Question

During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time t, in hours. At time t = 0 hours, the population is 300. At time t = 24 hours, the population is 1000. At what time t is the population 500?
(A) t = 2112
(B) t = 24 In 500 In 1000
(D) t = 300e (9)

Answer 1

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

$\frac{dP}{dt} = Pr$

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

$\frac{dP}{P} = r dt$

$\int \frac{dP}{P} = \int r dt$

$\ln{P} = rt + K$

Applying the exponential to both sides:

$P(t) = Ke^{rt}$

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

$P(t) = 300e^{rt}$

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

$e^{24r} = \frac{1000}{300}$

$e^{24r} = \frac{10}{3}$

$\ln{e^{24r}} = \ln{\frac{10}{3}}$

$24r = \ln{\frac{10}{3}}$

$r = \frac{\ln{\frac{10}{3}}}{24}$

$r = 0.05$

So

$P(t) = 300e^{0.05t}$

At what time t is the population 500?

This is t for which P(t) = 500. So

$P(t) = 300e^{0.05t}$

$500 = 300e^{0.05t}$

$e^{0.05t} = \frac{500}{300}$

$e^{0.05t} = \frac{5}{3}$

$\ln{e^{0.05t}} = \ln{\frac{5}{3}}$

$0.05t = \ln{\frac{5}{3}}$

$t = \frac{\ln{\frac{5}{3}}}{0.05}$

$t = 10.22$

The population is of 500 after 10.22 hours.