Chapter 4.6, Problem 14E

Population Growth These exercises use the population growth model.

14. Bacteria Culture The count in a culture of bacteria was 400 after 2 hours and 25,600 after 6 hours.

1. (a) What is the relative rate of growth of the bacteria population? Express your answer as a percentage.
2. (b) What was the initial size of the culture?
3. (c) Find a function that models the number of bacteria n(t) after t hours.
4. (d) Find the number of bacteria after 4.5 hours.
5. (e) After how many hours will the number of bacteria reach 50,000?

To determine

To find: The relative growth rate of the bacteria population if the bacteria population after $2\text{hours}$ is $400$ and $6\text{hours}$ is $25,600$.

Answer to Problem 14E

The relative growth rate is $103.97\%$.

Explanation of Solution

Formula used:

Exponential growth model (relative growth rate) is, $n\left(t\right)=n_0{e}^{rt}$ (1)

Where, $n_0$ is the initial size of the population, $r$ is relative growth rate and $n\left(t\right)$ is size of the population at time $t$.

Calculation:

Given that the bacteria population after $2\text{hours}$ is $400$.

Substitute $400$ for $n\left(t\right)$ and $2$ for $t$ in equation (1), and get the bacteria population at 2 hours.

Thus, the bacteria population at 2 hours is $400=n_0{e}^{2r}$.

The bacteria population after $6\text{hours}$ is $25,600$.

Thus, the bacteria population at 6 hours is $25600=n_0{e}^{6r}$.

Divide $25600=n_0{e}^{6r}$ by $400=n_0{e}^{2r}$ to find the relative growth rate of the bacteria population,

$\begin{array}{c}\frac{n_0{e}^{6r}}{n_0{e}^{2r}}=\frac{25600}{400}\\ e^{4r}=64\\ \ln \left(e^{4r}\right)=\ln \left(64\right)\left(\because \text{taking}\text{ln}\text{to}\text{each}\text{side}\right)\end{array}$

Simplify the above expression,

$\begin{array}{c}4r=\ln \left(64\right)\left(\because \ln \left(e^a\right)=a\right)\\ r=\frac{4.158883}{4}\\ =1.03972\\ \approx 103.97\%\end{array}$

Thus, the relative growth rate is $103.97\%$.

To determine

To find: The initial size of the culture of bacteria.

Answer to Problem 14E

The initial size of the bacteria population in the culture is $50$.

Explanation of Solution

Given that the bacteria population after $2\text{hours}$ is $25,600$.

From part (a), the relative growth rate is $103.97\%$.

Substitute $400$ for $n\left(t\right)$, $2$ for $t$ and $1.0397$ for $r$ in equation (1) to find the initial population of bacteria in the culture,

$\begin{array}{c}400=n_0{e}^{1.0397\left(2\right)}\\ n_0=\frac{400}{e^{2.0794}}\\ =\frac{400}{7.9997}\\ \approx 50.00188\end{array}$

Thus, the initial size of the bacteria population in the culture is $50$.

To determine

To find: The function for bacteria population.

Answer to Problem 14E

The function for bacteria population is $n\left(t\right)=50e^{1.0397t}$.

Explanation of Solution

From part (b), the initial population is $50$.

From part (a), the relative growth rate is $103.97\%$.

Substitute $50$ for $n_0$ and $1.0397$ for $r$ in equation (1) to find the bacteria growth function,

$\begin{array}{l}n\left(t\right)=n_0{e}^{rt}\\ n\left(t\right)=50e^{1.0397t}\end{array}$

Thus, the function for bacteria population is $n\left(t\right)=50e^{1.0397t}$.

To determine

To find: The number of bacteria after $4.5\text{hours}$.

Answer to Problem 14E

The bacteria population after $4.5\text{hours}$ is about $5,381$.

Explanation of Solution

From part (c), the population function is $n\left(t\right)=50e^{1.0397t}$.

Substitute $4.5$ for $t$ in $n\left(t\right)=50e^{1.0397t}$ to find the bacteria population after $4.5\text{hours}$,

$\begin{array}{c}n\left(4.5\right)=50e^{1.0397\left(4.5\right)}\\ =50e^{4.67865}\\ \approx 50\left(107.625\right)\left(\because e^{4.67865}\approx 107.625\right)\\ =5,381.25\end{array}$

Thus, the bacteria population after $4.5\text{hours}$ is about $5,381$.

To determine

To find: The time in which the bacteria population reach to $50,000$.

Answer to Problem 14E

The bacteria population reach to $50,000$ in $6.64\text{hours}$.

Explanation of Solution

From part (c), the population function is $n\left(t\right)=50e^{1.0397t}$.

Substitute $50,000$ for $n\left(t\right)$ in $n\left(t\right)=50e^{1.0397t}$ to find the time in which the bacteria population reaches to $50,000$,

$\begin{array}{c}50,000=50e^{1.0397t}\\ 1,000=e^{1.0397t}\\ \ln \left(1000\right)=\ln \left(e^{1.0397t}\right)\left(\because \text{Take}\text{ln}\text{to}\text{each}\text{side}\right)\\ \ln \left(1000\right)=1.0397t\left(\because \ln \left(e^a\right)=a\right)\end{array}$

Simplify the above expression,

$\begin{array}{c}t=\frac{\ln \left(1000\right)}{1.0397}\\ =\frac{6.907755}{1.0397}\\ =6.643989\\ \approx 6.64\end{array}$

Thus, the bacteria population reach to $50,000$ in $6.64\text{hours}$.