Author Archives: Lin McMullin

The Opposite of Negative

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Next year, for the first time in 15 years, I am going to be teaching high school full-time. While I have enjoyed writing and working primarily with teachers for the last 15 years, I’m looking forward to “going back to the classroom” as they say. It looks like I’ll be teaching BC calculus and Algebra 1 – two of my favorite classes. I’m very positive about that.

With that in mind I have been thinking of some of the things I want to be sure I get right in the Algebra 1 classes to get the kids off to a good start. So a few of my blogs in the coming year may be on Algebra 1 topics with the view of having students do things right from the start and not having to relearn things when they get to calculus.

So here is the first thing I want to be sure to work on: the m-dash also known as the minus sign.

According to Wikipedia:

The minus sign (−) has three main uses in mathematics:

  1. The subtraction operator: A binary operator to indicate the operation of subtraction, as in 5 − 3 = 2. Subtraction is the inverse of addition.
  2. Directly in front of a number and when it is not a subtraction operator it means a negative number. For instance −5 is negative 5.
  3. A unary operator that acts as an instruction to replace the operand by its opposite. For example, if x is 3, then −x is −3, but if x is −3, then −x is 3. Similarly, −(−2) is equal to 2.

Using the same symbol understandably can confuse beginning math students. I am not going to invent new symbols so I will just have to be careful with what I say and let the kids say. And I have to say it right , if I expect them to.

When used between two numbers or two expressions with variables the symbol means subtraction. That’s pretty easy to spot and understand in context.  But when used alone in front of something the minus sign means different things.

The m-dash may always be read “opposite.” So “–a” is read “the opposite of a” and not “negative a.”  Likewise, –5 is read “the opposite of five.”

There is only one instance where the m-dash may be read “negative.” When it is used in front of a number it indicates a negative number so      “–5” is also correctly read “negative five.” This is the only time the m-dash should be read “negative.”  Things like “–a” should always be read “the opposite of a” and never read “negative a.”

There was a time when the folks who write math books tried to make the distinction by using a slightly raised dash to indicate negative number so negative 3 was written  “–3.” This has carried over into calculators where the key marked “(–)” is used for “negative” and “opposite.” and is printed on the screen as a shorter and slightly raised dash. The subtraction key is only used for subtraction.

Oh, if it were only that simple. What do you do with –(–5)? Not really a problem the “opposite of the opposite of 5” and the “opposite of negative 5” are both 5.

I’ll know I’ve succeeded when everyone can get 100% on this little True-False quiz:

  1. The opposite of a number is a negative.
  2. –x < 0
  3. x > 0
  4. | x | = x
  5. |– x |= x

Answers are in my next post.

Revised 10-27-2018

A Standard Problem?

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Sometimes even the simplest problems can lead to interesting places. I got interested in this ordinary textbook problem a few months ago while observing an AB calculus class. Here is how my thinking went written as a series of exercises:

Original exercise: Find the point(s) on the parabola y = x2 that are closest to the point (0, 2).

The solution is straightforward:

Let z(x) be the distance from (0, 2) to the point (x, x2). Then \displaystyle z\left( x \right)=\sqrt{{{\left( x-0 \right)}^{2}}+{{\left( {{x}^{2}}-2 \right)}^{2}}}and setting the derivative equal to zero, we find the critical points at x = 0 a local maximum and x=-\sqrt{\tfrac{3}{2}} and x=\sqrt{\tfrac{3}{2}} both absolute minimums located symmetrically to the y-axis.

The students had some trouble understanding what was going on. I suggested they consider a point moving along the parabola starting high on the left side. As the point comes down the graph its distance from \left( 0,2 \right) gets shorter until it reaches some minimum length somewhere. Then the distance gets longer again until – where? Obviously, at the origin. Then symmetry takes over and the distance decreases again until it reaches a second minimum directly across the y-axis from the first point, after which it increases forever.

Exercise 1: Sketch the graph of the distance function, z(x), on top of the graph of the parabola. Give the exact coordinates of the maximum point without doing any computations. Discuss the y-coordinates of the minimum points.

It then occurred to me that there is nothing sacred about (0, 2). We could use any point on the y-axis, . Or could we? I pictured a circle centered at the point (0, a) on the y-axis tangent to the parabola (i.e. tangent to the tangent line of the parabola). The radius of such a circle is the minimum distance for that y-axis center point.

Exercise 2: If the point is below the origin then obviously the minimum distance is the distance directly up to the origin. There are places above the x-axis where the origin is the closest point. For what values of a is the origin the closest point? Use the circle mentioned above to explain how this is possible.

The distance graph that we drew above has a W shape similar to the shape of some 4th degree polynomials (note: z(x) is not a 4th degree polynomial, it is not even a polynomial). So next I drew the graph of z’(x), the first derivative of z; this graph is similar to a cubic polynomial (again it is not a polynomial)

Exercise 3: Without actually computing the derivative’s equation, sketch the graph of the derivative with the graph of the parabola and the distance function’s graph. Be sure the critical points are where they should be.

Exercise 4: Use a graphing program or a graphing calculator do draw the parabola, z(x) and z’(x) and use it to check your sketches.

Project: Using a suitable program (Winplot, Geogebra, etc.) set up an animation that will show all the features discussed above. It should show the parabola which does not move, the movable point (0, a), the circle tangent to the parabola, z(x) and z’(x). These last 3 should move with the slider for a.

At this point I noticed that the derivative changed concavity 4 times (up-down-up-down). The origin is one of 3 points of inflection. I wondered if there was anything interesting about the location of the points of inflection.

Exercise 5: Find the coordinates of the points of inflection of z’(x). Are they related to any of the features in the problem? (Use a CAS for this one.)

The solutions can be found here: Closest Point Problem Solution.

Looking Ahead

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Well the AP calculus exams are over. I hope all your students did well.

I’ll be cutting down the number of posts for the next few months. I will try to post once a week or so when I find something interesting. Remember if you click one of the RSS feeds or the “Follow this Blog” at the bottom of the right hand column you can get the postings without coming directly here.

In August I plan to go back to teaching high school for the first time in 15 years. I will be filling in for a friend who is going on sabbatical. That should be interesting and fun.

This blog was stared in August 2012 and I tried to do post on topics that would help a few weeks ahead of when I though most people would be teaching them. So you can go through the year monthly from the archives section on the right. Starting this coming August I hope to fill in the gaps from the first year by writing on topics I missed or expanding on the previous ones.

In the meanwhile, I’ll be working on the third edition of my book Teaching AP Calculus. I hope it will be out early next year.

Hope you have  relaxing and productive summer!

Today’s the Day

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For all the AP Calculus teachers out there, my best wishes for your students today as they take the big test. I hope they all do well and all your hard work and effort pays off.

Getting Ready for the Exam

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I think the idea of writing this blog came to me about this time last year when folks were looking for last-minute advice to give their students before the AP calculus exams. I had some ideas of my own and collected some from others. Here is a list in no particular order.

The review time

  • Concentrate your reviewing on the things you don’t know (yet). Try to pick up those details you are not too sure of.
  • Work as many actual AP problems as you can, but concentrate on the form and ideas. None of these questions will be on the test, but many very much like these will be.
  • With, or without your class, find one (or more) of the released exams and take it in one sitting with the time allowed for each section.  This is to get you used to the real timing and the fact that you may not finish one or more sections.

The day before the test

  • Take a good look at the various formulas you will need; be sure you have them memorized correctly.
  • Put fresh batteries in you calculator and be sure it is in radian mode.
  • Take the afternoon and evening off. Relax. Do something fun.
  • Get to bed early and get a good night’s sleep.
  • Have a good breakfast.
  • Bring a snack for the short break between the two sections of the test.
  • Get Psyched!

During the test

  • Don’t panic! There is no extra credit for 100%. You may miss quite a few points and still get a 5; and quite a few more and get a 3.
  • Concentrate on the things you know. If you don’t know a how to do a problem, go onto the next one.
  • Keep your eye on the clock. Just before the multiple-choice sections are over, bubble in anything you left blank – there is no penalty for guessing.
  • On the free-response section, do not do arithmetic or algebraic simplification – it is not required and simplifying a correct answer incorrectly will lose a point. And it wastes time.
  • Don’t get bogged down in a problem – if you are not getting anywhere, stop and go to the next part or next question.

Good Luck!

What’s a Mean Old Average Anyway?

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Students often confuse the several concepts that have the word “average” or “mean” in their title. This may be partly because not just the names, but the formulas associated with each are very similar, but I think the main reason may be that they are keying in on the word “average” rather than the full name.

Here are the three items. We will assume that the function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b):

1.  The average rate of change of a function over the interval is simply the slope of the line from one endpoint of the graph to the other.

 \displaystyle \frac{f\left( b \right)-f\left( a \right)}{b-a}

2. The mean (or average) value theorem say that somewhere in the open interval (a, b) there is a number c such that the derivative (slope) at x = c is equal to the average rate of change over the interval.

\displaystyle {f}'\left( c \right)=\frac{f\left( b \right)-f\left( a \right)}{b-a}

3. The average value of a function is literally the average of all the y-coordinates on the interval. It is the vertical side of a rectangle whose base extends on the x-axis from x = a to x =b and whose area is the same as the area between the graph and the x-axis and the function over the same interval.

\displaystyle \frac{\int_{a}^{b}{f\left( x \right)dx}}{b-a}

Notice that when you evaluate the integral, the result looks very much like the ones above. This formula is also called the mean value theorem for integrals or the integral form of the mean value theorem. No wonder people get confused.

The three are closely related. Consider a position-velocity-acceleration situation. The average rate of change of position (#1 above) is the average value of the velocity (#3) and somewhere the velocity must equal this number (#2). Similarly, the average rate of change of velocity (#1) is the average acceleration (#3) and somewhere in the interval the acceleration (derivative of velocity) must equal this number (#2).

These ideas are tested on the AP calculus exams sometimes in the same question. See for example 2004 AB 1 parts c and d.

So, help your students concentrate on the entire name of the concepts, not just the “average” part.

Proof

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When math books present a theorem, they almost always immediately present its proof. I tend to skip the proofs. I assume they are correct. I want to get on with the ideas in the text. Later I may come back and read through them. Is this a good thing to advise students to do? I don’t know.

There are reasons to read proofs. One reason is to help understand why a theorem is true, by seeing the reasoning that leads to the result. Another is to check the reasoning yourself. A third is to learn how to do proofs.

Learning to write original proofs is not usually one of the goals of a beginning calculus course. That comes later in a course with “analysis” in its title. There are many theorems that involve some one-off that rarely will be used again. I’m thinking of a proof like that of the sum of the limits is equal to the limit of the sums, where you add and subtract the same expression and this more complicated form allows you to group and factor the terms of the numerator and arrive at the result. Another example is in the Mean Value Theorem where you consider a new function that gives the vertical distance between a function and its secant line. These always bring the question, “How did you know to do that?”

If a student can accept things like that, then the proof is usually easy enough to follow. But I would never spend a lot of time making every student fight his or her way through each and every proof.

On this other hand, I would never just present a theorem and not give some explanation as to why it is true (and why it is important enough to mention). Unfortunately, I have seen teachers write the Fundamental Theorem of Calculus on the board and proceed to show how to use it to evaluate definite integrals, with no hint of why this important theorem is true. Sure kids can memorize it and use it, but it seems to me they should also have a hint as to why it is true.

Some theorems are easy to understand if explained in ways other than giving a proof. For an example of this, see my post of October 1, 2012 on the Mean Value Theorem. Almost every book will bail out on the Intermediate Value Theorem by claiming (quite rightly) that, “the proof is beyond the scope of this book,” or they give the proof in an appendix. But a simple drawing will convince you that it is true.

So my feeling is that you do not need to labor over a proof for every theorem, BUT, big BUT, you should provide a good explanation of why it is true.

This is important for all students and especially for young women. Jo Boaler writes

“As I interviewed more and more boys and girls, I noticed that the desire to know why was something that separated the girls from the boys. The girls were able to accept the method that were shown them and practice them, but they wanted to know why they worked, where they came from, and how they connected with other methods…. When they could not get access to the depth of understanding they wanted, the girls started to turn away from the subject…. Classes in which students discuss concepts, giving them access to a deep and connected understanding of math, are good for boys and girls. Boys may be willing to work in isolation on abstract rules, but such approaches do not give many students, girls or boys, access to the understanding they need. In addition, high-level work in mathematics, science and engineering is not about isolated, abstract rule following, but about collaboration and connection making.”

[Jo Boaler, What’s Math Got to Do with It? Helping Children Learn to Love Their Most Hated Subject – And Why It’s Important for America, © 2008 Penguin Group, New York. From Chapter 6]