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

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

`(dp)/(dt) \propto p\\\\(dp)/(dt) =kp\\\\ [k \texttt {is constant}]\\\\(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=(In2)/(5) \\\\\therefore k=(In2)/(5)`

Substitute
`k=(In2)/(5)` in
`p(t)=p_oe^(kt)` to get

`\large \boxed {p(t)=p_oe^{(In2)/(5)t }}`

Given that population of a community is 9000 at 3 years

substitute t = 3 in
`{p(t)=p_oe^{(In2)/(5)t }}`

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

Therefore, the initial population is 5937.8