Chapter 4.3, Problem 60PS

Solution

a.

Write the function for the number of bacteria after 1:00 p.m.

  $P\left(t\right)=30e^{0.116t}$

Given information:

The exponential function $P\left(t\right)=P_0{e}^{0.116t}$ where $P\left(t\right)$ is the population after $t$ hours and $P_0$ is the population.

At $1:00$ pm there are $30$ mycobacterium.

Concept Used:

Substitute the value of a $P\left(t\right)$ on the exponential function $P\left(t\right)=P_0{e}^{0.116t}$.

Calculation:

As per the given problem

Note that for this problem the starting population $P_0=30$

Now the working equation for this problem,

  $\begin{array}{l}P\left(t\right)=P_0{e}^{0.116t}\\ P\left(t\right)=30e^{0.116t}\end{array}$

Conclusion:

Hence, the equation is $P\left(t\right)=30e^{0.116t}$

b.

Graph the function

The graph is in the solution

Given:

The exponential function $P\left(t\right)=P_0{e}^{0.116t}$ where $P\left(t\right)$ is the population after $t$ hours and $P_0$ is the population.

At $1:00$ pm there are $30$ mycobacterium.

Concept Used:

Draw the graph Substitute the value of a $P\left(t\right)$ on the exponential function $P\left(t\right)=P_0{e}^{0.116t}$

.Graph:

  

Conclusion:

Hence, the graph is above in the solution.

c.

Write the population at 5:00 P.M.

  $=48$

Given:

The exponential function $P\left(t\right)=P_0{e}^{0.116t}$ where $P\left(t\right)$ is the population after $t$ hours and $P_0$ is the population.

  $P\left(t\right)=30e^{0.116t}$

Concept Used:

Substitute the value of a $P\left(t\right)$ on the exponential function $P\left(t\right)=P_0{e}^{0.116t}$.

Calculation:

As per the given problem

Note that for 5:00 P.M that would be $4$ hours later so $t=4$ . Going up and over from $4$ on the graph leads me to just under $48$

Now the working equation for this problem,

  $\begin{array}{l}P\left(t\right)=P_0{e}^{0.116t}\\ P\left(t\right)=30e^{0.116t}\end{array}$

Conclusion:

Hence, the population at 5:00 pm $48$.

d.

How to find the population at 3:45 P.M.

  $t=2\frac{3}{4}$

Given:

The exponential function $P\left(t\right)=P_0{e}^{0.116t}$ where $P\left(t\right)$ is the population after $t$ hours and $P_0$ is the population.

  $P\left(t\right)=30e^{0.116t}$

Concept Used:

Substitute the value of a $P\left(t\right)$ on the exponential function $P\left(t\right)=P_0{e}^{0.116t}$.

Calculation:

As per the given problem

Note that for 5:00 P.M that would be $4$ hours later so $t=4$ . Going up and over from $4$ on the graph leads me to just under $48$

Now the working equation for this problem,

  $\begin{array}{l}P\left(t\right)=P_0{e}^{0.116t}\\ P\left(t\right)=30e^{0.116t}\end{array}$

Graph the function

  

Now to find the population at $3:45$ , figure out the time . $2$ hours and $45$ minutes have passed.

For the equation, time needs to be in hours so this would convert to $2.75$ hours.

Now plug it in and do the math or a less $t=2\frac{3}{4}$ on the graph and read up and over.

Conclusion:

To find the population at 3:45 pm read answer above.