5128_ch04ansTE_pp652-661
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658 Additional Answers<br />
63. (a) y(0) 0 (b) y(L) 0<br />
(c) y(0) 0, so d 0. y(0) 0, so c 0.<br />
Then y(L) aL 3 bL 2 H and<br />
y(L) 3aL 2 2bL 0.<br />
H H<br />
Solving, a 2 L 3 and b 3 L 2, which gives the equation shown.<br />
64. (a) V(x) 3 2 a x<br />
<br />
2<br />
2 a2 <br />
a<br />
2 <br />
x<br />
2<br />
2<br />
(b) When a 4: r 4 6<br />
, h 4 3<br />
;<br />
3 3<br />
when a 5: r 5 6<br />
, h 5 3<br />
;<br />
3 3<br />
when a 6: r 26, h 23;<br />
when a 8: r 8 6<br />
, h 8 3<br />
<br />
3 3<br />
(c) h<br />
r 2<br />
a<br />
65. (a) The x- and y-intercepts of the line through R and T are x and f (x)<br />
a xf(x) respectively.<br />
The area of the triangle is the product of these two values.<br />
(b) Domain: (0, 10)<br />
2. (a) L(x) 4 5 x 9 5 <br />
(b) Differs from the true value in absolute value by less than 10 3<br />
3. (a) L(x) 2<br />
(b) Differs from the true value in absolute value by less than 10 2<br />
4. (a) L(x) x<br />
(b) Differs from the true value in absolute value by less than 10 2<br />
5. (a) L(x) x <br />
(b) Differs from the true value in absolute value by less than 10 3<br />
6. (a) L(x) x 2 <br />
(b) Differs from the true value in absolute value by less than 10 3<br />
7. f(0) 1. Also, f (x) k(1 x) k1 , so f (0) k.<br />
This means the linearization at x 0 is L(x) 1 kx.<br />
19. (a) dy (3x 2 3) dx<br />
(b) dy 0.45 at the given values<br />
2 2x2<br />
20. (a) dy (1 x2) 2 dx<br />
(b) dy 0.024 at the given values<br />
21. (a) dy (2x ln x x) dx<br />
(b) dy 0.01 at the given values<br />
1 2x2<br />
22. (a) dy (1 x2) 1 /2 dx<br />
(b) dy 0.2 at the given values<br />
23. (a) dy (cos x) e sin x dx<br />
(b) dy 0.1 at the given values<br />
24. (a) dy csc 1 3<br />
x <br />
cot 1 3<br />
x <br />
dx<br />
The vertical asymptotes at x 0 and x 10 correspond to horizontal<br />
or vertical tangent lines, which do not form triangles.<br />
(c) Height 15, which is 3 times the y-coordinate of the center of the<br />
ellipse.<br />
(d) Part (a) remains unchanged.<br />
The domain is (0, C) and the graph is similar.<br />
The minimum area occurs when x 2 3C 2<br />
. From this, it follows that<br />
4<br />
the triangle has minimum area when its height is 3B.<br />
Section 4.5<br />
Quick Review 4.5<br />
9.<br />
10.<br />
[0, 10] by [–100, 1000]<br />
[0, π] by [–0.2, 1.3]<br />
(b) dy 0.205525 at the given values<br />
25. (a) dy (x <br />
dx1) 2<br />
(b) dy 0.01 at the given values<br />
x y<br />
26. (a) dy 2 2 x<br />
dx<br />
(b) dy 0.0375 at the given values<br />
53. Since V 4 3 r3 , we have dV 4r 2 dr 4r 2 1<br />
16<br />
r 2<br />
.<br />
4<br />
The volume error in each case is simply r 2<br />
in 3 .<br />
4<br />
Sphere Type True Radius Tape Error Radius Error Volume Error<br />
Orange 2" 1/8" 1/16 " 1 in 3<br />
Melon 4" 1/8" 1/16 " 4 in 3<br />
Beach Ball 7" 1/8" 1/16 " 12.25 in 3<br />
54. Since A 4r 2 1 r<br />
, we have dA 8rdr 8r <br />
16 <br />
. 2<br />
The surface area error in each case is simply 2<br />
r in 2 .<br />
Sphere Type True Radius Tape Error Radius Error S. Area Error<br />
Orange 2" 1/8" 1/16 " 1 in 2<br />
Melon 4" 1/8" 1/16 " 2 in 2<br />
Beach Ball 7" 1/8" 1/16 " 3.5 in 2<br />
1<br />
64. If x 1 h, then f(x 1 ) 2h 1/2 and x 2 h h 1/2<br />
<br />
1<br />
<br />
h 2h h. If x 1 h,<br />
2h 1/2<br />
1<br />
then f(x 1 ) 2h 1/2 and x 2 h 2h h.<br />
Exercises 4.5<br />
1. (a) L(x) 10x 13<br />
(b) Differs from the true value in absolute value by less than 10 1