# Question

1. The population of a certain community is known to increase at a rate proportional to the number of people present at time $t$. If the population has doubled in 5 years, how long will it take to triple? To quadruple?

2. Suppose it is known that the population of the community in Problem 1 is 10,000 after 3 years. What was the initial population? What will be the population in 10 years?

Note - I need help with #2. I only included #1 because the first line of the second problem points to it.

The logistic equation demonstrated to us in class is

$$P =\frac{ak}{e^{-at}+bk}$$

where $a$ = birth rate and $b$ = death rate. I assumed that the death rate is zero given the scope of this problem and attempted to solve for $a$, but I cannot isolate the variable.

$$-3a = ln(\frac{ak}{10000})$$

Am I using the right equation? If not, then when is this one supposed to be used?

• How is "to increase at a rate proportional to the number of people present at time t" related to the logistic equation?
– Did
Commented Jun 20, 2015 at 20:17
• @Did yes I noticed that. So before the attempt shown above I simply used $$P(t) = rt$$. I found the rate to be \frac{2}{5}, however this did not give me the correct answer for #1. So I started thinking that maybe I needed the logisitc equation. Commented Jun 20, 2015 at 20:38
• P(t)=rt is not "to increase at a rate proportional to the number of people present at time t" but to increase at a constant rate independent of the number of people present at time t.
– Did
Commented Jun 20, 2015 at 21:09

Since the population increases at a rate proportional to the number of people present at time t (read: the rate of change of $P$ is proportional to $P$.)

$$\frac{\mathrm{d}P}{\mathrm{d}t} = kP$$

Separating the variables and integrating yields

$$\int \frac{1}{P} \, \mathrm{d}P = \int k \, \mathrm{d}t$$

So that we get

$$\ln P = kt + c$$ or equivalently, letting the initial population be $P_0$ so that when $t=0$, $P= P_0$ to find the arbitrary constant and re-arranging yields

$$P = P_0e^{kt}$$

We are given, that the population doubles ($2P_0$) in five years ($t=5$) so:

$$2P_0 = P_0 e^{5k}$$

Solving for $k$ yields

$$e^{5k} = 2 \implies k = \frac{\ln 2}{5}$$

Our population growth equation is then $$P = P_0 e^{\frac{\ln 2}{5}t}$$

So for the population to triple, we need to solve $3P_0 = P_0e^{kt}$ for $t$, where $k$ is what we found above. This yields

$$e^{kt} = 3 \implies kt = \ln 3 \implies t = \frac{5\ln 3}{\ln 2}$$

The same can be done to find the time it takes for the population to quadruple, that is solve $4P_0 = P_0e^{kt}$ for t.

For the second question, is we have that the population after three years ($t=3$) is $P = 10000$

then substituting this into our population growth equation gives

$$10000 = P_0e^{3k}$$

so that we solve for $P_0$ to get

$$P_0 = \frac{10000}{e^{3k}} = \frac{10000}{\exp{\left(\frac{3\ln 2}{5}\right)}}$$

• Thank you, I sincerely appreciate your time in helping me solve this! :) Commented Jun 22, 2015 at 1:36
• Glad you understood, no problem! :) Commented Jun 22, 2015 at 13:20