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Assignment #2 MTH301 (Fall 2013) Total marks: 10 Lecture # 21 to 35 Due Date February 10, 2014

Assignment #  2

 

MTH301 (Fall 2013)

          

           Total marks: 10

            Lecture # 21 to 35  

           Due Date: February 10, 2014

 

 

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Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

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

Question No. 3 

Quiz 2 relevant example

Determine whether or not the following differential equation is exact and solve it if it is exact:?

(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

Question no.2

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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