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Thread: MTH301 Assignment 5 Deadline 19 July 2010

  1. #1
    Senior Member viki's Avatar
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    Junior Member Death race's Avatar
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    any one plz help in 2nd question

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    Senior Member Awais's Avatar
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    what happened in 2nd question??? What u need to know??

  4. #4
    Senior Member viki's Avatar
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    22 Idea Solution For MTH301

    To find Int {y dx +x^2dy} where C is the curve x=t and y = (1/2) t^2.

    Solution:

    x= t. So dx = dt
    y =(1/2)t^2. Therefore dy = 1/2* 2tdt or dy = tdt. Substituting in the integral,
    Int (y dx+x^2dy) = Int { (1/2)t^2 dt +t^2 tdt)
    = (1/2)t^(2+1) / (2+1) + t^(3+1) / (3+1)
    = (1/6)t^3 + t^4/4
    =[ t^3(2+3t)/12]. Now for t= 0 to 2 the integral value is:
    = {2^3(2+3*2)/12 - 0 }
    = 64/12
    = 16/3
    ***************************

    Use Greenís theorem to evaluate the integral , where C is the circle
    and c is oriented counterclockwise.


    Green's theorem states that for a contour integral,
    int( L dx + M dy ) = dblint( dM/dx - dL/dy ) dx dy
    So in your problem,
    L = x^2 - y ; dL/dy = 2x dx/dy - 1
    M = x ; dM/dx = 1
    dx/dy = -y/x (since x^2 + y^2 = 4)
    dblint( 1 + 2y + 1 ) dx dy
    dblint( 2y + 2 ) dx dy
    int( 2xy + 2x ) dy | -2 <= x <= 2
    int( 8y + 8 ) dy
    ( 8y^2/2 + 8y ) | -2 <= y <= 2
    =0 + 32 = 32
    ***********************************
    Question 3rd
    If , determine curl of A at the point (1,4,-3).

    Curl of vector A = [ (i j k), (db/dbx , db/dy , db/dz), ( A1i , A2j, A3k)]
    Where vector A = A1i+A2j+A3k
    Curl of vector A = (dbA3/dby -dbA2/dbz)i + (dbA1/dbz - dbA3/dbx)j + (dbA2/dbx-dbA1/dy)]
    = i {db/dy(-x^3y^3*z^3) - db/dz(x^2+y^2)}
    + j {db/dbz(x^4-y^2z^2) - db/dbx (-x^3y^3z^3)}
    k { db/dbx ( x^2+y^2) - db/dy (x^4-y^2z^2)}
    = i { (-x^3*3y^2z^3) - ( 0) }
    +j { (-y^2*2z) - (-3x^2y^3z^3)}
    +k{ (2x) - (2yz^2}.
    =i( - 3x^3*y^2*z^3) +j(-2y^2z +3x^2*y^3*z^3)+k(2x-2yz^2)
    Threfore curl of A at (1, 4, -3) = i(-3*4^2(-3)^2) + J(-2*4^2*(-3)+3*4^3*(-3)^3)+k(2-2*3(-3)^2)
    = i(-432) +j(96 - 5184)+k(2-54)
    = (-432)i + (-5088)j + (-52) k is the Curl of A at (1,4,-3).


    Solution 2nd

    The mathematical curl of a vector function A = Ax i + Ay j + Az k is defined as:
    curl A = (dAz/dy - dAy/dz)i + (dAx/dz - dAz/dx)j + (dAy/dx - dAx/dy)k
    Calculate the above derivatives and plug in for the point (1,4,-3)
    dAz/dy = -3x^3y^2z^3 = -3(4)^2(-3)^3 = 1296
    dAy/dz = 0
    dAx/dz = -2y^2z = -2(4)^2(-3) = 96
    dAz/dx = -3x^2y^3z^3 = -3(1)^2(4)^3(-3)^3 = 5184
    dAy/dx = 2x = 2(1) = 2
    dAx/dy = -2yz^2 = -2(4)(-3)^2 = -72
    Curl A = (1296 - 0 )i + (96 - 5184)j + (2 + 72)k
    Curl A = 1296i - 5088j + 74k
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  5. #5
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    AOA. yar 1st and 2nd ans main mistakes hain.

  6. #6
    Junior Member Death race's Avatar
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    yes he is right 1st or 2nd me mistake he esy thanks viki bhai for help
    thnk u so much.

  7. #7
    Junior Member Death race's Avatar
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    2nd question me to ye puchna tha ky integration me koi values to nahi dalni or ye jo viki bhai ne ass di he wo thek he

  8. #8
    Senior Member viki's Avatar
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    Quote Originally Posted by Death race View Post
    2nd question me to ye puchna tha ky integration me koi values to nahi dalni or ye jo viki bhai ne ass di he wo thek he
    oooh main na just idea provide kiya hia main kon sa kahta hn ka ankhain band kar ka solution ly lo

    please chk kiya karian then solution khud banya karain thankx
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