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Calculus 3 Past Paper Final Exam ;Questions and Answers, guaranteed 100% Pass

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Calculus 3 Past Paper Final Exam ;Questions and Answers, guaranteed 100% PassCalculus 3 Past Paper Final Exam ;Questions and Answers, guaranteed 100% PassCalculus 3 Past Paper Final Exam ;Questions and Answers, guaranteed 100% PassCalculus 3 Past Paper Final Exam ;Questions and Answers, guaranteed 100% PassCalculus 3 Past Paper Final Exam ;Questions and Answers, guaranteed 100% PassCalculus 3 Past Paper Final Exam ;Questions and Answers, guaranteed 100% PassCalculus 3 Past Paper Final Exam ;Questions and Answers, guaranteed 100% Pass

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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) 004 M. Daniel . . . . . . . . . . . (12am)
002 E. Angel . . . . . . . . . . . . .(10am) 005 A. Gorokhovsky . . . . . . (1pm)
003 I. Mishev . . . . . . . . . . . . (11am)



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

, 1. (15 pt) Find the area of the region enclosed by the curve
1. (15 pt) Find the area of the region enclosed by the curve
x = 1 − t22, y = t(1 − t22), −1 ≤ t ≤ 1
x = 1 − t , y = t(1 − t ), −1 ≤ t ≤ 1
(see the picture below)
(see the picture below)

0.5




0 0.5 1




-0.5




I
By Green’s Theorem, A = −y dx with C the above curve oriented counter-clockwise. Thus
I Z 1 C Z 1
2 8
A= −y dx = −t(1 − t )(−2t)dt = 2 (t2 − t4 )dt = .
C −1 −1 15




2. (15 pt) Find the equation for the plane tangent to the paraboloid z = 2x2 + 3y 2 that is also
parallel to the plane 4x − 3y − z = 10



2. (15 pt) Find the equation for the plane tangent to the paraboloid z = 2x2 + 3y 2 that is also
parallel to the plane 4x − 3y − z = 10



Write G(x, y, z) = 2x2 + 3y 2 − z, so ∇G(x, y, z) = h4x, 6y, −1i. This tells us that h4x, 6y, −1i
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 4x − 3y − z = 10, we need h4x, 6y, −1i to be parallel to
h4, −3, −1i. This will certainly happen when x = 1 and y = − 21 . Solving for z in the equation
for our surface, we get that the tangent plane at (1, − 12 , 11
4
) is parallel to 4x − 3y − z = 10. At
this point the equation for the tangent plane will be given by 4(x − 1) − 3(y + 21 ) − (z − 114
) = 0.
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