Suppose that the number of bacteria in a culture at time (t) is given by:

Question

Suppose that the number of bacteria in a culture at time (t) is given by:

N=4000(30+(t)e^(-t/10))

(a) Find the largest and smallest number of bacteria in the culture during the time interval 0 ≤ t  ≤ 100.

Largest number of bacteria = ?

Smallest number of bacteria = ?

(b) At what time during the time interval in part (a) is the number of bacteria decreasing most rapidly?

t=?

Answer 1

a)To find the largest and smallest number of bacteria we take the derivative dN/dt to find the rate of change in number of bacteria. Using the chain rule, we find that:

dN/dt=-(400t-4000)e^(-t/10)

A minimum or maximum occurs when dN/dt=0, so set dN/dt=0 and solve for t. Looking at the interior of the first parentheses it is clear dN/dt=0 if t=10.

Now we must determine if this is a minimum or a maximum. Taking the derivative again we find

d²N/dt²= (40t-800)e^-(t/10)

Inputting t=10 we find that d²N/dt² is a negative number. A negative number means that at t=10, the slope of the equation is decreasing. This indicates that this number is a maximum.

If the slope were increasing, that means the actual value N in the original equation would continue to increase as t increased, indicating a minimum.

Knowing t=10 is a maximum plug into the original equation to find N=134,715.

To find the smallest number of bacteria we must go back to the dN/dt term. Looking at the time interval 0<=t<=100, we can see that no other times in this range will result in dN/dt =0.

Below t=10 we know that the number N is increasing because the slope is always positive beneath this point. Above t=10 we can see that the number N is decreasing because the slope always negative above this point. (NOTE: e^-(t/10) is always positive). This means that the two ends will have the relative minimums for the original equation.

Calculate the values of N for both t=0 and t=100 to find

N=120000 @t=0 and N=120018.16 @t=100.

You can also come to this conclusion by simple inspection of the original equation. As stated before

e^-(t/10) is always positive. Therefore, if the value of t is above 0, the term te^(-t/10) term will always be positive and will always increase the value above the initial 120,000 (unless t->infinite).

b) To find when the number of bacteria is decreasing most rapidly, we want the minimum slope. To do this we set d²N/dt²=0. Solving the second derivative we find that at t=20. To determine if this is the minimum or maximum we take the derivative once more

=−(4t−120)e^-(t/10). Plugging in t=20 we find that the value is positive. This indicates a minimum and thus the point where the number of bacteria is decreasing most rapidly.