Question

Population Growth The number of a certain type of bacteria increases continuously at a rate proportional to the number present. After 2 hours, there are 200 bacteria, and after 4 hours, there are 300 bacteria.

(a) Find an exponential model given the population p after t hours.
(b) How many becteria will there be after 7 hours?
(c) How long will it take for the population to double?

Answer 1

a)
`P(t) =(400)/(3) e^(0.2027325541 t)`

b)
`P(t=7) =(400)/(3) e^(0.2027325541*7)= 551.135 \approx 552`

c) For this case we need to satisfy this equation:

`2P_o = P_o e^(kt)`

If we divide both sides by
`P_o` we got:

`2 = e^(kt)`

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

`ln(2) = kt`

And the
`t = (ln(2))/(0.2027325541)=3.419 hours`

Explanation:

For this case the proportional model is given by the following differential equation:

`(dP)/(dt) =kP`

Where P is the population, t the time and k a proportional constant.

IWe can rewrite the differential equation like this:

`(dP)/(P) =kdt`

And if we integrate both sides we got:

`ln |P| = kt +c`

Part a

And if we apply exponentiation on both sides we got:

`P(t) = e^(kt) e^c = e^(kt) P_o = P_o e^(kt)`

So then our model is given by:

`P(t) = P_o e^(kt)`

Where
`P_o` is the initial population for t =0.

For this case we have some conditions given:

`P(2) = 200, P(4) = 300`

So then we have two equations:

`200 = P_o e^(2k)` (1)

`300 = P_o e^(4k)` (2)

From equation (1) we can solve for
`P_o` like this:

`P_o =(200)/(e^(2k))` (3)

And if we replace equation (3) into equation (2) we got:

`300 =(200)/(e^(2k)) e^(4k)= 200e^(2k)`

And we can divide both sides by 200 and we got:

`(3)/(2) = e^(2k)`

Now we can apply natural logs on both sides and we got:

`ln((3)/(2)) = 2k`

And then the value for k is:

`k = (ln((3)/(2)))/(2)= 0.2027325541`

And since we have the value for k we can find the value or
`P_o` like this:

`P_o =(200)/(e^(2* 0.2027325541))= (400)/(3)`

And with that our model is given by:

`P(t) =(400)/(3) e^(0.2027325541 t)`

Part b

For this case we just need to replace t=7 and see what we got:

`P(t=7) =(400)/(3) e^(0.2027325541*7)= 551.135 \approx 552`

Part c

For this case we need to satisfy this equation:

`2P_o = P_o e^(kt)`

If we divide both sides by
`P_o` we got:

`2 = e^(kt)`

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

`ln(2) = kt`

And the
`t = (ln(2))/(0.2027325541)=3.419 hours`