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Unformatted text preview: 166 Section 4.4 53. (a) According to the graph, y (0) 0. (b) According to the graph, y ( L)
(c) y(0) 0, so d c 2bx r
h 3ax 2 bx 2 and 3aL 2 and y ( L) 2bL bL 2 H 0, so we have two linear 2. and let m y 3aL
. Substituting into the first
2
3aL 3
aL 3
equation, we have aL 3
H, or
H, so
2
2
H
H
a 2 3 . Therefore, b 3 2 and the equation for y is
L
L
H
H
x3
x2
y(x) 2 3 x 3 3 2 x 2, or y(x) H 2
3
.
L
L
L
L m(0 equation gives b 2a x
and so the
2
2a x2
.
2 a2 height is h r2 a2 r 2h 3 x2 2a
2 3 x2 2a
2 a2 m(x x0) x0) m(x x0) a a 3 d2
(r
3 dr f (r)
= a2 a2 2
r 3 3 a2 2a 2r
3 a2
r(2a 2 2 2 a2 h
r a 6
3 r 2)(2r) mx0 m x0 have A(x) y2 2 2a 2
, which gives
3 2 a6
. Then
3 a2 mx0 a m a m
a2 m
a2
m f(x) 2
.
f (x) f (x) x To graph, let y1 3r 2) r2 mx0 mx0) (b) The domain is the open interval (0, 10). 2 The critical point occurs when r
r ( a2 r2
r (a m m a. 3r 3 2 . 1
(xintercept of line RT)(yintercept of line RT)
2
1 mx0 a 2
2 r 2) r 2 3a 2
a
3 r a m Substituting x for x0, f (x) for m, and f (x) for a, we
( 2r) r2 2r(a
a r 2, where 0 mx0 0, or x 2(Area of triangle ORT) . r 2) 1 r2 3 r2 mx0, (Area of triangle RST) m
f(r) a Let O designate the origin. Then (b) To simplify the calculations, we shall consider the
volume as a function of r:
volume a. The yintercept of this line is and the xintercept is the solution of Therefore,
V(x) f (x0) be the slope of line RT. Then the equation of line RT is equations in the two unknowns a and b. The second 54. (a) The base radius of the cone is r a3
, the relationship is
3 and h P has coordinates (x0, a), aL 3 2bx. Then y( L) 6
3 55. (a) Let x0 represent the fixed value of x at point P, so that c, so y (0) implies that
ax 3 0. Therefore, y(x) y (x) 0. 0. 3ax 2 Now y (x) a (c) Since r f (x) y3 A(x) f (x) 5 51 x2
,
100 NDER(y1), and
y2 x y1 2
y2 . The graph of the area function y3 A(x) is shown below.
2a
3 2 2 a
3 a 3
3 . Using a3
,
3 and h we may now find the values of r and h for the given
values of a.
[0, 10] by [ 100, 1000] When a 4: r when a 5: r when a 6: r when a 8: r 46
,h
3
56
,h
3 2 4 3
3 5 3
3 ; ; 6, h
6 8
3 2 3;
8 ,h 3
3 The vertical asymptotes at x 0 and x 10
correspond to
horizontal or vertical tangent lines, which do not form
triangles. Section 4.4 167 (c) Using our expression for the yintercept of the tangent line, the height of the triangle is
a mx f (x) f (x) x
1
2
1
2 5
5 x x2 100 2 100
x2
2 100 x2 100 x2 x x2 We may use graphing methods or the analytic method in part (d) to find that the minimum value of A(x) occurs at x 8.66.
Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the ycoordinate of the
center of the ellipse.
(d) Part (a) remains unchanged. The domain is (0, C). To graph, note that
f (x)
f (x) x2
C2 B1 B
B
C2 1
C2 x2 B
C B C2 x 2 and Bx ( 2x) C C2 . x2 Therefore, we have A(x) f (x) 2
f (x) f (x) x C2 C B
C B Bx x x2 Bx
C C2 Bx
C C2 x2 1
BCx C2 BCx C2 BCx C2 x2 1
x2 1 C2 x2)
Bx [Bx 2 (BC B C2 x 2)( [Bx 2 BC C 2 x2 x2 x C2 C2 [BC(C C2 x2 2 B(C 2 C2 BC(C x2 C2 x 2)]2 x 2)]2 x 2)]2 x 2)2
x2 (x C 2 A (x) B (BC x x2 2 C2 x 2)(2)(C C2 x x 2) C2 BC x 2 (C 2 C 2 x 2)
BC(C
22
x (C
x 2) 2x 2 C 2 x 2)
BC(C
22
x (C
x 2) [ BC(C
C 2 x 2)
22
x (C
x 2) ( C2 (C Cx2 2x 2 C2 Cx 2
C2 BC(C
x 2(C 2 C 2 x 2)
[Cx 2
x 2)3/2 BC 2(C
x 2(C 2 C 2 x 2)
(2x 2
x 2)3/2 x2 x2 C C(C 2 C2 C2 x 2) C C2 x 2) C2 (C x2 ( x2
C 2 C2 x2 x2 ) C2 C2 C2 x 2) x
C2 x2 x 2) C C2 x2 x 2 )2 x x 2] x2 (C 2 ) x2 x 2) C2 x 2(1) 168 Section 4.5 55. continued
To find the critical points for 0
2x 2
4x 4 C2 C 4C 2x 2 C4 C4 4x 4 3C 2x 2
3C 2) x2 C 2x 2 0 The minimum value of A(x) for 0
corresponding triangle height is
a C, we solve: 0 x 2(4x 2 C2 x mx f (x) C occurs at the critical point x C3
, or x 2
2 3C 2
. The
4 f (x) x B
C B x C2 x2 Bx2
C C2 x2 2
B 3C B B
C B BC
C2 B B
2 C2 3C 2
4 4 CC 2 3C 2
4 3BC 2
4
C2
2 3B
2 3B
This shows that the triangle has minimum area when its height is 3B. s Section 4.5 Linearization and Newton’s Method (pp. 220–232)
Exploration 1
1. f (x) Approximating with Tangent Lines 2x, f (1) 2, so an equation of the tangent line is y 1 2(x 1) or y 2x 1. 3. Since (y1 y2)(1) y1(1) y2(1) 1 1 0, this view shifts the action from the point (1, 1) to the point (1, 0). Also
(y1 y2) (1) y1 (1) y2 (1) 2 2 0. Thus the tangent line to y1 y2 at x 1 is horizontal (the xaxis). The measure of
how well y2 fits y1 at (1, 1) is the same as the measure of how well the xaxis fits y1 y2 at (1, 0).
4. These tables show that the values of y1 y2 near x
x 1. Here are two tables with Table 0.0001. 1 are close to 0 so that y2 is a good approximation to y1 near Section 4.5
Exploration 2
Grapher Using Newton’s Method on the 1–3. Here are the first 11 computations. 7. (a) x
x 1 169 0
1 (b) 2ex
2ex
x e
e 1
(e 1 0
1)
0.684 2e 8. f (x) 3x 2 4
f (1) 3(1)2 4
1
Since f (1)
2 and f (1)
1, the graph of g(x) passes
through (1, 2) and has slope 1. Its equation is
g(x)
1(x 1) ( 2), or g(x)
x 1.
x 2. dy
dx (x 1)(1 1) sin x) (x
(x 1)2 x x sin x 1
(x sin x
1)2 1 x 2x cos (x 2 2 2.069 2.1 2.072 2.2 2.003 2.3 9. f (x) cos x
f (1.5) cos 1.5
Since f (1.5) sin 1.5 and f (1.5) cos 1.5, the tangent
line passes through (1.5, sin 1.5) and has slope cos 1.5. Its
equation is y (cos 1.5)(x 1.5) sin 1.5, or
approximately y 0.071x 0.891 Quick Review 4.5
d2
(x
dx 1.9 2 1.3 1) 1.871 1.2 cos (x 2 1.8 1.1 dy
dx 1.7 1.688 1.0 1. 1.457 0.8
0.9 2. g(x) 0.7
4. Answers will vary. Here is what happens for x1 f (x) 1) cos x)(1)
cos x cos x (x 1) sin x
(x 1)2 [0, ] by [ 0.2, 1.3] 10. For x , and so f (4) 2 1
. Since
2 x3
1
, the tangent line passes through
2
1
1
(4, 1) and has slope . Its equation is y
(x 4) 1, or
2
2
1
y
x 1.
2 f (4) 3. 1 3, f (x)
1 and f (4) [ 2, 6] by [ 3, 3] x 0.567 4. [ 1, 7] by [ 2, 2]
[ 4, 4] by [ 10, 10] x 0.322 5. f (x) (x)( e x) (e x)(1) e x xe x
f (0) 1
The lines passes through (0, 1) and has slope 1. Its equation
is y x 1.
6. f (x) (x)( e x) ( e x)(1) e x xe x
f ( 1) e 1 ( e 1) 2e
The lines passes through ( 1, e 1) and has slope 2e.
Its equation is y 2e(x 1) ( e 1), or
y 2ex e 1. Section 4.5 Exercises
1. (a) f (x) 3x 2 2
We have f (2) 7 and f (2) 10.
L(x) f (2) f (2)(x 2)
7 10(x 2)
10x 13
(b) Since f (2.1) 8.061 and L(2.1) 8, the
approximation differs from the true value in absolute
value by less than 10 1. 170 Section 4.5
1 2. (a) f (x)
2 x 9 We have f ( 4)
L(x) x 8. (a) f (x) 2 9 4
.
5 5 and f ( 4) f ( 4) 4
x
5 x)6 (1
2 (b) f (x) 1 2 9
5 (b) Since f ( 3.9) 4.9204 and L( 3.9) 4.92, the
approximation differs from the true value by less than
10 3. 1 2[1 ( x)] 1 1/2 x (1 ( 4)) 4) ( x)]6 [1 1 6( x) 1
x
2 2[1 1 6x ( 1)( x)] 2x (c) f (x) f ( 4)(x 4
(x
5 5 x (2x) 2 (d) f (x) x) x2 2
21 x 2 1/2
2
x2
4 21 1 x2
22 x
2 1 21 2 3. (a) f (x) 1 x
We have f (1) 2 and f (1)
L(x) f (1) f (1)(x 1)
2 0(x 1)
2 0. (e) f (x) 41/3 1 (b) Since f (1.1) 2.009 and L(1.1) 2, the
approximation differs from the true value by less than
10 2.
4. (a) f (x) (f) f (x) x 1 2
3 1 2x We have f (0)
L(x) (b) Since f (0.1) 0.0953 and L(0.1) 0.1 the
approximation differs from the true value by less than
10 2.
5. (a) f (x) sec2 x
We have f ( ) 0 and f ( )
L(x) f ( ) f ( )(x
)
0 1(x
)
x f (0) f (0)(x
( 1)(x 2 x x 3
3 and f (0) 2 x 2 1 x 1. 0)
0) (b) Since f (0.1) 1.47063 and L(0.1) 1.47080, the
approximation differs from the true value in absolute
value by less than 10 3. x 6 3x cos x
1 1 and f (0)
f (0)(x 3
2 0) 3
x
2
1
x
2 1.
0.002)100 1 0.021 10 0.009)1/3 1 1.009 (1 1.009 1.003 9 10 6 (100)(0.002) 1 1
(0.009)
3 10 f (x) 5 0(x 1) 5 1 2/3
x
3 We have f (8)
L(x) 1.003; 8 2 and f (8) k1 k. 2
2 11. Center
1
f (x) 4x 4
We have f ( 1)
5 and f ( 1) 0
L(x) f ( 1) f ( 1)(x ( 1))
5
12. Center 2 7. f (x) k(1 x)
We have f (0) 1 and f (0)
L(x) f (0) f (0)(x 0)
1 k(x 0)
1 kx 2/3 1 1 1 10. (a) (1.002)100 (1
1.2;
1.002100 1.2 x2
2 2/3 1 The linearization is the sum of the two individual for 1 We have f (0) f (0)
1 (b) 1 41/3 1 linearizations, which are x for sin x and 1 1. (b) Since f (
0.1) 0.10033 and L(
0.1) 0.1, the
approximation differs from the true value in absolute
value by less than 10 3.
6. (a) f (x) 1 3x
34 3x 1/3
4
x
4 41/3 1 1 9. f (x) 1. 3x)1/3 1 1 We have f (0) 0 and f (0)
L(x) f (0) f (0)(x 0)
0 1x
x L(x) (4 f (8) f (8)(x 8) 1
.
12 2 1
(x
12 8) x
12 4
3 ...
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