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Solução_Calculo_Stewart_6e

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F.<br />

394 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS<br />

TX.10<br />

(b) Since k>0, there will be an exponential expansion ⇔ P 0 − m k > 0 ⇔ mkP0.<br />

(d) P 0 =8,000,000, k = α − β =0.016, m = 210,000 ⇒ m>kP 0 (= 128,000), so by part (c), the population was<br />

declining.<br />

15. (a) The term −15 represents a harvesting of fish at a constant rate—in this case, 15 fish/week. This is the rate at which fish<br />

are caught.<br />

(b) (c) From the graph in part (b), it appears that P (t) = 250 and P (t) =750<br />

are the equilibrium solutions. We confirm this analytically by solving the<br />

equation dP/dt =0as follows: 0.08P (1 − P/1000) − 15 = 0<br />

⇒<br />

0.08P − 0.00008P 2 − 15 = 0 ⇒<br />

−0.00008(P 2 − 1000P + 187,500) = 0<br />

⇒<br />

(P − 250)(P − 750) = 0 ⇒ P = 250 or 750.<br />

(d)<br />

For 0

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