finalsol_s07

# finalsol_s07 - MATH 2400: CALCULUS 3 MAY 9, 2007 FINAL EXAM...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 = I C - y dx with C the above curve oriented counter-clockwise. Thus A = I C - y dx = Z 1 - 1 - t (1 - t 2 )( - 2 t ) dt = 2 Z 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 ) = h 4 x, 6 y, - 1 i . This tells us that h 4 x, 6 y, - 1 i 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 h 4 x, 6 y, - 1 i to be parallel to h 4 , - 3 , - 1 i . 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.
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. We know Φ =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/13/2008 for the course MATH 1012 taught by Professor N/a during the Fall '07 term at Colorado.

### Page1 / 8

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online