Chapter 7.4, Problem 2E

a.

To determine

To calculate : the relative growth

Answer to Problem 2E

The relative growth is $\ln \left(8\right)\text{ or }2.07944$

Explanation of Solution

Given information : The question is

A cell of bacterium in a nutrient-broth medium divides into two cell every $20$ minutes. The initial population of a culture is $60$

Calculation : For the relative growth rate

  $\begin{array}{l}\frac{dy}{dt}=k\times y\\ y\left(0\right)=60\\ y\left(\frac{1}{3}\right)=2\times 60=120\end{array}$

Let $y\left(t\right)$ be the population of the bacteria and let $t$ be time in hours,

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{tk}\\ y\left(\frac{1}{3}\right)=60\times e^{\frac{1}{3}k}\\ 120=60\times e^{\frac{1}{3}k}\\ 2=e^{\frac{1}{3}k}\\ \ln \left(2\right)=\frac{1}{3}k\\ 3\ln \left(2\right)=k\\ \ln \left(2^3\right)=k\\ \ln \left(8\right)=k\end{array}$

Using the exponentially growth equation $y=y\left(0\right)\times e^{tk}$

Solve for $k$ using $y\left(0\right)=60\&y\left(\frac{1}{3}\right)=120$

So, $\ln \left(8\right)\text{ or }2.07944$

b.

To determine

To find : an expression for the number of cells after $t$ hours

Answer to Problem 2E

The number of cells after $t$ hours is $y\left(t\right)=60\times 8^t$

Explanation of Solution

Given information : The question is

A cell of bacterium in a nutrient-broth medium divides into two cell every $20$ minutes. The initial population of a culture is $60$

Calculation : the number of cells after $t$ hours

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{tk}\\ y\left(t\right)=60\times e^{\ln \left(8\right)\times t}\\ y\left(t\right)=60\times 8^t\end{array}$

c.

To determine

To find : the number of cells after $8$ hours

Answer to Problem 2E

The number of cells after $8$ hours is $1,006,632,960cells$

Explanation of Solution

Given information : The question is

A cell of bacterium in a nutrient-broth medium divides into two cell every $20$ minutes. The initial population of a culture is $60$

Calculation : The number of cells after $8$ hours is,

  $\begin{array}{l}y\left(t\right)=60\times 8^t\\ y\left(8\right)=60\times 8^8\\ y\left(8\right)=1,006,632,960cells\end{array}$

d.

To determine

To find : the rate of growth after $8$ hours

Answer to Problem 2E

The rate of growth after $8$ hours is $2,093,234,394.2481$

Explanation of Solution

Given information : The question is

A cell of bacterium in a nutrient-broth medium divides into two cell every $20$ minutes. The initial population of a culture is $60$

Calculation : The rate of growth after $8$ hours is,

  $\begin{array}{l}\frac{dy}{dt}=k\times y\\ \frac{dy}{dt}=\ln \left(8\right)\times 60\times 8^8\\ \frac{dy}{dt}=2,093,234,394.2481\end{array}$

e.

To determine

When will the population reach $20,000$ cells?

Answer to Problem 2E

The population will reach in $2.794\text{ hours}$

Explanation of Solution

Given information : The question is

A cell of bacterium in a nutrient-broth medium divides into two cell every $20$ minutes. The initial population of a culture is $60$

Calculation :

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ 20,000=60\times 8^t\\ \frac{1,000}{3}=8^t\\ \log 8\left(\frac{1,000}{3}\right)=t\\ t=2.794\text{ hours}\end{array}$