2021 AP Calculus AB FRQ Question 1: Bacteria Density in a Petri Dish

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2021.

Question 1

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Bacteria Density in a Petri Dish

Rate & Data from Tables

Hard

The density of a bacteria population in a circular petri dish at a distance

r

centimeters from the center of the dish is given by an increasing, differentiable function

f

, where

f(r)

is measured in milligrams per square centimeter. Values of

f(r)

for selected values of

r

are given in the table shown.

Values of f(r)

r (centimeters)	0	1	2	2.5	4
f(r) (mg per cm²)	1	2	6	10	18

Part AEasy2 points

Use the data in the table to estimate

f'(2.25)

. Using correct units, interpret the meaning of your answer in the context of this problem.

Answer

f'(2.25)\approx8

mg/cm² per cm

Full Solution & Work

Set up the difference quotient

Since

2.25

is the midpoint of

2

and

2.5

f'(2.25) \approx \frac{f(2.5)-f(2)}{2.5-2} = \frac{10-6}{0.5}=8

Interpret the answer

At a distance

r=2.25

cm from the center, the density of bacteria is **increasing at a rate of about 8 mg/cm² per cm**.

AP Scoring — 2 Points

**P1**: The difference quotient, correctly set up and evaluated.
**P2**: Correct interpretation with units, tied to

r=2.25

Part BMedium3 points

The total mass, in milligrams, of bacteria in the petri dish is given by the integral expression

2\pi\displaystyle\int_0^4 rf(r)\,dr

. Approximate the value of

2\pi\displaystyle\int_0^4 rf(r)\,dr

using a right Riemann sum with the four subintervals indicated by the data in the table.

Answer

\approx 269\pi \approx 845.088

Full Solution & Work

Set up the right Riemann sum for the inner integral

\int_0^4 rf(r)\,dr \approx (1)\big(1\cdot f(1)\big)+(1)\big(2\cdot f(2)\big)+(0.5)\big(2.5\cdot f(2.5)\big)+(1.5)\big(4\cdot f(4)\big)

Substitute values and compute

= (1)(2)+(1)(12)+(0.5)(25)+(1.5)(72) = 2+12+12.5+108=134.5

Multiply by 2π

2\pi\int_0^4 rf(r)\,dr \approx 2\pi(134.5)=269\pi\approx845.088 \text{ mg}

AP Scoring — 3 Points

**P1**: Correct form of the right Riemann sum, using

r\cdot f(r)

at each right endpoint with correct widths.
**P2**: Correctly computes the inner sum, 134.5.
**P3**: Correct final answer including the factor of

2\pi

269\pi\approx845.088

mg.

Part CMedium1 point

Is the approximation found in part (b) an overestimate or an underestimate of the total mass of bacteria in the petri dish? Explain your reasoning.

Answer

Overestimate.

Full Solution & Work

Reason about the integrand

f

is given to be an **increasing** function, and

r

is also increasing (it's the variable of integration), so the product

r\cdot f(r)

is increasing on

[0,4]

Apply the right-sum rule for increasing integrands

For an increasing integrand, a **right** Riemann sum uses the largest value on each subinterval, so it **overestimates** the true value of the integral.

AP Scoring — 1 Point

**P1**: "Overestimate," with reasoning that

f

(and hence

r f(r)

) is increasing, so the right Riemann sum overestimates.

Part DHard3 points

The density of bacteria in the petri dish, for

1\leq r\leq4

, is modeled by the function

g

defined by

g(r)=2-16\big(\cos(1.57\sqrt r)\big)^3

. For what value of

k

1<k<4

, is

g(k)

equal to the average value of

g(r)

on the interval

1\leq r\leq4

Answer

k\approx2.497

Full Solution & Work

Set up the average value

\text{Average value} = \frac{1}{4-1}\int_1^4 g(r)\,dr

Evaluate the average value

\approx 9.876

Solve g(k) = average value for k

g(k) = 9.876 \implies k \approx 2.497

AP Scoring — 3 Points

**P1**: Correct average-value setup

\frac13\int_1^4 g(r)\,dr

.
**P2**: Correctly evaluates the average value,

\approx9.876

.
**P3**: Correctly solves

g(k)=9.876

for

k\approx2.497

within

(1,4)

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