Applied Maths i u IV Solution

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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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


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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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



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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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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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


(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


7/33


8/33



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


9/33





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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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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(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



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


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


12/33



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


13/33


14/33



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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(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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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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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


18/33



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



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)


19/33



sin 2u u


2 2 22 2

2 22 2cos 2


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).


20/33



(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).


(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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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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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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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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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)


25/33


26/33



sin2u u

x y ux y

--------(2)

sin 2u u


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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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


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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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


2 2

u u u u ux xy y u

x x y y

2 2 2

2 2

2 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


(a)State Eulers theorem on homogeneous function.


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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


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


2 2 22 2

2 22 2cos 2


x x y y x y x y


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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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If a=0 ,then I=0 c=0Therefore, I=

( 1 + )

Now when a


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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


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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ii. Differentiate the equation 1 w.r.t to x and y respectively and add we willhave

sin2 ........................2u u


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)

Applied Maths i u IV Solution

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