Solution:
Given that the number of bacteria in a culture, N, at time t is expressed as
The rate of change is expressed by taking the derivative of the above relation with respect to t.
This gives:
[tex]N^{\prime}(t)=\frac{dN}{dt}=2000\left(-\frac{te^{-\frac{t}{20}}}{20}+e^{-\frac{t}{20}}\right)[/tex]
Thus, when the rate of change equals zero, we have
[tex]\begin{gathered} N^{\prime}(t)=0 \\ \Rightarrow2000\left(-\frac{te^{-\frac{t}{20}}}{20}+e^{-\frac{t}{20}}\right)=0 \\ thus\text{ we have} \\ -\frac{te^{-\frac{t}{20}}}{20}=-e^{-\frac{t}{20}} \\ multiply\text{ both sides by 20} \\ -te^{-\frac{t}{20}}=-20e^{-\frac{t}{20}} \\ cancel\text{ out similar factors,} \\ \therefore \\ t=20 \end{gathered}[/tex]
Hence, the rate of change of of the number of bacteria equals zero when
[tex]t=20[/tex]
