Thomas Calculus 13th [Solutions]

540 Chapter 7 Transcendental Functions
A k k
2 2
kt
0 5730 5730
39. Since the half life is 5730 years and A( t)
A0
e we have A
1
0e e ln(0.5) 5730k
k With 10% of the original carbon-14 remaining we have
ln(0.5) ln(0.5)
5730 t
0.1A A e 0.1 e
5730
ln(0.5)
5730 .
ln(0.5) (5730)ln(0.1)
5730 ln(0.5)
0 0
ln(0.1) t t 19,035 years (rounded to the nearest year).
t
40.
kt
T T T T e 180 40 (220 40) e , time in hours, k 4ln
7
4ln
9
s o s
4ln(9/7) t ln 6
4ln 9
7
k /4
9 7
70 40 (220 40) e t 1.78hr 107 min, the total time the time it took to cool
from 180 F to 70 F was 107 15 92 min
CHAPTER 7
ADDITIONAL AND ADVANCED EXERCISES
1.
t x x
t
t t 2x 2x t
2t
0 0 0
2 0 2
A ( t ) e dx e 1 e , V ( t ) e dx e 1 e
(a) lim A( t) lim 1 e 1
(b)
(c)
t
t
lim
t
2
( )
2 1 t
V t
e
lim
A( t) t
t 1 e 2
t
2
V ( t)
1 e t 1 e t 1 e
t
2 2
A( t) t
t
1 1
2
t 0 t 0 e t 0 e t 0
lim lim lim lim 1
e
t
2. (a) lim log ln 2
a 2 lim 0;
ln a
a 0 a 0
lim log ln 2
a 2 lim ;
ln a
a 1 a 1
lim log 2 lim ln 2
a
;
ln a
a 1 a 1
lim log 2 lim ln 2
a
0
a
a
ln a
4. In the interval x 2 the function
sin x
sin x 0 (sin x ) is not defined for all values in
that interval or its translation by 2 .
5. (a) g( x) h( x) 0 g( x) h( x ); also g( x) h( x) 0 g( x) h( x) 0 g( x) h( x) 0
g( x) h( x ); therefore h( x) h( x) h( x) 0 g( x) 0
f ( x) f ( x)
fE ( x) fO ( x) fE ( x) fO ( x) fE ( x) fO ( x) fE ( x) fO
( x)
(b)
f ( x);
2 2 2
f ( x) f ( x)
fE ( x) fO ( x) fE ( x) fO ( x) fE ( x) fO ( x) fE ( x) fO
( x)
2 2 2
(c) Part b such a decomposition is unique.
E
fO
( x)
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