# When will she pass me for the first time? [solution]

Recently, my dad posed the following question here:

My wife and I walk on a circular track, starting at the same point.  She does m laps in the time that it takes me to do n laps.  She walks faster than I do, so m > n.  After how many laps will she catch up with me again?

If you haven’t solved it yet, give it a crack. It’s a fun problem that has surprising depth.

Here’s my solution (in it, I refer to “mom” rather than “my wife” for obvious reasons!):

Since mom’s lap rate is $m$ laps per unit time, and dad’s lap rate is $n$ laps per unit time, in time $t$, mom goes $mt$ laps and dad goes $nt$ laps.

They meet whenever their distance (measured in laps) is separated by an integer number of laps $k$. That is, mom and dad meet when

$mt=nt+k, k\in\mathbb{Z}.$

This happens at time

$t=\frac{k}{m-n}.$

Mom will have gone

$mt=\frac{mk}{m-n}$

laps and dad will have gone

$nt=\frac{nk}{m-n}$

laps when they meet for the $k$th time.

And that’s it! That’s the general solution. This means that:

• At time $t=0$, dad and mom “meet” because they haven’t even started walking at all (they are $k=0$ laps apart).
• At time $t=\frac{1}{m-n}$, dad and mom meet for their first time after having started walking (they are $k=1$ lap apart). This is the answer to the problem as it was originally stated. Mom will have gone $mt=\frac{m}{m-n}$ laps and dad will have gone $nt=\frac{n}{m-n}$ laps when they meet for the first time.
• At time $t=\frac{2}{m-n}$, dad and mom meet for their second time (now $k=2$ laps apart).
• At time $t=\frac{k}{m-n}$, dad and mom meet for their $k$th time.

Here are two examples:

• If mom walks 15 laps in the time it takes dad to walk 10 laps, when they meet up for the first time, mom will have gone $\frac{m}{m-n}=3$ laps and dad will have gone $\frac{n}{m-n}=2$ laps.
• If mom walks 12 laps in the time it takes dad to walk 5 laps, when they meet up for the first time, mom will have gone $\frac{m}{m-n}=1\frac{5}{7}$ laps and dad will have gone $\frac{n}{m-n}=\frac{5}{7}$ laps.

Boom! Problem solved! 🙂

# Four ways to compute a probability

I have a guest blog post that appears on the White Group Mathematics blog here. (My first guest post!) Here’s a taste:

One thing I love about math, and particularly combinatorics and probability, is the fact that many methods exist for solving the same problem.

Each method may have its advantages. The advantage might be conceptual (as in “this makes most sense to me”) or the advantage might be computational (as in “this is the fastest way to do it”).

Discussing the merits of different methods is exactly what math class is for!

For example, check out this typical probability question that could appear in a Precalculus course:

The Texas Ranger pitching staff has 5 right-handers and 8 left-handers. If 2 pitchers are selected at random to warm up, what is the probability that at least one of them is a right-hander?

In fact, it’s one I use in my own Precalculus course and it generated a great class discussion. In teaching it this past year, I ended up showing students four ways to do the problem this year! Here they are…

For the epic conclusion of this post, visit White Group Mathematics. 🙂

# Composing power functions

I presented the following example in my Precalculus classes this past week and it bothered students:

Let $f(x)=4x^2$ and $g(x)=x^{3/2}$. Compute $f(g(x))$ and $g(f(x))$ and state the domain of each.

As usual, I’ll give you a second to think about it yourself.

..

..

Done yet?

.

$f(g(x))=4(x^{3/2})^2=4x^3, x\geq 0$

$g(f(x))=(4x^2)^{3/2}=8|x|^3, x\in\mathbf{R}$

The reason that the first one was unsettling, I think, is because of the restricted domain (despite the fact that the simplified form of the answer seems not to imply any restrictions).

The reason the second one was unsettling is because they had forgotten that $\sqrt{x^2}=|x|$. It seems to be a point lost on many Algebra 2 students.

# Integration by parts and infinite series

I was teaching tabular integration yesterday and as I was preparing, I was playing around with using it on integrands that don’t ‘disappear’ after repeated differentiation. In particular, the problem I was doing was this:

$\int x^2\ln{x}dx$

Now this is done pretty quickly with only one integration by parts:

Let $u=\ln{x}$ and $dv=x^2dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{x^3}{3}$. Rewriting the integral and evaluating, we find

$\int x^2\ln{x}dx = \frac{1}{3}x^3\ln{x}-\int \left(\frac{x^3}{3}\cdot\frac{1}{x}\right)dx$

$=\frac{1}{3}x^3\ln{x}-\int \frac{x^2}{3}dx$

$= \frac{1}{3} x^3 \ln{x} - \frac{1}{9} x^3 + c$.

But I decided to try tabular integration on it anyway and see what happened. Tabular integration requires us to pick a function $f(x)$ and compute all its derivatives and pick a function $g(x)$ and compute all its antiderivatives. Multiply, then insert alternating signs and voila! In this case, we choose $f(x)=\ln{x}$ and $g(x)=x^2$. The result is shown below.

$\int x^2\ln{x}dx = \frac{1}{3}x^3\ln{x}-\frac{1}{12}x^3-\frac{1}{60}x^3-\frac{1}{180}x^3-\cdots$

$= \frac{1}{3}x^3\ln{x} - x^3 \sum_{n=0}^\infty \frac{2}{(n+4)(n+3)(n+2)(n+1)} +c$

If I did everything right, then the infinite series that appears in the formula must be equal to $\frac{1}{9}$. Checking with wolframalpha, we see that indeed,

$\sum_{n=0}^\infty \frac{2}{(n+4)(n+3)(n+2)(n+1)} = \frac{1}{9}$.

Wow!! That’s pretty wild. It seemed like any number of infinite series could pop up from this kind of approach (Taylor series, Fourier series even). In fact, they do. Here are just three nice resources I came across which highlight this very point. I guess my discovery is not so new.

# Great NCTM problem

Yesterday I presented this problem from NCTM’s facebook page:

Solve for all real values of $x$:

$\frac{(x^2-13x+40)(x^2-13x+42)}{\sqrt{x^2-12x+35}}$

Don’t read below until you’ve tried it for yourself.

Okay, here’s the work. Factor everything.

$\frac{(x-8)(x-5)(x-7)(x-6)}{\sqrt{(x-5)(x-7)}}=0$

Multiply both sides by the denominator.

$(x-8)(x-5)(x-7)(x-6)=0$

Use the zero-product property to find $x=5,6,7,8$. Now check for extraneous solutions and find that $x=5$ and $x=7$ give you $\frac{0}{0}\neq 0$ and $x=6$ gives $x=\frac{0}{\sqrt{-1}}=\frac{0}{i}=0$. This last statement DOES actually hold for $x=6$ but we exclude it because it’s not in the domain of the original expression.The original expression has domain $(-\infty,5)\cup(7,\infty)$. We could have started by identifying this, and right away we would know not to give any solutions outside this domain. The only solution is $x=8$.

Does this seem problematic? How can we exclude $x=6$ as a solution when it (a) satisfies the equation and (b) is a real solution? This is why we had such a lively discussion.

But this equation could be replaced with a simpler equation. Here’s one that raises the same issue:

Solve for all real values of x:

$\frac{x+5}{\sqrt{x}}=0$

Same question: Is $x=-5$ a solution? Again, notice that it DOES satisfy the equation and it IS a real solution. So why would we exclude it?

Of course a line is drawn in the sand and many people fall on one side and many fall on the other. It’s my impression that high-school math curriculum/textbooks would exclude $x=-5$ as a solution.

Here’s the big question: What does it mean to “solve for all real values of x“? Let’s consider the above equation within some other contexts:

Solve over $\mathbb{Z}$:

$\frac{x+5}{\sqrt{x}}=0$

Is $x=-5$ a solution? No, I think we must reject it. If we try to check it, we must evaluate $\frac{0}{\sqrt{5}}$ but this expression is undefined because $\sqrt{5}\notin\mathbb{Z}$. Here’s another one:

Solve over $\mathbb{Z}_5$:

$\frac{x+5}{\sqrt{x}}=0$

Is $x=-5$ a solution? No. Now when we try to check the solution we get $\frac{0}{\sqrt{5}}=\frac{0}{\sqrt{0}}=\frac{0}{0}$ which is undefined.

The point is that, if we go back to the same question and ask about the solutions of $\frac{x+5}{\sqrt{x}}=0$ over the reals, and we check the solution $x=-5$, we must evaluate $\frac{0}{\sqrt{-5}}$ which is undefined in the reals.[1]

So in the original NCTM question, we must exclude $x=6$ for the same reason. When you test this value, you get $\frac{0}{i}$ on the left side which YOU may think is 0. But this is news to the real numbers. The reals have no idea what $\frac{0}{i}$ evaluates to. It may as well be $\frac{0}{\text{moose}}$.

There’s a lot more to say here, so perhaps I’ll return to this topic another time. Special thanks to all the other folks on facebook who contributed to the discussion, especially my dad who helped me sort some of this out. Feel free to comment below, even if it means bringing a contrary viewpoint to the table.

________________________

[1] This last bit of work, where we fix the equation and change the domain of interest touches on the mathematical concept of algebraic varieties, which I claim to know *nothing* about. If someone comes across this post who can help us out, I’d be grateful! 🙂

# Trapezoid Problem (take 2)

Am I blundering fool? You decide!

It turns out the trapezoid construction I posted earlier today is trivial. Thanks to Alexander Bogomolny for pointing out my error. The construction is quite easy (and it does not require the height), and I quote Alexander:

No, you do not need the height.

Imagine a trapezoid. Draw a line parallel to a side (not a base) from a vertex not on that side. In principle, there are two such lines. One of these is inside the trapezoid. This line, the other side (the one adjacent to the line) and the difference of the bases form a triangle that could be constructed with straightedge and compass by SSS. Next, extend its base and draw through its apex another base. That’s it.

So I redid my Geogebra Applet and posted it here. It’s not really worth checking out, though, since it’s indistinguishable from my previous applet. (In truth, you can reveal the construction lines and see the slight differences.) But I did it for my own satisfaction, just to get the job done correctly :-). Anyway, three cheers for mathematical elegance, and for Alexander Bogomolny*.

*check out Alexander’s awesome blog & site, a true institution in the online math community!

# Vi Hart’s Blog

It’s high time I gave a bit of press to Vi Hart’s Blog. If you haven’t checked it out, do so right away. It’s brilliant.  A number of people have pointed me to her blog, including one of my Calc students. Her little math videos are fresh, funny, and insightful. Denise, at Let’s Play Math, gave her some press too, which is what reminded me to finally make this post. Here’s the video Denise highlighted (the most recent of Vi’s creations):

This is particularly appropriate because there were a couple of us in our math department discussing this very question: In total, how many gifts are given during the 12 Days of Christmas song? It’s a nice problem, perfect for a Precalculus student. Or any student. Here’s a super nice explanation of how to calculate this total, posted at squareCircleZ. But before you go clicking that link, take out a piece of scrap paper and a pencil and figure it out yourself!

Here’s another nice video from Vi Hart:

You could spend a lot of time on her site. Here’s another awesome video. I’ll have to have my Precalculus class watch this one when we do our unit on sequences and series.

And you’ve got to love the regular polyhedra made with Smarties ,  right?

Plus, Vi Hart plays StarCraft, which is awesome too.  Back in the day, I really loved playing. I haven’t played in a while, and I certainly haven’t tried SC 2 yet, because then I’d never grade my students’ papers.

Bottom line is, you need to check out all the playful stuff Vi Hart is doing at her blog. Happy Wednesday everyone!