Solução_Calculo_Stewart_6e

F.
116 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
(d) P 0 (1920) = 3(0.0012937063)(1920) 2 +2(−7.061421911)(1920) + 12,822.97902
≈ 14.48 million/year [smaller than the answer in part (a), but close to it]
P 0 (1980) ≈ 75.29 million/year (smaller, but close)
(e) P 0 (1985) ≈ 81.62 million/year, so the rate of growth in 1985 was about 81.62 million/year.
27. (a) Using v = P
4ηl (R2 − r 2 ) with R =0.01, l =3, P =3000,andη =0.027,wehavev as a function of r:
v(r) = 3000
4(0.027)3 (0.012 − r 2 ). v(0) = 0.925 cm/s, v(0.005) = 0.694 cm/s, v(0.01) = 0.
(b) v(r) = P
4ηl (R2 − r 2 ) ⇒ v 0 (r) = P
Pr
(−2r) =−
4ηl 2ηl .
Whenl =3, P =3000,andη =0.027,wehave
v 0 (r) =−
3000r
2(0.027)3 . v0 (0) = 0, v 0 (0.005) = −92.592 (cm/s)/cm, and v 0 (0.01) = −185.185 (cm/s)/cm.
(c) The velocity is greatest where r =0(at the center) and the velocity is changing most where r = R =0.01 cm
(at the edge).
29. (a) C(x) = 1200 + 12x − 0.1x 2 +0.0005x 3 ⇒ C 0 (x) =12− 0.2x +0.0015x 2 $/yard, which is the marginal cost
function.
(b) C 0 (200) = 12 − 0.2(200) + 0.0015(200) 2 =$32/yard, and this is the rate at which costs are increasing with respect to
the production level when x =200.
C 0 (200) predicts the cost of producing the 201st yard.
(c) The cost of manufacturing the 201st yard of fabric is C(201) − C(200) = 3632.2005 − 3600 ≈ $32.20,whichis
approximately C 0 (200).
31. (a) A(x) = p(x)
x
⇒ A 0 (x) = xp0 (x) − p(x) · 1
x 2
= xp0 (x) − p(x)
x 2 .
A 0 (x) > 0 ⇒ A(x) is increasing; that is, the average productivity increases as the size of the workforce increases.
(b) p 0 (x) is greater than the average productivity ⇒ p 0 (x) >A(x) ⇒ p 0 (x) > p(x)
x
xp 0 (x) − p(x) > 0 ⇒ xp0 (x) − p(x)
x 2 > 0 ⇒ A 0 (x) > 0.
⇒ xp 0 (x) >p(x) ⇒
33. PV = nRT ⇒ T = PV
nR = PV
(10)(0.0821) = 1 (PV). Using the Product Rule, we have
0.821
dT
dt = 1
0.821 [P (t)V 0 (t)+V (t)P 0 (t)] = 1 [(8)(−0.15) + (10)(0.10)] ≈−0.2436 K/min.
0.821
35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is, dC
dt =0and dW dt
(b) “The caribou go extinct” means that the population is zero, or mathematically, C =0.
=0.
(c)Wehavetheequations dC
dt = aC − bCW and dW dt
= −cW + dCW .LetdC/dt = dW/dt =0, a =0.05, b =0.001,
c =0.05,andd =0.0001 to obtain 0.05C − 0.001CW =0 (1) and −0.05W +0.0001CW =0 (2). Adding10 times
(2) to (1) eliminates the CW-terms and gives us 0.05C − 0.5W =0 ⇒ C =10W . Substituting C =10W into (1)