LB 220 Review Problems for Midterm Exam III
1. (13.3 #24) Find fx and fy for f (x, y) = ln [(x + y)/(x − y)]. Answer:
fx = 1/(x + y) − 1/(x − y), fy = 1/(x + y) + 1/(x − y); (use properties of
logarithms to avoid the quotient rule)
Ry
Rx
2. (13.3 #40) Find fx and fy for f (x, y) = x (2t + 1) dt + y (2t − 1) dt.
Answer: fx = 2x + 1 − (2x − 1) = 2, fy = −(2y + 1) + (2y − 1) = −2
3. (13.3 #82) Find all points (x, y) such that fx (x, y) = 0 and fy (x, y) = 0
where f (x, y) = x2 − xy + y 2 − 5x + y. Answer: (3, 1)
4. (13.3 #100) Show that z = arctan (y/x) satisfies Laplace’s equation zxx +
zyy = 0. Answer: carefully compute the derivatives; here’s a fancy solution: z = θ, −π/2 < θ < π/2 in polar coordinates and this clearly satisfies
zrr + r−2 zθθ + r−1 zr = 0.
√
5. (13.3 #110) Find fx where
f (x) = x(sinh (y/z))(y 2 − 2 y − 1)z. Answer:
√
fx = (sinh (y/z))(y 2 − 2 y − 1)z.
6. (13.3 #130) True or false: if z = f (x)g(y), then zx + zy = f 0 (x)g(y) +
f (x)g 0 (y)? Answer: True
7. (13.4) True or false: a function f (x, y) is differentiable if it has a tangent
plane at each of the points in its domain? Answer: True
8. (13.4) True or false: if fx and fy exist and are continuous, then f is
differentiable. Answer: False
9. (13.5) True or false: if f (x, y), x(t), and y(t) are differentiable and x0 (t) >
0 and y 0 (t) > 0, then f 0 (t) > 0. Answer: False
10. (13.5) True or false: if (a, b) is a critical point of a differentiable function
f (x, y) and x(t) and y(t) are differentiable, then f 0 (t) = 0? Answer: True
11. (13.5 #4) Find wt where w = ln (y/x), x = cos t, and y = sin t. Answer:
wt = cot t sec2 t.
12. (13.5 #24) Find ws and wt where w = x2 + y 2 + z 2 , x = t sin s, y = t cos s,
z = st2 . Answer: ws = 2st4 , wt = 2t + 4s2 t3
13. (13.5 #28) Differentiate implicitly to find dy/dx where sec (xy)+tan (xy)+
5 = 0. Answer: dy/dx = −y/x
14. (13.6) Suppose the temperature at a point (x, y) is given by T (x, y) = x2 +
2y 2 . Determine how fast the temperature of a particle moving along the
trajectory x(t) = cos t, y(t) = sin(t) is changing when t = π/4. Answer: 1
15. (13.6 #14) Determine Du f (x, y) for f (x, y) = y/(x + y), where u is the
unit
√ vector which makes an angle of π/6 with the positive x-axis. Answer:
−( 3y + x)/[2(x + y)2 ]
1
16. (13.6 #30) Use the gradient to determine the directional derivative of
−−→
f (x, y) = sin (2x) cos y at√the point P (π, 0) in the direction of P Q where
Q(π/2, π). Answer: −2/ 5
17. (13.6 #40)
Determine the gradient of f (x, y, z) = xeyz at (2, 0, −4). An√
swer: 65
18. (13.6 #46) Find a unit vector u orthogonal to ∇f (3, 2), where f (x, y) =
3 − x/3 − y/2. Then compute Du√f (3, 2). What is the geometric meaning
of the result? Answer: u = (1/ 13)h3, −2i, zero, the function has zero
rate of change in a direction orthogonal to that of the gradient.
2
19. (13.6 #56) Find a unit
√ normal vector to the curve 3 = x − y at the point
(4, −1). Answer: (1/ 5)h1, 2i
2
2
2
20. (13.6) Find a unit normal vector
√ to the surface 9 = x + 2y + 3z at the
point (2, 1, 1, ). Answer: (1/ 17)h2, 2, 3i
21. (13.7 #40) Determine an equation of the tangent plane and the normal
line to the surface y ln (xz 2 ) = 2 at the point (e, 2, 1). Answer: (2/e)(x −
e)+(y −2)+4(z −1) = 0, (x−e)/(2/e) = (y −2)/1 = (z −1)/4 (symmetric
equations of the normal line)
22. (13.8) Determine where the function f (x, y) = x2 − xy + y 2 − 5x + y has
any relative extrema or saddle points. State whether any such points are
where f has a relative min, relative max, or saddle point. Answer: Critical
point at (3, 1), relative minimum
23. (13.8 #46) Let f (x, y) = x2 + xy. Let R = {(x, y) | |x| ≤ 2, |y| ≤ 1}.
Determine the absolute extreme values of f on R. Answer: absolute max
6, absolute min −1/4
24. (13.9 $12) Show that the rectangular box of maximum volume inscribed
in a sphere of radiuspr is a cube. Let V = 8xyz, where x2 + y 2 + z 2 = r2 .
Then f (x, y) = 8xy r2 − x2 − y 2 has a critical point when the
√volume is
maximum. Solving ∇f (x, y) = 0, we find that x = y = z = r/ 3.
R3Ry
25. (14.1 #26) Compute 1 0 4/(x2 + y 2 ) dx dy. Answer: π ln 3
26. (14.1 #68) Sketch the region of integration, switch the order of integration
and evaluate the following integral:
Z 4Z 2
3
dy dx
√ 2 + y3
0
x
Answer:
R 2 R y2
0
0
3/(2 + y 3 ) dx dy = ln 5
2