Solução_Calculo_Stewart_6e

F.
118 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
9. (a) If y(t) is the mass (in mg) remaining after t years, then y(t) =y(0)e kt =100e kt .
y(30) = 100e 30k = 1 (100) ⇒ 2 e30k = 1 2
⇒ k = −(ln 2)/30 ⇒ y(t) = 100e −(ln 2)t/30 = 100 · 2 −t/30
(b) y(100) = 100 · 2 −100/30 ≈ 9.92 mg
(c) 100e −(ln 2)t/30 =1 ⇒ −(ln 2)t/30 = ln 1
ln 0.01
100
⇒ t = −30
ln 2
≈ 199.3 years
11. Let y(t) be the level of radioactivity. Thus, y(t) =y(0)e −kt and k is determined by using the half-life:
y(5730) = 1 y(0) ⇒ 2 y(0)e−k(5730) = 1 y(0) ⇒ 2 e−5730k = 1 ⇒ −5730k =ln 1 ⇒ k = − ln 1 2
2 2
5730 = ln 2
5730 .
If 74% of the 14 C remains, then we know that y(t) =0.74y(0) ⇒ 0.74 = e −t(ln 2)/5730 ⇒ ln 0.74 = − t ln 2 ⇒
5730
5730(ln 0.74)
t = − ≈ 2489 ≈ 2500 years.
ln 2
13. (a) Using Newton’s Law of Cooling, dT
dt = k(T − T s), wehave dT = k(T − 75). Nowlety = T − 75, so
dt
y(0) = T (0) − 75 = 185 − 75 = 110,soy is a solution of the initial-value problem dy/dt = ky with y(0) = 110 and by
Theorem 2 we have y(t) =y(0)e kt =110e kt .
y(30) = 110e 30k =150− 75 ⇒ e 30k = 75 = 15
110 22
⇒ k = 1 15
30
ln ,soy(t) = 110e 30 1 t ln( 15
22 )
22 and
y(45) = 110e 45
30
ln( 15
22 ) ≈ 62 ◦ F.Thus,T (45) ≈ 62 + 75 = 137 ◦ F.
(b) T (t) =100 ⇒ y(t) =25. y(t) = 110e 30 1 t ln( 15
1
22 ) =25 ⇒ e 30
t ln( 15
22 ) =
25
110
⇒ 1 15 25
30
t ln
22
=ln110 ⇒
25
30 ln
110
t =
ln 15 ≈ 116 min.
22
dT
15. = k(T − 20). Lettingy = T − 20, wegetdy
dt dt = ky,soy(t) =y(0)ekt . y(0) = T (0) − 20 = 5 − 20 = −15, so
y(25) = y(0)e 25k = −15e 25k ,andy(25) = T (25) − 20 = 10 − 20 = −10,so−15e 25k = −10 ⇒ e 25k = 2 . Thus,
3
25k =ln
2
3 and k =
1
ln
2
25 3 ,soy(t) =y(0)e kt = −15e (1/25) ln(2/3)t .Moresimply,e 25k = 2 ⇒ e k =
2 1/25
⇒
3 3
e kt =
2 t/25
⇒ y(t) =−15 ·
2 t/25
.
3
3
(a) T (50) = 20 + y(50) = 20 − 15 ·
2 50/25
=20− 15 ·
2 2
=20− 20 =13.¯3 ◦ C
3
3
3
(b) 15 = T (t) =20+y(t) =20− 15 ·
2 t/25
3
⇒ 15 ·
2 t/25
3
=5 ⇒
2 t/25
3
= 1 3
⇒
(t/25) ln
2
3 =ln 1
⇒ t =25ln
1
3
3 ln 2
3 ≈ 67.74 min.
17. (a) Let P (h) be the pressure at altitude h. ThendP/dh = kP ⇒ P (h) =P (0)e kh =101.3e kh .
P (1000) = 101.3e 1000k =87.14 ⇒ 1000k =ln
87.14
⇒ k = 1 ln
87.14
⇒
101.3
1000 101.3
P (h) =101.3 e 1000 1 h ln( 87.14
101.3) ,soP (3000) = 101.3e
3ln( 87.14
101.3 ) ≈ 64.5 kPa.
(b) P (6187) = 101.3 e 6187 ln( 87.14
1000 101.3 ) ≈ 39.9 kPa