Chapter 7.5, Problem 6E

a.

To determine

To plot : the data and estimate the carrying capacity for the yeast population

Answer to Problem 6E

The carrying capacity of the population is

Explanation of Solution

Given information : The question is

Time (hours)	Yeast cells	Time (hours)	Yeast cells
0	18	10	509
2	39	12	597
4	80	14	640
6	171	16	664
8	336	18	672

Curve is converging to $680$ . So, $680$ is the carrying capacity of the population

The graph plot shown below

  

b.

To determine

To calculate : the initial relative growth rate

Answer to Problem 6E

The initial relative growth rate $k\approx 0.4$

Explanation of Solution

Given information : The question is

Time (hours)	Yeast cells	Time (hours)	Yeast cells
0	18	10	509
2	39	12	597
4	80	14	640
6	171	16	664
8	336	18	672

Calculation :

From the logistic equation

  $\begin{array}{l}P=\frac{M}{1+A{e}^{-kt}}\\ \text{where },A=\frac{M-P_0}{P_0}\\ M=800\\ A=\frac{M-P_0}{P_0}=\frac{800-18}{18}=43.44\\ P=\frac{800}{1+43.44e^{-kt}}\\ P\left(2\right)=39\\ P\left(2\right)=\frac{800}{1+43.44e^{-k\left(2\right)}}\\ 39=\frac{800}{1+43.44e^{-k\left(2\right)}}\\ 39\cdot 34.44e^{-2k}=761\\ e^{-2k}=0.4493\\ \ln e^{-2k}=\ln 0.4493\\ k=-\frac{\ln 0.4493}{2}\\ k=0.4\\ k\approx 0.4\end{array}$

c.

To determine

To calculate : the exponential and logistic model of the data

Answer to Problem 6E

In exponential model is $P\left(t\right)=18e^{0.4t}$ and the logistic model is $P=\frac{800}{1+43.44e^{-0.4t}}$

Explanation of Solution

Given information : The question is

Time (hours)	Yeast cells	Time (hours)	Yeast cells
0	18	10	509
2	39	12	597
4	80	14	640
6	171	16	664
8	336	18	672

Calculation :

The exponential model equation is $P\left(t\right)=P_0{e}^{kt}$

In part (b) $k=0.4$ . Therefore, the exponential model is

  $P\left(t\right)=18e^{0.4t}$

The logistic model equation

  $\begin{array}{l}P=\frac{M}{1+A{e}^{-kt}}\\ \text{where },A=\frac{M-P_0}{P_0}\\ M=800\\ A=43.44\\ k=0.4\\ P=\frac{800}{1+43.44e^{-0.4t}}\end{array}$

d.

To determine

To compare : the predicted values to observed values

Explanation of Solution

Given information : The question is

Time (hours)	Yeast cells	Time (hours)	Yeast cells
0	18	10	509
2	39	12	597
4	80	14	640
6	171	16	664
8	336	18	672

  

From the above graph it is observed that the red graph is the logistic model, and the black graph is the exponential model and blue dots are the observed dots

In the graph it can be seen that the red one is far and close to the blue dotes in comparison to the black graph, therefore the conclusion that logistic model is superior in this case

e.

To determine

To calculate : the number of yeast cells after $7$ hours

Answer to Problem 6E

The number of yeast cells after $7$ hours is $220$

Explanation of Solution

Given information : The question is

Time (hours)	Yeast cells	Time (hours)	Yeast cells
0	18	10	509
2	39	12	597
4	80	14	640
6	171	16	664
8	336	18	672

Calculation :

From part (b)

  $\begin{array}{l}P=\frac{800}{1+43.44e^{-0.4t}}\\ P\left(7\right)=P=\frac{800}{1+43.44e^{-0.4\left(7\right)}}\\ P\left(7\right)\approx 220\end{array}$