Chapter 7.4, Problem 4E

a.

To determine

To calculate : The relative growth rates express answer in percentage

Answer to Problem 4E

The relative growth rate is $104\%$

Explanation of Solution

Given information : The question is

A bacteria culture grows with constant relative growth rate. The bacteria count was $400$ after $2$ hours and $25,600$ after $6$ hours

Calculation : Let $k$ be the relative growth rate, which is given as constant

So,

  $\begin{array}{l}A\left(t\right)=A_0{e}^{kt}\\ A\left(2\right)=400\\ A\left(2\right)=A_0{e}^{2k}\\ 400=A_0{e}^{2k}\\ A_0=\frac{400}{e^{2k}}\\ A\left(t\right)=A_0{e}^{kt}\Rightarrow A\left(6\right)=25,600\\ A\left(6\right)=A_0{e}^{6k}\\ 25,600=A_0{e}^{6k}\\ \frac{25,600}{e^{6k}}=A_0\\ \frac{25,600}{e^{6k}}=\frac{400}{e^{2t}}\\ \frac{25,600}{400}=\frac{e^{6k}}{e^{2k}}\\ 64=e^{4k}\\ \ln \left(64\right)=\ln \left(e^{4k}\right)\\ \ln \left(64\right)=4k\\ k=\frac{\ln \left(64\right)}{4}\\ k=\frac{\ln \left(8\right)}{2}\approx 1.04\\ In\%\\ k=\frac{\ln \left(8\right)}{2}=104\%\end{array}$

b.

To determine

To calculate : the initial size of the culture

Answer to Problem 4E

The initial size of the culture is $50$

Explanation of Solution

Given information The question is

A bacteria culture grows with constant relative growth rate. The bacteria count was $400$ after $2$ hours and $25,600$ after $6$ hours

Calculation : The initial size of culture is the size of culture when $t=0$

  $\begin{array}{l}A\left(t\right)=A_0{e}^{kt}\\ A\left(0\right)=A_0\\ A\left(2\right)=400\Rightarrow k=\frac{\ln \left(8\right)}{2}\\ A\left(2\right)=A_0{e}^{2\times \frac{\ln \left(8\right)}{2}}=400\\ A_0{e}^{\ln \left(8\right)}=400\\ 8A_0=400\\ A_0=50\end{array}$

c.

To determine

To calculate : the expression for the number of bacteria after $t$ hours

Answer to Problem 4E

The number of bacteria after $t$ hours is $50e^{1.04t}$

Explanation of Solution

Given information : The question is

A bacteria culture grows with constant relative growth rate. The bacteria count was $400$ after $2$ hours and $25,600$ after $6$ hours

Calculation : The number of bacteria after $t$ hours is,

  $\begin{array}{l}{A}_0=50\\ k=1.04\\ A\left(t\right)=A_0{e}^{kt}\\ A\left(t\right)=50e^{1.04t}\end{array}$

d.

To determine

To calculate : the number of bacteria cells after $4.5$ hours

Answer to Problem 4E

The number of bacteria cells after $4.5$ hours is $5381$

Explanation of Solution

Given information : The question is

A bacteria culture grows with constant relative growth rate. The bacteria count was $400$ after $2$ hours and $25,600$ after $6$ hours

Calculation : from part (c)

The number of bacteria after $t$ hours is $50e^{1.04t}$

  $\begin{array}{l}A\left(t\right)=50e^{1.04t}\\ A\left(t\right)=50e^{0.5\times \ln 8\times 4.5}\\ A\left(t\right)=5381\end{array}$

e.

To determine

To calculate : the rate of growth after $4.5$ hours

Answer to Problem 4E

The rate of growth after $4.5$ hours is $5595$

Explanation of Solution

Given information : The question is

A bacteria culture grows with constant relative growth rate. The bacteria count was $400$ after $2$ hours and $25,600$ after $6$ hours

Calculation : Number of bacteria cells after $t$ hours

  $\begin{array}{l}A\left(t\right)=50e^{0.5\times \ln 8\times 4.5}\\ \frac{dA\left(t\right)}{dt}=50e^{0.5\times \ln 8\times t}\times 0.5\times \ln \left(8\right)\\ t=4.5\\ \Rightarrow 50e^{0.5\times \ln 8\times 4.5}\times 0.5\times \ln \left(8\right)=5595\end{array}$

So, the rate of growth after $4.5$ hours is $5595$ cells per hour

f.

To determine

When will the population reach $50,000$

Answer to Problem 4E

The population of bacteria will reach $50,000$ in $6.65hours$

Explanation of Solution

Given information : The question is

A bacteria culture grows with constant relative growth rate. The bacteria count was $400$ after $2$ hours and $25,600$ after $6$ hours

Calculation : Number of bacteria cells after $t$ hours

  $\begin{array}{l}A\left(t\right)=50e^{0.5\times \ln 8\times t}\\ 50000=50e^{0.5\times \ln 8\times t}\\ 1000=e^{0.5\times \ln 8\times t}\\ 1000={\left(e^{0.5\times \ln 8}\right)}^t\\ \ln \left(1000\right)=t\times \ln \left(e^{0.5\times \ln 8}\right)\\ \ln \left(1000\right)=t\times \left(0.5\times \ln 8\right)\\ t=\frac{\ln \left(1000\right)}{\left(0.5\times \ln 8\right)}=6.65hours\end{array}$