Category Archives: Integration Theory

Good Question 11 – Riemann Reversed

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Good Question 11 – or not. double-riemann

 

The question below appears in the 2016 Course and Exam Description (CED) for AP Calculus (CED, p. 54), and has caused some questions since it is not something included in most textbooks and has not appeared on recent exams. The question gives a Riemann sum and asks for the definite integral that is its limit. Another example appears in the 2016 “Practice Exam” available at your audit website; see question AB 30. This type of question asks the student to relate a definite integral to the limit of its Riemann sum. These are called reversal questions since you must work in reverse of the usual order. Since this type of question appears in both the CED examples and the practice exam, the chances of it appearing on future exams look good.

To the best of my recollection the last time a question of this type appeared on the AP Calculus exams was in 1997, when only about 7% of the students taking the exam got it correct. Considering that by random guessing about 20% should have gotten it correct, this was a difficult question. This question, the “radical 50” question, is at the end of this post.

Example 1

Which of the following integral expressions is equal to \displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( \sqrt{1+\frac{3k}{n}}\cdot \frac{1}{n} \right)} ?

There were 4 answer choices that we will consider in a minute.

The first key to answering the question is to recognize the limit as a Riemann sum. In general, a right-side Riemann sum for the function f on the interval [a, b] with n equal subdivisions, has the form:

\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( f\left( a+\frac{b-a}{n}\cdot k \right)\cdot \frac{b-a}{n} \right)}=\int_{a}^{b}{f\left( x \right)dx}

To evaluate the limit and express it as an integral, we must identify, a, b, and f. I usually begin by looking for \displaystyle \frac{b-a}{n}. Here \displaystyle \frac{b-a}{n}=\frac{1}{n} and from this conclude that b – a = 1, so b = a + 1.

Usually, you can start by considering a = 0 , which means that the \displaystyle \frac{b-a}{n}\cdot k becomes the “x.”. Then rewriting the radicand as \displaystyle 1+3\frac{1}{n}k=1+3\left( a+\frac{1}{n}\cdot k \right), it appears the function is \sqrt{1+3x} and the limit is \displaystyle \int_{0}^{1}{\sqrt{1+3x}}dx=\frac{14}{9}.

The answer choices are

(A)  \displaystyle \int_{0}^{1}{\sqrt{1+3x}}dx        (B)    \displaystyle \int_{0}^{3}{\sqrt{1+x}}dx      (C)    \displaystyle \int_{1}^{4}{\sqrt{x}}dx     (D)   \displaystyle \tfrac{1}{3}\int_{0}^{3}{\sqrt{x}}dx

The correct choice is (A), but notice that choices B, C, and D can be eliminated as soon as we determine that b = a + 1. That is not always the case.

Let’s consider another example:

Example 2: \displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( {{\left( 2+\frac{3}{n}k \right)}^{2}}\left( \frac{3}{n} \right) \right)}=

As before consider \displaystyle \frac{b-a}{n}=\frac{3}{n}, which implies that b = a + 3. With a = 0,  the function appears to be {{\left( 2+x \right)}^{2}} on the interval [0, 3], so the limit is \displaystyle \int_{0}^{3}{{{\left( 2+x \right)}^{2}}}dx=39

BUT

What if we take a = 2? If so, the limit is \displaystyle \int_{2}^{5}{{{x}^{2}}dx}=39.

And now one of the “problems” with this kind of question appears: the answer written as a definite integral is not unique!

Not only are there two answers, but there are many more possible answers. These two answers are horizontal translations of each other, and many other translations are possible, such as \displaystyle \int_{-25.65}^{-22.65}{{{\left( 27.65+x \right)}^{2}}dx}=39.

The same thing can occur in other ways. Returning to example 1,and using something like a u-substitution, we can rewrite the original limit as \displaystyle \frac{1}{3}\cdot \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( \sqrt{1+\frac{3k}{n}}\cdot \frac{3}{n} \right)}.

Now b = a + 3 and the limit could be either \displaystyle \frac{1}{3}\int_{0}^{3}{\sqrt{1+x}}dx=\frac{14}{9} or \displaystyle \frac{1}{3}\int_{1}^{4}{\sqrt{x}}dx=\frac{14}{9}, among others.

My opinions about this kind of question.

The real problem with the answer choices to Example 1 is that they force the student to do the question in a way that gets one of the answers. It is perfectly reasonable for the student to approach the problem a different way, and get a different correct answer that is not among the choices. This is not good.

The problem could be fixed by giving the answer choices as numbers. These are the numerical values of the 4 choices:(A) 14/9   (B) 14/3   (C)  14/3   (D)    2\sqrt{3}/3. As you can see that presents another problem. Distractors (wrong answers) are made by making predictable calculus mistakes. Apparently, two predictable mistakes give the same numerical answer; therefore, one of them must go.

A related problem is this: The limit of a Riemann sum is a number; a definite integral is a number. Therefore, any definite integral, even one totally unrelated to the Riemann sum, which has the correct numerical value, is a correct answer.

I’m not sure if this type of question has any practical or real-world use. Certainly, setting up a Riemann sum is important and necessary to solve a variety of problems. After all, behind every definite integral there is a Riemann sum. But starting with a Riemann sum and finding the function and interval does not seem to me to be of practical use.

The CED references this question to MPAC 1: Reasoning with definitions and theorems, and to MPAC 5: Building notational fluency. They are appropriate,and the questions do make students unpack the notation.

My opinions notwithstanding, it appears that future exams will include questions like these.

These questions are easy enough to make up. You will probably have your students write Riemann sums with a small value of n when you are teaching Riemann sums leading up to the Fundamental Theorem of Calculus.  You can make up problems like these by stopping after you get to the limit, giving your students just the limit, and having them work backwards to identify the function(s) and interval(s). You could also give them an integral and ask for the associated Riemann sum. Question writers call questions like these reversal questions since the work is done in reverse of the usual way.

Here is the question from 1997, for you to try. The answer is below.

riemann-reversal

 

 

 

 

Answer B. Hint n = 50

 

 

 

 

 

Revised 5-5-2022

 

Parts and More Parts

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At an APSI this summer the participants and I got to discussing the “tabular method” for integration by parts. Since we were getting far from what is tested on the BC Calculus exams, I ended the discussion and said for those that were interested I would post more on the tabular method this blog going farther than just the basic set up. So here goes.

Here are some previous posts on integration by parts and the tabular method

 Integration by Parts 1 discusses the basics of the method. This is as far as a BC course needs to go.

Integration by Parts 2 introduces the tabular method

Modified Tabular Integration presents a very quick and slick way of doing the tabular method without making a table. This is worth knowing.

There is also a video on integration by parts here. Scroll down to “Antiderivatives 5: A BC topic – Integration by parts.” The tabular method is discussed starting about time 15:16. There are several ways of setting up the table; one is shown here and a slightly different way is in the Integration by Parts 2 post above. There are others.

Going further with the tabular method.

The tabular method works well if one of the factors in the original integrand is a polynomial; eventually its derivative will be zero and you are done. These are shown in the examples in the posts above and Example 1 below. To complete the topic, this post will show two other things that can happen when using integration by parts and the tabular method.

First we look at an example with a polynomial factor and learn how to stop midway through. Why stop? Because often there will be no end if you don’t stop. There are ways to complete the integration as shown in the examples.

Example 1:  Find \displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx} by the tabular method (See Integration by Parts 2 for more detail on how to set the table up)

Capture

Adding the last column gives the antiderivative:

\displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx}=4{{x}^{3}}\sin \left( x \right)+12{{x}^{2}}\cos \left( x \right)-24x\sin \left( x \right)-24\sin \left( x \right)+C

Now say you wanted to stop after 12{{x}^{2}}\cos \left( x \right). Example 2 shows why you want (need) to stop. In Example 1 you will have

\displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx}=4{{x}^{3}}\sin \left( x \right)+12{{x}^{2}}\sin \left( x \right)+\int_{{}}^{{}}{-24x\cos \left( x \right)}dx

The integrand on the right is the product of the last column in the row at which you stopped and the first two columns in the next row, as shown in yellow above.

Example 2 Find \displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx}

Tabular 4

As you can see things are just repeating the lines above sometimes with minus signs. However, if we stop on the third line we can write:

\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx={{e}^{x}}\sin \left( x \right)}+{{e}^{x}}\cos \left( x \right)-\int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx}

The integral at the end is identical to the original integral.  We can continue by adding the integral to both sides:

\displaystyle 2\int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx={{e}^{x}}\sin \left( x \right)}+{{e}^{x}}\cos \left( x \right)

Finally, we divide by 2 and have the antiderivative we were trying to find:

\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx=\tfrac{1}{2}{{e}^{x}}\sin \left( x \right)}+\tfrac{1}{2}{{e}^{x}}\cos \left( x \right)+C

In working this type of problem you must be aware of that the original integrand showing up again can happen and what to do if it does. As long as the coefficient is not +1, we can proceed as above. The same thing happens if we do not use the tabular method. (If the coefficient is +1 then the other terms on the right will add to zero and you need to make different choices for u and dv.)

Reduction Formulas.

Another use of integration by parts is to produce formulas for integrals involving powers. An integral whose integrand is of less degree than the original, but of the same form results. The formula is then iterated to continually reduce the degree until the final integral can be integrated easily.

Example 3: Find \displaystyle \int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}

Let u={{x}^{n}},\ du=n{{x}^{n-1}}dx,\ dv={{e}^{x}}dx,\ v={{e}^{x}}

\displaystyle \int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}={{x}^{n}}{{e}^{x}}-n\int_{{}}^{{}}{{{x}^{n-1}}{{e}^{x}}dx}

This is a reduction formula; the second integral is the same as the first, but of lower degree. Here is how it is used.  At each step the integrand is the same as the original, but one degree lower. So the formula can be applied again, three more times in this example.

Tabular 7

Most textbooks have a short selection of reduction formulas.

Final Thoughts.

Back in the “old days”, BC (before calculators), beginning calculus courses spent a lot of time on the topic of “Techniques of Integration.” This included integration by parts, algebraic techniques, techniques known as trig-substitutions, and others. Mathematicians and engineers had tables of integrals listing over a thousand forms and students were taught how to use the tables and distinguish between similar forms in the tables. (See the photo below from the fourteenth edition of the CRC tables (c) 1965.) Current textbooks often contain such sections still.

Today, none of this is necessary. CAS calculators can find the antiderivatives of almost any integral. Websites such as WolframAlpha are also available to do this work.

I’m not sure why the College Board recently expanded slightly the list of types of antiderivatives tested on the exams. Certainly a few of the basic types should be included in a course, but what students really need to know is how to write the integral appropriate to a problem, and what definite and indefinite integrals mean. This, in my opinion, is far more important than being able to crank out antiderivatives of increasingly complicated expressions: let technology do that – or buy yourself an integral table. Just saying … .

tabular130

Trapezoids – Ancient and Modern

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The other day, in the course of about 10 minutes, I came across two interesting things about Trapezoidal approximations that I thought I would share with you.

cuneifirm tablet

Cuneiform writing

The first was a link to a story about how the ancient Babylonian astronomers sometime between 350 and 50 BCE used trapezoids to, in effect, find the area under a velocity-time graph tracking Jupiter’s motion. This was an NPR story based on a January 2016 Science magazine article in which the author, Mathieu Ossendrijver discusses his work deciphering cuneiform tablets written over 1,400 years before the technique showed up in Europe.

The second was a question asked on the AP Calculus bulletin board. A teacher asked, “Can someone please help me answer this question a student posed the other day. We were comparing left, right and midpoint and trapezoidal approximations. He asked since the trapezoidal calculation is the best estimate what is the use of LRAM and RRAM?” Here is an expanded form of my answer.

There are several things to consider here.

  1. First, if all you need is an estimate of the area or integral of a continuous function then a Trapezoid sum is certainly better than the left Riemann sum (left-RΣ) or the right-RΣ. Better, yes, the “best” maybe not: midpoint sums are about as good and parabola sums (Simpson’s Rule) are better.

2. Another reason to do left RΣ and right RΣ with small values of n is simply to give students practice in setting up Riemann sums so that they will be familiar with them when they move on to finding their limits and getting ready to define definite integrals.

3. A RΣ for a function f on a closed interval [a, b] is formed by partitioning the interval into subintervals and taking exactly one function value from each closed subinterval, multiplying the value by the width of that subinterval and adding these results. You may pick the function value any way you want – left end, middle, right end, any place at random in the subintervals and someplace else in the next subinterval. One way is to pick the smallest function value in each subinterval; this gives a RΣ called the lower RΣ. Likewise, you could pick the largest value in each subinterval; this gives the upper RΣ. Now it is true that

lower RΣ ≤ (any/all other RΣs) ≤ upper RΣ

Then as you add more partition points (n approaches infinity, or Δx approaches 0, etc.) the lower sum increases and the upper sum decreases. The series of lower sums is increasing and bounded above (by the upper sum) and therefore converges to its least upper bound. The upper sum decreases forming a decreasing series that is bounded below and therefore converges to its greatest lower bound.

If the lower RΣ and the upper RΣ approach the same value, then ALL the other RΣs approach that same value by the Squeeze theorem. This value is then defined as the definite integral of f from a to b.

In most AP calculus course, the textbooks do not deal with upper and lower sums. Instead, they deal with left RΣ and right RΣ on intervals on which f is only increasing (or only decreasing). In this case the lower RΣ = left RΣ and the upper RΣ = the right RΣ (or the other way around for decreasing functions).

So, this is why you need the left RΣ and right RΣ; not so much to approximate, but to complete the theory leading to the definite integral.

Good Question 8 – or not?

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Seattle rainToday’s question is not a good question. It’s a bad question.

But sometimes a bad question can become a good one.

This one leads first to a discussion of units, then to all sorts of calculus.

Here’s the question a teacher sent me this week taken from his textbook:

The normal monthly rainfall at the Seattle-Tacoma airport can be approximated by the model R=3.121+2.399\sin \left( 0.524t+1.377 \right), where R is measured in inches and t is the time in months, t = 1 being January. Use integration to approximate the normal annual rainfall.  Hint: Integrate over the interval [0,12].

Of course, with the hint it’s not difficult to know what to do and that makes it less than a good question right there. The answer is \displaystyle \int_{0}^{12}{R(t)dt=37.4736} inches. You could quit here and go on to the next question, but …

Then a student asked. “If R is in inches shouldn’t be in units of the integral be inch-months, since the unit of an integral is the unit of the integrand times the units of the independent variable?”  Well, yes, they should. So, what’s up with that?

Also, the teacher figured that the integral of a rate is an amount and our answer is an amount, so why isn’t the integrand a rate?

The only answer I could come up with is that the statement “R is measured in inches” is incorrect; R should be measured in inches /month. The opening phrase “normal monthly rainfall” also seems to point to the correct units for R being inches/month.

Problem solved; or maybe does this lead to a different concern?

The teacher pointed out that R(6) = 0.7658 inches is a reasonable answer for the amount of rain in June whereas \displaystyle \int_{0}^{6}{R(t)dt=}20.4786 is not.

If R is a rate, then the amount of rain that falls in June (t = 6) is given by \displaystyle \int_{5}^{6}{R(t)dt}=0.9890.

From here on we will assume that R is a rate with units of inches/month. Here are the individual monthly rates calculated with a CAS. Ques 8 a

The total amount of rainfall (second line above) appears be R(1) + R(2) + R(3) + … +R(12) = 37.4742. This is very close to the amount calculated by integration.

The slight difference of 0.0006 is not a round off error.

Remember, behind every definite integral there is a Riemann sum!

Again, the units are the problem. Why does the sum of the monthly rates seem to give the total amount?  The reason is that the terms of the sequence above are actually the values of a right-side Riemann sum of the rate, R(t), over the interval [0,12] with 12 equal subdivisions of width 1 (month) each with the 1’s left out as 1’s often are. Therefore, their sum should come close to the total yearly rainfall, but it is really just an approximation of it.

The actual total for any month, n, is given by \displaystyle \int_{n-1}^{n}{r(t)}dt. For example the amount of rain that falls in June is given by \displaystyle \int_{5}^{6}{R(t)dt}=0.9890 inches.

Here is the sequence of the actual monthly rainfall values in inches, and their sum.

Ques 8 b

This agrees with the integral. Why? Because one of  the properties of integrals tell us that \displaystyle \sum\limits_{n=1}^{12}{\int_{n-1}^{n}{r(t)dt}}=\int_{0}^{12}{r(t)dt}.

Another instructive thing with this integral is this: The function R=3.121+2.399\sin \left( 0.524t+1.377 \right) is periodic with a period of  \frac{2\pi }{0.524}\approx 11.9908\approx 12. So the sine function takes on (almost) all its values in a year, as you would expect. Since the sine values all but cancel each other out

\displaystyle \int_{0}^{12}{3.121+2.399\sin \left( 0.524t+1.377 \right)dt}\approx \int_{0}^{12}{3.121dt=3.121\left( 12-0 \right)=37.452}. Close!

The total rainfall divided by 12 is \frac{37.452}{12}=3.121 this must be close to the average rainfall each month. The average rainfall is \displaystyle \frac{1}{12}\int_{0}^{12}{R\left( t \right)dt}=3.1228 inches. Close, again!

So, there you have it. Is this a good question or not? We considered all these concepts while working not just with an equation but with numbers from a poorly stated problem:

  • Reading and interpreting words.
  • Unit analysis
  • Integration by technology
  • Realizing that a pretty good approximation is not correct, due again to units.
  • A Riemann sum approximation in a real situation that comes very close to the value by integration
  • Using a property of a periodic function to greatly simplify an integral
  • Finding average value two ways

So, it turned out to be a sunny day in Seattle.seattle sun

.

November 2015

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Here is the list of “November Topics,” that is what AP classes usually consider from mid-November into December. There has been a lot of discussion about inverses this month at the AP Calculus Community. While not the most read on this blog, the series on inverses may be helpful in considering all the ins and outs of inverses.

The four featured posts on the first page are the most popular from this month. Speed with 3563 hits this year and 6157 hits since it first appear is one of the most popular overall. “Open or Closed?” is another poplar post.

Thinking ahead into December, the first posts on integration are here and will continue into December. (As I’ve mentioned I try to post a few weeks ahead of where most people are now, so you have some time to read and plan.)

October 13, 2014 Extremes without Calculus

November 2, 2012 Open or Closed?

November 5, 2012 Inverses

November 7, 2012 Writing Inverses

November 9, 2012 The Range of the Inverse

November 12, 2012 The Calculus of Inverses

November 14, 2012 Inverses Graphically and Numerically

November 16, 2012 Motion Problems: Same Thing, Different Context.

November 19, 2012 Speed

April 17, 2013 The Ubiquitous Particle Motion Question  

September 16, 2014 Matching Motion

November 21, 2012 Derivatives of Exponential Functions

November 26, 2012 Integration Itinerary

November 18, 2012 Antidifferentiation

November 30, 2012 The Old Pump

Good Question 7 – 2009 AB 3

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Another in my occasional series on Good Questions to teach from. This is the Mighty Cable Company question from the 2009 AB Calculus exam, number 3

This question presented students with a different situation than had been seen before. It is a pretty standard “in-out” question, except that what was going in and out was money. Students were told that the Mighty Cable Company sold its cable for $120 per meter. They were also told that the cost of the cable varied with its distance from the starting end of the cable. Specifically, the cost of producing a portion of the cable x meters from the end is 6\sqrt{x} dollars per meter. Profit was defined as the difference between the money the company received for selling the cable minus the cost of producing the cable.  Steel Wire Rope 3

Students had a great deal of trouble answering this question. (The mean was 1.92 out of a possible 9 points. Fully, 36.9% of students earned no point; only 0.02% earned all 9 points.) This was probably because they had difficulty in interpreting the question and translating it into the proper mathematical terms and symbols. Since economic problems are not often seen on AP Calculus exams, students needed to be able to use the clues in the stem:

  • The $120 per meter is a rate. This should be deduced from the units: dollars per meter.
  • The cost of producing the portion of cable x meters from one end cable is also a rate for the same reason. In economics this is called the marginal cost; the students did not need to know this term.
  • The profit is an amount that is a function of x, the length of the cable.

Part (a): Students were required to find the profit from the sale of a 25-meter cable. This is an amount. As always, when asked for an amount, integrate a rate. In this case integrate the difference between the rate at which the cable sells and the cost of producing it.

\displaystyle P(25)=\int_{0}^{25}{\left( 120-6\sqrt{x} \right)dx}=\$2500

or

\displaystyle P(25)=120(25)-\int_{0}^{25}{6\sqrt{x}\ dx}=\$2500

Part (b): Students were asked to explain the meaning of \displaystyle \int_{25}^{30}{6\sqrt{x}dx} in the context of the problem. Since the answer is probably not immediately obvious, here is the reasoning involved.

This is the integral of a rate and therefore, gives the amount (of money) needed to manufacture the cable. This can be found by a unit analysis of the integrand: \displaystyle \frac{\text{dollars}}{\text{meter}}\cdot \text{meters}=\text{dollars} .

Let C be the cost of production, so \displaystyle \frac{dC}{dx}=6\sqrt{x}, and therefore, \displaystyle \int_{25}^{30}{6\sqrt{x}dx}=C\left( 30 \right)-C\left( 25 \right) by the Fundamental Theorem of Calculus (FTC).

Therefore, \displaystyle \int_{25}^{30}{6\sqrt{x}dx} is the difference in dollars between the cost of producing a cable 30 meters long, C(30), and the cost of producing a cable 25 meters long, C(25). (Another acceptable is that the integral is the cost in dollars of producing the last 5 meters of a 30 meter cable.)

Part (c): Students were asked to write an expression involving an integral that represents the profit on the sale of a cable k meters long.

Part (a) serves as a hint for this part of the question. Here the students should write the same expression as they wrote in (a) with the 25 replaced by k.

\displaystyle P(k)=\int_{0}^{k}{\left( 120-6\sqrt{x} \right)dx}

or

\displaystyle P(k)=120k-\int_{0}^{k}{6\sqrt{x}\ dx}

Part d: Students were required to find the maximum profit that can be earned by the sale of one cable and to justify their answer. Here they need to find when the rate of change of the profit (the marginal profit) changes from positive to negative.

Using the FTC to differentiate either of the answers in part (c) or by starting fresh from the given information:

\displaystyle \frac{dP}{dx}=120-6\sqrt{x}

\displaystyle \frac{dP}{dx}=0 when x = 400 and P(400)= $16,000.

Justification: The maximum profit on the sale of one cable is $16,000 for a cable 400 meters long. For 0<x<400,{P} '(x)>0 and for x>400,{P} '(x)<0 therefore, the maximum profit occurs at x = 400. (The First Derivative Test).

Once students are familiar with in-out questions, this is a good question to challenge them with. The actual calculus is not that difficult or unusual but concentrating on the translation of the unfamiliar context into symbols and calculus ideas is different. Show them how to read the hints in the problem such as the units.

Steel cable or steel wire rope as it is called also has some interesting geometry in its construction. You can find many good illustrations of this, such as the ones below, by Googling “steel wire rope.”

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Good Question 6: 2000 AB 4

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2000 AB 4 Water tankAnother of my favorite questions from past AP exams is from 2000 question AB 4. If memory serves it is the first of what became known as an “In-out” question. An “In-out” question has two rates that are working in opposite ways, one filling a tank and the other draining it.

In subsequent years we saw a question with people entering and leaving an amusement park (2002 AB2/BC2), sand moving on and off a beach (2005 AB 2), another tank (2007 AB2), an oil leak being cleaned up (2008 AB 3), snow falling and being plowed (2010 AB 1), gravel being processed (2013 AB1/BC1), and most recently water again flowing in and out of a pipe (2015 AB1/BC1). The in-between years saw rates in one direction only but featured many of the same concepts.

The questions give rates and ask about how the quantity is changing. As such, they may be approached as differential equation initial value problems, but there is an easier way. This easier way is that a differential equation that gives the derivative as a function of a single variable, t, with an initial point \left( {{t}_{0}},y\left( {{t}_{0}} \right) \right) always has a solution of the form

y\left( t \right)=y\left( {{t}_{0}} \right)+\int_{{{t}_{0}}}^{t}{{y}'\left( x \right)dx}.

This is sometimes called the “accumulation equation.” The integral of a rate of change {y}'\left( t \right) gives the net amount of change over the interval of integration [{{t}_{0}},t]. When this is added to the initial amount the result is an expression that gives the amount at any time t.

In a motion context, this same idea is that the position at any time t, is the initial position plus the displacement:

\displaystyle s\left( t \right)=s\left( {{t}_{0}} \right)+\int_{{{t}_{0}}}^{t}{v\left( x \right)dx} where v\left( t \right)={s}'\left( t \right)

The scoring standard gave both forms of the solution. The ease of the accumulation form over the differential equation solution was evident and subsequent standards only showed this one.

2000 AB 4

The question concerned a tank that initially contained 30 gallons of water. We are told that water is being pumped into the tank at a constant rate of 8 gallons per minute and the water is leaking out at the rate of \sqrt{t+1} gallons per minute.

Part a asked students to compute the amount of water that leaked out in the first three minutes. There were two solutions given. The second solves the problem as an initial value differential equation:

Let L(t) be the amount that leaks out in t minutes then

\displaystyle \frac{dL}{dt}=\sqrt{t+1}

L\left( t \right)=\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+C

L\left( 0 \right)=\frac{2}{3}{{\left( 0+1 \right)}^{3/2}}+C=0 since nothing has leaked out yet, so C = -2/3

L\left( t \right)=\frac{2}{3}{{\left( t+1 \right)}^{3/2}}-\frac{2}{3}

L\left( 3 \right)=\frac{14}{3}

The first method, using the accumulation idea takes a single line:

\displaystyle L\left( 3 \right)=\int_{0}^{3}{\sqrt{t+1}dt}=\left. \frac{2}{3}{{\left( t+1 \right)}^{3/2}} \right|_{0}^{3}=\frac{2}{3}{{\left( 4 \right)}^{3/2}}-\frac{2}{3}{{\left( 1 \right)}^{3/2}}=\frac{14}{3}

I think you’ll agree this is easier and more direct.

Part b asked how much water was in the tank at t = 3 minutes.  We have 30 gallons to start plus 8(3) gallons pumped in and 14/3 gallons leaked out gives 30 + 24 – 14/3 = 148/3 gallons.

This part, worth only 1 point, was a sort of hint for the next part of the question.

Part c asked students to write an expression for the total number of gallons in the tank at time t.

Following part b the accumulation approach gives either

\displaystyle A\left( t \right)=30+8t-\int_{0}^{t}{\sqrt{x+1}dx}  or

\displaystyle A\left( t \right)=30+\int_{0}^{t}{\left( 8-\sqrt{x+1} \right)dx}.

The first form is not a simplification of the second, but rather the second form is treating the difference of the two rates, in minus out, as the rate to be integrated.

The differential equation approach is much longer and looks like this:

\displaystyle \frac{dA}{dt}=8-\sqrt{t+1}

A\left( t \right)=8t-\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+C

A\left( 0 \right)=30=8(0)-\frac{2}{3}{{\left( 0+1 \right)}^{3/2}}+C, so C=\frac{92}{3}

A\left( t \right)=8t-\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+\frac{92}{3}

Again, this is much longer. In recent years when asking student to write an expression such as this, the directions included a phrase such as “write an equation involving one or more integrals that gives ….” This pretty much leads students away from the longer differential equation initial value problem approach.

Part d required students to find the time when in the interval 0\le t\le 120 minutes the amount of water in the tank was a maximum and to justify their answer. The usual method is to find the derivative of the amount, A(t), set it equal to zero, and then solve for the time.

{A}'\left( t \right)=8-\sqrt{t+1}

Notice that this is the same regardless of which of the three forms of the expression for A(t) you start with. Thus, an excellent example of the Fundamental Theorem of Calculus used to find the derivative of a function defined by an integral. Or you could just start here without reference to the forms above: the overall rate in the rate in minus the rate out.

{A}'\left( t \right)=0 when t = 63

This is a maximum by the First Derivative Test since for 0 < t < 63 the derivative of A is positive and for 63 < t <120 the derivative of A is negative.

There is an additional idea on this part of the question in the Teaching Suggestions below.

I like this question because it is a nice real (as real as you can hope for on an exam) situation and for the way the students are led through the problem. I also like the way it can be used to compare the two methods of solution.  Then the way they both lead to the same derivative in part d is nice as well. I use this one a lot when working with teachers in workshops and summer institutes for these very reasons.

Teaching Suggestions

  • Certainly, have your students work through the problem using both methods. They need to learn how to solve an initial value problem (IVP) and this is good practice. Additionally, it may help them see how and when to use one method or the other.
  • Be sure the students understand why the three forms of A(t) in part c give the same derivative in part d. This makes an important connection with the Fundamental theorem of Calculus.
  • Like many good AP questions part d can be answered without reference to the other parts. The question starts with more water being pumped in than leaking out. This will continue until the rate at which the water leaks out overtakes the rate at which it is being pumped in. At that instant the rate “in” equals the rate “out” so you could start with 8=\sqrt{t+1}. After finding that t = 63, the answer may be justified by stating that before this time more water is being pumped in than is leaking out and after this time the rate at which water leaks out is greater than the rate at which it is pumped in, so the maximum must occur at t = 63.
  • And as always, consider the graph of the rates.

2000 AB 4

I used this question as the basis of a lesson in the current AP Calculus Curriculum Module entitled Integration, Problem Solving and Multiple Representations © 2013 by the College Board. The lesson gives a Socratic type approach to this question with a number of questions for each part intended to help the teacher not only work through this problem but to bring out related ideas and concepts that are not in the basic question. The module is currently available at AP sponsored workshops and AP Summer Institutes. Eventually, it will be posted at AP Central on the AB and BC Calculus Home Pages.