Multivariable Calculus 9th Edition by James Stewart solution manual

36
¤
CHAPTER 1 FUNCTIONSAND MODELS
25.  = |cos|. Start with the graph of  = cos, shrink horizontally by a factor of , and reflect all the parts of the graph
below the ­axis about the ­axis.
26.  =

√
 − 1
.
Start with the graph of  =
√
, shift 1 unit downward, and then reflect the portion of the graph below the
­axis about the ­axis.
27. This is just like the solution to Example 4 except the amplitude of the curve (the 30 ◦ N curve in Figure 9 on June 21) is
14 − 12 = 2. So the function is () = 12 + 2sin 
2
365 ( − 80)
 . March 31 is the 90th day of the year, so the model gives
(90) ≈ 1234 h. The daylight time (5:51 AM to 6:18 PM ) is 12 hours and 27 minutes, or 1245 h. The model value differs
from the actual value by
1245−1234
1245
≈ 0009, less than 1%.
28. Using a sine function to model the brightness of Delta Cephei as a function of time, we take its period to be 54 days, its
amplitude to be 035 (on the scale of magnitude), and its average magnitude to be 40. If we take  = 0 at a time of
average brightness, then the magnitude (brightness) as a function of time  in days can be modeled by the formula
() = 40 + 035sin 
2
54 
 .
29. The water depth () can be modeled by a cosine function with amplitude
12 − 2
2
= 5 m, average magnitude
12 + 2
2
= 7 m,
and period 12 hours. High tide occurred at time 6:45 AM ( = 675 h), so the curve begins a cycle at time  = 675 h (shift
6.75 units to the right). Thus, () = 5cos
 2
12 ( − 675)

+ 7 = 5cos
 
6 ( − 675)

+ 7, where  is in meters and  is the
number of hours after midnight.
30. The total volume of air  () in the lungs can be modeled by a sine function with amplitude
2500 − 2000
2
= 250 mL, average
volume
2500 + 2000
2
= 2250 mL, and period 4 seconds. Thus,  () = 250sin
2
4
 + 2250 = 250sin

2  + 2250, where
 is in mL and  is in seconds.
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3 NEW FUNCTIONSFROM OLD FUNCTIONS
¤
37
31. (a) To obtain  = (||), the portion of the graph of  = () to the right of the ­axis is reflected about the ­axis.
(b)  = sin|| (c)  =
 ||
32. The most important features of the given graph are the ­intercepts and the maximum
and minimum points. The graph of  = 1() has vertical asymptotes at the ­values
where there are ­intercepts on the graph of  = (). The maximum of 1 on the graph
of  = () corresponds to a minimum of 11 = 1 on  = 1(). Similarly, the
minimum on the graph of  = () corresponds to a maximum on the graph of
 = 1(). As the values of  get large (positively or negatively) on the graph of
 = (), the values of  get close to zero on the graph of  = 1().
33. () =
√ 25 −  2
is defined only when 25 −  2 ≥ 0 ⇔  2 ≤ 25 ⇔ −5 ≤  ≤ 5, so the domain of  is [−55].
For () =
√  + 1, we must have  + 1 ≥ 0
⇔  ≥ −1, so the domain of  is [−1∞).
(a) ( + )() =
√ 25 −  2
+
√  + 1. The domain of  +  is found by intersecting the domains of  and : [−15].
(b) ( − )() =
√ 25 −  2
−
√  + 1. The domain of  −  is found by intersecting the domains of  and : [−15].
(c) ()() =
√ 25 −  2
·
√  + 1 = √ − 3
−  2 + 25 + 25. The domain of  is found by intersecting the domains of 
and : [−15].
(d)
 


() =
√ 25 −  2
√  + 1 =

25 −  2
 + 1
. Notice that we must have  + 1 6= 0 in addition to any previous restrictions.
Thus, the domain of  is (−15].
34. For () =
1
 − 1 , we must have  − 1 6= 0
⇔  6= 1. For () =
1

− 2, we must have  6= 0.
(a) ( + )() =