60t
2. Let N (t) be the number of cells after t hours. N (t) = 60 · 2( 20 ) = 60 · 23t = 60 · 8t .
(a)
1 dN
N dt
=
1
60·8t
· (60 ln 8 · 8t ) = ln 8 hr−1 .
(b) N (t) = 60 · 8t cells.
(c) N (8) = 60 · 88 cells.
(d) N 0 (8) = 60 ln 8 · 88 cells/hr.
20000
60
(e) 20000 = N (t) = 60 · 8t ⇒ ln
= t ln 8 ⇒ t =
ln 1000−ln 3
ln 8
≈ 2.793 hrs.
3. Let N (t) be the number of cells after t hours.
(a) N (1) = 100 · ekt |t=1 = 420 ⇒ ek = 4.2 ⇒ k = ln 4.2 ⇒ N (t) = 100 · 4.2t cells.
(b) N (3) = 100 · 4.23 ≈ 7409 cells.
(c) N 0 (3) = 100 ln 4.2 · 4.23 ≈ 10632 cells/hr.
(d) 10000 = N (t) = 100 · 4.2t ⇒ t =
ln 100
ln 4.2
≈ 3.2 hrs.
5. Let P (t) be the population in year t.
(a) P (1750) = 790 ⇒ P (t) = 790ek(t−1750) , P (1800) = 980 ⇒ 980 = 790e50k ⇒ k =
ln 98−ln 79
≈ 0.00431. P (1900) ≈ 1508 million, P (1950) ≈ 1871 million.
50
(b) P (1850) = 1260 ⇒ P (t) = 1260ek(t−1850) , P (1900) = 1650 ⇒ 1650 = 1260e50k ⇒
126
k = ln 165−ln
≈ 0.00539. P (1950) ≈ 2160 million.
50
(c) P (1900) = 1650 ⇒ P (t) = 1650ek(t−1900) , P (1950) = 2560 ⇒ 2560 = 1650e50k ⇒
165
k = ln 256−ln
≈ 0.00878. P (2000) ≈ 3970 million. World Wars v.s. improve50
ment in medical care.
7. Let C(t) be the concentration after t seconds.
(a) C(t) = Ce−0.0005t .
(b) 0.9 = e−0.0005t ⇒ t =
ln 0.9
−0.0005
≈ 211 s.
t
9. (a) 100 · 0.5( 30 ) mg.
100
(b) 100 · 0.5( 30 ) ≈ 9.92 mg.
t
(c) 100 · 0.5( 30 ) = 1 ⇒
t
30
ln 0.5 = ln 0.01 ⇒ t = 30 lnln0.01
≈ 199.3 years.
0.5
t
11. 0.5( 5730 ) = 0.74 ⇒ t = 5730 ·
ln 0.74
ln 0.5
≈ 2489 years.
13. (a) 75+(185−75)e0.5k = 150 ⇒ k = 2(ln 75−ln 110) ≈ −0.766 ⇒ 75+110e0.75(−0.766) ≈
137o F.
(b) 75 + (185 − 75)e−0.766t = 100 ⇒ t =
1
ln 25−ln 110
−0.766
≈ 1.93 hrs≈ 116 minutes.
25k
15. (a) 20 + (5 − 20)e( 60 ) = 10 ⇒ k =
13.3o C.
(b) 20 − 15e−0.973t = 15 ⇒ t =
16.
60
(ln 10 − ln 15)
25
− ln 3
−0.973
≈ −0.973 ⇒ 20 − 15e(
50·−0.973
60
)≈
≈ 1.129 hrs≈ 67.74 minutes.
= k(T − T0 )|T =70 = k(70 − 20) = −1 ⇒ k = −0.02 ⇒ T = 20 + 75e−0.02t .
70 = 20 + 75e−0.02t ⇒ t = −50(ln 50 − ln 75) ≈ 20.3 minutes.
dT
|
dt T =70
2. (a)
dA
dt
dA
dt
=
d(
V
πr 2
d(πr2 )
dt
= 2πr dr
.
dt
3.
dA
dt
= 2π · 30 · 1 = 60π m2 /sec.
√
= 2a da
= 2 · 16 · 6 = 48 cm2 /sec.
dt
5.
dh
dt
=
7.
dy
dt
= (2x + 1)−0.5 dx
.
dt
(b)
(a)
dt
dy
dt
)
=
1 dV
π·52 dt
=
3
25π
m/min.
= (9)−0.5 · 3 = 1.
(b) 5 = (25)−0.5 dx
⇒
dt
dx
dt
= 25.
10. xy 0 + x0 y = 0 ⇒ 4(−3) + 2x0 = 0 ⇒ x0 = 6.
2