2025 AP Calculus AB FRQ Question 1: Invasive Species Spread Rate

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2025.

Question 1

Calculator OK

Invasive Species Spread Rate

Rate, Average Value & Optimization

Hard

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) = \dfrac{38}{25 + t^2}

. (Calculator should be in radian mode.)

Part AEasy2 points

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.

Answer

\dfrac{1}{4-0}\int_0^4 C(t)\,dt = \dfrac{1}{4}(11.112896) = 2.778224 \approx 2.778

acres

Full Solution & Work

Apply the average value formula

The average value of

C

[0,4]

is:

\frac{1}{4-0}\int_0^4 C(t)\,dt

Evaluate with a calculator

\int_0^4 C(t)\,dt = 11.112896

\frac{1}{4}(11.112896) = 2.778224

From

t=0

t=4

weeks, the average number of acres affected was **2.778 acres**.

AP Scoring — 2 Points

**P1**: Correct average-value setup

\frac{1}{4}\int_0^4 C(t)\,dt

(division by 4 can be shown implicitly via the correct answer).
**P2**: Correct answer 2.778, accurate to three decimal places.

Part BMedium2 points

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 \leq t \leq 4

. Show the setup for your calculations.

Answer

t = 2.154298 \approx 2.154

weeks

Full Solution & Work

Find the average rate of change of C on [0,4]

\frac{C(4)-C(0)}{4-0} = 1.282008

(since

C(0)=0

and

C(4) = 7.6\arctan(0.8) = 5.128031

)

Set the instantaneous rate equal to it and solve

C'(t) = \frac{38}{25+t^2} = 1.282008

Solving with a calculator:

t = 2.154298

AP Scoring — 2 Points

**P1**: Uses the correct average rate of change

\frac{C(4)-C(0)}{4-0}

(or an equivalent expression like

1.282

).
**P2**: Correct answer

t = 2.154

, supported by the equation

C'(t) = 1.282

Common mistake: Presenting only "$t = 1.282$" without the supporting equation earns neither point — the average rate of change and the equation $C'(t) = \text{average rate}$ must both appear.

Part CEasy2 points

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.

Answer

\lim_{t \to \infty} C'(t) = \lim_{t \to \infty} \dfrac{38}{25+t^2} = 0

Full Solution & Work

Write the limit of the rate function

The rate of change is

C'(t)

, so the end behavior is described by:

\lim_{t \to \infty} C'(t) = \lim_{t \to \infty} \frac{38}{25+t^2}

Evaluate

t \to \infty

, the denominator

25+t^2 \to \infty

while the numerator stays fixed at 38, so:

\lim_{t \to \infty} \frac{38}{25+t^2} = 0

AP Scoring — 2 Points

**P1**: Correct limit expression — either

\lim_{t\to\infty} C'(t)

\lim_{t\to\infty} \frac{38}{25+t^2}

. Writing

\lim_{t\to\infty} C(t)

instead does **not** earn this point (that describes the acres, not the rate).
**P2**: Correct value 0.

Common mistake: A very common slip: writing $\lim_{t\to\infty} C(t)$ instead of $C'(t)$. The question asks about the *rate of change*, so it must be the derivative inside the limit.

Part DHard3 points

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 \leq t \leq 36

. At what time

t

, for

4 \leq t \leq 36

, does

A

attain its maximum value? Justify your answer.

Answer

t = 11.442

(or

11.441

) weeks

Full Solution & Work

Differentiate A using FTC

By the Fundamental Theorem of Calculus:

A'(t) = C'(t) - 0.1\ln t

Set A'(t) = 0 and solve

A maximum on

[4,36]

occurs where

A'(t)=0

or at an endpoint.

A'(t) = C'(t) - 0.1\ln t = 0 \implies C'(t) = 0.1\ln t

Solving numerically:

t = 11.441700

Candidates test

\begin{array}{c|c} t & A(t) \\\hline 4 & 5.128031 \\ 11.4417 & 7.316978 \\ 36 & 1.743056 \end{array}

The largest value of

A(t)

occurs at

t = 11.442

AP Scoring — 3 Points

**P1**: Considers

A'(t) = 0

, i.e. sets

C'(t) - 0.1\ln t = 0

(equivalently

C'(t)=0.1\ln t

). Just presenting

t=11.4417

without this equation does **not** earn the point.
**P2**: A complete justification — either a candidates-test table evaluating

A(4)

A(11.442)

A(36)

, or a sign analysis of

A'(t)

around

t=11.442

.
**P3**: Correct final answer

t = 11.442

(or 11.441), supported by the work above.

Common mistake: A first- or second-derivative test alone (a *local* argument) does not earn the justification point on a closed-interval max/min question — AP requires a *global* argument: either compare $A$ at all candidates (endpoints + critical points), or show $A'$ changes from $+$ to $-$ specifically at $t=11.442$ and note it's the only critical point in the interval.

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