Interpreting average and instantaneous change in a spread model

Question

An invasive species of plant appears in a fruit grove at time $t=0$ and begins to spread. The function $C$ defined by $C(t)=7.6\arctan(0.2t)$ models the number of acres in the fruit grove affected by the species $t$ weeks after the species appears. It can be shown that $C'(t)=\frac{38}{25+t^2}$.

(Note: Your calculator should be in radian mode.)

A. Find the average number of acres affected by the invasive species from time $t=0$ to time $t=4$ weeks. Show the setup for your calculations.

B. Find the time $t$ when the instantaneous rate of change of $C$ equals the average rate of change of $C$ over the time interval $0\le t\le 4$. Show the setup for your calculations.

C. Assume that the invasive species continues to spread according to the given model for all times $t>0$. Write a limit expression that describes the end behavior of the rate of change in the number of acres affected by the species. Evaluate this limit expression.

D. At time $t=4$ weeks after the invasive species appears in the fruit grove, measures are taken to counter the spread of the species. The function $A$, defined by $A(t)=C(t)-\int_4^t 0.1\ln(x)\,dx$ models the number of acres affected by the species over the time interval $4\le t\le 36$. At what time $t$, for $4\le t\le 36$, does $A$ attain its maximum value? Justify your answer.

Acemy AI · Accepted Answer

**Key takeaway**: This problem tests how a modeled accumulation function, its derivative, and a related antiderivative-style adjustment work together on an interval.

## Part A: Average number of acres affected
The function
$$
C(t)=7.6\arctan(0.2t)
$$
models the number of acres affected after $t$ weeks. Since the prompt asks for the **average number of acres affected from $t=0$ to $t=4$**, compute the average value of the function on that interval:
$$
\frac{1}{4-0}\int_0^4 C(t)\,dt
=\frac14\int_0^4 7.6\arctan(0.2t)\,dt.
$$
That setup is the correct one because “average number of acres affected” refers to the average value of $C$, not the average rate of change.

Using integration by parts or a calculator, this evaluates to approximately
$$
\frac14\int_0^4 7.6\arctan(0.2t)\,dt \approx 1.92.
$$
So the average number of acres affected is about **1.9 acres**.

## Part B: When instantaneous rate equals average rate
First find the average rate of change of $C$ on $[0,4]$:
$$
\frac{C(4)-C(0)}{4-0}.
$$
Since $C(0)=7.6\arctan(0)=0$, we have
$$
\text{average rate} = \frac{7.6\arctan(0.8)}{4} \approx 1.25.
$$
Now set the derivative equal to this value:
$$
C'(t)=\frac{38}{25+t^2}=1.25.
$$
Solve:
$$
38=1.25(25+t^2)
$$
$$
38=31.25+1.25t^2
$$
$$
6.75=1.25t^2
$$
$$
t^2=5.4
$$
$$
t\approx 2.3.
$$
So the instantaneous rate equals the average rate at about **$t=2.3$ weeks**.

## Part C: End behavior of the rate of change
The limit expression for the end behavior of the rate of change is
$$
\lim_{t\to\infty} C'(t)=\lim_{t\to\infty}\frac{38}{25+t^2}.
$$
As $t$ grows, the denominator increases without bound, so the fraction approaches 0:
$$
\lim_{t\to\infty}\frac{38}{25+t^2}=0.
$$
This means the spread slows down over time, with the rate of change eventually approaching zero acres per week.

## Part D: Maximum of the adjusted model
The adjusted function is
$$
A(t)=C(t)-\int_4^t 0.1\ln(x)\,dx, \qquad 4\le t\le 36.
$$
Differentiate:
$$
A'(t)=C'(t)-0.1\ln(t)=\frac{38}{25+t^2}-0.1\ln(t).
$$
To find a maximum on a closed interval, check critical points and endpoints. For $t\ge 4$, $\ln(t)$ is positive and the first term $\frac{38}{25+t^2}$ is decreasing, so the derivative becomes negative after the starting point. In fact, at $t=4$,
$$
A'(4)=\frac{38}{41}-0.1\ln 4<0.
$$
Since $A'(t)$ is already negative and stays negative on the interval, $A$ decreases for $4\le t\le 36$. Therefore, the maximum value occurs at the left endpoint:
$$
\boxed{t=4}.
$$

---
**Pitfalls the pros know** 👇
A common mistake is to confuse the average value of the function with the average rate of change. In Part A, you must average $C(t)$ itself, which means integrating $C$ and dividing by the interval length. In Part B, you use the secant slope $\frac{C(4)-C(0)}{4}$, not the derivative at an endpoint. Another subtle error is forgetting that the derivative formula is already given, so there is no need to differentiate $7.6\arctan(0.2t)$ again unless you are checking the model. For Part D, students often set $A'(t)=0$ but then ignore endpoint testing; because the domain is closed, the maximum could occur at $t=4$ or $t=36$ even if no interior critical point exists.

**What if the problem changes?**
If the interval in Part B were changed to $0\le t\le 8$, the average rate of change would become $\frac{C(8)-C(0)}{8}$, which is smaller because $C(t)$ grows more slowly as $t$ increases. The equation $C'(t)=\text{average rate}$ would likely still have a solution, but the solution would shift to a larger value of $t$ because $C'(t)=\frac{38}{25+t^2}$ decreases as $t$ increases. If the adjusted model in Part D had been $A(t)=C(t)+\int_4^t 0.1\ln(x)\,dx$, then the added term would increase the function instead of reducing it, and the maximum might move away from the left endpoint.

`Tags`: average value of a function, instantaneous rate of change, closed interval maximum

Question

Interpreting average and instantaneous change in a spread model

Expert Verified Solution

Part A: Average number of acres affected