#$&* course Mth 277 Question: `q001. Find the Jacobian d(x,y,z)/d(u,v,w) when x = 2u - v, y = 2v + 2w, z = v - w.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Let R be the parallelogram with vertices (0,0), (1,4), (4,6), (4,2). Sketch and decribe the corresponding region after the transformation u = x^2, v = x+ y. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: New vertices (0,0),(1,5),(16,10),(16,6) As you can see, the parallelogram also has a vertice at the origin, but is obviously much larger than the original parallelogram. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!