Question

The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial population P0 has doubled in 5 years. Suppose it is known that the population is 9,000 after 3 years. What was the initial population P0? (Round your answer to one decimal place.)

Answer 1

Answer:

Step-by-step explanation:

Let P be the population of the community

So the population of a community increase at a rate proportional to the number of people present at a time

That is

$\frac{dp}{dt} \propto p\\\\\frac{dp}{dt} =kp\\\\ [k \texttt {is constant}]\\\\\frac{dp}{dt} -kp =0$

Solve this equation we get

$p(t)=p_0e^{kt}---(1)$

where p is the present population

p₀ is the initial population

If the  initial population as doubled in 5 years

that is time t = 5 years

We get

$2p_o=p_oe^{5k}\\\\e^{5k}=2$

Apply In on both side to get

$Ine^{5k}=In2\\\\5k=In2\\\\k=\frac{In2}{5} \\\\\therefore k=\frac{In2}{5}$

Substitute $k=\frac{In2}{5}$  in $p(t)=p_oe^{kt}$ to get

$\large \boxed {p(t)=p_oe^{\frac{In2}{5}t }}$

Given that population of a community is 9000 at 3 years

substitute t = 3 in ${p(t)=p_oe^{\frac{In2}{5}t }}$

$p(3)=p_oe^{3 (\frac{In2}{5}) }\\\\9000=p_oe^{3 (\frac{In2}{5}) }\\\\p_o=\frac{9000}{e^{3(\frac{In2}{5} )}} \\\\=5937.8$

Therefore, the initial population is 5937.8