## Calculus: Early Transcendentals 9th Edition

(a) The relative growth rate is $104\%$ (b) $P(0) = 50$ (c) $P(t) = 50~e^{1.03972~t}$ (d) 5382 cells (e) $5596~cells/h$ (f) 6.6 hours
(a) We can find the value of $k$: $P(6) = P(0)e^{6k} = 25,600$ $P(2) = P(0)e^{2k} = 400$ $\frac{P(6)}{P(2)} = e^{6k-2k} = \frac{25,600}{400}$ $e^{4k} = 64$ $4k = ln(64)$ $k = \frac{ln(64)}{4}$ $k = 1.03972$ The relative growth rate is $104\%$ (b) We can find the initial size: $P(t) = P(0)~e^{kt}$ $P(2) = P(0)~e^{2k} = 400$ $P(0) = \frac{400}{e^{(2)( 1.03972)}}$ $P(0) = 50$ (c) $P(t) = P(0)~e^{kt}$ $P(t) = 50~e^{1.03972~t}$ (d) We can find the number of bacteria cells after 4.5 hours: $P(t) = 50~e^{1.03972~t}$ $P(4.5) = 50~e^{(1.03972)(4.5)}$ $P(4.5) = 5382$ (e) We can find the rate of growth after 4.5 hours: $\frac{dP}{dt} = k~P(4.5)$ $\frac{dP}{dt} = (1.03972)(5382)$ $\frac{dP}{dt} = 5596~cells/h$ (f) We can find the time $t$ when $P(t) = 50,000$: $P(t) = (50)~e^{1.03972~t} = 50,000$ $e^{1.03972~t} = 1000$ $1.03972~t = ln(1000)$ $t = \frac{ln(1000)}{1.03972}$ $t = 6.6~h$