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 MTH301 LATEST ASIGNMENT SOLUTION JULY 2010

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PostSubject: MTH301 LATEST ASIGNMENT SOLUTION JULY 2010   Tue Jul 20, 2010 9:21 pm

Quote :






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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