My Favorite One-Liners: Part 80

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today’s awful pun comes courtesy of Math With Bad Drawings. Suppose we need to solve for $x$ in the following equation:

$2^{2x+1} = 3^{x}$ .

Naturally, the first step is taking the logarithm of both sides. But with which base? There are two reasonable options for most handheld scientific calculators: base-10 and base- $e$ . So I’ll tell the class my preference:

I’m organic; I only use natural logs.

My Favorite One-Liners: Part 79

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

I’ll use today’s quip when there are multiple reasonable ways of solving a problem. For example,

Two fair dice are rolled. Find the probability that at least one of the rolls is a six.

This can be done by directly listing all of the possibilities:

$11 \qquad 12 \qquad 13 \qquad 14 \qquad 15 \qquad 16$

$21 \qquad 22 \qquad 23 \qquad 24 \qquad 25 \qquad 26$

$31 \qquad 32 \qquad 33 \qquad 34 \qquad 35 \qquad 36$

$41 \qquad 42 \qquad 43 \qquad 44 \qquad 45 \qquad 46$

$51 \qquad 52 \qquad 53 \qquad 54 \qquad 55 \qquad 56$

$61 \qquad 62 \qquad 63 \qquad 64 \qquad 65 \qquad 66$

Of these 36 possibilities, 11 have at least one six, so the answer is $11/36$ .

Alternatively, we could use the addition rule:

$P(\hbox{first a six or second a six}) = P(\hbox{first a six}) + P(\hbox{second a six}) - P(\hbox{first a six and second a six})$

$= P(\hbox{first a six}) + P(\hbox{second a six}) - P(\hbox{first a six}) P(\hbox{second a six})$

$= \displaystyle \frac{1}{6} + \frac{1}{6} - \frac{1}{6} \times \frac{1}{6}$

$= \displaystyle \frac{11}{36}$ .

Another possibility is using the complement:

$P(\hbox{at least one six}) = 1 - P(\hbox{no sixes})$

$= 1 - P(\hbox{first is not a six})P(\hbox{second is not a six})$

$= 1 - \displaystyle \frac{5}{6} \times \frac{5}{6}$

$= \displaystyle \frac{11}{36}$

To emphasize that there are multiple ways of solving the problem, I’ll use this one-liner:

There are plenty of ways to skin a cat… for those of you who like skinning cats.

When I was a boy, I remember seeing some juvenile book of jokes titled “1001 Ways To Skin a Cat.” A recent search for this book on Amazon came up empty, but I did find this:

My Favorite One-Liners: Part 78

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

One of my pet peeves is students who don’t give me a reasonable amount of time to grade their final exams and compute their grade for the semester. Usually, “reasonable” means “tomorrow morning.” (However, sometimes I have to give multiple final exams on the same day, and there just isn’t enough time in the day to compute grades for all of my classes in that amount of time.)

So, to head this off, I’ll announce to my students when they can expect me to finish grading the finals so that it’s safe to ask for their grade for the semester; usually the answer is “9:00 tomorrow morning.” And, to make sure that no one bugs me before then, I’ll give the following playful admonition:

Anyone who asks me for their grade before 9:00 tomorrow morning gets an automatic F in the course.

My Favorite One-Liners: Part 77

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

At the end of every semester, instructors are often asked “What do I need on the final to make a ___ in the course?”, where the desired course grade is given. (I’ve never done a survey, but A appears to be the most desired course grade, followed by C, D, and B.) Here’s the do-it-yourself algorithm that I tell my students, in which the final counts for 20% of the course average.

Let $F$ be the grade on the final exam (as I write a big F on the chalkboard). [groans] After all, final starts with $F$ , and it’s important to assign variable names that make sense.

Also, let $D$ be the up-to-date course average prior to the final. [more groans]

This gives us the course average. Just to be nice, let’s call that $A$ . [sighs of relief]

So $A = 0.2F + 0.8D$ .

More seriously, here’s a practical tip for students to determine what they need on the final to get a certain grade (hat tip to my friend Jeff Cagle for this idea). It’s based on the following principle:

If the average of $x_1, x_2, \dots x_n$ is $\overline{x}$ , then the average of $x_1 + c, x_2 + c, \dots, x_n + c$ is $\overline{x} + c$ . In other words, if you add a constant to a list of values, then the average changes by that constant.

As an application of this idea, let’s try to guess the average of $78, 82, 88, 90$ . A reasonable guess would be something like $85$ . So subtract $85$ from all four values, obtaining $-7, -3, 3, 5$ . The average of these four differences is $(-7-3+3+5)/4 = -0.5$ . Therefore, the average of the original four numbers is $85 + (-0.5) = 84.5$ .

So here’s a typical student question: “If my average right now is an $88$ , and the final is worth $20\%$ of my grade, then what do I need to get on the final to get a $90$ ?” Answer: The change in the average needs to be $+2$ , so the student needs to get a grade $+2/0.2 = +10$ points higher than his/her current average. So the grade on the final needs to be $88 + 10 = 98$ .

Seen another way, we’re solving the algebra problem

$88(0.8) + x(0.2) = 90$

Let me solve this in an unorthodox way:

$88(0.8) + x(0.2) = 88 + 2$

$88(0.8) + x(0.2) = 88(0.8+0.2) + 2$

$88(0.8) + x(0.2) = 88(0.8) + 88(0.2) + 2$

$x(0.2) = 88(0.2) + 2$

$x = \displaystyle \frac{88(0.2)}{0.2} + \frac{2}{0.2}$

$x = 88 + \displaystyle \frac{2}{0.2}$

This last line matches the solution found in the previous paragraph, $x = 88 + 10 = 98$ .

My Favorite One-Liners: Part 76

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a problem that might arise in trigonometry:

Compute $\cos \displaystyle \frac{2017\pi}{6}$ .

To begin, we observe that $\displaystyle \frac{2017}{6} = 336 + \displaystyle \frac{1}{6}$ , so that

$\cos \displaystyle \frac{2017\pi}{6} = \cos \left( \displaystyle 336\pi + \frac{\pi}{6} \right)$ .

We then remember that $\cos \theta$ is a periodic function with period $2\pi$ . This means that we can add or subtract any multiple of $2\pi$ to the angle, and the result of the function doesn’t change. In particular, $-336\pi$ is a multiple of $2 \pi$ , so that

$\cos \displaystyle \frac{2017\pi}{6} = \cos \left( \displaystyle 336\pi + \frac{\pi}{6} \right)$

$= \cos \left( \displaystyle 336\pi + \frac{\pi}{6} - 336\pi \right)$

$= \cos \displaystyle \frac{\pi}{6}$

$= \displaystyle \frac{\sqrt{3}}{2}$ .

Said another way, $336\pi$ corresponds to $336/2 = 168$ complete rotations, and the value of cosine doesn’t change with a complete rotation. So it’s OK to just throw away any even multiple of $\pi$ when computing the sine or cosine of a very large angle. I then tell my class:

In mathematics, there’s a technical term for this idea; it’s called $\pi$ throwing.

My Favorite One-Liners: Part 75

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

The $\delta-\epsilon$ definition of a limit is often really hard for students to swallow:

$\forall \epsilon > 0 \exists \delta > 0 \forall x (0 < |x - c| < \delta \Rightarrow |f(x) - L| < \epsilon)$

To make this a little more palatable, I’ll choose a simple specific example, like $\lim_{x \to 2} x^2 = 4$ , or

$\forall \epsilon > 0 \exists \delta > 0 \forall x (0 < |x - 2| < \delta \Rightarrow |x^2 - 4| < \epsilon)$

I’ll use one of the famous lines from “Annie Get Your Gun”:

Anything you can do, I can do better.

In other words, no matter how small a $\delta$ they give me, I can find an $\epsilon$ that meets the requirements of this limit.

My Favorite One-Liners: Part 74

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

After presenting the Fundamental Theorem of Calculus to my calculus students, I make a point of doing the following example in class:

$\displaystyle \int_0^4 \frac{1}{4} x^2 \, dx$

Hopefully my students are able to produce the correct answer:

$\displaystyle \int_0^4 \frac{1}{4} x^2 \, dx = \displaystyle \left[ \frac{x^3}{12} \right]^4_0$

$= \displaystyle \frac{(4)^3}{12} - \frac{(0)^3}{12}$

$= \displaystyle \frac{64}{12}$

$= \displaystyle \frac{16}{3}$

Then I tell my students that they’ve probably known the solution of this one since they were kids… and I show them the classic video “Unpack Your Adjectives” from Schoolhouse Rock. They’ll watch this video with no small amount of confusion (“How is this possibly connected to calculus?”)… until I reach the 1:15 mark of the video below, when I’ll pause and discuss this children’s cartoon. This never fails to get an enthusiastic response from my students.

If you have no idea what I’m talking about, be sure to watch the first 75 seconds of the video below. I think you’ll be amused.

My Favorite One-Liners: Part 73

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today’s entry is courtesy of Season 1 of The Simpsons. I’ll tell this joke just after introducing derivatives to my calculus students. Here is some dialogue from the episode “Bart The Genius”:

Teacher: So y = r cubed over 3. And if you determine the rate of change in this curve correctly, I think you’ll be pleasantly surprised.
[The class laughs except for Bart who appears confused.]
Teacher: Don’t you get it, Bart? Derivative dy = 3 r squared dr over 3, or r squared dr, or r dr r. Har-de-har-har! Get it?

For a more detailed listing of mathematical references, I highly recommend http://www.simpsonsmath.com (or http://mathsci2.appstate.edu/~sjg/simpsonsmath/), maintained by Dr. Sarah J. Greenwald of Appalachian State University and Dr. Andrew Nestler of Santa Monica College.

My Favorite One-Liners: Part 72

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

In calculus, the Intermediate Value Theorem states that if $f$ is a continuous function on the closed interval $[a,b]$ and $y_0$ is any number between $f(a)$ and $f(b)$ , then there is at least one point $c \in [a,b]$ so that $f(c) =y_0$.

When I first teach this, I’ll draw some kind of crude diagram on the board:

In this picture, $f(a)$ is less than $y_0$ while $f(b)$ is greater than $y_0$ . Hence the one-liner:

I call the Intermediate Value Theorem the Goldilocks principle. After all, $f(a)$ is too low, and $f(b)$ is too high, but there is some point in between that is just right.

My Favorite One-Liners: Part 71

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Some of the algorithms that I teach are pretty lengthy. For example, consider the calculation of a $100(1-\alpha)\%$ confidence interval for a proportion:

$\displaystyle \frac{\hat{p} + \displaystyle \frac{z_{\alpha/2}^2}{2n}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } - z_{\alpha/2} \frac{\sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z_{\alpha/2}^2}{4n^2}}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } < p < \displaystyle \frac{\hat{p} + \displaystyle \frac{z_{\alpha/2}^2}{2n}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } + z_{\alpha/2} \frac{\sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z_{\alpha/2}^2}{4n^2}}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} }$ .

Wow.

Proficiency with this formula definitely requires practice, and so I’ll typically give a couple of practice problems so that my students can practice using this formula while in class. After the last example, when I think that my students have the hang of this very long calculation, I’ll give my one-liner to hopefully boost their confidence (no pun intended):

By now, you probably think that this calculation is dull, uninteresting, repetitive, and boring. If so, then I’ve done my job right.