Query128

#$&*

course Phy 241

4pm 12/10/2011

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what

you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

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Question: `q001. Find the Jacobian d(x,y,z)/d(u,v,w) when x = 2u - v, y = 2v + 2w, z = v - w.

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Your solution:

x = 2u - v

dx/du = 2

dx/dv = -1

dx/dw = 0

y = 2v + 2a

dy/du = 0

dy/dv = 2

dy/dw = 2

z = v - w

dz/du = 0

dz/dv = 1

dz/dw = -1

Matrix form

dx/du , dx/dv , dx/dw

dy/du , dy/dv , dy/dw

dz/du , dz/dv , dz/dw

=

2 , 1 , 0 | 2 , -1

0 , 2 , 2 | 0 , 2

0 , 1 , -1 | 0 , 1

= [2*(2)*(-1) + 1*2*0 + 0*0*1] - [0*2*0 + 1*2*2 + -1*0*-1]

= -4 - 4

= -8

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q002. Let R be the parallelogram with vertices (0,0), (1,4), (4,6), (4,2). Sketch and decribe the corresponding region after the

transformation u = x^2, v = x+ y.

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Your solution:

Assume region lines in the positive uv plane

u = x^2

x = sqrt(u)

0 = sqrt(u)

u = 0

v = x + y

y = v - x

0 = v - sqrt(u)

v = 0

At (x,y) = (0,0)

(u,v) = (0,0)

x = sqrt(u)

1 = sqrt(u)

u = 1

y = v - sqrt(u)

4 = v - 1

v = 5

At (x,y) = (1,4)

(u,v) = (1,5)

x = sqrt(u)

4 = sqrt(u)

u = 2

y = v - sqrt(u)

6 = v - 4

v = 10

At (x,y) = (4,6)

(u,v) = (2,10)

x = sqrt(u)

4 = sqrt(u)

u = 2

y = v - sqrt(u)

2 = v - 4

v = 6

At (x,y) = (4,2)

(u,v) = (2,6)

looks like a triangle with a hypotenuse from (0,0) to (2,10) and lengths of the triangle from the origin (0,0) to (2,6) and (2,6) to (2,10)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q003. Under the transformation u = 1/5(2x + y) and v = 1/5(x - 2y) the region D which is a square in the xy-plane with vertices (0,0), (1,-2) is

mapped onto a square in the uv-plane. Use this information to find the integral of cos(2x + y)*sin(x - 2y) with respect to V over D.

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Your solution:

u = 1/5(2x + y)

v = 1/5(x - 2y)

dx/du = 2/5

dx/dv = 1/5

dy/du = 1/5

dy/dv = -2/5

(2/5)*(-2/5) - (1/5)*(1/5)

= (-4/25) - (1/25)

= -1/5

(0,0), (1, -2), (1,0), (0,-2)

u = 1/5(2x + y)

u(0,0) = 0

u(1,-2) = 0

u(1,0) = 2/5

u(0,-2) = -2/5

v = 1/5(x - 2y)

v(0,0) = 0

v(1,-2) = 4/5

v(1,0) = 1/5

v(0,-2) = 4/5

x = v + 2u

y = u - 2v

(0,0), (1, -2), (1,0), (0,-2)

becomes

(0,0) , (0, 4/5) , (2/5 , 1/5) , (-2/5 , 4/5)

Int[cos(2x + y)*sin(x - 2y)dA

= Int[cos(5u)*sin(5v)(1/5)dv, 0, 4/5]du, -2, 2]

= Int[-1/25cos(5u)*cos(4) + 1/25cos(5u)*cos(0)]du, -2, 2]

= Int[0.0661cos(5u)] du, -2, 2

= 0.0132sin(10) - 0.0132sin(-10)

= -0.0144 units

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

-.01439 seems alittle off, but I'll stick with it because 1/5 and 4/5 is not very much, either

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Self-critique rating:

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Question: `q004. A rotation of the xy-plane through the fixed angle theta is given by

x = u cos(theta) - v sin(theta), y = u sin(theta) + v cos(theta).

Compute the Jacobian of this transformation.

Let E denote the ellipse x^2 + xy + y^2 = 9. Use a rotation of pi/4 to obtain an integral which is equivalent to the double integral of y with respect to V over E.

Evaluate the integral found in the previous step.

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Your solution:

x = u cos(theta) - v sin(theta)

y = u sin(theta) + v cos(theta)

dx/du*dy/dv - dy/du*dx/dv

= [-u*sin(theta) + cos(theta)] * [-v*sin(theta) + cos(theta)] - [u*cos(theta) + sin(theta)] * [-vcos(theta) - sin(theta)]

= [uvsin^2(theta) - usin(theta)cos(theta) + -vsin(theta)cos(theta) + cos^2(theta)] - [-uvcos^2(theta) - usin(theta)cos(theta) -vsin(theta)cos(theta) - sin^2(theta)]

= uvsin^2(theta) + uvcos^2(theta) + cos^2(theta) + sin^2(theta)

= uv + 1

x^2 + xy + y^2 = 9

r = 3

theta = pi/4

Int[dr * d(theta)

Not sure what to do next.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

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The partial derivatives of x and y with respect to u are cos(theta) and sin(theta).

The partial derivatives of x and y with respect to v are -sin(theta) and cos(theta).

The Jacobian is therefore

det([cos theta, sin theta; -sin theta, cos theta]) = 1

If theta = pi / 4 we have

x = sqrt(2) / 2 ( u - v ) and y = sqrt(2) / 2 ( u + v ), so our equation becomes (sqrt(2) / 2 * ( u - v ) ) ^ 2 + sqrt(2) / 2 * (u - v) * sqrt(2) / 2 * (u + v) + (sqrt(2) / 2 * (u + v) ) ^ 2 = 9

1/2 (u^2 - 2 u v + v^2) + 1/2 (u^2 - v^2) + 1/2 (u^2 + 2 u v + v^2) = 9

3/2 u^2 + 1/2 v^2 = 9

This is the equation of an ellipse with semimajor and semiminor axes 3 sqrt(2) and sqrt(6).

The region within this ellipse can be described as

-sqrt(6) <= u < = sqrt(6)

-sqrt(18 - 3 u^2) < = v <= sqrt(18 - 3 u^2)

leading directly to the integral.

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Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#