Tangent Lines and Rates of Change — Question 3

Question 3

A population of bacteria in a lab experiment is modeled by the function $$P(t) = 200e^{0.3t},$$ where $t$ is the time in hours since the start of the experiment and $P(t)$ is the number of bacteria.

(a) Estimate the average rate of growth of the population from $t = 2$ to $t = 5$.

(b) Find the instantaneous rate of growth at $t = 2$.

(c) Sketch the population function along with the secant and tangent lines.

Question 3 - Solution

We are given: $$P(t) = 200e^{0.3t}$$

(a) Average rate of change on $[2, 5]$: $$P(2) = 200e^{0.6} \approx 200(1.8221) = 364.42$$ $$P(5) = 200e^{1.5} \approx 200(4.4817) = 896.34$$ $$\text{Average rate} = \frac{896.34 - 364.42}{5 - 2} \approx \frac{531.92}{3} \approx 177.31 \text{ bacteria/hr}$$

(b) Instantaneous rate at $t = 2$: $$P'(t) = 200 \cdot 0.3e^{0.3t} = 60e^{0.3t}$$ $$P'(2) = 60e^{0.6} \approx 60(1.8221) = 109.33 \text{ bacteria/hr}$$

(c) Interpretation: The population grows faster over the longer interval than it does exactly at $t = 2$, illustrating exponential acceleration.