#$&*
course Mth 277
4/19 3
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:
For our 3x3 matrix of d(x,y,z)/d(u,v,w) we get
|2 -1 0|
|0 2 2|
|0 1 -1|
Which has a determinant of
[(-4)+(-2)+0] - [0+4+0] = -10
So Jacobian is -10.
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Given Solution:
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Self-critique (if necessary):
I believe I have done this correctly.
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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:
For u = x^2 and v = x+ y, we get another parallelogram with vertices at (0,0),(1,5),(16,6),(16,10) which has covers approx. 4.5 times more units than the original parallelogram.
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@& The transformation in the preceding question had a Jacobian matrix consisting of constant numbers. That is because x, y and z are all linear combinations of u, v and w, so all the derivatives are constants.
Thus the transformation in the first question was a linear transformation. Among other things, linear transformations map straight lines to straight lines.
That Jacobian was -10, which tells us that any region of the u-v plane (e.g., say, a parallelogram) would map onto a region of the x-y plane which has 10 times the area, and on which (due to the negative sign) the direction in which the region is traversed is the reverse of the direction in which the original region is traversed.
In this case we have among other things u = x^2. u is not linear in x, so the transformation is not linear. The Jacobian matrix would not consist of just constants. Also we don't expect straight lines to always map to straight lines, so the image of the mapping would not be a parallelogram.
However your estimate of about 4.5 times the area looks about right.*@
Given Solution:
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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:
square in the xy-plane with vertices (0,0), (1,-2) has the boundries 0<=x<=1 and -2<=y<=0
Solving for x and y w/ u = 1/5(2x + y) and v = 1/5(x - 2y) we get the system of equations
(1/5)( 2x + y) = u
(1/5)(x - 2y) = v
We get
x =2u+v
y =u - 2v
So our Jacobian is
(2*(-2)) - ((-1)*1) =|- 5|
We now can look at the integral
Int(Int(cos(2x + y)*sin(x - 2y) dy dx) = Int(Int(cos(5u)*sin(5v)*|-5| du dv)
Need some help, can’t quite figure out what the boundries are . When x = 0 u = (1/2)v and when y = 0 v = (1/2)u. I see the the x(triangular region) formed by the 2 lines centered at (0,0) but can’t put it together.
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Given Solution:
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Self-critique (if necessary):
@& It doesn't appear to me that the square maps to a square, since (0, 0) maps to (0, 0), (1, 0) mapes to (2/5 , 1/5), (0, -2) maps to (-2/5, 4/5) and (1, -2) maps to (0, 1).
What should happen is that the region does map to a square in the u-v plane. Since the transformation is linear, the Jacobian is constant. So the ratio of the area of a region of the u-v plane to that of the corresponding region in the x-y plane will be the Jacobian.
The region in the u-v plane corresponding to the given square in the x-y plane is a rectangle with sides sqrt(5) / 5 and 2 sqrt(5) / 5, so its area is 2/5.
The area of the original square was 2. The area ratio is therefore 1/5.
Noting that the vertices of the mapped region are traversed counterclockwise when the original vertices are traversed counterclockwise, we conclude that the Jacobian is -1/5.
If you calculate the Jacobian you confirm this, getting det( [ 2/5, 1/5; 1/5, -2/5 ] ) = - 1/5.
This means that we could find the integral of cos(2 x + y) * sin(x - 2 y) in the x-y plane by finding the integral of cos(u) sin(v) in the u-v plane, the multiplying the result by -5.
The region is a little inconvenient (it would have to be broken into 3 subregions) but the boundaries are all straight lines, so it wouldn't be really difficult.
To integrate cos(2x + y) * sin(x - 2 y) in x and y would take a little work. It probably turns out to be easier to do the integration in the u-v system.*@
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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:
Jacobian is
[cos(`theta)* cos(`theta)] - [-sin(`theta)* sin(`theta)]
= cos(`theta)^2 + sin(`theta)^2 = 1
Next we find what transformation gives us
(u cos(theta) - v sin(theta))^2 + (u cos(theta) - v sin(theta))( u sin(theta) + v cos(theta)) + (u sin(theta) + v cos(theta))^2 = 9
=[u(cos^2 + sin^2) + v(cos^2+sin^2) + (2uv*(sin(`theta)cos(`theta)) - 2uv*(cos(`theta)sin(`theta))]
@& A rotation of pi/4 is provided by
x = u cos(pi/4) - v sin(pi/4) = sqrt(2) / 2 ( u - v)
y = ... = sqrt(2 ) / 2 ( u + v).
With this transformation the equation becomes
1/2 ( u - v) ^2 + 1/2 (u - v) ( u + v) + 1/2 (u + v)^2 = 9,
or
3/2 u^2 + 1/2 v^2 = 9
This ellipse is centered at the origin and is symmetric with respect to both u and v axes. Its semi-axis in the u direction is sqrt(6); in the v direction it is 3 sqrt(2)
In the first quadrant it is bounded above by the curve
u = sqrt( 18 - 3 u^2)
so its area in this quadrant is
integral(sqrt(18 - 3 u^2) du, u from 0 to sqrt(6) ) = 3 sqrt(3) pi / 2 (approximately 8.2).
The area of the entire ellipse is 4 times this, 6 sqrt(3) pi or about 33.
Since the Jacobian is 1, the area of the ellipse in the x-y plane is the same.
The ellipse in the x-y plane, when rotated pi / 4 or 45 deg in the positive direction, is the ellipse in the u-v plane. So if the ellipse in the u-v plane is rotated 45 degrees in the negative direciton, we will have the ellipse in the x-y plane.
*@
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Given Solution:
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&#Your work looks good. See my notes. Let me know if you have any questions. &#