5128_ch04ansTE_pp652-661

658 Additional Answers
63. (a) y(0) 0 (b) y(L) 0
(c) y(0) 0, so d 0. y(0) 0, so c 0.
Then y(L) aL 3 bL 2 H and
y(L) 3aL 2 2bL 0.
H H
Solving, a 2 L 3 and b 3 L 2, which gives the equation shown.
64. (a) V(x) 3 2 a x
2
2 a2
a
2
x
2
2
(b) When a 4: r 4 6
, h 4 3
;
3 3
when a 5: r 5 6
, h 5 3
;
3 3
when a 6: r 26, h 23;
when a 8: r 8 6
, h 8 3
3 3
(c) h
r 2
a
65. (a) The x- and y-intercepts of the line through R and T are x and f (x)
a xf(x) respectively.
The area of the triangle is the product of these two values.
(b) Domain: (0, 10)
2. (a) L(x) 4 5 x 9 5
(b) Differs from the true value in absolute value by less than 10 3
3. (a) L(x) 2
(b) Differs from the true value in absolute value by less than 10 2
4. (a) L(x) x
(b) Differs from the true value in absolute value by less than 10 2
5. (a) L(x) x
(b) Differs from the true value in absolute value by less than 10 3
6. (a) L(x) x 2
(b) Differs from the true value in absolute value by less than 10 3
7. f(0) 1. Also, f (x) k(1 x) k1 , so f (0) k.
This means the linearization at x 0 is L(x) 1 kx.
19. (a) dy (3x 2 3) dx
(b) dy 0.45 at the given values
2 2x2
20. (a) dy (1 x2) 2 dx
(b) dy 0.024 at the given values
21. (a) dy (2x ln x x) dx
(b) dy 0.01 at the given values
1 2x2
22. (a) dy (1 x2) 1 /2 dx
(b) dy 0.2 at the given values
23. (a) dy (cos x) e sin x dx
(b) dy 0.1 at the given values
24. (a) dy csc 1 3
x
cot 1 3
x
dx
The vertical asymptotes at x 0 and x 10 correspond to horizontal
or vertical tangent lines, which do not form triangles.
(c) Height 15, which is 3 times the y-coordinate of the center of the
ellipse.
(d) Part (a) remains unchanged.
The domain is (0, C) and the graph is similar.
The minimum area occurs when x 2 3C 2
. From this, it follows that
4
the triangle has minimum area when its height is 3B.
Section 4.5
Quick Review 4.5
9.
10.
[0, 10] by [–100, 1000]
[0, π] by [–0.2, 1.3]
(b) dy 0.205525 at the given values
25. (a) dy (x
dx1) 2
(b) dy 0.01 at the given values
x y
26. (a) dy 2 2 x
dx
(b) dy 0.0375 at the given values
53. Since V 4 3 r3 , we have dV 4r 2 dr 4r 2 1
16
r 2
.
4
The volume error in each case is simply r 2
in 3 .
4
Sphere Type True Radius Tape Error Radius Error Volume Error
Orange 2" 1/8" 1/16 " 1 in 3
Melon 4" 1/8" 1/16 " 4 in 3
Beach Ball 7" 1/8" 1/16 " 12.25 in 3
54. Since A 4r 2 1 r
, we have dA 8rdr 8r
16
. 2
The surface area error in each case is simply 2
r in 2 .
Sphere Type True Radius Tape Error Radius Error S. Area Error
Orange 2" 1/8" 1/16 " 1 in 2
Melon 4" 1/8" 1/16 " 2 in 2
Beach Ball 7" 1/8" 1/16 " 3.5 in 2
1
64. If x 1 h, then f(x 1 ) 2h 1/2 and x 2 h h 1/2
1
h 2h h. If x 1 h,
2h 1/2
1
then f(x 1 ) 2h 1/2 and x 2 h 2h h.
Exercises 4.5
1. (a) L(x) 10x 13
(b) Differs from the true value in absolute value by less than 10 1