Math 251
Sample Exam 1 Answers
1. C
2. D
3. E
4. E
5. B
6. D
7.
3x2
6xy 2 − 3x2 y
∂ 2g
= 6x ln(x − y) +
+
.
∂x ∂y
x−y
(x − y)2
∂f
∂g
∂f
∂f
∂f
∂g
=
(t+3u, teu )+eu (t+3u, teu ) and
= 3 (t+3u, teu )+teu (t+3u, teu ).
∂t
∂x
∂y
∂u
∂x
∂y
√
−→
−→
9. (a) We have QR = h2, 3, −3i, and so |QR| = 22.
−→ −→
−→
(b) The area will be 21 |QP × QR|. We find QP = h4, −2, 1i, and then calculate
√
−→
1 −→
1
461.
|
QP
×
QR|
=
2
2
−→ −→
−→ −→
(c) Yes, the angle is obtuse. The reason is that QP · QR = −1, but also QP · QR =
−→ −→
cos(∠P QR)|QP | |QR|. This means that cos(∠P QR) must be negative, so ∠P QR is
obtuse.
8.
10. (a) The equation of the tangent plane at (2, −1, 9) is 3(x − 2) − (y + 1) − (z − 9) = 0.
(b) To find such points, we need to find points where fx (x, y) = 0 and fy (x, y) = 0
simultaneously. This amounts to solving the equations 2x + y = 0 and −5 + x − 2y = 0.
The only solution is x = 1, y = −2, and so the point we are looking for is (1, −2, 8).