# My Favorite One-Liners: Part 88

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 the first few weeks of my calculus class, after introducing the definition of a derivative,

$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$,

I’ll use the following steps to guide my students to find the derivatives of polynomials.

1. If $f(x) = c$, a constant, then $\displaystyle \frac{d}{dx} (c) = 0$.
2. If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$.
3.  If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$.
4. If $f(x) = x^n$, where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$.
5. If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let $A(r) = \pi r^2$. Notice I’ve changed the variable from $x$ to $r$, but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.)

What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$.

Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Generally, students start waking up even though it’s near the end of class. I continue:

Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$. Does this remind you of anything? (Students answer: the volume of a sphere.)

What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$.

Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

By now, I’ve really got my students’ attention with this unexpected connection between these formulas from high school geometry. If I’ve timed things right, I’ll say the following with about 30-60 seconds left in class:

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.) Class is dismissed.

If you’d like to see the answer, see my previous post on this topic.

# My Favorite One-Liners: Part 87

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.

When teaching Calculus I, I use the following mantra throughout the semester. I heard this from my calculus instructor back in 1984, and I repeat it for my own students:

There are two themes of calculus: approximating curved things by straight things, and passing to limits.

For example, to find a derivative, we approximate a curved function by a straight tangent line and then pass to a limit. Later in the semester, to find a definite integral, we approximate the area under a curve by the sum of a bunch of straight rectangles and then pass to a limit.

For further reading, I’ll refer to this series of posts on what I typically do on the first day of my calculus class.

# My Favorite One-Liners: Part 86

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.

To get students comfortable with $i = \sqrt{-1}$, I’ll often work through a quick exercise on the powers of $i$:

$i^1 = i$

$i^2 = -1$

$i^3 = -i$

$i^4 = 1$

$i^5 = i$

Students quickly see that the powers of $i$ are a cycle of length 4, so that $i^5 = i \cdot i \cdot i \cdot i \cdot i$ is the same thing as just $i$. So I tell my students:

There’s a technical term for this phenomenon: aye-yai-yai-yai-yai.

# My Favorite One-Liners: Part 85

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 one-liner is one that I’ll use when I want to discourage students from using a logically correct and laboriously cumbersome method. For example:

Find a polynomial $q(x)$ and a constant $r$ so that $x^3 - 6x^2 + 11x + 6 = (x-1)q(x) + r$.

Hypothetically, this can be done by long division:

However, this takes a lot of time and space, and there are ample opportunities to make a careless mistake along the way (particularly when subtracting negative numbers). Since there’s an alternative method that could be used (we’re dividing by something of the form $x-c$ or $x+c$, I’ll tell my students:

Yes, you could use long division. You could also stick thumbtacks in your eyes; I don’t recommend it.

Instead, when possible, I guide students toward the quicker method of synthetic division:

# My Favorite One-Liners: Part 84

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.

Every once in a while, I’ll show my students that there’s a difficult way to do a problem that I don’t want them to do for homework. For example, here’s the direct derivation of the mean of the binomial distribution using only Precalculus; this would make an excellent homework problem for the Precalculus teacher who wants to torture his/her students:

$E(X) = \displaystyle \sum_{k=0}^n k {n \choose k} p^k q^{n-k}$

$= \displaystyle \sum_{k=1}^n k {n \choose k} p^k q^{n-k}$

$= \displaystyle \sum_{k=1}^n k \frac{n!}{k!(n-k)!} p^k q^{n-k}$

$= \displaystyle \sum_{k=1}^n \frac{n!}{(k-1)!(n-k)!} p^k q^{n-k}$

$= \displaystyle \sum_{k=1}^n \frac{n (n-1)!}{(k-1)!(n-k)!} p^k q^{n-k}$

$= \displaystyle \sum_{i=0}^{n-1} \frac{n (n-1)!}{i!(n-1-i)!} p^{i+1} q^{n-1-i}$

$= \displaystyle np \sum_{i=0}^{n-1} \frac{(n-1)!}{i!(n-1-i)!} p^i q^{n-1-i}$

$= \displaystyle np(p+q)^{n-1}$

$= np \cdot 1^{n-1}$

$=np.$

However, that’s a lot of work, and the way that I really want my students to do this, which is a lot easier (and which will be used throughout the semester), is by writing the binomial random variable as the sum of indicator random variables:

$E(X) = E(I_1 + \dots + I_n) = E(I_1) + \dots + E(I_n) = p + \dots + p = np$.

So, to reassure my students that they’re going to be asked to reproduce the above lengthy calculation, I’ll tell them that I wrote all that down for my own machismo, just to prove to them that I really could do it.

Since my physical presence exudes next to no machismo, this almost always gets a laugh.

# My Favorite One-Liners: Part 83

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 from calculus:

Let $f(x) = x^2 e^{3x}$. Find $f''(x)$.

We begin by finding the first derivative using the Product Rule:

$f'(x) = 2x e^{3x} + 3x^2 e^{3x}$.

Next, we apply the Product Rule again to find the second derivative:

$f''(x) = (2 e^{3x} + 6x e^{3x}) + (6x e^{3x} + 9x^2 e^{3x})$.

At this point, before simplifying to get the final answer, I’ll ask my students why the $6x e^{3x}$ term appears twice. After a moment, somebody will usually volunteer the answer: the first term came from differentiating $x^2$ first and then $e^{3x}$ second, while the other term came from differentiating $e^{3x}$ first and then $x^2$ second. Either way, we end up with the same term.

I then tell my class that there’s a technical term for this: Oops, I did it again.

While on the topic, I can’t resist also sharing this (a few years ago, this was shown on the JumboTron of Dallas Mavericks games during timeouts):

# My Favorite One-Liners: Part 81

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 hypothetically could appear in Algebra II or Precalculus:

Find the solutions of $x^4 + 2x^3 + 10 x^2 - 6x + 65 = 0$.

While there is a formula for solving quartic equations, it’s extremely long and hence is not typically taught to high school students. Instead, the techniques that are typically taught are the Rational Root Test and (sometimes, depending on the textbook) Descartes’ Rule of Signs. The Rational Root Test constructs a list of possible rational roots (in this case $\pm 1, \pm 5, \pm 13, \pm 65$) to test… usually with synthetic division to accomplish this as quickly as possible.

The only problem is that there’s no guarantee that any of these possible rational roots will actually work. Indeed, for this particular example, none of them work because all of the solutions are complex ($1 \pm 2i$ and $2 \pm 3i$). So the Rational Root Test is of no help for this example — and students have to somehow try to find the complex roots.

So here’s the wisecrack that I use. This wisecrack really only works in Texas and other states in which the state legislature has seen the wisdom of allowing anyone to bring a handgun to class:

What do you do if a problem like this appears on the test? [Murmurs and various suggestions]

Shoot the professor. [Nervous laughter]

It’s OK; campus carry is now in effect. [Full-throated laughter.]

# 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 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, $x = 88 + 10 = 98$.