Chapter 4.6, Problem 8E

a.

To determine

Find the projected population after $24$ hours for the following conditions.

Answer to Problem 8E

  $\boxed{392}$

Explanation of Solution

Given information:

It is observed that a certain bacteria culture has a relative growth rate of $12\%$ per hour, but in the presence of an antibiotic the relative growth rate is reduced to $5\%$ per hour. The initial number of bacteria in the culture is $22$ . Find the projected population after $24$ hours for the following conditions.

No antibiotic is present, so the relative growth rate is $12\%$ .

Calculation:

Consider exponentially growth model,

  $n\left(t\right)=n_0{e}^{rt}$

Initial growth rate , $r=12\%=0.12$

  $n_0=22$

The projected population is $\boxed{n\left(t\right)=22e^{0.12t}}$

Now find population in $24$ hours,

Put $t=24$

  $\begin{array}{l}n\left(24\right)=22e^{0.12\times 24}\\ =22\left(17.8\right)\\ =\boxed{392}\end{array}$

Hence, the solution is, $\boxed{392}$

b.

To determine

Find the projected population after $24$ hours for the following conditions.

Answer to Problem 8E

  $\boxed{73}$

Explanation of Solution

Given information:

It is observed that a certain bacteria culture has a relative growth rate of $12\%$ per hour, but in the presence of an antibiotic the relative growth rate is reduced to $5\%$ per hour. The initial number of bacteria in the culture is $22$ . Find the projected population after $24$ hours for the following conditions.

An antibiotic is present in the culture, so the relative growth rate is reduced to 5%.

Calculation:

Consider exponentially growth model,

  $n\left(t\right)=n_0{e}^{rt}$

Initial growth rate, $r=5\%=0.05$

  $n_0=22$

The projected population is $\boxed{n\left(t\right)=22e^{0.05t}}$

Now find population in $24$ hours,

Put $t=24$

  $\begin{array}{l}n\left(24\right)=22e^{0.05\times 24}\\ =22\left(3.32\right)\\ =\boxed{73}\end{array}$

Hence, the solution is, $\boxed{73}$