Math 113 - Review Worksheet 2 answers - 10/7/2015
1. (a) It exists and equals −1/6.
(b) It does not exist.
(c) It exists and equals 0.
2. The z-value is increasing as we move in the positive x or y directions, and so fx > 0 and
fy > 0. The cross-sections cut parallel to the x-axis and y-axis are concave down, and so
fxx < 0 and fyy < 0. [ The surface is the plot of f (x, y) = (x2 − 2x)(y 2 − 3y). ]
3. (x2 y 2 z 2 + 3xyz + 1)exyz .
4. (a) z = 2 + 2(x − 1) + 2(y − 1).
√
√ 2
π 2
1
√
(b) z = 2 + 2 − 8 (x − 1) −
√
2
2
y−
π
4
.
(c) z = (e + 1) + 2e(x − 1) + (e + 2)(y − 1).
(d) 4(x − 4) + 7(y − 7) − 8(z − 8) = 0.
5. The approximate value is 1.
6. We have
∂u
∂t
= 9.
√
7. We have D~u f (1, 1) = 6 2.
√ √
8. We have ∇f (0, 0) = h− 2, 2 2i.
9. There are four critical points: (0, 1), (0, −1), (−2, 1) and (2, −1). The first two are saddle
points, the third is a local max, and the fourth is a local min.
10. At the point on the paraboloid that is closest to the plane, the normal vector of the tangent
plane to the paraboloid will be parallel to the normal vector of x + y + z = 1. The answer is
(−1/2, −1/2, 11/2).
11. The direction ~vold = h 24
, − 18
, −12i works.
5
5
12. Use the equality of mixed partials.
13. There are nine real critical points (Maple sometimes thinks there is a really small imaginary
part, but the solutions are in fact real). The points (−1.185, 0.6854), (0.6854, −1.185) are local
mins, (0.9222, 0.9222), (−0.4399, −0.4399), (−1.232, −1.232), and (−0.3984, 1.960) are local
maxima, and the points (1.960, −0.3984), (−1.880, −0.6811) and (−0.6811, −1.880) are saddle
points.
R1
14. (a) f (t) = 0 Gx (x, t) dt = G(1, t) − G(0, t).
(b) f 0 (t) =
d
dt
[G(1, t) − G(0, t)] = Gy (1, t) − Gy (0, t).
1
2
R1
∂
(c) f 0 (t) = 0 Gyx (x, t) dx. Since G is a polynomial Gyx = Gxy and Gxy = ∂y
(x2 + y)10000 =
10000(x2 + y)9999 . We get
Z 1
Z 1
Z 1
10000
0
Gyx (x, 0) dx =
Gxy (x, 0) dx =
10000x19998 =
.
f (0) =
19999
0
0
0
15. fxx (0, 0) − 2fxy (0, 0) + fyy (0, 0) + 6fx (0, 0) + 4fy (0, 0).
16. The limit exists and equals zero. A proper proof needs to use the squeeze theorem or polar
coordinates.
√
17. The minimum area is 6.