PRACTICE PROBLEMS FOR EXAM 2
1. Let f (x, y) have continuous second partial derivatives, and let x = st and y = est .
(a) Find ∂x/∂t and ∂y/∂t.
(b) Find ∂f /∂t in terms of ∂f /∂x, ∂f /∂y, s and t.
(c) Find ∂ 2 f /∂t2 in terms of ∂ 2 f /∂x2 , ∂ 2 f /∂x∂y, ∂ 2 f /∂y 2 , ∂f /∂x, ∂f /∂y, s and t.
2. Consider the function f (x, y) = 3x2 − xy + y 3 .
(a) Find the rate of change of f at (1, 2) in the direction of v = 3i + 4j.
(b) In what direction (unit vector) does f decrease at (1, 2) at the maximum rate ?
What is this maximum rate of change?
(c) In what directions is the rate of change of f at (1, 2) equal to zero? Your answer
should be a pair of opposite unit vectors.
3. Suppose the gradient ∇f (2, 4) of a function f (x, y) has length equal to 5. Is there
a direction u such that the directional derivative Du f at the point (2, 4) is 7? Explain
your answer.
4. Find the tangent plane to the ellipsoid x2 +4y 2 = 169−9z 2 at the point P = (3, 2, 4).
5. Find the points on the surface (ellipsoid) x2 + 2y 2 + 4z 2 + xy + 3yz = 1 where the
tangent plane is parallel to the xz plane.
6. Find all the critical points of f (x, y) = x2 + y 2 /2 + x2 y and apply the second
derivative test to each of them.
7. Find the absolute maximum and minimum values of the function f (x, y) = (x −
1)2 + (y − 1)2 in the rectangular domain D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2}. Justify
your answer.
8. Find the maximum of f (x, y) = xy restricted to the curve (x + 1)2 + y 2 = 1. Give
both the coordinates of the point and the value of f .
9. Find the dimensions of a rectangular box of maximum volume such that the sum of
the lengths of its 12 edges is a constant C.
RR
10. Compute D (3x + 1) dxdy where D is the region in the first quadrant bounded
by the parabolas y = x2 and y = (x − 1)2 and the y–axis.
2
11. Change the order of integration in the following iterated integral :
Z
Z √ 4−y2
2
f (x, y) dx dy.
−2
0
12. (a) Find the area of the region enclosed by the cardioid given in polar coordinates
by r = 1 + cos(θ). (b) Use polar coordinates to evaluate
√
Z
√
2Z
0
4−y 2
y
1
dx dy.
1 + x2 + y 2
SOLUTIONS
1. (a) xt = s, yt = s est ;
(b) ft = s fx + s est fy ;
(c) ftt = s2 est fy + s2 fxx + 2s2 est fxy + s2 e2st fyy .
2. (a) u =
3
4
v
= i + j,
|v|
5
5
∇f
(b) −
(1, 2) =
|∇f |
(Du f )(1, 2) = (∇f )(1, 2) · u = 56/5.
4
11
−√
, −√
.
137
137
(c)
11
4
11
4
√
, −√
, −√
,√
.
137
137
137
137
3. (a) No, because |(Du f )(2, 4)| = |(∇f )(2, 4)||u| cos θ where θ is the angle between
u and (∇f )(2, 4). If |(Du f )(2, 4)| = 7 then 7 = 5 cos θ which is impossible because
cos θ ≤ 1.
∂f
∂f
4. 3x + 8y + 36z = 169.
5. The points where
=
= 0, i.e.
∂x
∂z
4
3
2
4
3
2
√ , −√ , √
−√ , √ , − √
and
.
19
19 2 19
19
19 2 19
6. (0, 0) is a local minimum, (1, −1) and (−1, −1) are saddles.
7. Minimum 0, maximum 2.
9. C/12, C/12, C/12.
Z
11.
0
2Z
√
√
√
8. Maximum f (−3/2, − 3/2) = 3 3/4.
10. 3/8.
4−x2
√
− 4−x2
f (x, y) dy dx.
12. (a) 3π/2. (b)
π
ln 5.
8