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Assignment # 2
MTH301 (Fall 2013)
Total marks: 10
Lecture # 21 to 35
Due Date: February 10, 2014
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Qno 1:
0.643 or 3/28
Qno 2: confuse from last portion plzz if any one have solved do share
Qno3:
it is exact z= x2ey+ycosx2+ycosx
this is my ans do share ur ans as well.....
asslm o alikum to all
we are facing promblem in this assignment plz help in this assigment of mth301-2
(y^2+3)dx+(2xy-4)dy=0
Let M(x,y) = (y^2+3) and N(x,y) = (2xy-4)
To find if they are exact, we take the partial derivative of M(x,y) in respect to y:
partial d/dy [y^2+3] = 2y
Then we take the partial derivative of N(x,y) respect to x:
partial d/dx [2xy-4] = 2y
Since the partial derivatives equal each other, the diff eqn is exact.
To find the solution, we first integrate M(x,y) respect to x:
1) integral (y^2 +3)dx = xy^2 + 3x + h
where h is a constant in respect to x.
To find what h is we find the partial derivative of this equation in respect to y:
2) partial d/dy [xy^2 + 3x + h] = 2xy + h'
where h' = dh/dy
Set this equation equal to N(x,y) and integrate
3) 2xy + h' = 2xy -4
dh/dy = -4
integral dh = integral -4dy
h = -4y + c
Substituting -4 + c for h in the equation we found in step 1) will give us the general solution:
xy^2 + 3x - 4y = c
any one can help us plz
mth301
Relevant example to solve qus no 3
Evaluate the line integral with respect to arclength:
f(x,y,z)=1/(y^3)
C=r(t)=<ln(t), t, 2> where t is greater than 1 and less than e
f(x,y,z) = y^(-3) and
r(t) = <ln(t),t,2> = <x(t),y(t),z(t)>; hence,
F(x,y,z) = ∇f = <0,-3(y^(-4)),0> = <0,-3(y(t)^(-4)),0> = <0,-3(t^(-4)),0> and
r'(t) = <1/t,1,0>;hence,
∫[1,e] ∇f•r' dt = ∫[1,e] 0∙(1/t) + (-3)(t^(-4))∙(1) + 0∙0 dt = (-3)((t^(-3)/(-3)) | [1,e] = t^(-3) | [1,e] = e^(-3) - 1.
Question # 1
Use polar co-ordinates to evaluate the double integral, where R is the sector in the first quadrant bounded by .
0.643 or 3/28
is me kia direct yehi answer aa rha hai koi solution nahi hai?
Question # 2
Determine whether the following differential is exact? If so, find z.
it is exact z= x2ey+ycosx2+ycosx
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