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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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### Replies to This Discussion

plz mujy es qustion k solv smja den

2nd ques main z ko find hi nai kya....

thanks sath me correction b kar deti tu achai hota.

yes 2nd Question is wrong 1st is also some mistake

Question # 2

Determine whether the following differential is exact? If so, find z.

it is exact z= x2ey+ycosx2+ycosx

Get help from here for quiz 3

Evaluate the line integral ∫C x ds, where C is the given curve (0,0) to (1,1)
along the parabola y=x^2 followed by the curve from (1,1) to (0,0) along the
curve y=x.

since ds is the infinitesimal arclength, we can say:
ds^2 = dx^2 + dy^2
ds = √(dx^2 + dy^2)
ds/dt = √(dx^2 + dy^2)/dt = √((1/dt^2)(dx^2 + dy^2)) = √((dx/dt)^2 + (dy/dt)^2)
ds = √((dx/dt)^2 + (dy/dt)^2) dt

since we changed the variable that we integrate with respect to, we have to change the limits of integration and change the integrand into a function of t.

first of all, we must split up the line integral between the y = x^2 part and the y = x part.

for the y = x^2 part, we can parametrize this curve with: x = t, y = t^2, and so the x in the integrand will simply be t. for the limits of integration, since the curve y = x^2 goes from x = 0 to x = 1, t will range from t = 0 to t = 1, and so that will be the limits of integration. now, we can integrate this:

(0 to 1) ∫ [x * √((dx/dt)^2 + (dy/dt)^2)] dt
= (0 to 1) ∫ [t * √(1 + 4t^2)] dt <--- use u-substitution
= [(1/12)√(1 + 4t^2)^3] | (0 to 1)
= (1/12)√(5)^3 - (1/12)

now, we have to find the line integral over y = x. we can parametrize this as x = t, y = t, where t ranges from 1 to 0:

(1 to 0) ∫ [x * √((dx/dt)^2 + (dy/dt)^2)] dt
= (1 to 0) ∫ [t * √(1 + 1)] dt
= (1 to 0) ∫ [√(2)t] dt
= (1/√(2))t^2 | (1 to 0)
= -(1/√(2))

now, we add the results we got from the two line integrals:

(1/12)√(5)^3 - (1/12) - (1/√(2))
≈ 0.148

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

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thanks 