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| Subject: MTH301 LATEST ASIGNMENT SOLUTION JULY 2010 Tue Jul 20, 2010 9:21 pm | |
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MTH301 IDEA SOL BY Mr.Sepharian,,,,,,,,,,,,,
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 Green’s 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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