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

    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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    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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    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

    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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    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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    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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    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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    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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    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)

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    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

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    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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    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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    (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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    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)

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    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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    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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    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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    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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    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

    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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    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)