MA005: Calculus I

Suppose we set up a machine to count the number of bacteria growing on a petri plate (Fig. 7). At first there are few bacteria so the population grows slowly. Then there are more bacteria to divide so the population grows more quickly. Later, there are more bacteria and less room and nutrients available for the expanding population, so the population grows slowly again. Finally, the bacteria have used up most of the nutrients, and the population declines as bacteria die. 

The population graph can be used to answer a number of questions.

(a) What is the bacteria population at time $\mathrm{t}=3$ days?

(Answer: about 500 bacteria)

(b) What is the population increment from $\mathrm{t}=3$ to $\mathrm{t}=10$ days?

(about 4000 bacteria)

(c) What is the rate of population growth from $\mathrm{t}=3$ to $\mathrm{t}=10$ days? (Fig. 7)

Solution: The rate of growth from $\mathrm{t}=3$ to $\mathrm{t}=10$ is the average change in population during that time:

$\begin{aligned} \text { average change in population } &=\frac{\text { change in population }}{\text { change in time }}=\frac{\Delta \text { population }}{\Delta \text { time }} \\ &=\frac{4000 \text { bacteria }}{7 \text { days }} \approx 570 \text { bacteria/day.} \end{aligned}$

This is the slope of the secant line through the two points $(3,500)$ and $(10,4500)$.

(d) What is the rate of population growth on the third day, at $\mathrm{t}=3$?

Solution: This question is asking for the instantaneous rate of population change, the slope of the line which is tangent to the population curve at (3,500). If we sketch a line approximately tangent to the curve at (3,500) and pick two points near the ends of the tangent line segment (Fig. 8), we can estimate that instantaneous rate of population growth is approximately $\text{320 bacteria/day}$.

$\begin{aligned}  \text { Average population growth rate } &=\frac{\Delta \text { population }}{\Delta \text { time }} \\ &=\text { slope of the secant line through } 2 \text { points. } \\ \\ \text { Instantaneous population growth rate }& =\text { slope of the line tangent to the graph. }\end{aligned}$

Practice 4: Approximately what was the average change in population between $\mathrm{t}=9$ and $\mathrm{t}=13$ ? Approximately what was rate of population growth at $\mathrm{t}=9$ days?

The tangent line problem, the instantaneous velocity problem and the instantaneous growth rate problem are all similar. In each problem we wanted to know how rapidly something was changing at an instant in time, and each problem turned out to be finding the slope of a tangent line. The approach in each problem was also the same: find an approximate solution and then examine what happened to the approximate solution over shorter and shorter intervals. We will often use this approach of finding a limiting value, but before we can use it effectively we need to describe the concept of a limit with more precision.