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finalsol_s07 - MATH 2400 CALCULUS 3 MAY 9 2007 FINAL EXAM I...

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MATH 2400: CALCULUS 3 MAY 9, 2007 FINAL EXAM I have neither given nor received aid on this exam. Name: 001 E. Kim . . . . . . . . . . . . . . . . (9am) 002 E. Angel . . . . . . . . . . . . . (10am) 003 I. Mishev . . . . . . . . . . . . (11am) 004 M. Daniel . . . . . . . . . . . (12am) 005 A. Gorokhovsky . . . . . . (1pm) If you have a question raise your hand and remain seated. In order to receive full credit your answer must be complete , legible and correct . Show all of your work, and give adequate explanations. DO NOT WRITE IN THIS BOX! Problem Points Score 1 15 pts 2 15 pts 3 15 pts 4 15 pts 5 20 pts 6 15 pts 7 30 pts 8 15 pts 9 15 pts 10 30 pts 11 15 pts TOTAL 200 pts
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1. (15 pt) Find the area of the region enclosed by the curve x = 1 - t 2 , y = t (1 - t 2 ) , - 1 t 1 (see the picture below) 1. (15 pt) Find the area of the region enclosed by the curve x = 1 - t 2 , y = t (1 - t 2 ) , - 1 t 1 (see the picture below) 0 0.5 1 -0.5 0.5 2. (15 pt) Find the equation for the plane tangent to the paraboloid z = 2 x 2 + 3 y 2 that is also parallel to the plane 4 x - 3 y - z = 10 By Green’s Theorem, A = C - y dx with C the above curve oriented counter-clockwise. Thus A = C - y dx = 1 - 1 - t (1 - t 2 )( - 2 t ) dt = 2 1 - 1 ( t 2 - t 4 ) dt = 8 15 . 2. (15 pt) Find the equation for the plane tangent to the paraboloid z = 2 x 2 + 3 y 2 that is also parallel to the plane 4 x - 3 y - z = 10 Write G ( x, y, z ) = 2 x 2 + 3 y 2 - z , so G ( x, y, z ) = 4 x, 6 y, - 1 . This tells us that 4 x, 6 y, - 1 is a normal vector for the tangent plane to our surface at the point ( x, y, z ). For the tangent plane at a point to be parallel to 4 x - 3 y - z = 10, we need 4 x, 6 y, - 1 to be parallel to 4 , - 3 , - 1 . This will certainly happen when x = 1 and y = - 1 2 . Solving for z in the equation for our surface, we get that the tangent plane at (1 , - 1 2 , 11 4 ) is parallel to 4 x - 3 y - z = 10. At this point the equation for the tangent plane will be given by 4( x - 1) - 3( y + 1 2 ) - ( z - 11 4 ) = 0.
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3. (15 pt) Find the flux of F ( x, y, z ) = ( x + y ) i + ( y + z ) j + ( z + x ) k across the portion of the plane x + y + z = 1 in the first octant oriented by unit normals with positive components.
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