Michael Corral: Vector Calculus

190 Appendix A: Answers and Hints to Selected Exercises
parallel to c (c) Hint: Think of the
functions as position vectors.
15. Hint: Theorem 1.16
Section 1.9 (p. 63)
1. 3π√ 5 2
2
3.
((
t by
27s+16
2
27 (133/2 −8) 5. Replace
) 2/3−4 )/
9 6. Hint: Use
Theorem 1.20(e), Example 1.37, and
Theorem 1.16 7. Hint: Use Exercise 6.
9. Hint: Use f ′ (t) = ‖f(t)‖T, differentiate
that to get f ′′ (t), put those expressions
into f ′ (t) × f ′′ (t), then write
T ′ (t) in terms of N(t). 11. T(t) =
√
1
(−sint,cost,1), N(t) = (−cost,−sint,0),
2
B(t)= 1 √
2
(sint,−cost,1),κ(t)=1/2
Chapter 2
Section 2.1 (p. 70)
1. domain: 2 , range: [−1,∞) 3. domain:
{(x,y) : x 2 +y 2 ≥ 4}, range: [0,∞)
5. domain: 3 , range: [−1,1] 7. 1
9. does not exist 11. 2 13. 2 15. 0
17. does not exist
Section 2.2 (p. 74)
1. ∂f ∂f ∂f
∂x
= 2x,
∂y
= 2y 3.
∂x = x(x2 +y+4) −1/2 ,
∂f
∂y
= 1 2 (x2 + y+4) −1/2 ∂f
5.
∂x
= ye xy + y,
∂f
∂y = xexy + x 7. ∂f
∂x = 4x3 , ∂f
∂y = 0
9. ∂f
∂x = x(x2 +y 2 ) −1/2 , ∂f
∂y = y(x2 +y 2 ) −1/2
11. ∂f
∂x = 2x
3 (x2 +y+4) −2/3 ,
∂f
∂y = 1 3 (x2 +y+4) −2/3 13. ∂f
∂x =−2xe−(x2 +y 2) ,
∂f
∂y =−2ye−(x2 +y 2 )
15. ∂f
∂x = ycos(xy),
∂f
∂y = xcos(xy) 17. ∂2 f
= 2, ∂2 f
= 2,
∂x 2 ∂y 2
∂ 2 f
∂x∂y = 0 19. ∂2 f
= (y+4)(x 2 +y+4) −3/2 ,
∂x 2
∂ 2 f
=− 1 ∂y 2 4 (x2 +y+4) −3/2 ,
∂ 2 f
∂x∂y =−1 2 x(x2 +y+4) −3/2
21. ∂2 f
= y 2 e xy , ∂2 f
= x 2 e xy ,
∂x 2 ∂y 2
∂ 2 f
∂x∂y = (1+ xy)exy +1 23. ∂2 f
∂ 2 f
= 0,
∂y 2
∂ 2 f
=−y −2 ,
∂y 2
∂ 2 f
∂x∂y = 0 25. ∂2 f
∂ 2 f
∂x∂y = 0
Section 2.3 (p. 77)
∂x 2 = 12x 2 ,
∂x 2 =−x −2 ,
1. 2x+3y−z−3=0 3.−2x+y−z−2=0
5. x+2y=z 7. 1 2 (x−1)+ 4 9 (y−2)+ √
11
12 (z−
2 √ 11
3
)=0 9. 3x+4y−5z=0
Section 2.4 (p. 82)
x
1. (2x,2y) 3. ( √ , y
√ )
x 2 +y 2 +4 x 2 +y 2 +4
5. (1/x,1/y) 7. (yzcos(xyz),xzcos(xyz),xycos(xyz))
9. (2x,2y,2z) 11. 2 √ 2 13.
1 √3
15. √ 3 cos(1) 17. increase: (45,20),
decrease: (−45,−20)
Section 2.5 (p. 88)
1. local min. (1,0); saddle pt. (−1,0)
3. local min. (1,1); local max. (−1,−1);
saddle pts. (1,−1),(−1,1) 5. local min.
(1,−1); saddlept. (0,0) 7. localmin. (0,0)
9. local min. (−1,1/2) 11. width = height
= depth=10 13. x=y=4, z=2
Section 2.6 (p. 95)
2. (x 0 ,y 0 ) = (0,0) : → (0.2858,−0.3998);
(x 0 ,y 0 )=(1,1) :→(1.03256,−1.94037)