Solução_Calculo_Stewart_6e

F.
SECTION TX.10 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 115
(c) The circumference is C(r) =2πr = A 0 (r). Thefigure suggests that if ∆r is small,
then the change in the area of the circle (a ring around the outside) is approximately equal
to its circumference times ∆r. Straightening out this ring gives us a shape that is approximately
rectangular with length 2πr and width ∆r, so∆A ≈ 2πr(∆r). Algebraically,
∆A = A(r + ∆r) − A(r) =π(r + ∆r) 2 − πr 2 =2πr(∆r)+π(∆r) 2 .
So we see that if ∆r is small, then ∆A ≈ 2πr(∆r) and therefore, ∆A/∆r ≈ 2πr.
15. S(r) =4πr 2 ⇒ S 0 (r) =8πr ⇒
(a) S 0 (1) = 8π ft 2 /ft (b) S 0 (2) = 16π ft 2 /ft (c) S 0 (3) = 24π ft 2 /ft
As the radius increases, the surface area grows at an increasing rate. In fact, the rate of change is linear with respect to the
radius.
17. The mass is f(x) =3x 2 , so the linear density at x is ρ(x) =f 0 (x) =6x.
(a) ρ(1) = 6 kg/m (b) ρ(2) = 12kg/m (c) ρ(3) = 18 kg/m
Since ρ is an increasing function, the density will be the highest at the right end of the rod and lowest at the left end.
19. The quantity of charge is Q(t) =t 3 − 2t 2 +6t +2,sothecurrentisQ 0 (t) =3t 2 − 4t +6.
(a) Q 0 (0.5) = 3(0.5) 2 − 4(0.5) + 6 = 4.75 A
(b) Q 0 (1) = 3(1) 2 − 4(1) + 6 = 5 A
The current is lowest when Q 0 has a minimum. Q 00 (t) =6t − 4 < 0 when t< 2 3 . So the current decreases when t< 2 3 and
increases when t> 2 3 . Thus, the current is lowest at t = 2 3 s.
21. (a) To find the rate of change of volume with respect to pressure, we first solve for V in terms of P .
PV = C ⇒ V = C P ⇒ dV
dP = − C P 2 .
(b) From the formula for dV/dP in part (a), we see that as P increases, the absolute value of dV/dP decreases.
Thus, the volume is decreasing more rapidly at the beginning.
(c) β = − 1 V
dV
dP = − 1 − C
=
V P 2
C
(PV)P =
C
CP = 1 P
23. In Example 6, the population function was n =2 t n 0. Since we are tripling instead of doubling and the initial population is
400, the population function is n(t) =400· 3 t . The rate of growth is n 0 (t) =400· 3 t · ln 3, so the rate of growth after
2.5 hours is n 0 (2.5) = 400 · 3 2.5 · ln 3 ≈ 6850 bacteria/hour.
1860 − 1750
25. (a) 1920: m 1 =
1920 − 1910 = 110
2070 − 1860
=11, m2 =
10 1930 − 1920 = 210
10 =21,
(m 1 + m 2 )/ 2 = (11 + 21)/2 =16million/year
4450 − 3710
1980: m 1 =
1980 − 1970 = 740
10 =74, m 5280 − 4450
2 =
1990 − 1980 = 830
10 =83,
(m 1 + m 2)/ 2 = (74 + 83)/2 =78.5 million/year
(b) P (t) =at 3 + bt 2 + ct + d (in millions of people), where a ≈ 0.0012937063, b ≈−7.061421911, c ≈ 12,822.97902,
and d ≈−7,743,770.396.
(c) P (t) =at 3 + bt 2 + ct + d ⇒ P 0 (t) =3at 2 +2bt + c (in millions of people per year)