## Calculus: Early Transcendentals 8th Edition

(a) $P(t) = 200~e^{1.17557~t}$ (b) The number of bacteria after 4 hours is $~~22,040$ (c) The rate of growth after 4 hours is $~~25,910~cells/hour$ (d) The population reaches $10,000$ after $3.33~hours$
(a) $\frac{dP}{dt} = kP$ We know that the solution to this equation is: $P(t) = P(0)e^{kt}$ Note that $P(0) = 200$ We can find $k$: $P(0.5) = 200~e^{0.5k} = 360$ $e^{0.5k} = \frac{360}{200}$ $0.5k = ln(1.8)$ $k = \frac{ln(1.8)}{0.5}$ $k = 1.17557$ Therefore: $P(t) = 200~e^{1.17557~t}$ (b) We can find the number of bacteria after $4~hours$: $P(t) = 200~e^{1.17557~t}$ $P(4) = 200~e^{(1.17557)(4)}$ $P(4) = 22,040$ The number of bacteria after 4 hours is $~~22,040$ (c) We can find the rate of growth after 4 hours: $\frac{dP}{dt} = kP$ $\frac{dP}{dt} = (1.17557)(22,040)$ $\frac{dP}{dt} = 25,910$ The rate of growth after 4 hours is $~~25,910~cells/hour$ (d) We can find the time $t$ when the population reaches $10,000$: $P(t) = 200~e^{1.17557~t} = 10,000$ $e^{1.17557~t} = 50$ $1.17557~t = ln(50)$ $t = \frac{ln(50)}{1.17557}$ $t = 3.33~hours$ The population reaches $10,000$ after $3.33~hours$