Question

A bacteria culture grows with a constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count is 50,000.

(a) Find the initial population. P(0) = 80 )ãbacteria
(b) Find an expression for the population after t hours. r(t) = 180( 125(6 Plt) =180(125(2))-
(c) Find the number of cells after 7 hours. (Round your answer to the nearest integer.) P(7)=72.358- bacteria
(d) Find the rate of growth after 7 hours. (Round your answer to the nearest integer.) P(7) 2x bacteria/hour
(e) When will the population reach 200,000? (Round your answer to one decimal place.) hours

Answer 1

a) P(0) = 80

b)
`P(t) = 80(2.2361)^t`

c) 22,363 cells.

d) The rate of growth after 7 hours is of 18,000 bacteria per hour.

e) 9.7 hours.

Explanation:

A bacteria culture grows with a constant relative growth rate.

This means that the population is given by:

`P(t) = P(0)(1+r)^t`

In which P(0) is the initial population and r is the growth rate, as a decimal.

After 2 hours there are 400 bacteria and after 8 hours the count is 50,000.

This means that in 6 hours, the population went from 400 bacteria to 50,000 bacteria. We use this to find r. So

`50000 = 400(1+r)^6`

`(1+r)^6 = (50000)/(400)`

`(1+r)^6 = 125`

`\sqrt[6]{(1+r)^6} = \sqrt[6]{125}`

`1 + r = 125^{(1)/(6)}`

`1 + r = 2.2361`

So

`P(t) = P(0)(2.2361)^t`

(a) Find the initial population. P(0)

We have that P(2) = 400. We use this to find P(0). So

`P(t) = P(0)(2.2361)^t`

`400 = P(0)(2.2361)^2`

`P(0) = (400)/((2.2361)^2)`

`P(0) = 80`

So

`P(t) = 80(2.2361)^t`

(b) Find an expression for the population after t hours.

`P(t) = 80(2.2361)^t`

(c) Find the number of cells after 7 hours.

This is P(7). So

`P(7) = 80(2.2361)^7 = 22363`

22,363 cells.

(d) Find the rate of growth after 7 hours.

We have to find the derivative when t = 7. So

`P(t) = 80(2.2361)^t`

`P^(\prime)(t) = 80ln(2.2361)(2.2361)^t`

`P^(\prime)(7) = 80ln(2.2361)(2.2361)^7 = 18000`

The rate of growth after 7 hours is of 18,000 bacteria per hour.

(e) When will the population reach 200,000?

This is t for which
`P(t) = 200000`. So

`P(t) = 80(2.2361)^t`

`200000 = 80(2.2361)^t`

`(2.2361)^t = (200000)/(80)`

`(2.2361)^t = 2500`

`\log{(2.2361)^t} = \log{2500}`

`t\log{2.2361} = \log{2500}`

`t = \frac{\log{2500}}{\log{2.2361}}`

`t = 9.7`

So 9.7 hours.