Multivariable Calculus 9th Edition by James Stewart solution manual

34. cos −1  = 2 ⇒ cos(cos −1 ) = cos2 ⇒  = cos2 ≈ −0416
35. tan −1 (3 2 ) =

4
⇒ tan(tan −1 (3 2 )) = tan

4
⇒ 3 2 = 1 ⇒  2 =
1
3
⇒  = ±
1
√ 3 ≈ ±0577
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.
66
¤
CHAPTER 1 FUNCTIONSAND MODELS
36. ln − 1 = ln(5 + ) − 4 ⇒ ln − ln(5 + ) = −4 + 1 ⇒ ln

5 + 
= −3 ⇒  ln((5+)) =  −3 ⇒

5 + 
=  −3 ⇒  = 5 −3 +  −3 ⇒  −  −3 = 5 −3 ⇒ (1 −  −3 ) = 5 −3 ⇒  =
5 −3
1 −  −3
or, multiplying by
 3
 3
, we have  =
5
 3 − 1
≈ 0262.
37. (a) The half­life of the virus with this treatment is eight days and 24 days is 3 half­lives, so the viral load after 24 days is
520  1
2
 1
2
 1
2

= 520
 1
2
 3
= 65 RNA copies/mL.
(b) The viral load is halved every 8 days, so  () = 520  1
2
 8 .
(c)  =  () = 520  1
2
 8
⇒

520
=
 1
2
 8
= 2 −8 ⇒ log 2


520

= log 2

2 −8

= −

8
⇒
 = ( ) − 8log 2


520

. This gives the number of days  needed after treatment begins for the viral load to be reduced
to  RNA copiesmL.
(d) Using the function from part (c), we have (20) = −8log 2

20
520

= −8 ·
ln
1
26
ln2
≈ 376 days.
38. (a) The population would reach 900 in about 44 years.
(b)  =
100,000
100 + 900 −
⇒ 100 + 900 − = 100,000 ⇒ 900 − = 100,000 − 100 ⇒
 − =
100,000 − 100
900
⇒ − = ln
 1000 − 
9

⇒  = −ln
 1000 − 
9

, or ln

9
1000 − 

;
this is the time required for the population to reach a given number .
(c)  = 900 ⇒  = ln

9 · 900
1000 − 900

= ln81 ≈ 44 years, as in part (a).
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.
PRINCIPLES OF PROBLEM SOLVING
1. By using the area formula for a triangle,
1
2
(base)(height), in two ways, we see that
1
2 (4)() =
1
2 ()(), so  =
4

. Since 4 2 +  2 =  2 ,  =
√  2
− 16, and
 =
4 √  2 − 16

.
2. Refer to Example 1, where we obtained  =
 2 − 100
2
. The 100 came from
4 times the area of the triangle. In this case, the area of the triangle is
1
2 ()(12) = 6. Thus,  =
 2 − 4(6)
2
⇒ 2 =  2 − 24 ⇒
2 + 24 =  2 ⇒ (2 + 24) =  2 ⇒  =
 2
2 + 24 .
3.
 4 − | + 1|  = 3
⇒ 4 − | + 1| = −3 (Equation 1) or 4 − | + 1| = 3 (Equation 2).
If  + 1  0, or   −1, then | + 1| = −( + 1) = − − 1. If  + 1 ≥ 0, or  ≥ −1, then | + 1| =  + 1.
We thus consider two cases,   −1 (Case 1) and  ≥ −1 (Case 2), for each of Equations 1 and 2.
Equation 1, Case 1: 4 − | + 1| = −3 ⇒ 4 − (− − 1) = −3 ⇒ 5 + 1 = −3 ⇒
5 = −4 ⇒  = − 4
5
which is invalid since   −1.
Equation 1, Case 2: 4 − | + 1| = −3 ⇒ 4 − ( − 1) = −3 ⇒ 3 − 1 = −3 ⇒
3 = −2 ⇒  = − 2
3 , which is valid since  ≥ −1.
Equation 2, Case 1: 4 − | + 1| = 3 ⇒ 4 − (− − 1) = 3 ⇒ 5 + 1 = 3 ⇒
5 = 2 ⇒  =