Multivariable Calculus 9th Edition by James Stewart solution manual

  | 10 
− 100 ≥ 0  =
  | 10 
≥ 10 2  = { |  ≥ 2} = [2∞).
(b) The sine and exponential functions have domain , so () = sin(  − 1) also has domain .
19. Use  =   with the points (16) and (324). 6 =  1
  =
6


and 24 =  3 ⇒ 24 =
 6


 3 ⇒
4 =  2 ⇒  = 2 [since   0] and  =
6
2
= 3. The function is () = 3 · 2  .
20. Use  =   with the points (−13) and
 1
4
3
 . From the point (−13), we have 3 =  −1 , hence  = 3. Using this and
the point
 1
4
3
 , we get
4
3
=  1 ⇒
4
3
= (3) ⇒
4
9
=  2 ⇒  =
2
3
[since   0] and  = 3( 2
3 ) = 2. The
function is () = 2( 2
3 )
 .
21. If () = 5  , then
( + ) − ()

=
5 + − 5 

=
5  5  − 5 

=
5   5  − 1 

= 5 
 5 
− 1


.
22. Suppose the month is February. Your payment on the 28th day would be 2 28−1 = 2 27 = 134,217,728 cents, or
$1,342,177.28. Clearly, the second method of payment results in a larger amount for any month.
23. 2 ft = 24 in, (24) = 24 2 in = 576 in = 48 ft. (24) = 2 24 in = 2 24 (12 · 5280) mi ≈ 265 mi
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SECTION 1.4 EXPONENTIALFUNCTIONS
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47
24. We see from the graphs that for  less than about 18, () = 5   () =  5 , and then near the point (18171) the curves
intersect. Then ()  () from  ≈ 18 until  = 5. At (53125) there is another point of intersection, and for   5 we
see that ()  (). In fact,  increases much more rapidly than  beyond that point.
25. The graph of  finally surpasses that of  at  ≈ 358.
26. We graph  =   and  = 1,000,000,000 and determine where
  = 1×10 9 . This seems to be true at  ≈ 20723, so    1×10 9
for   20723.
27. (a) (b) Using a graphing calculator, we obtain the exponential
curve () = 3689301(106614)  .
(c) Using the TRACE and zooming in, we find that the bacteria count
doubles from 37 to 74 in about 1087 hours.
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48
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CHAPTER 1 FUNCTIONSAND MODELS
28. Let  = 0 correspond to 1900 to get the model  =   , where   808498 and   101269. To estimate the population in
1925, let  = 25 to obtain   111 million. To predict the population in 2020, let  = 120 to obtain   367 million.
29. (a) Three hours represents 6 doubling periods (one doubling period is 30 minutes). Thus, 500 · 2 6 = 32,000.
(b) In  hours, there will be 2 doubling periods. The initial population is 500,
so the population  at time  is  = 500 · 2 2 .
(c)  =
40
60
=
2
3
⇒  = 500 · 2 2(23) ≈ 1260
(d) We graph  1 = 500 · 2 2 and  2 = 100,000. The two curves intersect at
 ≈ 382, so the population reaches 100,000 in about 382 hours.
30. (a) Let  be the initial population. Since 18 years is 3 doubling periods,  · 2 3 = 600 ⇒  =
600
8
= 75. The initial
squirrel popluation was 75.
(b) A period of  years corresponds to 6 doubling periods, so the expected squirrel population  years after introduction
is  = 75 · 2 6 .
(c) Ten years from now will be 18 + 10 = 28 years from introduction. The population is estimated to be
 = 75 · 2 286 ≈ 1905 squirrels.
31. Half of 760 RNA copies per mL, corresponding to  = 1, is 380 RNA copies per mL. Using the graph of  in Figure 11, we
estimate that it takes about 35 additional days for the patient’s viral load to decrease to 38 RNA copies per mL.
32. (a) The exponential decay model has the form () =   1
2
 15 , where  is the
number of hours after midnight and () is the BAC. We are given that
(0) = 014, so  = 014, and the model is () = 014  1