#### Answer

(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 $3.33~hours$

#### Work Step by Step

(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 $3.33~hours$