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.

Answer 2

By finding the population as a function of time, we will see that the population will be 500 after 10.2 hours.

How to find the population equation.

Let's say that the population equation is p(t).

We know that the rate of change is proportional to the population, so we have the differential equation:

p'(t) = a*p(t).

This means that p(t) is an exponential function of the form:

p(t) = A*e^(a*t)

Now we know that:

p(0) = A*e^(a*0) = A = 300

And:

P(24) = 300*e^(a*24) = 1000

           e^(a*24) = 1000/300

           a*24 = ln(1000/300)

            a = ln(1000/300)/24 = 0.05

Then the population equation is:

p(t) = 300*e^(0.05*t)

Now we want to find the value of t such that the population is 500, so we need to solve:

p(t) = 300*e^(0.05*t) = 500

         e^(0.05*t) = 500/300

             0.05*t =   ln(500/300)

                      t =   ln(500/300)/0.05 = 10.2

So the population will be 500 at t = 10.2 hours.

If you want to learn more about differential equations, you can read:

https://brainly.com/question/18760518