+ia L(cosat+isinat) = L(e lat ) = --.- ==-)-S -lU s- + a 2
(s> 0) from (a)
627
Laplace Transforms
Equating real and imaginary parts we get L(sinat) =
a s 2 ,L(cosat) = 2 2 +a s +a
2 S
(c)
Proof:
L(sinhat) =
2
S
.
a 2 and L(coshat) = -a
(e _eOI
L(smhat)=L
s 2
S
I]
from (a)
s+a
a
Thus (sinhat) =
2
S
-a
2
similarly we can show that L(coshat) = L(tn) =
I(n + 1)
----;+]
s
laD
1
-----
2 s-a
(d)
(s>
2 -(I
1 01 1 -01 =-L(e )--L(e ) 222
1[1
- -
ol
..... (4)
s s
2
2
•••.
(5)
-(I
[(n + I) > 0 and s> 0]
Proof:
OOfe-P.pnd =T(n+l) o
S
'P
n+1
S
11+1
00
Since T(n+I)= fe-xxndx
and
T(n+ I)
=
n!
ifn is a +ve integer
o
n!
:. L(tn) = ----;;-+J s From (6) we have
1 L(I)= -, s
L(t) =
..... (6)
1 -2 ,
s
Solved Examples
8.1.6
Find the Laplace transform of 1. 2t3 + cos4t + e-2t 2. e2t + 4t3 - sin2tcos3t 3. cos(wt + ~)
4. 3t2 + e- t + sin 32t 5. cos32t
Engineering Mathematics - I
628
Sol: I.
2.
6 s 1 12 s 1 + - - =-+ +-L(2t3 + cos4t + e-2t ) = 2x-+ S4 S2 + 16 S + 2 S4 S2 + 16 S + 2 L(e2t ) + 4L(t3 ) - L(sin2tcos3t)
I
6 (5
1)
S - 2 + 4x7 - S2 + 25 - n 2 + 1 [.: sin2tcos3t = 3.
1
"2 (sin5t -
sint)]
L cos(wt + 13) = L[coswt.cosj3 - sinwtsinj3]
=cosfJL(coswt)-sinfJL(sinwt)= cos fJ 4.
S
2
+w L[3t2+e-t+sin 32t] = 3L(t2 ) + L(e-t ) + L (sin 32t)
2
.
-sm
fJ
S
2
S
2 1 3 3 =3.-+--+ - -?-S3 s+1 2(S2+4) 2s-+36 2
1
3
1
3
1
=-+--+-------S3 S + 1 2 S2 + 4 2 S2 + 36 1
4" L[3sin2t -
(.: L(sin32t) = 5.
sin6t]
L(cos 2t) = L
[cos6t + 3cos2t] 4
1
(s 3S) S2 + 36 + S2 + 4
3
=
4"
Exercise 8(A) Find the Laplace transform of 1.
4t2 + sin3t + e-2t
8 3 1 Ans. - + - - + - S3 S2 +9 s+2
2.
(sin2t - cos2tf
Ans. S
3.
4. 5.
1
4
S2 + 16
Ans. !(!+_S_) 2 S S2 -64
w +w
2
629
Laplace Transforms
6.
Ans. - - + ? ? 2 s s- + 16b-
162 Ans. -----::,---------:--(S2 -81)(.~·2 -9)
7.
8.
sinhat- sinbt
An s.
8.1.7
b
a -?::-------::-)
s- -(r
S(S2 - 28) Ans. -----:----::--2 (S2 - 36)(5 - 4)
9.
10.
s)
1(1
cos 2 (2bt)
coshat - cosbt
Ans.
s 2 S
s 2
-(I
Examples on properties 8.1.4 Find the Laplace transform of
2. t 3 e-4t
1. e-btcosat 4. e-3t sin3t
5. e-2t[2cos3t-3sin3t]
Sol. S
1.
L( cosat)
=)
s- +0
2
(from 8.1.4( I) first shifting theorem)
2.
6 L(t3 ) = 4
s
By shifting theorem 8.1.4(1) 6 L(e-4t. t 3) = - - (s
3.
+ 4)4
. 3 L(sm3t) = -2-S +9 by 8.1.4(1), 2t
L( e sin3t)
=
(5 _
3 2)2 + 9.
Engil~eering
630
4.
Mathematics - I
e-3t .sin3t
[~(3Sin 1 - Sin3/)] = ~ L(sin I) - ~ L(sin 3/) 4 4 4
L(sin3t) =- L 3
1
1
3
By 8.1.4(1) -3t .
L(e
5.
-
~
sm3t) - 4·
(s+
I 3)2
+I
e-2 t(2cos3t - 3sin3t) s L(2cos3t) = 2.-2s +9
9
~9
L(3sin3t) =
8
+
By shifting theorem 8.1.4( I)
2L(e-2tcos3t) - 3L(e-2t sin3t) 2(s + 2) = (s
8.1.8 SoL
9 + 2)2 + 9 - (8 + 2)2 + 9 3s
If L[f(t)] = 'L[f(t)]
~9 . 8
+
find L[f(3t)]
38 +9
= S2
s
I By 8.1.4 (3), L[f(3t)]
=
3")
38
3(8)2 - +9 3
8
2
+27
Exercise 8(8) Find the Laplace Transform of 1.
Ans. 4cosa
2
8-3 +sin-----,:----(s - 3)2 + 16 (8 - 3)2 + 16
2. 3.
cosh2t.cos2t
Ans
~[ 8-2 + __8_+_2_] . 2 (8 - 2)2 + 4 (8 + 2)2 + 4
Laplace Transforms
4.
631
coshat sinat
Ans.
a(s2 + 2( 2) S4
+ 2a 4
S(S2 -
5.
cos3t.cosh4t
Ans. S4
6.
e- 21(2cos3t-3sin3t)
7)
+ 625 -14s 2
2s-5
Ans. S2
+4s+ 13 S3
7.
Ans.
coshat cosat
S4
8.
elsin4t.cos2t
Ans.
Ans.
Ans.
8.1.9
+ 4a 4
3 (s _1)2 +36
5s 2 -3s+2 S3
6 n+ 1 (8 + a)"+1 + (.\'2 + 1)(s2 + a)
Property (4) : Effect of multiplication by t If L[f(t)l = F(s) then L[tf(t)] = -FI(s) 00
Proof:
We know that F(s) = Je-"f(t)dt. o differentiating with respect to s 00
a
00
F(s)= J-[e-"f(t)]dt= Je-"'f(t)dt o 0
as
00
:. -FI(s)
=
Je-" [if(t)]dl = L[if(t)] o
8.1.10 Examples I.
d 1 _ 1 - -3- L(Ie -2')_ --------_o_ L( e-2')_ s+2' ds s + 2 (s + 2)2
2,
d d ( 2 2S2 L(lcosat)=(-I)-[L(cosal)]=-2 = 2 2 2 ds ds s + a (s + a )
3.
L(t sin I) = __d d..
s)
(_I_) = S2
+1
---,-2_s---,(S2 + 1)2
632
4.
Engineering Mathematics - I
a
L(atsinat-cosat)=aL(tsinat)-L(COsat)=a[-dd ( 2S S +a 2as]
s = a [ (S2 +a2f~ - (S2 +a 2)
5.
2)]- s 2+a 2 S
2 2a s s = (S2 +a 2 )2 - (S2 +l?)
L(1+tet)3 = L[I+3et+3t2e2t+t3e3t] = L(I) + 3L(tet)+ 3L(t2e 2t) + L(t3e3t) 2 3 d 1 d 1 d ( 1 ) '1 3 6 6 =~-3 ds (s-I) +3 ds 2 (s-2)ds 3 s-2 =2+ (s-1)2 + (S-2)3 + (s-3)4 1
6.
7.
d
d [
=-
41'
d [ 2 ds S2 -85+20
J
] by shifting property
45 -16
= (S2 -8s+20)2
2
8.
2
L(te4t sin2t) = -L[e sm 2/1 = - 2 ds ds (s-4) +4
2
l d d [ s -1 ] L(t2etcos4t) = ds2 L(e cos4/) = d~2 (s _1)2 + 16 by shifting theorem
=_~[
S2-2S-15]= ds (S2 - 2s + 17)2
3 2 (2s -6s -90s+94) (S2 - 2s + 17)3
Exercise 8(C) Find the Laplace transform of the following:
I
H~
Ie-"'sjn2.
Ans.
2.
3t te- sin3t
6s-18 Ans. (S2 _ 6s + 18)2
3.
Ans,
(s
(S2
3)' -
(X~~::15;),]
1.
2s+4 + 4s + 5)2
633
Laplace Transforms
4.
2tsin3t-3tcos3t
Ans.
5.
tetsin2tcost
Ans.
8.1.11
Property (5) If L[f(t)] = f(s), then L
9+ 12s -3s 2 2 2 (s +9) 35-3 2
?
(s - 2s + 10)-
U.f(t)]
=
J/(5) ds ,
00
Proof:
f(s)= fe-'I/(t)dt Integrating bothsides w.r.t 's' between the limits s, integration on right hand side (R.H.S),
'1Je-.\II(t)dS]dt = J-[e~\1 l(t)] "tL
J/(S)ds
=
0
8.1.12
dt
0
and changing the order of
= Je-\I
I~t) dr
0
Examples Find the Laplace Transforms of the following
1.
~(1-cosal)
2.
I
Sol.1.
00
00
~(e-I sinal) I
1
L(I-cosat) = L(l) - L(cosat) = - S
(
)K
s 2 S
+a
2
] [
]
[-1
-1
l I s 1 2 L -(I-cosat) = - - 2 2 ds= logs--log(s2+a) t 2 , s s +a
2.
L(e-tsinat) =
00
0
(2? 1
1 s + a=-Iog 2 2 s
a 2
(s+l) +a
2
I ISlOat) . ] a2 (S+I)]OO 1t s+1 _I(S+I) L -(e= oof 2ds= tan - =--tan --=cot -[t s (s + 1) + a a ,2 a a
Engineering Mathematics - I
634
3.
L(1-e2 t)
=
L(I) - L(e2t)
1 1 s s-2
= ---
X
21] I I)ds= [logs-log(s-2)Js100 =Iog-(s - 2) I L[-(1-e ) = ---t s s-2 s s 4.
L(e-3tsin2t) =
L(!et
3/
Sin2t)= } s
1t =-=tan 2
5.
L(eal )
2 (S+3)2 +4
---::---
_1(S+3) - - =cot _1(S+3) -2
[
2
1 s-a
I L(e ht )= - -
= --,
s-b
1 al -e bl] L [ -(e ) = t
=
22 ds=tan- I (S+3)00 (s + 3) + 4 2 s
s-a log( - )] s-b
00
s
XI s
100 - - - -1) - ds= [log(s-a)-Iog(s-bh,. s-a s-b
s-a s-b ) ( =-IOg(-)=IOgs-b s-a
Exercise 8(0) Find the Laplace transfonn ofthe following:
1 (hI]
1.
1 -(1-cost) t
Ans. 2"log - s -
2.
I _I • -e SlOt t
Ans. coc 1 (s+l)
3.
1 - (-at e - e-hI) t
Ans.log C+b) -s+a
4.
!
(e-3tsin2t)
S-3) Ans. coC I ( -2-
5.
!
(sin2t)
Ans.
t t
1 (.'+4]
"4 log
~
635
Laplace Transforms
6.
(S2+b 2)
I - (coat - cosbt)
I Ans. -log
2
t
2
S
+(1
2
(S2 +9) --2-
I - (I - cos3t) t
I Ans. -log 2
8.
!
Ans.
8.2.1
Theorem: Iff(t) is continous ¥ t ~ 0 and of exponential order, say 0" and has a derivative ~(t) which is piecewise continuous on every finite interval [0, N] for each N > 0, then the Lapalce transform of the derivative ~(t) exists for s > 0" and
7.
(sin3t)
t
S
_) 4"I [3cot
S-
cot
_)
"3s]
L{~(t)} = sLf(t) - f(O) 00
L(f(t)]
Proof:
=
je-'I f(t)dt o 00
L(~(t)]
=
je-'I fl(t)dt
o
Integration by parts gives
L(~(t)] =
00
[e-.'I f(1)1 + s je-'I f(t)dt
o Since f(t) is of an exponential order, the integrand in first integral on the R.H.S. is zero at the upper limit when s> 0' and iff(O) at the lower limit. Thus, we have 00
L( ~(t)]
= 0 - f(O) + S
je
-;t
f(t)dt =' sLf(t) - f(O)
o
If Lt f(t)
= f(O+)
exists, but is not equal to f(O), which mayor may not exist then
1-)0
L(~(t)]
Note:
= sLf(t) - f(O+).
Iff(t) and its derivatives ~(t), ~I(t) .............. fI- 1 (t) are continous ¥ t ~ 0 and are functions of exponential order for some M > 0 and 0', and the derivative fI(t) is Piecewise continous on every finite interval in the range t ~ 0, then the Laplace transform offl(t) exists when s> 0' and is given by L[fI(t)] = s"[Lf(t) - s" - If(O) - s" - 2 ~(O) ............... _tn - 1(0)] If
n = 2, L [~I(t)]
=
s2[L(f(t»] - sf(O) -~(O) and so on.
636
8.2.2
Engineering Mathematics - I
Example Prove that L(cosat) =
Sol.
S2
+ a2
We can show this with the help of above theorem Let f(t) = cosat => f(O) = 1 then [I(t) [l1(t)
= -asint => fl(O) = 0
= -a2cosat
Hence, we have L[fll(t)]
= s2Lf(t) - sf(O) - fl(O)
:. L[-a 2cosat]
= s2L(cosat) - s - 0 2 s2L(cosat) + a L(cosat) = s
:. L( cosat)
=
s2L(cosat) - s
s
=
2
2
+a Theorem: Iff is Piecewise continous on [0, N] for each N > 0 and is of exponential S
8.2.3
I I order cr, then L ff(u)du = -L[/(/)] v s > CT, o s I
Proof.
Let «p(t)
= ff(u)du. Then we have «p1(t) = f(t) and f(O) = 0; o
L[~(t)l =s.L[~(t)l-~(O) =s[L(~(t»l =SL[!f(U)du1 :. L Iff(u)du [ o 8.2.4
III
= -L[¢I(/)] = -[L(f(/»] s
s
Examples
Find the Laplace transforms of I
I. fsin 2pdp o
Sol.
I
2. fp cosh p dp o
2 1. L(sin2t) = Z--4 = f(s) , say s + 112 fsin 2pdp =- f(s) = --=-2-o S s(s +4)
r
I
3.
I
•
mp
o p
dp
rp
.
sm P
4. je . - - d'P o P
637
Laplace Transforms
2. 2
I
)
I (s + (1-) then L pcosh pdp = - I(s) =- ) ) o 8 (8- -a-)s
J
3.
. L(smt)=
I ~I ;
s +
L
(I .
-SlOt
)
t
J-2s +1
00
=
I
ds
0\
p dOp=-Ijo() I -I L 'JSin -s =-cot s o P s s
4.
I L(sint)= -2-1 s +
I ... L( etsint) = --.,,---(s _1)2 + I
ds2 I L [ -ee' sint) ] = ooJ t (s -I) +1 ,
'f
sin p 1 1 1 :. L e P --dp = - I(s) = -cot- (s -1) o P s s
Exercise 8(E) Find the Laplace Transform of I
I
I. Jcos2pdp
3 2. Jpe- sin 4 pdp
0
Ans. I.
0
s S(S2
+ 4)
2.
0
8(s + 3) S(S2
I
Jp sin 3pdp
I
3.
+ 6s + 25)2
I _I 3. -cot s s
Engineering Mathematics - I
638
Laplace Transforms of Periodic Functions
8.2.5
A function f(t) in said to be periodic with period p>O iff(t+p) = f(t) for all t>O Since periodic appear in many practical problems and in most cases are more complicated than single sine or cosine functions, we find it useful to establish the following result: I I' L([f(t)] = I-e- Ps fe-Slf(t)dt o
Proof:
Let f(t) be a periodic function of period 'p' so that f(t) = f(t+p) = f(t+2p) = ...... . I'
00
L[f(t)] = fe -si f(t)dt o
21'
= fe -s/ f(t)dt + o
fe -si f(t)dt + ............... . I'
writing t = u+p 21'
I'
now fe-ollf(t)dt= fe-o,(u+p)f(u+ p)du 0
I'
I'
= e- Ps fe-OVlf(t + p)dt
(change u to t)
o I'
f -vlfCt )dt = e -1" Je
[.: f(t+p)=f(t)]
0
o 31'
I'
similarly, fe-oIl f(t)dt
= e- 21'S fe-ollf(t)dt
and so on
0
21'
I'
L[f(t)] = (1+e-ps+e-2ps + .................... 00) fe-ollf(t) o The expression within parantheses is a G.P. with c.r. = e-PS :. L[f(t)] = 1_:-1'$ Je-Slf(t)dt o
639
Laplace Transforms
Solved Problems
8.2.6
Find the Laplace Transform of
2.
t - for 0 < t < P and f(t) = f(t+p) P f(t) = 1 for 0 < t < a and f(t) = -1, a < t < 2a and f(t) is periodic with period 2a
Sol.
Since f(t) is periodic with period p.
l.
f(t)
=k
L[f(l)] =
2.
P
P
1. Je-s1f(l)dt=
l-e- sp
o
1
l-e-"P
k
Je-"I.~dt
0
p
f(t) is peirodic with peirod 2a
L[f(t)] =
1 2a si 1 [a 2a. Je- f(t)dt = Je-sl.l.dt + Je-·'I (-I)dt 1 _ e- 2as 1- e-2as 0 0 0
_ 1 [{_e-SI}a -1_e-2a, - s - 0
-
1
SI {_e- }2aj_l 1 (l -a,)2 - s - 0 - s l_e-2as ' -e
1-
]_ 1[/f - e-'T --tanh 1 [-as ] - 1s [11+- ee -2as --. -2as s 2
--.
S
8.2.7
~
e 2 +e
-~ 2
Find the Laplace transform of the rectangular wave shown in the figure: f(t)
1s
o
b 2b
3b
Rectangular wave
640
Engineering Mathematics - I
Sol:
The period of the given function is '2b'. I
Hence L[f(t)]=
2h. -2h, Je-Mj(/)dt=
l-e
hs
-
e 2 -e
0
I
[ h .S1 2b -2h.\ J1.e- j(t)dt+ J(-I)e-S1dt
l-e
0
1
b
-bs
-
2
I bs =-tanhs 2
8.2.8
What is the Laplace transform of the staircase function (fig(a)) given by f( t) = n+ I, np < t « n+ 1)p for n = 0, 1, 2 ............ .
Sol.
The easient way of getting the Laplace transform of the staircase function is to consider it as the difference between the two functions shown in Fig(b) the transform ofthe linear function (t + p) can easily be found by using first shifting, P second shifting and change of scale properties. Thus, we have
f(t)
o
f(t)
P 2p
3p
(a) Staircse function (b) solution of the stair case function (c) Saw-tooth function.
Laplace Transforms
641
t The function fl(t) = - is called a saw-tooth function (fig(c» and its transform is p obtained as shown below.
L[ft(t)] =
1 P 1 1 [ -,I ]P _, fe-'I~d'= _.- _e_(_st_1) 1- e Ip p 1- e "P p S2 o
0
1- (1 + ps)e - P' ps2(1-e-"P) This can be simplified to
L[ I' (I)] JI
-
~[1 + ps P S2
1
p s(1-e- PS )
Thus from (1) and (2), we get
L[f(t)]=~(~+P)_~(I+PS_ p- 1 p s- s P S2 s(1-e P") 1
Exercise 8(F) 1.
If Lf(t) = (s), Prove that L[~II(t)] = s3(s) - s2 (s) - sl(s) - 1I(s)
L[~I(t)] = Tan-I (+), f(O) =,2 and ~(O) = -1 find L[f(t)]
2.
If
3.
A periodic function f(t) of period 2a is defind by f(t) = t, = 2a - t,
for 0 .'S t .'S a, for a < t < 2a
7
show that L[f(t)] = 1 tanh
(as) 2
..... (2)
642
Engineering Mathematics - I
Iff(t) = t2,
4.
°
< t < 2 and f(t+2)
=
f(t)
find L[f(t)].
5.
Iff(s) = L(f(t»), Prove that
F~) and hence generalise the result.
L[Sdl SJ'(U)duj == o 0 s
8.2.9
Laplace transform of the unit step function The unit step function orthe Heaviside function is defined as H(t-a) =
°
H(t-a)
lift < a
=
iftO
The unit step function is the basic building block of certain functions whose knowledge greately increases the usefulness of the Laplace transformation. It is easy to show that the Laplace transform of H(t-a) is given by e- U \ L(H(t-a»==s a
00
00
L[H(t - a) == fe-vI H(t - a)dt = fe-VI O.dl + fe-·\/l.d' o 0 a 00
-e -sl s o
e-as s
8.2.10
. 1 In Particular, when a=O, we get L(H(t» = s Heaviside shift theorem: If Lf(t) = f(s), then L[f(t-a)H(t-a) = e-as f(s).
Proof:
By defnition, we have a
00
00
L[f(t - a)H(t - a») == fe- si f(t - a)H(t - a)dt == fe- si f(t - a)O.dt + fe- si f(t - a)l.dt o 0 0 00
00
== fe -s(p+a) f(p )dp == e -sa fe -sp (p )dp o 0
(consider p = t-a)
643
Laplace Transforms
8.2.1
Examples
Find the L.T. of the function f(t)
1.
=1 =0
} if 0 < t < 1t if1tO) then fl(t) = fit) (v- t::::O) Some of the important properties of the inverse laplace transforms are analogous to the corresponding properties of Laplace transforms.
1.
2.
=
Linearity Property: If kl' k2 are any constants and F I(s) and Fis) are the Laplace transforms of fl(t) and fit) respectives, then L-I[kIFI(s) + k2Fis)] = klfl(t) + k/i t). First shifiing property: If L- I [F(s)] = f(t), then
jor t > a for t < a 3.
Change of scale property: if L-' [F(s)] = f(t), then
4.
Inverse Laplace transforms of derivatives: If L-'[F(s)] = f(t) then
5.
Inverse Laplace transfonn of integrals: If L-'[F(s)] = f(t) then
6.
Multiplication by's': If L-'[F(s)] L-[sF(s)]
7.
=
=
f(t) and f(O)
fI(t)
Division by's': If L-I [F(s)]
=
f(t) then
=
0, then
Engineering Mathematics - I
646
F(S»)
. result can be extended to L- I ( - This s" I I
=
I
ff..·················ff(u)du". 00
0
(the proof is left to the student)
8.3.2
Table of inverse transforms L- I [F(s)]
L[f(t)] L(1)
I
s
L(e-at)
L(tn )
L- I
= -
1
L- I
= --
s+a
n!
1-1
= S"+I
(~)
=
1
(_I ) s+a ( I) + I)! =
-
S'1+1
-
e-at
(n
t"
if 'a' is a positive integer L(sinat) =
L(cosat)
rl (
s =
L(sinhat)
L(coshat) 8.3.3
S2
a +a 2
=
=
I
2) =
~a sinat
L-I ( 2 S
2)
a _a 2
L_ I ( 21
2)=~sinhat a
s _a 2
L- I (
S2
+a 2
S2
S2
2
s +a
s +a
s -a
S2
=
cosat
s 2 ) coshat _a
Using the above results inverse Laplace transforms of some simple functions can b~ obtained as below. Find the inverse Laplace transform of:
2s+3
I.
3.
3 6.s+ 4
2s+ I 7. (s + 1)
• S2
I
5. 4s-5
3s+5 2 9s -25
2 --
+9
1_ 4._ 4s+5 s+3
8·~4 s +
647
Laplace Transforms
Sol.
From the above table, we get
I.
2. =
3.
I L- (
3~+5
9s- - 25
2cos3t + sin3t
)=L-I(
3s )+L-I( 2 5 9s 2 - 25 9s - 25
)=~COSh(~)+~sinh(~) 3
3
3
3
4.
5.
6.
L- (_3_) = 3rl_1_ = 3es+4 s+4
7.
-2L-I(_s ) + L-I(_l )_" . L-I 2s+1 1 1 2 - .... cost+smt (s- + 1) s- + 1 s +1
8.
L
I
-I( S+3) -I( -1--
s- + 4
=L
41
-I(
s ) +L s- + 4
-1--
3 ) =cos2t+-sm21 3 . s- + 4 2
-1--'
Exercise 8(H) Find the invere Laplace transform of:
2
1
S
S3
1
1. -+-+-S
+4
4s+ 15 5. 16s 2 _ 25
Engineering Mathematics - I
648
[2
Ans.
1.2+ - + e-4t 2
2. 3cos4t + sin4t
3. 2cosh3t+sinh3t
,4
4.1 + t 2 + 24
Examples 8.3.4 Find the inverse Laplace transform of: 1.
6s-4 --"-2- - -
s -4s+ 20
2. (s _ 3)3
3. (s _ 2)4
4.
s-2 + -=-2- - (s-l) s -4s+5 I
2
Sol.
4 L- I
(
.
1 ) + L-I (s - 2) (S-2)4 (s2-4s+5)
-I(
=L
-l(
1 ') ) + L (s-2t
= L-I(
S-2) (S-2)2 + 12
=
I ) + L- I ( (s - 2) ) (s-2i s2-4s+4+1
') et.t + e~t.cost
Exercise 8(1) Find the inverse Laplace transform of:
s+4 S2 -4s+ 13
(i)----
(ii)
s +6 S2 -Ss +25
(iii)
4s+ 12 S2 +Ss + 16
-=----
649
Laplace Transforms
B.3.5
Method of partial fractions: Whenever possible we exress the given function F(s) into the sum of linear or quadratic partial fractions as F(s) =
A Bs+C r + ? 2 r and then use the standard results to find the inverse L.T. (s + a) (s- + a )
Examples (I) Find Sol.
4] for all s>O, s +s
r l [ S3 +
By partial fraction expansion, we have s+ 1 A Bs+C ---=-+---
S(S2 + I)
S
S2
+1
s+1 = A(s2+1) + (Bs+c)s = As2 + A + Bs2 + cs comparing coefficients of different powers of s from both sides O=A+B )=
c
:. I = A, B =-1 .
s+1 1 -s+1 =-+-+ I) S S2 + 1
.. S(S2
By the linearity property of L-I, we have
- - = L-l(I) - - L-1(-s- ) + L-1(-I- ) L-l(S+l) S3 + s ' S S2 + 1 S2 + I
. =I -cost+smt
2S2 -6s+5
2.
If L[f(t)] =
Sol.
By partial fractions we have
S3
+ 6s2 + lIs -6 find [f(t)]
J(s) = S3
2S2 -6s+5 =_A_+_B_+_C_ -6s 2 +Ils-6 (s-l) (s-2) (s-3)
:. 2S2 -6s + 5 = A(s -2)(s -3)+ B(s -l)(s -3)+C(s -l)(s -2) put s = I we get A= 1/2
I =2A put s
=
B=-1
2
650
Engineering Mathematics - I
Similarly s = 3 gives 5/2
C
=
. ..
L-I[f'( S.)]_ I L- I - I - L- I ( -I- ) +5 L- I ( -A- )_ 1 1 -e 21 +-e 5 31 ---e
.
2
(s-I)
8-2
2
8-3
2
2
3s+ I )( 2 I ' find f(t) s-I s + )
3.
If Lf(t) = (
Sol.
By partial fractions we have
j(s)=~+ Bs+C
s -1 S2 + I :. Solving for A, B C
we have A = 2, B = -2, C = -I L- I [j(s)] = 2C l _1__ 2CI _s_ + L- I _)_1_ S -1 S2 + I s- + I =
4.
If L[f(t)]
Sol.
j(s) =
2e t - 2cost + sint
s+2 (s + 3)(s + 1)3 find f(t)
=
s+2 (s+3)(s+I)3
=~_I__ ~_I_+~ I +1 1 8s+3 8s+1 4(8+1)2 2(8+1)3
(after expressing the function in partial fractions)
:.rl[f(8)]=~rl-I--~rl-I-+ 1 8
8+38
?
8+14(8+1)-
+ I 1 2(8+1)3
=~e-31 - ~e-I -t ~/e-I +t 2e-1 = ~[(2t2 + 21 -1)e-' + e-31 ] 8
8
5.
If L[f(t)]
Sol:
Let f(s) :.S2
=
=
4
(S2
8
(S2 + 2s - 42) + 28 + 5)(8 2 + 2s + 2) find f{t)
as+b 82 + 28 + 5 +
S2
c8+d + 28 + 2
+ 28 -4 =(a8 +b)(8 2 + 28 + 2)+ (cs + d)(S2 + 2s + 5)
Equating the coefficients of like powers of s we get a = 0, b = 3, c = 0, d = -2.
Laplace Transforms
651
3 :·f(s)= s2+2s+5
-I(
:.L
f(s)]=3L
-I(
2 s2+2s+2
-I[
12 ? ) -2L (s+l) +2-
If L[f(t)]
Sol.
By change of scale property we have
-I[
L
2 (s+I)2+12
12 ? (s+l) +1-
1
2s 4s2 + 16 ' find f(t)
6.
=
3 (s+I)2+22
since L- I [
2s
] =-cos 1 21 4s +16 2 2
2 S
=
]
s +16
cos4t
e- 5• 7.
If L[f(t)]
Sol.
We know
=
(s _ 2)4 find f(t)
rl[
1
(s - 2)4
1 L-I[_I] = e 21
S4
=
e2/.~3! = ~t3e21 6
Thus by second shifting property
L- I [
e -5s (S-2)4
1={Ig
-(I - 5)3e2(t - 5)
ift>5 if 1 < 5
Exercise 8(J) Determine the inverse Laplace transform of each of the following functions. (i)
(v)
3 s+2 s-5 S2 +6s+ 13
(ii)
s s2+2s+6
(iii)
(vi)
e -2 .. s(s + 1)
(vii)
2s+ 1 s(s + 1) S2 (S2 + 9)(S2 + 16)
(iv)
10g(~) s+2
Engineering Mathematics - I
652
2S2 + s -I 0
(viii)
-(ix) log (I + Sl2 )
(s - 4)(S2 + 2s + 2)
(xii)
(xi)
s(s + I)(s + 2)(s + 3) 2
8
(xiii) 8
+ lOs + 13
-I 0(8
2
-
(xiv)
5s + 6)
8+29
(xv)
(xvi)
(S+4)(S2 +9)
(xvii)
48+4
(8
4
+ 64)
s (s4+s2+1) s
2
(8 + 1)(s2 +4)
5s+3 (s -1)(s2 + 2s + 5) S2 + 1
(xviii)
(8 _1)2(8 + 2)
(x)
s3
+3s 2 +2s
. ) -1 (-2/ e -e -3/) ( IV t
Ans.
(v) e-3tcos2t - 4e-3tsin2t
(vi) 4(1 - e-2t)(t - 2)
(viii) e4t - e-t(cos2t+2sint)
2 (ix) -(I-cost)
t
. I I _/ 1 -2/ 1 -3/ -.l--e +-e --e 6 2 2 6
(XI)
(xiii) 12et - 37e2t + 25e3t
(xviii)
.! - 22
et
+
~ 2
e-2l
(x)
(xiv)
(vii)
I
"8
(~Sin4t -%Sin3t)
sin2t sinh2t
. J32(. TJ5 t.sln'ht]2"
(XII)
"3I
Sin
(cost - cos2t)
(xv) e--4t - cos3t +
"35
sin3t
Laplace Transforms
8.4
653
Convolution Theorem If L[f(t)] = F(s) and L[g(t)] = G(s) the inverse transfonn of F(s). G(s) denoted by H(t), is called the convolution of f and g is defind as given below. I
H(t) = (f*g)(t) = f(t-u)g(u)dll,vtzO o
..... ( I)
The convolution operation * has the following properties. (f*g)(t) = (g*f)*t),
8.4.1
v t
~
0 (commutative)
[f*(g*h)](t)
=
[(f*g)*h](t)
v- t
~
0 (associative)
[1'*(g+h)](t)
=
(f*g)(t) + (f*h)(t)
v- t
~
0 (dstributive)
Theorem: Let f(t) and get) be the inverse transforms of F(s) and G(s) respectively, and satify the hypothesis of the existence theorem (8.10.2) then the the inverse transform h(t) of the product F(s) G(s) is the convolution of f(t) and get). I
h(t) = (f*g)(t) = fl(t-u)g(ll)du o I
i.e.
L-'[F(s)G(s)]
=
fl(t - u)g(u)du
..... (i)
o
Proof:
It is sufficient if we show that F(s) G(s)
Now F(s)G(s)
~
L[
1f
(l- u)g(u)du
~ [fe" f( v)dv] =
1
[V·
g(u)du ]
rrllo
Ile-(U+V)S J(v)g(u)dvdu = le-S(V+U) J(V)dV]g(U)dU' 00
where the integration is extended over the first quadrant (u ~ 0, v .:=:: 0) in the uv-plane. We now introduce a new variable 't' in the inner integral of the last expression by taking v = t-u so that u + v = t and dv = dt (u being fixed during this integration). Thus
F(s)G(s) =
rrllule->t
J(t - U)dt]g(U)dU =
Ile->t J(t -u)g(u)dtdu 0
U
Engineering Mathematics - I
654
al
a
s, I
F(s)G(s)= fTIe-SI!(t-u)g(u)du]dt= fe- jlr{t-u)g(u)du}tt 00
0
0
8.4.2 Examples (I)
Evaluate
~ r' [ s2(s1+ I )2l using the convolution theorem
Sol.
we have
L-'C, )~I
and
r'((S~I)' )~/e-'
By the convolution theorem =L-
I
[
2(
2
U
U
1 y]= J(ue- -{t-u}du= J(ut-u )e- )du s s+1 0 0
= [(ut -u2)e-u - (t-2u)e-u + (-2)(-e-U) = t-e-t + 2e- 1 + t-2 (2)
~valuate L-
Sol.
Let F(s) =
I
[
2 S 2 2] using the convolution theorem.
(s +2 )
s s ' G(s) = - 2 - and Let s + s +4
~4
_I s f(t) = L-I [F(s)] = L (-2-) = cos2t s +4
1 g(t) = L-l[G(s)] = r-I (-2-_) = ! sin2t s +4 2
f(u).;: cos2u CIld g(t-u) =
2"1 sin2(t -u)
~s)
= F(s) G(s)
655
Laplace Transforms
I I C l [tp(S)] = JCOS 2u -sin 2(1 - u) du
2
o I
~ JSin 2t + sin 2(t -
=
2u )dll
40
=~[sin2t.U+~COS2(t-2U]1 ~tsin2u 4
(3)
Evaluate L-
Sol.
Let F(s) = ¢i...s)
I
[
2
~ 1using convolution theorem.
s s+4 ~
G(s) =
I
I
vs+4 F(s).G(s)
=
04
.. f(t) = L- [F(s)] =
r
(s + 4)
1
- and choose s
~
= e-4t
CI_1_ =e-41
s~
t~ = r;; r41
e-
1
2
2
1 =1 s
g(t) = L-1[G(s)] = L- 1 (-)
e --4u rand g(t-u) = 1
f(u) =
'\I1tU
By convolution theorem I
-4u
L-1[$(s)] = Jer- du o '\I 1tU
where 4u
=
.xl
=
1 r
Je-
2!i
x2
dx
'\I 1t 0
2du =xdx
J
I
(where erf(t)
= 2r e- x '\I1t 0
(erft stands for error function)
2
dx
'\I1tl
656
Engineering Mathematics - I
Exercise 8(K) Using the convolution theorem find the inverse Laplace transforms of
(v) log«s+2» S +1
(vi)
1
?
(s + 1)(s- + 1)
(ii)
(iv)
e
I o
-/I
-e u
-211
du
1
"3 (2sin2t-sint)
(iii)
1
"3 tsint
I f4 -cost-smt . ] (vi) -Le 2
8.4.3 Applications to differential equations Suppose we wish to find the particular solution of the differential equation
lay" + /3y' + yy = J(t)
..... (1)
=
which satisfiestheinitial conditionsy(O) Yo and yeO) = Yo Taking the Lapa1ce transform on bothsides of (l) and using its linearity, it follows that
aL(y") + /3L(y') + yL(y) =
L[J(t)]
but L(y") = s2L(y) -sy(O) _yl(O)
L(y') = sL(y) - yeO) a[s2L(y) - sy(O) - yl(O)] + /3[sL(y) - yeO)] + yL(y) and hence (as 2 + /3s + y)L(y) - L[f(t)] + (as + /3)Yn + ayJ
= L[f(t)]
=> L(y) = L[/(t)] + (as + /3)yo + aYb as 2 + /3s +y The function L[f(t)] is a specific function of s since f(t) is known; and since a, /3, y. }'0 and y J are constants, L(y) is completely known as a function of s. The inverse Laplace transform of L(y) will be the solution of the given diff. eqn. (Here it is assumed that the functions f( t), y, y' and y" must have Laplace transforms).
657
Laplace Transforms
Solved Examples
8.4.4 2
' d th ' 0f d2 y + -dy - 2Y = smt ' w h'IC h satls 'files t h'" , e soiutlOn e Imtla I cond'ItlOI1S y = 0, Fm dt dt Sol.
y' = 0 when t = 0 Applying the Laplace transforms to both sides of the given equation and by linearity, we have L(yW) + L(y') - 2L(y) = L(sin t)
1 As we know that L(sint) = -2-1 ;s > 0 s + [s2L(y) -sy(O) - y' (0)] + [sL(y) - yeO)] -2 L(y) Given conditions yeO) = 0, y' (0) = 0
=> s2L(y) + sL(y) -2 L(y) = L(y)=
1 ?
(s- + S - 2)(S2 + 1)
1 -2-
s +1
=-----:---
(s -IXs + 2)(S2 + 1)
we have the partial fraction expansion
1 (s -1)(s + 2)(S2 + I)
A B C . +D s -I s + 2 S2 + I
- - - - - - - : - - = - - + - - + -":--
where
A
=.! B =-=! ,C =-=! and D = - 3 6'
15
10
10
Therefore we get •
1 1 yet) = L- 1 - - - - - - 1 (s -1)(s + 2)(S2 + 1)
as the solution to the given initial value problem,
1
= -2-1 s +
Engineering Mathematics - I
658
8.4.5 Example Solve the initial value problem y"-3y'+2y=e 31 .y=l,y'=0 wheret=O Sol.
Applying Laplace transform to both sides of the differential equation, we get [s2Ly(t) - sy(O) - y' (0)] -3 [sLy(t) - yeO)] +2 L(t)
=
L(e3!) = -
1
s-3
Applying the given conditions, we get (s2 - 3s + 2) Ly(t) = s-3 + -
1
s-3
s-J
I
:. L[y(t)] = (S2 _ 3s + 2) + (s - 3)(S2 - 3s + 2)
1
s-3
:. L(y(t)] = (s _ 3)(S2 _. 3s + 2) + (8
2-
3s + 2)
1
8-3
(8-1)(s-2)(s-3)
(s-I)(8-2)
--------+------
.... (1)
To find yet), we expand each term on the right hand side of (I) in partial functions, Thus ABC ------=----+--(s-I)(s-2)(s-3) s-1 s-2 s-3 1 = A(s-2) (s-3) + B(s-l) (s-3) + C(s-I)(s-2) I
substituting s = I gives A = 2" ;similarly substituting s = 2 gets B = -1 and substituting I s = 3 gives c ="2
hence
- - -1- - - = (s-I)(s-2)(s-3)
1
2(8-1)
Similarly we can also write s-3 D E ----=-+-(s-I)(s-2) s-1 s-2
--- +---s-2
2(8-3)
659
Laplace Transforms
s-3 = D (s-2) R(s-I) Taking s =: 2, we get E =:-1 similarly s = I we get D = 2 L[f(t)] =
Hence
I
I
I
s-2
2{s-3)
--+
2(s-l)
I
I
s-I
s-2
+---
221 - - - + -,----- 2( s - I ) s - 2 2{s - 3)
-
Example 8.4.6 Solve yll + 3yl + 2y = 2t3 + 2t + 2 with yeO) = 2, yl(O)
Sol.
Applying Laplace transform to both sides of the given equation we get L(yll) + 3L(yl) + 2L(y) = 2L(t2) + 2L(t) + 2 ? I 4 2 2 [s-Ly(t) - sy(O) - yeO)] + 3 [sL[y(t)] - yeO)] + 2Ly(t) = 3 + -2 + -
_
s
s
s
using the given conditions (1) gives 2
(
3 2
()
s + s+ )Ly t
=
2(2+s+s2) 3
2
+ s+
6
3
=
2(S4 +3s +S2 +s+2)
S
By partial fractions, yet) = yet)
3L-I(~) -2r
=3 -
L\ ) 2L-ILI
1
2t + t 2 - e-2t
3
S
+
3) -
L-Ie ~ 2)
.... (1)
660
Engineering Mathematics - I
Exercise 8(1)
Using Laplace transform method solve the following differential equations with the given conditions.
Ans.l. y
I.
yll + 2yl - 3y = 0 at, y = 0,
yl = 4 when x = 0
2.
yl = 0 when t = 0
3.
yll + 4y = 4t at y = I, yll - 2yl + Y = et at y = 2,
4.
yll + k2y = coskt at y = 0, yl =
5. 6.
yll- y = 1 at y = -I, yl = I when t = 0 yll + 2yl + 3y = 3e-t sint; yeO) = yl(O) = 0
7.
xii + 4xl = -8t given x(O) = 0, xl(O) = 0
8.
yll + 9y = 18t given yeO) = 0, yl(O) = 0
9.
yll + Y = sint given yeO) = -I, yl(O) =
= eX -
-I
1
8 +"2
t - t2 +
I
"2
when t = 0
"21
1 I I 2. y= - t- - sint + cost 3. y = 2et - tet + -2 t2et 4 8
e-3x
1. k 1 . 4. Y = 2k Sill t + "2 tSll1t 7. x =
yl = -I when t = 0
I
"8
5. y = -1 + et - cosht
e-4t 8. y = 2t -
2
"3
sin3t
6. y = 3e-tsint
