Good Question 16: 2018 BC 2(b)

In this post we look at another part of the AP Calculus BC exam. Good Question 15 discussed the unusual units in 2018 BC 2(a). In this post we look at 2018 BC 2(b) where units help us find the correct integral to answer the question.

How do you answer a question of a type you’ve never seen before? I expect that’s what many of the students taking the 2018 AP Calculus exam were asking when they got to BC 5. If you’ve never done a density question how do you handle this one? 

The question concerns density. Density gives you how much of something exists in a certain length, area, or volume.  Density questions have appeared on the exam now and then, most recently 2008 AB 92 (which really isn’t recent, but then there are a lot of questions we never see). I have a blog post about the density here with several examples. In that post the alternate solution to example 3 explained how I used a unit analysis to find the answer; I used a similar approach here.

2018 BC 2 (b)

The stem of the question tells us that at a depth of meters, 0 < h < 30, the number of plankton in a cubic meter of sea water is modeled by p\left( h \right)=0.2{{h}^{2}}{{e}^{{-0.0025{{h}^{2}}}}} million cells per cubic meter. Part (b) asks for the number of million of plankton in a column of water whose horizontal cross sections have a constant area of 3 square meters.

If the density were constant, then it is just a matter of multiplying the volume of the column times the constant density. Alas, the density is not constant; it varies with the depth. What to do?

Since an amount is asked for, you usually look around for a rate to integrate. Density is a kind of rate: the units are millions of cells per cubic meter. You need to integrate something concerning the density so that you end up with millions of cells; something that will “cancel” the cubic meters.

Consider a horizontal slice thru the column at depth h meters. While I’m not sure plankton is a good topping for pizza, you could picture this as a rather large pizza box whose sides are \sqrt{3} meters long and whose height is  \Delta h meters. This box has a volume of 3 \Delta h cubic meters. For small values of \Delta h the number of million plankton in the box is nearly constant, so at depth hi , there are p(hi) million plankton per cubic meter or {3p\left( {{{h}_{i}}} \right)\Delta h} million plankton in the box.

Notice how the units of the individual quantities combine to assure you the final quantity has the correct units:

\displaystyle (3\text{ square meters)}\cdot \left( {p\left( {{{h}_{i}}} \right)\text{ }\frac{{\text{million plankton}}}{{\text{cubic meters}}}} \right)\left( {\Delta h\text{ meters}} \right)=3p\left( {{{h}_{i}}} \right)\Delta h\text{ million plankton}

Now to find the amount in the column of water we can add up a stack of “pizza boxes.” The sum is \sum\limits_{{i=1}}^{n}{{3p\left( {{{h}_{i}}} \right)\Delta h}}. Now, if we take thinner boxes by letting \Delta h\to 0, we are looking at a Riemann sum. And calculus gives us the answer.

\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{3p\left( {{{h}_{i}}} \right)\Delta h}}=\int_{0}^{{30}}{{3p\left( h \right)dh}}\approx 1,675 million plankton in the column of water (rounded to the nearest million as directed in the question.)

Previous Good Questions can be found under the “Thru the Year” tab on the black navigation bar at the top of the page, or here.


Good Question 15: 2018 BC 2(a)

My choices for the Good Question series are somewhat eclectic. Some are chosen because they are good, some because they are bad, some because I learned something from them, some because they can be extended, and some because they can illustrate some point of mathematics. This question and the next, Good Question 16, are in the latter group. They both concern units. They both are taken from this year’s AP calculus BC exam; both are suitable for AB classes. In this question 2018 BC 2(a) has some unusual units and in the next 2018 BC 2(b) the units help you figure out what to do. Part (c) concerns an improper integral and pard (d) is about parametric equation, neither of these are AB topics. 

2018 BC 2(a)

2018 BC 2 gave an equation that modeled the density p(h) of plankton in a sea in units of millions of cells per cubic meter, as a function of the depth, h, in meters.  Specifically, p\left( h \right)=0.2{{h}^{2}}{{e}^{{-0.0025{{h}^{2}}}}} for 0\le h\le 30. Part (a) asked for the value of {p}'\left( {25} \right) and also asked students “Using correct units, [to] interpret the meaning of {p}'\left( {25} \right) in the context of the problem.”


This was a calculator active question, so the computation is easy enough: {p}'\left( {25} \right)=-1.17906

Now units of the derivative are always very easy to determine; this should be automatic. The derivative is the limit of a difference quotient, so its units are the units of the numerator divided by the units of the denominator. In this case that’s millions of cells per cubic meter per meter of depth.

While “millions of cells per meter to the fourth power” is technically correct and will probably earn credit, what is a meter to the fourth power?

It is similar to the better-known situation with velocity and acceleration. I never liked the idea of saying the acceleration is so many meters per square second. What’s a square second? Are there round seconds? Acceleration is the change in velocity in meters per second per second; that is, at a particular time the velocity is changing at the rate of so-many meters per second each second

Returning to the question, a cubic meter (volume) and a meter of depth (linear) are not things that you should combine. The notational convenience of writing meters to the fourth power hides the true meaning. So, a better interpretation is “At depth of  25 meters, the number cells is decreasing at the rate of 1.179 million cells per cubic meter per meter of depth.” or “The number of cells changing at the rate of -1.179 million cells per cubic meter per meter of depth.”

Had the model been given using volume units such as millions of cells per liter, then the units of the derivative would be millions of cells per liter per meter. That makes more sense.

But what does it mean?

Let’s look at the graph of the derivative. The window is 0 < h < 30 and –2.5 < p(x) < 2.5

It means, that as we pass thru that thin (thickness \Delta h\to 0) film of water 25 meters down, there are approximately 1.179 million cells per cubic meter less than in the thin film right above it and more than in the thin film right below it.

For reference, p\left( {25} \right)\approx 26.2014 million cells per cubic meter. Of course, that thin (thickness \Delta h\to 0) film of water has very little volume; it is kind of difficult to think of a cubic meter exactly 25 meters below the surface (maybe a cube extending from 24.5 meters to 25.5 meters?). As \Delta h\to 0 does a cubic meter approach a square meter?

The cubic meter above h = 25 has \displaystyle \int_{{24}}^{{25}}{{p\left( h \right)(1)dh}}=26.763 cells and the cubic meter below has 25.586 million cells. This is a decrease of 1.1767 million cells. So, the derivative is reasonable.

(To make the units of \displaystyle \int_{{24}}^{{25}}{{p\left( h \right)(1)dh}} correct, I included a factor of 1 square meter, this multiplied by p(h) million cells per cubic meter and by dh in meters give a result of millions of cells. More on why this is necessary in Good Question 16 on density.)

Previous Good Questions can be found under the “Thru the Year” tab on the black navigation bar at the top of the page, or here.

Density Functions

Density, as an application of integration, has snuck onto the exams. It is specifically not mentioned in the “Curriculum Framework” chapter of the 2016 Course and Exam Description, there is one example in the 2020 CED There is an example (#12 p. 58) in the AB sample exam question section of 2020 Course and Exam Description. The first time this topic appeared was in the 2008 AB Calculus exam. There was a hint in the few years before that with a question in the old Course Description book. Both questions will be discussed below. The idea is that students are supposed to understand integration well enough to apply their knowledge to a new situation (density). 

The Mathematics

A density function gives the amount of something per unit of length, area, or volume, for example

  • The density of a metal rod may be given in units of grams per centimeter.
  • The density of the population of a city may be given in units of people per square mile. (See map at end.)
  • The density of a container of substance may be given in pounds per cubic foot.

The density can be used to find the amount. In each example, notice that the length, area, or volume of the region is multiplied by the density to find the amount.

Example 1: A 10 cm rod with a constant density of 3 g/cm has a mass of 10\text{ cm}\cdot \frac{3\text{ grams}}{\text{cm}}=30\text{ grams}

In other situations, the density is not constant and is given by some function. Suppose our metal rod of length b has a density of \rho \left( x \right) grams/cm where x is measured from one end of the rod. To find the total mass we think of cutting the rod in the very small pieces. (Think partition: each piece has a length of \Delta x in which the density is nearly constant say \rho \left( x \right).) The sum of the mass of these pieces is the Riemann sum \sum\limits_{i=1}^{n}{\rho \left( {{x}_{i}} \right)\Delta x}. The limit of this expression as \Delta x\to 0 gives the total mass in grams: \displaystyle M=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\rho \left( {{x}_{i}} \right)\Delta x}=\int_{0}^{b}{\rho \left( x \right)dx} Notice that \sum\limits_{i=1}^{n}{\Delta x} is the length of the rod. This is multiplied by the density to find the mass.

Example 2: The next example is from the old Course Description book.

A city is built around a circular lake that has a radius of 1 mile. The population density of the city is f\left( r \right) people per square mile, where r is the distance from the center of the lake, in miles. Which of the following expressions gives the number of people who live within 1 mile of the lake?

(A) 2\pi \int_{0}^{1}{rf\left( r \right)dr}                 (B) 2\pi \int_{0}^{1}{r\left( 1+f\left( r \right) \right)dr}

(C) 2\pi \int_{0}^{2}{r\left( 1+f\left( r \right) \right)dr}      (D) 2\pi \int_{1}^{2}{rf\left( r \right)dr}

(E) 2\pi \int_{1}^{2}{r\left( 1+f\left( r \right) \right)dr}

We need to partition the region so that each piece has a close to a constant density. Thin rings around the lake will accomplish this. A ring, if straightened out is similar to a rectangle of length 2\pi {{r}_{i}} where {{r}_{i}} is the distance from the center of the lake (this is the circumference of the ring), the width of this ring (rectangle) is \Delta r. In this ring (rectangle) the population density is people per square mile, so the population in the ring is f\left( {{r}_{i}} \right) approximated by multiplying the area by the density: 2\pi {{r}_{i}}f\left( {{r}_{i}} \right)\Delta r. Adding these gives a Riemann sum whose limit gives the total population:

\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{2\pi {{r}_{i}}f\left( {{r}_{i}} \right)\Delta r}=2\pi \int_{1}^{2}{rf\left( r \right)dr}        Answer (D).

The limits of integration are from the edge of the lake, r = 1 to r = 2 (“one mile from the lake”). Another way to look at this is that \sum\limits_{i=1}^{n}{2\pi {{r}_{i}}\Delta r} is the area of the city; this is multiplied by the population density to find the population.

This type of density situation is called a radial density function.

Notice that the answer looks like the formula for volume by cylindrical shells; this is not quite an accident. The rings around the center are like the shells used when finding volume. It is the units that are different.

Example 3: From the 2008 AB Calculus exam #92.


A city located beside a river has a rectangular boundary as shown in the figure above. The population density of the city at any point along a strip x miles from the river’s edge is f\left( x \right) people per square mile. Which of the following expressions gives the population of the city?

(A) \int_{0}^{4}{f\left( x \right)dx}          (B) 7\int_{0}^{4}{f\left( x \right)dx}          (C) 28\int_{0}^{4}{f\left( x \right)dx}

(D) \int_{0}^{7}{f\left( x \right)dx}          (E) 4\int_{0}^{7}{f\left( x \right)dx}

A thin vertical strip of the city {{x}_{i}} miles to the right of the river has an area of 7\Delta x. The population in each such strip is found by multiplying the area by the density function; this gives 7f\left( {{x}_{i}} \right)\Delta x. These are then added forming a Riemann sum, etc.

\text{Population}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{7f\left( {{x}_{i}} \right)\Delta x}=7\int_{0}^{4}{f\left( x \right)dx}         Answer (B)

Alternative solution: When I first saw this question, not having thought about density for quite a while, I answered it by doing a unit analysis. Since unit analysis is a good thing for students to understand I’ll outline my thinking next.

We are looking for the population so the answer must be in units of “people.” The density function is in units of “people/square mile” (given). Both x and dx are in units of “miles” and the “7” also has units of “miles.” Therefore, the only choice that gives “people” is the one that multiplies the 7, the dx and the density function. This eliminates (A) and (D). The 28 in (C) must be square miles, making the overall units “people-miles” which is not what we’re going for. Finally, choice (E) is eliminated since the 7 and the dx are not in the same direction. This leaves (B).

Example 4: A volume problem adapted from Calculus by Hughes-Hallett, Gleason, et al.

The density of air h meters above the earth’s surface is \rho \left( h \right)=1.25{{e}^{-0.00012h}}\text{ kg/}{{\text{m}}^{3}}. Find the mass of a column of air 25 km high with a square base 3 meters on a side sitting on the surface of the earth.

At any height,  h meters above the earth the volume of a thin slice of the column of air is {{3}^{2}}\Delta h. The mass of this slice is {{3}^{2}}\rho \left( {{h}_{i}} \right)\Delta h. The sum of these slices gives a Riemann sum whose limit gives the total volume:

\displaystyle M=\underset{\Delta h\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{9\rho \left( {{h}_{i}} \right)\Delta h}=9\int_{0}^{25,000}{1.25{{e}^{-0.00012h}}dh}\approx 89,082\text{ kg}\text{.}

For other examples see 2018 BC 2, and 2021 AB 1, Good Question 15 and Good Question 16

Check the index of your textbook for density problems. Calculus by Hughes-Hallett Gleason, et al and Calculus by Rogawski (2nd edition) have good exercises and examples. My advice is not to make too big a deal of this, but if you have time, you can take a look. Should this kind of question appear on the free-response section I would guess that the question will be carefully worded so that students who never saw this kind of question would have a good chance of answering it.

The changing population density of Sydney, Australia in persons per hectare. Note the date changes in the key at the lower left.


Revised and updated 8-20-2019

Coming soon:

  • Jan 17th, Every Day series
  • Jan 24th, Logistic Growth – Real and Simulated
  • Jan 31st, The Logistic Equation
  • Feb 7th, Graphing Taylor Polynomials
  • Feb 14th,  Geometric Series – Far Out