Query128Mth

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course Mth 277

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:

2 -1 0

0 2 2

0 1 -1

[(-4)+(-2)+0] - [0+4+0] = -10

-10

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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:

New vertices (0,0),(1,5),(16,10),(16,6)

As you can see, the parallelogram also has a vertice at the origin, but is obviously much larger than the original parallelogram.

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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)

cos(2x + y)*sin(x - 2y)

dx/du = 2/5

dx/dv = 1/5

dy/du = 1/5

dy/dv = -2/5

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]

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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:

xu cos(theta) - v sin(theta)

yu sin(theta) + v cos(theta)

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

uv + 1

x^2 + xy + y^2 = 9 r=sqrt(9)=3

x=sqrt(2)/2( u - v )

y=sqrt(2)/2( u + v )

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

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Good.

To complete the solution:

The Jacobian is 1, as is easily enough verified, so the region enclosed by this curve in the uv plane will be equal to the ellipse in the xy plane.

u can vary from -sqrt( 2/3 * 9) = - sqrt(6) to sqrt(6).

For a given u, v = can vary from - sqrt( 18 - 3 u^2 ) to sqrt( 18 - 3 u^2 ).

In u, v coordinates we obtain the integral

int( int( dv, - sqrt(18 - 3 u^2), sqrt(18 - 3 u^2)) du, -sqrt(6), sqrt(6) )

which is 6 pi sqrt(3).

Comparing this with a circle of radius 9, we find that this is a little greater than the area of the circle.

The ellipse has major axis sqrt(18) = sqrt(3) * sqrt(6) and minor axis sqrt(6). A formula for the area of an ellipse is

A = pi * semimajor axis * semiminor axis

so the area should be

A = pi * sqrt(6) * sqrt(3) * sqrt(6) = 6 pi sqrt(3).

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Given Solution:

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Self-critique (if necessary):OK

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&#Good work. See my notes and let me know if you have questions. &#