# A bacteria culture initially contains 100 cells and grows at a rate proportional to its size....

## Question:

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 330.

(a) Find an expression for the number of bacteria after t hours.

(b) Find the number of bacteria after 2 hours. (Round your answer to the nearest whole number.)

(c) Find the rate of growth after 2 hours. (Round your answer to the nearest whole number.)

(d) When will the population reach 10,000? (Round your answer to one decimal place.)

## Exponential Growth:

To solve this problem, we will be using an exponential model. The model will be using will be of the type given below.

$$P(t)=P_o e^{kt}$$

Our first step will be to complete the model by finding the rate of growth k.

a) As the bacteria population is growing proportional to its size, we can use the following exponential model to find the population after t hours.

$$N(t)=100e^{kt}$$

Here, N(t) is the population after t hours when the initial population, 100, is growing at a rate of k per hour.

We have to find the value of k to complete the model. We can do this as we know that the population rose to 330 in 1 hour.

\begin{align} &N(1)=100e^{k}=330\\ &e^{k}=3.3\\ &k=\ln3.3\\ \therefore &N(t)=100e^{t\ln3.3} \end{align}

b) The population after 2 hours will be:

\begin{align} N(2)&=100e^{2\ln3.3}\\&=1089 \end{align}

c) The rate of growth of N(t) after any time t will be given by its derivative:

\begin{align} N'(t)&=\frac{\mathrm{d} }{\mathrm{d} t}\left (100e^{t\ln3.3} \right )\\ &=83.7575e^{t\ln 3.3} \end{align}

At t=2, this will be:

{eq}\begin{align} N'(2)&=83.7575e^{2\ln 3.3}\\ &=912.119 \end{align} {/eq}

d) The time after which the population will reach 10000 can be found as follows.

\begin{align} &100e^{t\ln3.3}=10000\\ &e^{t\ln3.3}=100\\ &t\ln3.3=\ln 100\\ &t\approx 3.857\,\text{hours} \end{align}