Question

In a lab experiment, 80 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to double every 16 hours. Write a function showing the number of bacteria after tt hours, where the hourly growth rate can be found from a constant in the function. Round all coefficients in the function to four decimal places. Also, determine the percentage of growth per hour, to the nearest hundredth of a percent.

Answer 1

The hourly growth rate is of 4.43%.

The function showing the number of bacteria after t hours is
`P(t) = 80(1.0443)^t`

Explanation:

Equation of population growth:

The equation for the population after t hours 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.

The conditions are such that the number of bacteria is able to double every 16 hours.

This means that
`P(16) = 2P(0)`. We use this to find r.

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

`2P(0) = P(0)(1+r)^(16)`

`(1+r)^(16) = 2`

`\sqrt[16]{(1+r)^(16)} = \sqrt[16]{2}`

`1 + r = 2^{(1)/(16)}`

`1 + r = 1.0443`

`r = 1.0443 - 1`

`r = 0.0443`

The hourly growth rate is of 4.43%.

80 bacteria are placed in a petri dish.

This means that
`P(0) = 80`.

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

`P(t) = 80(1+0.0443)^t`

`P(t) = 80(1.0443)^t`

The function showing the number of bacteria after t hours is
`P(t) = 80(1.0443)^t`