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Thomas Calculus 13th [Solutions]

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Section 3.11 Linearization and Differentials 197<br />

62.<br />

4<br />

3<br />

8 0.06t<br />

(1<br />

3 2<br />

t )<br />

C ( t) 0.06 e , where t is measured in hours. When the time changes from 20 min to 30 min, t in<br />

hours changes from 1 1<br />

3<br />

to<br />

2<br />

, so the differential estimate for the change in C is<br />

1 1 1 1 1<br />

C<br />

3 2 3 6<br />

C<br />

3<br />

0.584 mg/mL.<br />

3<br />

63. The relative change in V is estimated by<br />

dV / dr 4kr 4 r<br />

V<br />

r 4 r r . If the radius increases by 10%, r changes to<br />

kr<br />

1.1r and r 0.1 r . The approximate relative increase in V is thus 4(0.1 r<br />

r<br />

) 0.4 or 40%.<br />

1/2<br />

3/2 3/2<br />

64. (a) T 2 L dT 2 L 1 g dg Lg dg<br />

g<br />

2<br />

(b) If g increases, then dg 0 dT 0. The period T decreases and the clock ticks more frequently. Both the<br />

pendulum speed and clock speed increase.<br />

3/2<br />

2<br />

2<br />

(c) 0.001 100(980 ) dg dg 0.977 cm/sec the new g 979 cm/sec<br />

65. (a) i. Q( a) f ( a)<br />

implies that b 0 f ( a).<br />

ii. Since Q ( x) b1 2 b2<br />

( x a), Q ( a) f ( a ) implies that b 1 f ( a).<br />

f ( a)<br />

iii. Since Q ( x) 2 b2<br />

, Q ( a) f ( a ) implies that b2 .<br />

2<br />

f ( a)<br />

In summary, b 0 f ( a),<br />

b 1 f ( a ), and b2 .<br />

2<br />

1<br />

2 2<br />

3 3<br />

(b) f ( x) (1 x) ; f ( x)<br />

1(1 x) ( 1) (1 x) ; f ( x)<br />

2(1 x) ( 1) 2(1 x)<br />

Since<br />

f (0) 1, f (0) 1, and f (0) 2, the coefficients are b 2<br />

0 1, b1 1, b 2 1. The quadratic<br />

2<br />

2<br />

approximation is Q( x) 1 x x .<br />

(c)<br />

As one zooms in, the two graphs quickly become<br />

indistinguishable. They appear to be identical.<br />

(d)<br />

1<br />

2 3<br />

g( x) x ; g ( x)<br />

1 x ; g ( x) 2x<br />

Since g(1) 1, g (1) 1, and g (1) 2, the coefficients are b 2<br />

0 1, b1 1, b 2 1. The quadratic<br />

2<br />

2<br />

approximation is Q( x) 1 ( x 1) ( x 1) .<br />

As one zooms in, the two graphs quickly become<br />

indistinguishable. They appear to be identical.<br />

(e)<br />

1/2<br />

1/2 3/2<br />

h( x) (1 x) ; h ( x)<br />

1 (1 x) ; h ( x) 1 (1 x)<br />

2 4<br />

1<br />

Since h(0) 1, h (0) 1 1<br />

4<br />

, and h (0) , the coefficients are b 1 1<br />

2 4<br />

0 1, b1 , b<br />

2 2 . The quadratic<br />

2 8<br />

2<br />

approximation is Q( x)<br />

1 x x .<br />

2 8<br />

Copyright<br />

2014 Pearson Education, Inc.

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