A Calculus Journey

Posted on November 9, 2021 by Lin McMullin

I think that the path leading up to and including the Fundamental Theorem of Calculus (FTC) is one of the most beautiful walks in mathematics. I have written several posts about it. You will soon be ready to travel that path with your students. (I always try to post on topics shortly before most teachers will get to them, so that you have some time to consider them and work the ideas you like into your lessons.)

Here is an annotated list of some of the posts to guide you on your journey.

Working Towards Riemann Sums gives the preliminary definitions you will need to define and discuss Riemann sums.

Riemann Sums defines the several Riemann sums often used in the calculus left-side sums, right-side sums, midpoint sums and the trapezoidal sums. “The Area Under a Curve” in the iPad app A Little Calculus is a great visual display of these and shows what happens as you use more subintervals.

The Definition of the Definite Integral gives the definition of the definite integral as the limit of any Riemann sum. As with any definition, there is nothing to prove or argue about here. The thing to remember is that the limit of the Riemann sum and the definite integral are the same thing. Behind any definite integral is a Riemann sum. The advantage of the definition’s integral notation is that it shows the interval involved which the Riemann sum does not. (Any Riemann sum may be represented by many definite integrals. See Good Question 11 – Riemann Reversed.)

Foreshadowing the FTC is an example of how a definite integral may be evaluated. It is long and has a lot of notation, so you may not want to use this.

The Fundamental Theorem of Calculus is where the path leads. This post develops the FTC based on the other “big” idea of the calculus: the Mean Value Theorem. (I think the form here is preferable to the usual book notation that uses F(x) and its derivate f (x).)

Y the FTC? Tries to answer the question of what’s so important about the FTC. Example 1: The verbal interpretation of the FTC (the integral of a rate of change is the net amount of change over the interval.) will soon be used in many practical applications. While example 2 shows how the FTC allows one to evaluate a definite integral and, therefore the Riemann sum it represents, by evaluating a function whose derivative is the integrand (its antiderivative).

More About the FTC presents examples leading up to the other form of the FTC: the derivative of the integral is the integrand).

At this point you may go in the direction of learning how to find antiderivatives or working on applications. (See Integration itinerary.)

Bon Voyage!

The Old Pump

Posted on November 5, 2021 by Lin McMullin

A tank is being filled with water using a pump that is old and slows down as it runs. The table below gives the rate at which the pump pumps at ten-minute intervals. If the tank initially has 570 gallons of water in it, approximately how much water is in the tank after 90 minutes?

Elapsed time (minutes)	0	10	20	30	40	50	60	70	80	90
Rate (gallons / minute)	42	40	38	35	35	32	28	20	19	10

And so, integration begins.

Ask your students to do this problem alone. When they are ready (after a few minutes) collect their opinions. They will not all be the same (we hope, because there is more than one reasonable way to approximate the amount). Ask exactly how they got their answers and what assumptions they made. Be sure they always include units (gallons). Here are some points to make in your discussion – points that we hope the kids will make and you can just “underline.”

1. Answers between 3140 and 3460 gallons are reasonable. Other answers in that range are acceptable. They will not use terms like “left-sum”, “right sum” and “trapezoidal rule” because they do not know them yet, but their explanations should amount to the same thing. An answer of 3300 gallons may be popular; it is the average of the other two, but students may not have gotten it by averaging 3140 and 3460.
2. Ask if they think their estimate is too large or too small and why they think that.
3. Ask what they need to know to give a better approximation – more and shorter time intervals.
4. Assumptions: If they added 570 + 42(10) + 40(10) + … +19(10) they are assuming that the pump ran at each rate for the full ten minutes and then suddenly dropped to the next. Others will assume the rate dropped immediately and ran at the slower rate for the 10 minutes. Some students will assume the rate dropped evenly over each 10-minute interval and use the average of the rates at the ends of each interval (570 + 41(10) + 39(10) + … 14.5(10) = 3300).
5. What is the 570 gallons in the problem for? Well, of course to foreshadow the idea of an initial condition. Hopefully, someone will forget to include it and you can point it out.
6. With luck someone will begin by graphing the data. If no one does, you should suggest it; (as always) to help them see what they are doing graphically. They are figuring the “areas” of rectangles whose height is the rate in gallons/minute and whose width is the time in minutes. Thus the “area” is not really an area but a volume (gal/min)(min) = gallons). In addition to unit analysis, graphing is important since you will soon be finding the area between the graph of a function and the x-axis in just this same manner.

Follow up: Flying to Integrationland

Be sure to check the “Thoughts on ‘The Old Pump'” in the comments section below.

Revised from a post of November 30, 2012.

Unit 6 – Integration and Accumulation of Change

Posted on November 2, 2021 by Lin McMullin

Unit 6 develops the ideas behind integration, the Fundamental Theorem of Calculus, and Accumulation. (CED – 2019 p. 109 – 128). These topics account for about 17 – 20% of questions on the AB exam and 17 – 20% of the BC questions.

Topics 6.1 – 6.4 Working up to the FTC

Topic 6.1 Exploring Accumulations of Change Accumulation is introduced through finding the area between the graph of a function and the x-axis. Positive and negative rates of change, unit analysis.

Topic 6.2 Approximating Areas with Riemann Sums Left-, right-, midpoint Riemann sums, and Trapezoidal sums, with uniform partitions are developed. Approximating with numerical methods, including use of technology are considered. Determining if the approximation is an over- or under-approximation.

Topic 6.3 Riemann Sums, Summation Notation and the Definite Integral. The definition integral is defined as the limit of a Riemann sum.

Topic 6.4 The Fundamental Theorem of Calculus (FTC) and Accumulation Functions Functions defined by definite integrals and the FTC.

Topic 6.5 Interpreting the Behavior of Accumulation Functions Involving Area Graphical, numerical, analytical, and verbal representations.

Topic 6.6 Applying Properties of Definite Integrals Using the properties to ease evaluation, evaluating by geometry and dealing with discontinuities.

Topic 6.7 The Fundamental Theorem of Calculus and Definite Integrals Antiderivatives. (Note: I suggest writing the FTC in this form $displaystyle int_{a}^{b}{{{f}'left( x right)}}dx=fleft( b right)-fleft( a right)$ because it seem more efficient then using upper case and lower case f.)

Topics 6.5 – 6.14 Techniques of Integration

Topic 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation. Using basic differentiation formulas to find antiderivatives. Some functions do not have closed-form antiderivatives. (Note: While textbooks often consider antidifferentiation before any work with integration, this seems like the place to introduce them. After learning the FTC students have a reason to find antiderivatives. See Integration Itinerary

Topic 6.9 Integration Using Substitution The u-substitution method. Changing the limits of integration when substituting.

Topic 6.10 Integrating Functions Using Long Division and Completing the Square

Topic 6.11 Integrating Using Integration by Parts (BC ONLY)

Topic 6.12 Integrating Using Linear Partial Fractions (BC ONLY)

Topic 6.13 Evaluating Improper Integrals (BC ONLY) Showing the work requires students to show correct limit notation.

Topic 6.14 Selecting Techniques for Antidifferentiation This means practice, practice, practice.

Timing

The suggested time for Unit 6 is 18 – 20 classes for AB and 15 – 16 for BC of 40 – 50-minute class periods, this includes time for testing etc.

Previous posts on these topics include:

Introducing Integration

Integration Itinerary

The Old Pump and Flying to Integrationland Two introductory explorations

Working Towards Riemann Sums

Riemann Sums

The Definition of the Definite Integral

Foreshadowing the FTC

The Fundamental Theorem of Calculus

More About the FTC

Y the FTC?

Area Between Curves

Under is a Long Way Down

Properties of Integrals

Trapezoids – Ancient and Modern On Trapezoid sums

Good Question 9 – Riemann Reversed Given a Riemann sum can you find the Integral it converges to? A common and difficult AP Exam problem

Adapting 2021 AB 1 / BC 1

Adapting 2021 AB 4 / BC 4

Accumulation

Accumulation: Need an Amount?

Good Question 7 – 2009 AB 3

Good Question 8 – or Not? Unit analysis

AP Exams Accumulation Question A summary of accumulation ideas.

Graphing with Accumulation 1

Graphing with Accumulation 2

Accumulation and Differential Equations

Painting a Point

Techniques of Integrations (AB and BC)

Antidifferentiation

Why Muss with the “+C”?

Good Question 13 More than one way to skin a cat.

Integration by Parts – a BC Topic

Integration by Parts 1

Integration by Part 2

Parts and More Parts

Good Question 12 – Parts with a Constant?

Modified Tabular Integration

Improper Integrals and Proper Areas

Math vs the Real World Why $displaystyle int_{{-infty }}^{infty }{{frac{1}{x}}}dx$ does not converge.

Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

2019 CED – Unit 1: Limits and Continuity

2019 CED – Unit 2: Differentiation: Definition and Fundamental Properties.

2019 CED – Unit 3: Differentiation: Composite , Implicit, and Inverse Functions

2019 CED – Unit 4 Contextual Applications of the Derivative Consider teaching Unit 5 before Unit 4

2019 – CED Unit 5 Analytical Applications of Differentiation Consider teaching Unit 5 before Unit 4

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions

2019 CED Unit 10 Infinite Sequences and Series

Implicit Differentiation of Parametric Equations

Posted on October 31, 2021 by Lin McMullin

I’ve never liked memorizing formulas. I would rather know where they came from or be able to tie it to something I already know. One of my least favorite formulas to remember and explain was the formula for the second derivative of a curve given in parametric form. No longer.

If $\displaystyle y=y(t)$ and, $\displaystyle x=x(t)$ then the traditional formulas give

$\displaystyle \frac{{dy}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$ , and

$\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$

It is that last part, where you divide by $\displaystyle {\frac{{dx}}{{dt}}}$ , that bothers me. Where did the $\displaystyle {\frac{{dx}}{{dt}}}$ come from?

Then it occurred to me that dividing by $\displaystyle {\frac{{dx}}{{dt}}}$ is the same as multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$

It’s just implicit differentiation!

Since $\displaystyle \frac{{dy}}{{dx}}$ is a function of t you must begin by differentiating the first derivative with respect to t. Then treating this as a typical Chain Rule situation and multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$ gives the second derivative. (There is a technical requirement here that given $\displaystyle x=x(t)$ , then its inverse $\displaystyle t={{x}^{{-1}}}\left( x \right)$ exists.)

In fact, if you look at a proof of the formula for the first derivative, that’s what happens there as well:

$\displaystyle \frac{d}{{dx}}y(t)=\frac{{dy}}{{dt}}\cdot \frac{{dt}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$

The reason you do it this way is that since x is given as a function of t, it may be difficult to solve for t so you can find dt/dx in terms of x. But you don’t have to; just divide by dx/dt which you already know.

Here is an example for both derivatives.

Suppose that $\displaystyle x={{t}^{3}}-3$ and $\displaystyle y=\ln \left( t \right)$

Then $\displaystyle \frac{{dy}}{{dt}}=\frac{1}{t}$ and $\displaystyle \frac{{dx}}{{dt}}=3{{t}^{2}}$ and $\displaystyle \frac{{dt}}{{dx}}=\frac{1}{{3{{t}^{2}}}}$

Then $\displaystyle \frac{{dy}}{{dx}}=\frac{1}{t}\cdot \frac{{dt}}{{dx}}=\frac{1}{t}\cdot \frac{1}{{3{{t}^{2}}}}=\frac{1}{3}{{t}^{{-3}}}$

And $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\left( {\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)} \right)\cdot \frac{{dt}}{{dx}}=\left( {-{{t}^{{-4}}}} \right)\cdot \left( {\frac{1}{{3{{t}^{2}}}}} \right)=-\frac{1}{{3{{t}^{6}}}}$

Yes, it’s the same thing as using the traditional formula, but now I’ll never have to worry about forgetting the formula or being unsure how to explain why you do it this way.

Revised: Correction to last equation 5/18/2014. Revised: 2/8/2016. Originally posted May 5, 2014.

Open or Closed?

Posted on October 26, 2021 by Lin McMullin

About this time of year, you find someone, hopefully one of your students, asking, “If I’m finding where a function is increasing, is the interval open or closed?”

Do you have an answer?

This is a good time to teach some things about definitions and theorems.

The place to start is to ask what it means for a function to be increasing. Here is the definition:

A function is increasing on an interval if, and only if, for all (any, every) pairs of numbers x₁ < x₂ in the interval, f(x₁) < f(x₂).”

(For decreasing on an interval, the second inequality changes to f(x₁) > f(x₂). All of what follows applies to decreasing with obvious changes in the wording.)

Notice that functions increase or decrease on intervals, not at individual points. We will come back to this in a minute.
Numerically, this means that for every possible pair of points, the one with the larger x-value always produces a larger function value.
Graphically, this means that as you move to the right along the graph, the graph is going up.
Analytically, this means that we can prove the inequality in the definition.

For an example of this last point consider the function f(x) = x². Let x₂= x₁ + h where h > 0. Then in order for f(x₁) < f(x₂) it must be true that

$\displaystyle {{x}_{1}}^{2}<{{\left( {{{x}_{1}}+h} \right)}^{2}}$

$\displaystyle 0<{{\left( {{{x}_{1}}+h} \right)}^{2}}-{{x}_{1}}^{2}$

$\displaystyle 0<{{x}_{1}}^{2}+2{{x}_{1}}h+{{h}^{2}}-{{x}_{1}}^{2}$

$\displaystyle 0<2{{x}_{1}}h+{{h}^{2}}$

This can only be true if $\displaystyle {{x}_{1}}\ge 0$ Thus, x² is increasing only if $\displaystyle {{x}_{1}}\ge 0$

Now, of course, we rarely, if ever, go to all that trouble. And it is even more trouble for a function that increases on several intervals. The usual way of finding where a function is increasing is to look at its derivative.

Notice that the expression $\displaystyle 0<{{x}_{1}}^{2}+2{{x}_{1}}h+{{h}^{2}}-{{x}_{1}}^{2}$ looks a lot like the numerator of the original limit definition of the derivative of x² at x = x₁, namely $\displaystyle 0<{{\left( {{{x}_{1}}+h} \right)}^{2}}-{{x}_{1}}^{2}$ . If h > 0, where the function is increasing the numerator is positive and the derivative is positive also. Turning this around we have a theorem that says, If $\displaystyle {f}'\left( {{{x}_{1}}} \right)>0$ for all x in an interval, then the function is increasing on the interval. That makes it much easier to find where a function is increasing, we simplify find where its derivative is positive.

There is only a slight problem in that the theorem does not say what happens if the derivative is zero somewhere on the interval. If that is the case, we must go back to the definition of increasing on an interval or use a different method. For example, the function x³ is increasing everywhere, even though its derivative at the origin is zero.

Let’s consider another example. The function sin(x) is increasing on the interval $\displaystyle [-\tfrac{\pi }{2},\tfrac{\pi }{2}]$ (among others) and decreasing on $\displaystyle [\tfrac{\pi }{2},\tfrac{3\pi }{2}]$ . It bothers some that $\displaystyle \tfrac{\pi }{2}$ is in both intervals and that the derivative of the function is zero at x = $\displaystyle \tfrac{\pi }{2}$ . This is not a problem. Sin( $\displaystyle \tfrac{\pi }{2}$ ) is larger than all the other values is both intervals, so by the definition, and not the theorem, the intervals are correct.

It is generally true that if a function is continuous on the closed interval [a,b] and increasing on the open interval (a,b) then it must be increasing on the closed interval [a,b] as well.

Returning to the first point above: functions increase or decrease on intervals not at points. You do find questions in books and on tests that ask, “Is the function increasing at x = a.” The best answer is to humor them and answer depending on the value of the derivative at that point. Since the derivative is a limit as h approaches zero, the function must be defined on some interval around x = a in which h is approaching zero. So, answer according to the value of the derivative on that interval.

You can find more on this here.

Case Closed.

Slightly revised from a posted published on November 2, 2012.

Extreme Values

Posted on October 22, 2021 by Lin McMullin

Every function that is continuous on a closed interval must have a maximum and a minimum value on the interval. These values may all be the same (y = 2 on [-3,3]); or the function may reach these values more than once (y = sin(x)).

If the function is defined on a closed interval, then the extreme values are either (1) at an endpoint of the interval or (2) at a critical number. This is known as the Extreme Value Theorem. Thus, one way of finding the extreme values is to simply find the value of the function at the endpoints and the critical points and compare these to find the largest and smallest. This is called the Candidates’ Test or the Closed Interval Test. It is a good one to “play” with: do some sketches of the different situation above; discuss why the interval must be closed.

On an open or closed interval, the shapes can change if the first derivative is zero or undefined at the point where two shapes join. In this case the point is a local extreme value of the function – a local maximum or minimum value. Specifically:

If the first derivative changes from positive to negative, the shape of the function changes from increasing to decreasing and the point is a local maximum. If the first derivative changes from negative to positive, the shape of the function changes from decreasing to increasing and the point is a local minimum.

This is a theorem called the First Derivative Test. By finding where the first derivative changes sign and in which direction it changes (positive to negative, or negative to positive) we can locate and identify the local extreme value precisely.

Another way to determine if a critical number is the location of a local maximum or minimum is a theorem called the Second Derivative Test.

If the first derivative is zero (and specifically not if it is undefined) and the second derivative is positive, then the graph has a horizontal tangent line and is concave up. Therefore, this is the location of a local minimum of the function.

Likewise, if the first derivative is zero at a point and the second derivative is negative there, the function has a local maximum there.

If both the first and second derivatives are zero at a point, then the second derivative test cannot be used, for example y = x⁴ at the origin.

The mistake students make with the second derivative test is in not checking that the first derivative is zero. If “justify your answer” is required, students should be sure to show that the first derivative is zero as well as the sign of the second derivative.

In the case where both the first and second derivatives are zero at the same point the function changes direction but not concavity (e.g. f (x) = x⁴at the origin), or changes concavity but not direction (e.g. f (x) = x³at the origin).

This is a revised version of a post published on October 22, 2012

Reading the Derivative’s Graph

Posted on October 19, 2021 by Lin McMullin

A very typical calculus problem is given the equation of a function, to find information about it (extreme values, concavity, increasing, decreasing, etc., etc.). This is usually done by computing and analyzing the first derivative and the second derivative. All the textbooks show how to do this with copious examples and exercises. I have nothing to add to that. One of the “tools” of this approach is to draw a number line and mark the information about the function and the derivative on it.

A very typical AP Calculus exam problem is given the graph of the derivative of a function, but not the equation of either the derivative or the function, to find all the same information about the function. For some reason, students find this difficult even though the two-dimensional graph of the derivative gives all the same information as the number line graph and, in fact, a lot more.

Looking at the graph of the derivative in the x,y-plane it is easy to determine the important information. Here is a summary relating the features of the graph of the derivative with the graph of the function.

Feature	the function
${y}'$ > 0	is increasing
${y}'$ < 0	is decreasing
${y}'$ changes – to +	has a local minimum
${y}'$ changes + to –	has a local maximum
${y}'$ increasing	is concave up
${y}'$ decreasing	is concave down
${y}'$ extreme value	has a point of inflection

Here’s a typical graph of a derivative with the first derivative features marked.

Here is the same graph with the second derivative features marked.

The AP Calculus Exams also ask students to “Justify Your Answer.” The table above, with the columns switched does that. The justifications must be related to the given derivative, so a typical justification might read, “The function has a relative maximum at x = -2 because its derivative changes from positive to negative at x = -2.”

Conclusion	Justification
y is increasing	${y}'$ > 0
y is decreasing	${y}'$ < 0
y has a local minimum	${y}'$ changes – to +
y has a local maximum	${y}'$ changes + to –
y is concave up	${y}'$ increasing
y is concave down	${y}'$ decreasing
y has a point of inflection	${y}'$ extreme values

For notes on asymptotes see Asymptotes and the Derivative and Other Asymptotes.

Originally posted on October 26, 2012, and my single most viewed post over the years.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Author Archives: Lin McMullin

A Calculus Journey

The Old Pump