Question

Consider a population p of field mice that grows at a rate proportional to the current population, so that dp dt = rp. (Note: Remember that, as in the text, t is measured in months, not days. One month is 30 days.) (a) Find the rate constant r if the population doubles in 210 days. (Round your answer to four decimal places.) r = (b) Find r if the population doubles in N days. r = ?

Answer 1

a)
`r = (ln(2))/(210)=0.00330070086`

b)
`r = (ln(2))/(N)`

Explanation:

For this case we assume the followin differential equation:

`(dp)/(dt)= rp`

Where is is the consttant growth/decay rate , p represent the population and the the time.

For this case we can rewrite this expression like this:

`(dp)/(p)= rdt`

And now we can apply integrals on both sides like this:

`\int (dp)/(p) r\int dt`

`ln |p|= rt+C`

If we apply exponential on both sides we got:

`p(t) = e^(rt) *e^c = p_o e^(rt)`

And from the previous equation
`p_o` represent the initial population.

Part a

For this case we are assuming that the population doubles in t=210 so then we can set the following equation:

`2p_o = p_o e^(210r)`

We can cancel
`p_o` in both sides and we got:

`2 = e^(210r)`

We can apply natural log on both sides and we got:

`ln (2) = 210 r`

`r = (ln(2))/(210)=0.00330070086`

Part b

For this case we are assuming that the population doubles in t=N so then we can set the following equation:

`2p_o = p_o e^(Nr)`

We can cancel
`p_o` in both sides and we got:

`2 = e^(Nr)`

We can apply natural log on both sides and we got:

`ln (2) = N r`

`r = (ln(2))/(N)`