An Exploration in Differential Equations

This is an exploration based on the AP Calculus question 2018 AB 6. I originally posed it for teachers last summer. This will make, I hope, a good review of many of the concepts and techniques students have learned during the year. The exploration, which will take an hour or more, includes these topics:

  • Finding the general solution of the differential equation by separating the variables
  • Checking the solution by substitution
  • Using a graphing utility to explore the solutions for all values of the constant of integration, C
  • Finding the solutions’ horizontal and vertical asymptotes
  • Finding several particular solutions
  • Finding the domains of the particular solutions
  • Finding the extreme value of all solutions in terms of C
  • Finding the second derivative (implicit differentiation)
  • Considering concavity
  • Investigating a special case or two

I also hope that in working through this exploration students will learn not so much about this particular function, but how to use the tools of algebra, calculus, and technology to fully investigate any function and to find all its foibles.

The exploration is here in a PDF file. Here are the solutions.

As always, I appreciate your feedback and comments. Please share them with me using the reply box below.

The College Board is pleased to offer a new live online event for new and experienced AP Calculus teachers on March 5th at 7:00 PM Eastern.

I will be the presenter.

The topic will be AP Calculus: How to Review for the Exam:  In this two-hour online workshop, we will investigate techniques and hints for helping students to prepare for the AP Calculus exams. Additionally, we’ll discuss the 10 type questions that appear on the AP Calculus exams, and what students need know and to be able to do for each. Finally, we’ll examine resources for exam review.

Registration for this event is $30/members and $35/non-members. You can register for the event by following this link:









Karl Weierstrass (1815 – 1897) was the mathematician who (finally) formalized the definition of continuity. Included in that definition was the epsilon-delta definition of limit. This definition has been pulled out, so to speak, and now is usually presented on its own. So, which came first – continuity or limit? The ideas and situations that required continuity could only be formalized with the concept of limit. So, looking at functions that are or are not continuous helps us understand what limits are and why we first need them.

In the ideal world, students would have plenty of work with continuous and not continuous functions before starting the calculus. The vocabulary and notation, if not the formal definitions, would be used as early as possible. Then when students got to calculus, they would know the ideas and be ready to formalize the ideas.

The Intermediate Value Theorem (IVT) is an important property of continuous functions.

Using the definition of continuity to show that a function is or is not continuous at a point is a common question of the AP exams, as is the IVT.

Continuity The definition of continuity.

Continuity Should continuity come before limits?

From One Side or the Other One-sided limits and one-sided differentiability

How to Tell Your Asymptote from a Hole in the Graph  From the technology series. Showing holes and asymptotes on a graphing calculator.

Fun with Continuity Defined everywhere and continuous nowhere. Continuous only at a single point.

Theorems The Intermediate Value Theorem (IVT) and suggestions on teaching theorems.

Intermediate Weather  Using the IVT

Right Answer – Wrong Question Continuity or continuity “on its domain”?






Revised from a post of August 22, 2017


Limits – They Make the Calculus Work.

In an ideal world, I would like to have all students study limits in their pre-calculus course and know all about them when they get to calculus. Certainly, this would be better than teaching how to calculate derivatives in pre-calculus (after all derivatives are calculus, not pre-calculus).

Limits are the foundation of the calculus. Continuity, an important property of functions, depends on limits. All derivatives and all definite integrals are limits. For AP Calculus students need a good intuitive understanding of limits, what they mean, and how to find them The formal (delta-epsilon) definition is not tested and need not be taught, however, do not feel that you have to avoid it. If your students can handle it, let them try.

Here are a few of my previous posts on limits.

Why Limits?

Finding Limits  How to … and the use of “infinity” vs “DNE”

Dominance  Finding limits the easy way.

Deltas and Epsilons Not tested on the AP Exams; here’s why.

Asymptotes The graphical manifestation of limits at or equal to infinity.

Next Week: Continuity

Update of my post of August 15, 2017.


In an ideal world, I would like to have all students study limits in their precalculus course and know all about them when they get to calculus. Certainly, this would be better than teaching how to calculate derivatives in precalculus (after all derivatives are calculus, not pre… ). Here are a few of my previous posts on limit.

Why Limits?

Finding Limits  How to … and the use of “infinity” vs “DNE”

Dominance  Finding limits the easy way.

Deltas and Epsilons Not tested on the AP Exams; here’s why.

Asymptotes The graphical manifestation of limits at or equal to infinity.

Next Week: Continuity


Is this going to be on the exam?

confused-teacherRecently there was a discussion on the AP Calculus Community bulletin board regarding whether it was necessary or desirable to have students do curve sketching starting with the equation and ending with a graph with all the appropriate features – increasing/decreasing, concavity, extreme values etc., etc. – included. As this is kind of question that has not been asked on the AP Calculus exam, should the teacher have his students do problems like these?

The teacher correctly observed that while all the individual features of a graph are tested, students are rarely, if ever, expected to put it all together. He observed that making up such questions is difficult because getting “nice” numbers is difficult.

Replies ran from No, curve sketching should go the way of log and trig tables, to Yes, because it helps connect f. f ‘ and f ‘’, and to skip the messy ones and concentrate on the connections and why things work the way they do. Most people seemed to settle on that last idea; as I did. As for finding questions with “nice” numbers, look in other textbooks and steal borrow their examples.

But there is another consideration with this and other topics. Folks are always asking why such-and-such a topic is not tested on the AP Calculus exam and why not.

The AP Calculus program is not the arbiter of what students need to know about first-year calculus or what you may include in your course. That said, if you’re teaching an AP course you should do your best to have your students learn everything listed in the 2019 Course and Exam Description book and be aware of how those topics are tested – the style and format of the questions. This does not limit you in what else you may think important and want your students to know. You are free to include other topics as time permits.

Other considerations go into choosing items for the exams. A big consideration is writing questions that can be scored fairly.  Here are some thoughts on this by topic.

Curve Sketching

If a question consisted of just an equation and the directions that the student should draw a graph, how do you score it? How accurate does the graph need to be? Exactly what needs to be included?

An even bigger concern is what do you do if a student makes a small mistake, maybe just miscopies the equation? The problem may have become easier (say, an asymptote goes missing in the miscopied equation and if there is a point or two for dealing with asymptotes – what becomes of those points?) Is it fair to the student to lose points for something his small mistake made it unnecessary for him to consider? Or if the mistake makes the question so difficult it cannot be solved by hand, what happens then? Either way, the student knows what to do, yet cannot show that to the reader.

To overcome problems like these, the questions include several parts usually unrelated to each other, so that a mistake in one part does not make it impossible to earn any subsequent points. All the main ideas related to derivatives and graphing are tested somewhere on the exam, if not in the free-response section, then as a multiple-choice question.

(Where the parts are related, a wrong answer from one part, usually just a number, imported into the next part is considered correct for the second part and the reader then can determine if the student knows the concept and procedure for that part.)


A big topic in derivative applications is optimization. Questions on optimization typically present a “real life” situation such as something must be built for the lowest cost or using the least material. The last question of this type was in 1982 (1982 AB 6, BC 3 same question). The question is 3.5 lines long and has no parts – just “find the cost of the least expensive tank.”

The problem here is the same as with curve sketching. The first thing the student must do is write the equation to be optimized. If the student does that incorrectly, there is no way to survive, and no way to grade the problem. While it is fair to not to award points for not writing the correct equation, it is not fair to deduct other points that the student could earn had he written the correct equation.

The main tool for optimizing is to find the extreme value of the function; that is tested on every exam. So here is a topic that you certainly may include the full question in you course, but the concepts will be tested in other ways on the exam.

The epsilon-delta definition of limit

I think the reason that this topic is not tested is slightly different. If the function for which you are trying to “prove” the limit is linear, then \displaystyle \delta =\frac{\varepsilon }{\left| m \right|} where m is the slope of the line – there is nothing to do beside memorize the formula. If the function is not linear, then the algebraic gymnastics necessary are too complicated and differ greatly depending on the function. You would be testing whether the student knew the appropriate “trick.”

Furthermore, in a multiple-choice question, the distractor that gives the smallest value of must be correct (even if a larger value is also correct).

Moreover, finding the epsilon-delta relationship is not what’s important about the definition of limit. Understanding how the existence of such a relationship say “gets closer to” or “approaches” in symbols and guarantees that the limit exists is important.

Volumes using the Shell Method

I have no idea why this topic is not included. It was before 1998. The only reason I can think of is that the method is so unlike anything else in calculus (except radial density), that it was eliminated for that reason.

This is a topic that students should know about. Consider showing it too them when you are doing volumes or after the exam. Their college teachers may like them to know it.

Integration by Parts on the AB exam

Integration by Parts is considered a second semester topic. Since AB is considered a one-semester course, Integration by Parts is tested on the BC exam, but not the AB exam. Even on the BC exam it is no longer covered in much depth: two- or more step integrals, the tabular method, and reduction formulas are not tested.

This is a topic that you can include in AB if you have time or after the exam or expand upon in a BC class.

Newton’s Method, Work, and other applications of integrals and derivatives

There are a great number of applications of integrals and derivatives. Some that were included on the exams previously are no longer listed. And that’s the answer right there: in fairness, you must tell students (and teachers) what applications to include and what will be tested. It is not fair to wing in some new application and expect nearly half a million students to be able to handle it.

Also, remember when looking through older exams, especially those from before 1998, that some of the topics are not on the current course description and will not be tested on the exams.

Solution of differential equations by methods other than separation of variables

Differential equations are a huge and important area of calculus. The beginning courses, AB and BC, try to give students a brief introduction to differential equations. The idea, I think, is like a survey course in English Literature or World History: there is no time to dig deeply, but the is an attempt to show the main parts of the subject.

While the choices are somewhat arbitrary, the College Board regularly consults with college and university mathematics departments about what to include and not include. The relatively minor changes in the new course description are evidence of this continuing collaboration. Any changes are usually announced two years in advance. (The recent addition of density problems unannounced, notwithstanding.) So, find a balance for yourself. Cover (or better yet, uncover) the ideas and concepts in the course description and if there if a topic you particularly like or think will help your students’ understanding of the calculus, by all means include it.

PS: Please scroll down and read Verge Cornelius’ great comment below.

Happy Holiday to everyone. There is no post scheduled for next week; I will resume in the new year. As always, I like to hear from you. If you have anything calculus-wise you would like me to write about, please let me know and I’ll see what I can come up with. You may email me at


Good Question 11 – Riemann Reversed

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 ba = 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


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.






Answer B. Hint n = 50






Revised 5-5-2022


From One Side or the Other.

Recently, a reader wrote and suggested my post on continuity would be improved if I discussed one-sided continuity. This, along with one-sided differentiability, is today’s topic.

The definition of continuity requires that for a function to be continuous at a value x = a in its domain \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) and that both values are finite. That is, the limit as you approach the point in question be equal to the value at that point. This limit is a two-sided limit meaning that the limit is the same as x approaches a from both sides. That definition is extended to open intervals, by requiring that for a function to be continuous on an open interval, that it is continuous at every point of the interval

How do you check every point? One way is to prove the limit in the definition in general for any point in the open interval. Another is to develop a list of theorems that allow you to do this. For example, f(x) = x can be shown to be continuous at every number. Then sums, differences, and products, of this function allows you to extend the property to polynomials and then other functions.

But what about a function that has a domain that is not the entire number line? Something like f\left( x \right)=\sqrt{4-{{x}^{2}}},-2\le x\le 2. Here, f\left( -2 \right)=f\left( 2 \right)=0; the function is defined at the endpoints. A look at the graph shows a semi-circle that appears to contain the endpoints (–2, 0) and (2, 0). The function is continuous on the open interval (–2, 2) but cannot be continuous under the regular definition since the limit at the endpoints does not exist. The limit does not exist because the limit from the left at the left-endpoint, and the limit from the right at the right endpoint do not exist. What to do?


What is done is to require only that the one-sides limits from inside the domain exist. Here they do:\underset{x\to -2+}{\mathop{\lim }}\,f\left( x \right)=0  and the \underset{x\to 2-}{\mathop{\lim }}\,f\left( x \right)=0 and since the limits equal the values we say the function is continuous on the closed interval [–2, 2]. In general, when you say a function is continuous on a closed interval, you mean that the one-sided limits from inside the interval exist and equal the endpoint values.

You can determine that the limits exist by finding them as in the example above. Another way is to realize that if a < b < c < d and the function is continuous on the open (or closed) (a, d) then it is continuous on the closed interval [b c].

Why bother?

The reason we take this trouble is because for some reason the proof of the theorem under consideration requires that the endpoint value not only exist but hooks up with the function to make it continuous. Thus, the continuity on a closed interval is included in the hypotheses of theorem where this property is required. For example, the Intermediate Value Theorem would not work on a function that had no endpoints for f\left( c \right) to be between. Also, the Mean Value Theorem requires you to find the slope between the endpoints, so the endpoint needs to be not only defined, but attached to the rest of the function.

One-sided differentiability

The definition of the derivative at a point also requires a two-sided limit to exist at the point. Most of the early theorems in calculus require only that the function be differentiable on an open interval.

Is it possible to define differentiability at the endpoint of an interval? Yes. It’s done in the same way by using a one-sided limit. If x = a is the left endpoint of an interval, then the derivative from the right at that point is defined as

\displaystyle {f}'\left( a \right)=\underset{h\to 0+}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}.

By letting h approach 0 only from the right, you never consider values outside the interval. (At the right endpoint a similar definition is used with h\to 0-).

So why don’t we ever see one-sided derivatives? Because the theorems do not need them to prove their result. Hypotheses of theorems should be the minimum requirements needed, so if there is no need for the function to be differentiable at an endpoint, this is not listed in the hypotheses. This makes the hypothesis less restrictive and, therefore, covers more situations.

One theorem, beyond what is usually covered in beginning calculus, where endpoint differentiability is needed is Darboux’s Theorem. Darboux’s Theorem is sometime called the Intermediate Value Theorem for derivatives. It says that the derivative takes on all values between the derivatives at the endpoints, and thus needs the one-sided derivatives at the endpoints to exist. Interestingly, Darboux’s Theorem does not require the function to be continuous on the open interval between the endpoints.