MTH301 Assignment 5 Deadline 19 July 2010

[viki]

[SIZE="4"]MTH301 Assignment 5

Due Date 19 July 2010[/SIZE]

  mth301 5.JPG (Size: 24.64 KB / Downloads: 90)

[Death race]

any one plz help in 2nd question

[Awais]

what happened in 2nd question??? What u need to know??

[viki]

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

[bc090400375]

AOA. yar 1st and 2nd ans main mistakes hain.

[Death race]

yes he is right 1st or 2nd me mistake he esy thanks viki bhai for help
thnk u so much.

[Death race]

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

[viki]

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