Answers to Math 261 SP15 Exam 2
The following are just the final numerical (or otherwise) answers for Math 261 SP15 Exam 2.
Your solutions on the exam should include full details to allow for partial credit.
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hours.
1. (a)
∂f
∂y
=
∂f ∂r
∂r ∂y
+
∂f ∂s
.
∂s ∂y
(b) eπ.
2. (a) F (e, 1, 0) = 0 (so the point is indeed on the surface).
(b) 1e x + y + ez = 2.
3. (a) L(x, y) = 4 + 4(x − 1) + 2(y − 2).
(b) 12 (4.4)(0.3)2 .
2
= −144 < 0, so saddle point.
4. (a) fxx fyy − fxy
2
= 432 > 0 and fxx = 12, so local minimum.
(b) (2, 1). fxx fyy − fxy
5. (1, 2).
6. (a) Answer is the sketch of a triangle with corners (0, 0), (0, 4), and (2, 4).
Z 4 Z y/2
(b)
ex/y dxdy.
0
0