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Applied Maths i u IV Solution
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7/31/2019 Applied Maths i u IV Solution
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APPLIED MATHS I
Sol u t i on :CSVTU Ex a m i n a t i on
Papers
Depa r tm en t of Ma th em a ticsDIMAT
PARTIAL DIFFERENTI ATI ON
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 2
APPLIED MATHS ITime Allowed : Three hours
Maximum Marks : 80
Minimum Pass Marks : 28Note : Solve any two parts from each question. All questions carry equal marks.
UNIT IV
PARTIAL DIFFERENTI ATI ON
SOLUTION (Nov-Dec-2005)
(a) If 1 2tan / u y x prove that 2 2 22 2 22 22 sin .sin 2u u ux xy y u ux x y y
.
Ans: 1 2tan / u y x 2tan / u y x
Let tan u z2
2
2/ . .
y yz y x x x
x x
So, z is homogeneous function of order 1.
By Eulers formulaz z
x y nz
x y
z zx y z
x y
------------(1)
(tan ) (tan )tan
u ux y u
x y
2 2sec sec tanu u
x u y u ux y
2tan cosu u
x y u ux y
sin22
u u ux yx y
----------(2)
sin2
2
u u ux y x y x y
x y x y x y
2 2 22 2
2 22 cos 2
u u u u u u ux xy y x y u x y
x x y y x y x y
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 3
2 2 22 2
2 2
sin 2 sin 22 cos 2 .
2 2
u u u u ux xy y u
x x y y
2 2 22 2
2 2
sin 2 sin 22 cos 2 .
2 2
u u u u ux xy y u
x x y y
2 2 2
2 2
2 2sin22 . cos 2 1
2u u u ux xy y u
x x y y
2 2 2
2 2 2
2 2
sin 22 . 2sin
2
u u u ux xy y u
x x y y
2 2 22 2 2
2 22 sin .sin 2
u u ux xy y u u
x x y y
(Proved)
(b)Divide 24 into three parts such that the continued product of the first, squareof second and cube of third may be maximum.
Ans: Let , ,x y z are three parts of 24.
24x y z and2 3
p xy z So, 2 3p xy z and 24x y z
2 3 3 2 2
2 3 2 3 2 3
00 0
0 2 0 3 0
6 6 0 6 3 0 6 2 0
pp p
y yx x z z
y z xyz xy z
xy z x xy z y xy z z
Now, 6 3 2 6 3 2x y z x y z
3 , 2z x y x
24 2 3 24 6 24 4x y z x x x x x 4, 8, 12x y z (Ans)
(c)Using the method of differentiation under integral sign evaluate1
2
0
tan, 0
(1 )
axdx a
x x
.
Ans: Let1
2
0
tan( )
(1 )
axF a dx
x x
2
22
2 2 2 2 2 2 2 2 2
0 0 0
1
11 1 1 1 1'( )(1 ) 1 1 1 1 1
a
aaF a xdx dx dxx x a x x a x x a x
2
2 2 2 2 2
0 0
1'( )
1 1 1 1
dx a dxF a
a x a a x
2 11
2 200
1 tan'( ) . tan .
1 1
a axF a x
a a a
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 4
2 2
1'( ) . 0 . 0
1 2 1 2
aF a
a a
2
1 1'( ) .
2 1 2 1
aF a
a a
So, ( ) .log 12F a a C -------------- (1)
But a = 0, F(a) = 0. 0C
( ) .log 12
F a a
1
2
0
tanlog(1 )
(1 ) 2
axdx a
x x
(Proved)
SOLUTION (Apr-May-2006)
(a)If 3 31tan x yux y
prove that sin 2u u
x y ux y
and
2 2 22 2
2 22 sin 4 sin 2
u u ux xy y u u
x x y y
.
Ans:3 3
1tanx y
ux y
3 3
tanx y
ux y
Let tan u z 3
3
3 32
1
1
yx
xx y yz x
yx y xx
x
So, z is homogeneous function of order 2.
By Eulers formulaz z
x y nzx y
2z z
x y zx y
------------(1)
(tan ) (tan )2tan
u ux y u
x y
2 2sec sec 2 tanu u
x u y u ux y
22 tan cosu u
x y u ux y
2 sin cosu u
x y u ux y
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 5
sin2u u
x y ux y
(proved)--------(2)
sin 2u u
x y x y x y ux y x y x y
2 2 2
2 2
2 22 2cos 2u u u u u u ux xy y x y u x y
x x y y x y x y
2 2 22 2
2 22 sin 2 2cos 2 .sin 2
u u ux xy y u u u
x x y y
2 2 22 2
2 22 sin 4 sin 2
u u ux xy y u u
x x y y
(Proved)
(b)Find the values of , ,x y z for which 5( , , )2 4
xyzf x y z
x y z
is a maximum
subject to the condition 8xyz .Ans: Here 8xyz 8xyz
And5
2 4
xyzu
x y z
0 0 0u u u
x x y y z z
2 2 2
40 80 1600 0 0
2 4 2 4 2 4yz xz xy
x y z x y z x y z
2 2 2
40 80 160
2 4 2 4 2 4
x y zxyz xyz xyz
x y z x y z x y z
So,
2 2 2
40 80 160
2 4 2 4 2 4
x y z
x y z x y z x y z
40 80 160x y z 2 4x y z
As 8xyz , 4, 2, 1x y z (Ans).
(c) If cosh .cos , sinh .sinu a x y v a x y then show that2( , ) 1 (cosh 2 cos 2 )
( , ) 2
u va x y
x y
.
Ans: cosh .cos , sinh .sinu a x y v a x y
sinh .cos , cosh .sin
cosh .sin , sinh .cos
u va x y a x y
x x
u va x y a x y
y x
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 6
Now,( , )
( , )
u u
x yu v
v vx y
x y
sinh cos cosh sin( , )cosh sin sinh cos( , )
a x y a x yu va x y a x yx y
2 2 2 2 2 2( , ) sinh cos cosh sin( , )
u va x y a x y
x y
2 2 2 2 2( , )
sinh cos cosh sin( , )
u va x y x y
x y
2 2 2( , ) 1 cos 2 1 cos 2sinh cosh( , ) 2 2
u v y ya x x
x y
2
2 2 2 2( , ) sinh sinh cos 2 cosh cosh cos 2
( , ) 2
u v ax x y x x y
x y
2
2 2 2 2( , ) sinh cosh cos 2 sinh cosh( , ) 2
u v ax x y x x
x y
2( , )
cosh 2 cos 2( , ) 2
u v ax y
x y
(Proved)
2 2
2 2
sinh cosh cosh 2
sinh cosh 1
x x x
x x
SOLUTION (Nov-Dec-2006)
(a)If x y zx y z c , show that 2 1( log )z x exx y
at x y z .
Ans: x y zx y z c
log log log logx x y y z z c
log log log logz z x x y y c
Differentiate with respect to x partially we get
(1 log ) (1 log )z
z xx
(1 log )
(1 log )
z x
x z
Differentiate with respect to y partially we get2
2
1 1(1 log )
(1 log )
z zx
x z z y
2
2
(1 log ) 1 (1 log )
(1 log ) (1 log )
z x y
x z z z
2
3
(1 log )(1 log )
(1 log )
z x y
x z z
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 8
2 22 1
2 22 2
log(1 ) 1 log(1 )'( ) log(1 ) tan
1 12 1 1F
2 1
2 2
log(1 ) tan'( )
2 1 1F
2 1
2 2
1 log(1 ) tan( )
2 1 1F d d
12
2 2 2 2
1 2 tan( ) log(1 )
2 1 1 1 1
d dF d
1 12 1
2 2
1 tan tan( ) log(1 ) tan
2 1 1F d d
2 11( ) log(1 ) tan
2F C ---------- (1)
But at 0, ( ) 0 0F C
So equation (1) is 2 11
( ) log(1 ) tan2
F
2 1
2
0
log(1 ) 1log(1 ) tan
1 2
xdx
x
---------- (2) (Ans)
By putting 1 in equation (2) we get1
1
2
0
log(1 ) 1log2. tan 1
1 2
xdx
x
1
2
0
log(1 ) 1
log2.1 2 4
x
dxx
1
2
0
log(1 )
log21 8
x
dxx
(Proved)
SOLUTION (May-June-2007)
(a)If 3 31tan x yux y
prove that sin 2u u
x y ux y
Ans:3 3
1tanx y
ux y
3 3
tan
x y
u x y
Let tan u z 3
3
3 32
1
1
yx
xx y yz x
yx y xx
x
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 9
So, z is homogeneous function of order 2.
By Eulers formulaz z
x y nzx y
2z z
x y zx y
(tan ) (tan )2tan
u ux y u
x y
2 2sec sec 2 tanu u
x u y u ux y
22 tan cosu u
x y u ux y
2 sin cosu u
x y u ux y
sin2u u
x y ux y
(Proved)
(b)Find the volume of the greatest rectangular parallelepiped inscribed in theellipsoid whose equation is
2 2 2
2 2 21
x y z
a b c .
Ans: Let the edges of the parallelepiped be 2x, 2y, 2z which are parallel to axes.
Then its volume 8V xyz
Now we have to find maximum value of V subject to the condition that2 2 2
2 2 21 0
x y z
a b c
So, let
2 2 2
2 2 28 1x y z
F xyz a b c
22 2
2 22
2 22
22 28 08 0 8 0
2 228 88
F yF x F zxzyz zx
y bx a x c
x zyxyz xyzxyz
a cb
2 2 2
2 2 2
2 2 2x y z
a b c
2 2 2
2 2 2
1
3
x y z
a b c
, ,3 3 3
a b cx y z
At x = 0, the parallelepiped is just a rectangular sheet. So its volume = 0.
So, volume is maximum at , ,3 3 3
a b cx y z
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 10
So, maximum volume8
83 3
abcxyz (Ans).
(c) If sin cos , sin sin , cosx r y r z r then show that 2( , , ) sin( , , )
x y zr
r
.
Ans: sin cos , sin sin , cosx r y r z r
sin cos , cos cos , sin sinx x x
r rr
--------- (1)
sin sin , cos sin , sin cosy y y
r rr
--------- (2)
cos , sin , 0z z z
rr
--------- (3)
Now,( , , )
( , , )
x x x
r
x y z y y y
r r
z z z
r
sin cos cos cos sin sin( , , )
sin sin cos sin sin cos( , , )
cos sin 0
r rx y z
r rr
r
2 2
2 2
( , , )sin cos 0 sin cos cos cos (0 sin cos cos )
( , , )
sin sin ( sin sin cos sin )
x y zr r r
r
r r r
2 3 2 2 2 2 2 3 2 2 2 2( , , ) sin cos sin cos cos sin sin sin cos sin( , , )
x y zr r r r
r
2 3 2 2 2 2 2 2( , , )
sin cos sin sin cos cos sin( , , )
x y zr r
r
2 3 2 2( , , ) sin sin cos( , , )
x y zr r
r
2 2 2( , , )sin sin cos( , , )
x y zrr
2( , , ) sin( , , )
x y zr
r
(Proved)
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 11
SOLUTION (Nov-Dec-2007)
(a)State Eulers theorem on homogeneous function.Ans: If u be homogeneous function of degree n in x & y, then
u ux y nu
x y
.
(b)If 3 31tan x yux y
prove that2 2 2
2 2
2 22 2cos3 .sin
u u ux xy y u u
x x y y
.
Ans:3 3
1tanx y
ux y
3 3
tanx y
ux y
Let tan u z 3
3
3 3
2
1
1
yx
xx y yz xyx y x
xx
So, z is homogeneous function of order 2.
By Eulers formulaz z
x y nzx y
2z z
x y zx y
------------(1)
(tan ) (tan )2tan
u ux y u
x y
2 2sec sec 2 tanu u
x u y u ux y
22 tan cosu u
x y u ux y
2 sin cos
u ux y u u
x y
sin2u u
x y ux y
--------(2)
sin 2u u
x y x y x y ux y x y x y
2 2 22 2
2 22 2cos 2
u u u u u u u
x xy y x y u x yx x y y x y x y
2 2 22 2
2 22 sin 2 2cos 2 .sin 2
u u ux xy y u u u
x x y y
2 2 22 2
2 22 sin 4 sin 2
u u ux xy y u u
x x y y
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 12
2 2 22 2
2 22 2cos3 .sin
u u ux xy y u u
x x y y
(Proved)
(c)A rectangular box open at top, is to have a given capacity 32 cu unit. Find thedimensions of the box requiring least material for its construction.
Ans: Let , ,x y z be the edges of the box.
Given volume of the box = 32.
32xyz and surface area =
Let ( 2 2 ) ( 32)F xy yz zx xyz
2 0F
y z yzx
-----------(1)
2 0F
x z zxy
----------(2)
2 2 0F
y x xyz
----------(3)
(2) (3)x y 2 2 0zx zy x y --------(4)
(3) (4)y z 2 0 2xy xz y z ---------(5)
So, 2x y z
Given that 32xyz
2 .2 . 32z z z 3
4 32 2z z So, 4, 4, 2x y z (Ans)
(d)Using the method of differentiation under integral sign evaluate1
2
0
tan, 0
(1 )
axdx a
x x
.
Ans: Let1
2
0
tan( )
(1 )
axF a dx
x x
2 2 2
0
1 1'( )
(1 ) 1F a xdx
x x a x
2 2 2
0
1 1'( )
1 1F a dx
x a x
2
22
2 2 2
0
1
11'( )
1 1
a
aaF a dx
x a x
2
2 2 2 2 2
0 0
1'( )1 1 1 1
dx a dxF aa x a a x
2 11
2 200
1 tan'( ) . tan .
1 1
a axF a x
a a a
2 2
1'( ) . 0 . 0
1 2 1 2
aF a
a a
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 14
2 2 2 2 2
32 32 2 2
r x y x y
x rx y
Similarly,r y
y r
and
2 2
2 3
r x
y r
Now equation (1) becomes2 22 2 2 2
2 2 3 3''( ). '( )
u u x y y xf r f r
x y r r r r
2 2 2 2 2 2
2 2 2 3''( ). '( )
u u x y x yf r f r
x y r r
2 2
2 2
1''( ).1 '( ).
u uf r f r
x y r
as 2 2 2r x y
So,2 2
2 2
1( ) ( )
u uf r f r
x y r
.
(c)Find the maxima and minima of 2 2 2u x y z subject to the condition2 2 2 1ax by cz and 0lx my nc .
Ans: 2 2 2u x y z 2 2 2 1ax by cz
lx my nc
Then
1 21 2 1 2
1 2 1 21 2
00 0
2 2 0.....(1) 2 2 0.....(3)2 2 0.....(2)
uu u
y y yx x x z z zx ax l z cz ny by m
(1).x+(2).y+(3).z2 2 2 2 2 2
1 22( ) 2( ) ( ) 0x y z ax by cz lx my nz
1 12 2 0u u ---------(4)
Now putting in (1), (2), (3) we get
2 2 2, ,2 2 2 2 2 2
l m nx y z
au bu cu
Now putting in 0lx my nc we get,2 2 2
2 2 2 02 2 2 2 2 2
l m n
au bu cu
2 2 2
02 2 2 2 2 2
l m n
au bu cu
So, stationary value of u satisfies2 2 2
02 2 2 2 2 2
l m n
au bu cu
.
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 15
(d)Prove that / 2 2 2 2 20
log( cos sin ) log2
d
.
Ans:/2
2 2 2 2
0
( ) log( cos sin )F d
--------- (1)
Then by method of differentiation under integral sign, we get
/ 2
2 2 2 2
0
'( ) log( cos sin )F d
/ 2 2
2 2 2 2
0
2 cos'( )
cos sinF d
/ 2
2 2 2
0
'( ) 2n
dF
ta
----------- (2)
Let2
2 2 2 2tan sec
sec (1 tan ) t
dt dt dt t dt d d
So, (1) becomes
2 2 2 2
0
'( ) 2( )( )
dtF
t t
2 2 2 2 2 2
0
2 1 1'( )F dt
t t
1 1
2 2
0
2 1 1'( ) tan tan
t tF
2 2
2 1 1'( )
2 2
F
'( )F
( )F d C
( ) logF C ----------- (3)Putting 0 in (1) we get
/ 2 /2 /2
0 0 0
(0) 2 log( sin ) 2log 2 logsinF d d d
1(0) 2 log 2 log2 2 2
F
1(0) log log
2F
Putting 0 in (3) we get (0) logF C
So, 1
log log log2
C
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 16
1log
2C
Putting the value of C in (3) we get
1
( ) log log log2 2
F
(Proved).
SOLUTION (Dec-Jan-2008-2009)
(a)If y yz xx x
then:2 2 2
2 2
2 22 ____ .
u u ux xy y
x x y y
Ans: 0 (Zero).
(b)If8 8 8
2 3sin
x y zu
x y z
, show that 3tan 0
u u ux y z u
x y z
.
Ans:8 8 8
2 3sin
x y zu
x y z
Let8 8 8
2 3sin
x y zt u
x y z
3
8 84
1 2 3,
1
y zx x
y zx x
x y zt x
x xx
So, z is homogeneous function of order (n = -3)
So, by Eulers formulat t t
x y z ntx y z
sin sin sin 3sinu u ux y z ux y z
cos cos cos 3sinu u u
x u y u z u ux y z
3tanu u u
x y z ux y z
3tan 0u u u
x y z ux y z
(Proved)
(c)A rectangular box open at top, is to have volume of 32 cu. Cm. Find thedimensions of the box requiring least material for its construction.
Ans: Let , ,x y z be the edges of the box.
Given volume of the box = 32.
32xyz and surface area =
Let ( 2 2 ) ( 32)F xy yz zx xyz
2 0F
y z yzx
-----------(1)
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 17
2 0F
x z zxy
----------(2)
2 2 0F
y x xyz
----------(3)
(2) (3)x y 2 2 0zx zy x y --------(4)
(3) (4)y z 2 0 2xy xz y z ---------(5)
So, 2x y z
Given that 32xyz
2 .2 . 32z z z 3
4 32 2z z So, 4, 4, 2x y z (Ans)
(d)Using the method of differentiation under integral sign prove that1
2
0
tanlog(1 ), 0
(1 ) 2
axdx a a
x x
.
Ans: Let1
2
0
tan( )(1 )
axF a dxx x
2 2 2
0
1 1'( )
(1 ) 1F a xdx
x x a x
2 2 2
0
1 1'( )
1 1F a dx
x a x
2
22
2 2 2
0
1
11'( )
1 1
a
aaF a dx
x a x
2
2 2 2 2 2
0 0
1'( )
1 1 1 1
dx a dxF a
a x a a x
2 11
2 200
1 tan'( ) . tan .
1 1
a axF a x
a a a
2 2
1'( ) . 0 . 0
1 2 1 2
aF a
a a
2
1 1'( ) .
2 1 2 1
aF a
a a
So, ( ) .log 12
F a a C
-------------- (1)
But a = 0, F(a) = 0. 0C
( ) .log 12
F a a
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 18
1
2
0
tanlog(1 )
(1 ) 2
axdx a
x x
(Proved)
Solution (Apr-May-2009)
(a)Fill up the blank:If z is homogeneous function of degree n in x and y then,2 2 2
2 2
2 22 ____ .
u u ux xy y
x x y y
Ans:2 2 2
2 2
2 22 ( 1) .
u u ux xy y n n z
x x y y
(b)If 3 31tan x yux y
, prove that
(i) sin 2u ux y ux y
(ii) 2 2 22 22 2
2 2cos3 .sinu u u
x xy y u ux x y y
.
Ans:3 3
1tanx y
ux y
3 3
tanx y
ux y
Let tan u z 3
3
3 32
1
1
yx
xx y yz x
yx y x
x x
So, z is homogeneous function of order 2.
By Eulers formulaz z
x y nzx y
2z z
x y zx y
------------(1)
(tan ) (tan )2tan
u ux y u
x y
2 2sec sec 2 tanu u
x u y u u
x y
22 tan cosu u
x y u ux y
2 sin cosu u
x y u ux y
sin2u u
x y ux y
(proved)--------(2)
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 19
sin 2u u
x y x y x y ux y x y x y
2 2 22 2
2 22 2cos 2
u u u u u u ux xy y x y u x y
x x y y x y x y
2 2 22 2
2 22 sin 2 2cos 2 .sin 2
u u ux xy y u u u
x x y y
2 2 22 2
2 22 sin 4 sin 2
u u ux xy y u u
x x y y
2 2 22 2
2 22 2cos3 .sin
u u ux xy y u u
x x y y
(Proved)
(c)Find the volume of the greatest rectangular parallelepiped that can beinscribed in the ellipsoid:
2 2 2
2 2 21
x y z
a b c
.
Ans: Let the edges of the parallelepiped be 2x, 2y, 2z which are parallel to axes.
Then its volume 8V xyz
Now we have to find maximum value of V subject to the condition that2 2 2
2 2 21 0
x y z
a b c
So, let2 2 2
2 2 28 1
x y zF xyz
a b c
22 2
2 22
2 22
22 28 08 0 8 0
2 228 88
F yF x F zxzyz zx
y bx a x c
x zyxyz xyzxyz
a cb
2 2 2
2 2 2
2 2 2x y z
a b c
2 2 2
2 2 2
1
3
x y z
a b c
, ,3 3 3
a b cx y z
At x = 0, the parallelepiped is just a rectangular sheet. So its volume = 0.
So, volume is maximum at , ,3 3 3
a b cx y z
So, maximum volume8
83 3
abcxyz (Ans).
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 20
(d)Prove that2
1 1 2
0
1tan 2 tan log( 1)
2
ad x
dx a a ada a
. Verify your result by
direct integration.
Ans:
2 22
1 2 1
2 20 02
1tan tan 0
1
a ad x x d a
dx dx a
xda a a da aa
2 2
1 1
2 2
0 0
tan 2 tana a
d x xdx dx a a
da a x a
22
1 2 2 1
00
1tan log( ) 2 tan
2
aad x
dx x a a ada a
2
1 2 4 2 1
0
1tan log( ) log( ) 2 tan
2
ad x
dx a a a a ada a
2
2 41 1
2
0
1tan log 2 tan2
a
d x a adx a ada a a
2
1 1 2
0
1tan 2 tan log 1
2
ad x
dx a a ada a
(Proved).
SOLUTION (Nov-Dec-2009)
(a) Explain Jacobian of wvu ,, with respect to zyx ,, .Ans: - Jacobian of wvu ,, with respect to zyx ,, is
z
w
y
w
x
w
z
v
y
v
x
v
z
u
y
u
x
u
zyx
wvu
),,(
),,(
(b) If 3 3 3log 3u x y z xyz , show that2
2
9
( )u
x y z x y z
.
Ans: - 3 3 3log 3u x y z xyz 2 2 2
3 3 3 3 3 3 3 3 3
3 3 3 3 3 3, ,
3 3 3
u x yz u y xz u z xy
x x y z xyz y x y z xyz z x y z xyz
2 2 2
3 3 3 3 3 3 3 3 3
3 3 3 3 3 3
3 3 3
x yz y xz z xyu
x x x x y z xyz x y z xyz x y z xyz
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 21
2 2 2
3 3 3
3 3 3 3 3 3
3
x yz y xz z xyu
x x x x y z xyz
2 2 2
2 2 2
3
( )
x y z xy yz zxu
x x x x y z x y z xy yz zx
3
( )u
x x x x y z
2
u ux x x x x x x x x
23
( )u
x x x x x x x y z
2
2 2 2
3 3 3
( ) ( ) ( )u
x x x x y z x y z x y z
2
2
9
( )u
x y z x y z
(Proved)
(c) Show that the rectangular solid of maximum volume that can be inscribed ina given sphere is a cube.
Ans: - Let the equation of the sphere is 2 2 2 2x y z a --------------- (1)
Let , ,x y z be the sides of rectangular solid inscribed in the sphere
So, volume of the solid V xyz 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 4( )u V x y z x y a x y a x y x y x y
2 2 3 2 4 2 2 4 2 32 4 2 , 2 2 4u u
a xy x y xy a x y x y x y
x y
For maximum 0, 0u u
x y
2 2 3 2 4 2 2 4 2 32 4 2 0, 2 2 4 0a xy x y xy a x y x y x y
2 2 2 2 2 2 2 22 2 0, 2 2 0xy a x y x y a x y
2 2 2 2 2 22 0, 2 0a x y a x y 2 2 2 2 2 22 , 2x y a x y a ----------------(2)
2 2 0x y x y
Putting in (2) we get 3
a
x y ---------------(3)
Putting values of x and y in (1) we get3
az -------------(4)
From (3) and (4)3
ax y z .
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 22
Now,2
2 2 2 2 4
22 12 2
ur a y x y y
x
22 3 34 8 8
us a xy x y xy
x y
2
2 2 4 2 22
2 2 12ut a x x x yy
At3
ax y ,
4 4 48 4 8, ,
9 9 9
a a ar s t
So,8
2 48 081
art s and 0r so volume V is maximum when
3
ax y z
So, the volume is maximum when rectangular solid is a cube. (Proved)
(d) Evaluate the integral:1
0
1log
x dxx
by applying differentiation under integral
sign 0 .
Ans: Let1
0
1( )
log
xF dx
x
1
0
1( )
log
xF dx
x
11 1 1
0 0 0
log 1( )
log 1 1
x x xF dx x dx
x
1( ) log( 1)
1F d c
------------------ (1)
Form (1) when 0 , 0)( F
Putting in (2) we get 0c .So, ( ) log( 1)F
1
0
1log( 1)
log
xdx
x
(Ans).
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 23
Solution (May-June-2010)
(a) If zyxwzyxvyxu 32,,22 find),,(
),,(
zyx
wvu
.
Ans: - 4100)13(2)23(2
321
111
022
),,(
),,(
xx
x
z
w
y
w
x
w
z
v
y
v
x
v
zu
yu
xu
zyx
wvu.
(b) The temperature T at any point ),,( zyx in space is 2),,( kxyzzyxT where kis a constant. Find the highest temperature on the surface of the sphere
2222 azyx .
Ans: - Here 2),,( kxyzzyxT and 0),,( 2222 azyxzyx
),,(),,(),,( zyxzyxTzyxF
)(),,( 22222 azyxkxyzzyxF
xkyzx
F22
, ykxz
y
F22
, zkxyz
z
F22
As, 0
x
F, 0
y
F, 0
z
F
022 xkyz , 022 ykxz , 022 zkxyz
z
kxyz
y
kxz
x
kyz
2
2
22
22
2222
2
2
2
2
2
2
2 2
)(2
4
2
2
22 a
f
zyx
kxyz
z
kxyz
y
kxyz
x
kxyz
2,
2,
22,
4,
4
22
22
22 az
ay
ax
az
ay
ax
Maximum temperature =8222
),,(42
2 kaaaakkxyzzyxT (Ans).
(c) Find the maximum and minimum distances of the point (3, 4, 12) from thesphere 1222 zyx .
Ans: -Let P(x, y, z) be any point on the sphere and A(3, 4, 12) the given point so that
),,()12()4()3()( 2222 zyxfzyxAP ---------------- (1)
Wehave to find the maximum and minimum values of ),,( zyxf subject to the
condition
01),,( 222 zyxzyx .
Now, ),,(),,(),,( zyxzyxfzyxF
1)12()4()3( 222222 zyxzyx
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 24
xxx
F2)3(2
, yy
y
F2)4(2
, zz
z
F2)12(2
As, 0
x
F, 0
y
F, 0
z
F
02)3(2 xx
, 02)4(2 yy
, 02)12(2 zz
fzyx
zyx
z
z
y
y
x
x
222
222 )12()4()3(1243
So,f
zf
yf
x
1
12
1
12,
1
4
1
4,
1
3
1
3
As, 1222 zyx
11
12
1
4
1
3222
fff
16912
f
14,12131 ff So, Maximum value = 14 and minimum value = 12.
(d) Using the method of differentiation under the sign of integration, prove that)4/(
0
22
2cos
exdxe
x .
Ans: - Let
0
cos)(2
xdxeF x
00
sinsin)(22
xdxxedxxxeF xx
0 00
)(sinsin)(22
dxxdx
ddxxedxxexF
xx
00
cos22
.sin)(
22
xdxee
xFxx
)(2
)00()(
FF
2)(
)(
F
Fon integrating
CF ln4
))(ln(2
4
2
)(
CeF ----------------------(1)
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 26
sin2u u
x y ux y
--------(2)
sin 2u u
x y x y x y ux y x y x y
2 2 22 2
2 22 2cos 2u u u u u u ux xy y x y u x y
x x y y x y x y
2 2 2
2 2
2 22 sin 2 2cos 2 .sin 2
u u ux xy y u u u
x x y y
2 2 2
2 2
2 22 sin 4 sin 2
u u ux xy y u u
x x y y
(c) Prove that of all rectangular parallelepiped of given surface, cube has themaximum volume.
Ans.Let Sides of rectangular parallelepiped are 2x, 2y, 2z.
Let R be the radius of the sphere.
Volume of the solid = 8V xyz ,2 2 2 2x y z R
So,2 2 2 2 , 8x y z R V xyz
2 2 2
00 0
8 2 0 8 2 0 8 2 0
8 2 8 2 8 2
VV V
y yx x z z
yz x xz y xy z
xyz x xyz y xyz z
2 2 22 2 2x y z 2 2 2x y z x y z
So, Rectangular parallelepiped is a cube. (Proved).
(d) Evaluate 12
0
tan, 0
(1 )
axdx a
x x
by using the method of differentiation under integral
sign.
Ans: Let1
20
tan
( ) (1 )
ax
F a dxx x
2
22
2 2 2 2 2 2 2 2 2
0 0 0
1
11 1 1 1 1'( )
(1 ) 1 1 1 1 1
a
aaF a xdx dx dx
x x a x x a x x a x
2
2 2 2 2 2
0 0
1'( )
1 1 1 1
dx a dxF a
a x a a x
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 27
2 11
2 200
1 tan'( ) . tan .
1 1
a axF a x
a a a
2 2
1'( ) . 0 . 0
1 2 1 2
aF a
a a
2
1 1'( ) .2 1 2 1
aF aa a
So, ( ) .log 1
2F a a C
--------------(1)
But a = 0, F(a) = 0. 0C
( ) .log 12
F a a
1
2
0
tanlog(1 )
(1 ) 2
axdx a
x x
(Proved)
SOLUTION (Apr-May-2011)a) Define maxima and minima for function of two variables.b) If
2
4.nr
tt e
,find that value of n will make 22
1r
r r r t
c) 21tan ,yu prove that x
2 2 22 2 2
2 22 sin sin 2 .
u u ux xy y u u
x x y y
Ans: 1 2tan / u y x 2tan / u y x
Let tan u z2
2
2/ . .
y yz y x x x
x x
So, z is homogeneous function of order 1.
By Eulers formula
z zx y nz
x y
z z
x y zx y
------------(1)
(tan ) (tan )tan
u ux y u
x y
2 2sec sec tan
u ux u y u u
x y
2tan cos
u ux y u u
x y
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 28
sin2
2
u u ux y
x y
----------(2)
sin2
2
u u ux y x y x y
x y x y x y
2 2 22 2
2 22 cos 2u u u u u u ux xy y x y u x y
x x y y x y x y
2 2 2
2 2
2 2
sin 2 sin 22 cos 2 .
2 2
u u u u ux xy y u
x x y y
2 2 2
2 2
2 2
sin 2 sin 22 cos 2 .
2 2
u u u u ux xy y u
x x y y
2 2 2
2 2
2 2
sin22 . cos 2 1
2
u u u ux xy y u
x x y y
2 2 22 2 2
2 2
sin 2
2 . 2sin2
u u u u
x xy y ux x y y
2 2 2
2 2 2
2 22 sin .sin 2
u u ux xy y u u
x x y y
(Proved)
d) By the method of differentiation under integral sign, prove that1
0
1log( 1), 1
log
xdx
x
.
Ans: Let
1
0
1( )
log
xF dx
x
1
0
1( )
log
xF dx
x
11 1 1
0 0 0
log 1( )
log 1 1
x x xF dx x dx
x
1( ) log( 1)
1F d c
------------------ (1)Form (1) when 0 , 0)( F
Putting in (2) we get 0c .
So, ( ) log( 1)F
SOLUTION (Nov-Dec-2011)
(a)State Eulers theorem on homogeneous function.
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 29
Ans. If u be homogeneous function of degree n in x & y, thenu u
x y nux y
.
(b)If3 3
1tanx y
ux y
prove that uu
y
uy
yx
uxy
x
ux 2sin4sin2
2
22
2
2
22
2u u
x y Sin ux y
2u u
x y Sin ux y
3 3
1tanx y
ux y
3 3
tanx y
ux y
Let tan u z
3
3
3 32
1
1
yx
xx y yz x
yx y xx
x
So, z is homogeneous function of order 2.
By Eulers formula
z zx y nz
x y
2z z
x y zx y
(tan ) (tan )2tan
u ux y u
x y
2 2sec sec 2 tan
u ux u y u u
x y
22 tan cos
u ux y u u
x y
2 sin cos
u ux y u u
x y
sin2u u
x y ux y
--..AnsDifferentiate the equation 1 w.r.t to x and y respectively and add we will have
sin2 ........................2u u
x y x y x y ux y x y x y
2 2 22 2
2 22 2cos 2
u u u u u u ux xy y x y u x y
x x y y x y x y
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 30
2 2 22 2
2 22 sin 2 2cos 2 .sin 2
u u ux xy y u u u
x x y y
2 2 2
2 2
2 22 sin 4 sin 2
u u ux xy y u u
x x y y
Prove that of all rectangular parallelepiped of given surface, cube has the maximum
volume.
Evaluate1
2
0
tan, 0
(1 )
axdx a
x x
by using the method of differentiation under integral sign.
Ans:-Here1
2
0
tan, 0
(1 )
axdx a
x x
So ,by differentiating under the integral sign ,we have =
( 1 +
= () . =
()
=
[ ()
]
=
[ ]
=
. = .
=
.
Which on integration w.r.t to a gives
I= . + = ( 1 + ) + .
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 31
If a=0 ,then I=0 c=0Therefore, I=
( 1 + )
Now when a
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UNIT IV( I Semester)
Depar tm ent o f M athemat ics, DIM AT Page 32
2
3
( )
(1 log )(1 log )
(1 log )x y z
z x x
x x x
2
( )
1
(1 log )x y z
z
x x x
21
( )
1 1log
(log log ) logx y z
zx ex
x x e x x ex
(Proved)
(c) If 3 31tan x yux y
prove that
i. 2u ux y Sin ux y
2u u
x y Sin u
x y
3 31tan
x yu
x y
3 3
tanx y
ux y
Let tan u z
3
3
3 32
1
1
yx
xx y yz x
yx y xx
x
So, z is homogeneous function of order 2.
By Eulers formula
z zx y nz
x y
2z z
x y zx y
(tan ) (tan )
2tanu u
x y ux y
2 2sec sec 2 tan
u ux u y u u
x y
22 tan cos
u ux y u u
x y
2 sin cos
u ux y u u
x y
sin 2u u
x y ux y
--..1Ans:
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UNIT IV( I Semester)
ii. Differentiate the equation 1 w.r.t to x and y respectively and add we willhave
sin2 ........................2u u
x y x y x y ux y x y x y
2 2 22 2
2 22 2cos 2u u u u u u u
x xy y x y u x yx x y y x y x y
2 2 22 2
2 22 sin 2 2cos 2 .sin 2
u u ux xy y u u u
x x y y
2 2 2
2 2
2 22 sin 4 sin 2
u u ux xy y u u
x x y y
iii. 2 2 22 22 2
2 4 sin 2u u u
x xy y Sin u ux x y y
(d)Divide 24 into three parts such that the continued product of the first, square ofsecond and cube of third may be maximum
Ans: Let , ,x y z are three parts of 24.
24x y z and2 3p xy z
So,2 3p xy z and 24x y z
2 3 3 2 2
2 3 2 3 2 3
00 0
0 2 0 3 0
6 6 0 6 3 0 6 2 0
pp p
y yx x z z
y z xyz xy z
xy z x xy z y xy z z
Now, 6 3 2 6 3 2x y z x y z 3 , 2z x y x
24 2 3 24 6 24 4x y z x x x x x 4, 8, 12x y z (Ans)