# Doodling in Math

A few years ago, I blogged about this fun little doodle that students often make — and how another teacher and I found out the equation that “bounds” the figure. I honestly can’t remember if I ever posted how I got the answer. If I did and this is a repeat, apologies.

Tonight I wanted to see if I could re-derive it like I did before — and lo and behold I did. I’m curious if any of you have done it the way I did it, or if there are other ways you’ve learned to approach this problem. (There is a student who I had last year who created this amazing 3-d version of this using the edges of a cube and some string. I love the idea of asking — for this 3-d figure — what surface is generated by the intersections of these strings.)

We start out by having these lines which form a family of curves. But of course we’re not graphing all the lines. If we were, we’d get something more dense like this.

The main idea of what I’m going to do to find that curve… I’m going to pick two of those lines which are infinitely close to each other and find their point of intersection. That point of intersection will lie on the curve. (That’s the big insight in this solution.) But I’m not going to pick two specific lines — but instead keep things as general as possible. Thus when I find that point of intersection for those two lines, it will give me all the points of intersection for all the lines.

Watch.

First we pick two arbitrary lines.

We’ll have one line move down on the y-axis $k$ units (and thus over on the x-axis $k$ units). And the second line will be moved down on the y-axis just a tiny bit more (down an additional $e$ units). Yes, we are going to have that tiny bit, that $e$, eventually go to zero.

The two lines we have are:

$y=\frac{k-1}{k}(x-k)=\frac{k-1}{k}x-(k-1)$

$y=\frac{k+e-1}{k+e}(x-(k+e))=\frac{k+e-1}{k+e}x-(k+e-1)$

A little bit of algebra is needed to find the point of intersection. Setting the y-values equal:

$\frac{k-1}{k}x-(k-1)=\frac{k+e-1}{k+e}x-(k+e-1)$

And then doing some basic algebra:

$k^2+ke=x$

Now solving for $y$ we get:

$y=\frac{k-1}{k}(k)(k+e)-(k-1)$

$y=k^2+ke-2k-e+1$

So the point of intersection is:

$(k^2+ke, k^2+ke-2k-e+1)$

Here’s the kicker… Remember we wanted the two lines to be infinitely close together, right? So that means that we want $e$ to go to zero. Thus, our point of intersection of these infinitely close lines will be:

$(k^2, k^2-2k+1)$ or $(k^2,(k-1)^2)$.

Beautiful! And recall that we picked the lines arbitrarily. By varying $0\leq k \leq 1$ and plotting $(k^2,(k-1)^2)$, we can get any two lines on our doodle.

But I want an equation.

Simple. We know that $x=k^2$. Thus $x=\sqrt{k}$.*

Since $y=(k-1)^2$, we have $y=(\sqrt{x}-1)^2$

Let’s graph it to check.

Huzzah!!! And we’re done!

I wonder if I can do something similar with this cardioid:

I think I must (for funsies) do some investigation of “envelopes” this summer. I mean, Tina at Drawing on Math even introduces conics with these envelopes!

An extension for you. Do something with this 3d string-art.

*Of course you might be wondering why I don’t say $x=\pm \sqrt{k}$. Since $k$ is between 0 and 1, we know that $x$ must be positive.

# An Animal Problem

One of our math club leaders gave out this problem as the final problem of math club for the year. I had never seen it before, and after she handed it out, a number of math teachers were in a tizzy about finding the solution. So instead of planning for classes, we enjoyed working on this problem. But we got it! HUZZAH!

Here’s the problem:
In how many ways, counting ties, can eight horses cross the finishing line?

So we fully understand the problem, let me list all possibilities for three horses: Adam, Beatrice, and Candy. No, wait, those are better names for unicorns:

1st: A   2nd: B   3rd: C

1st: B   2nd: A   3rd: C

1st: A   2nd: C   3rd: B

1st: C   2nd: A   3rd: B

1st: B   2nd: C   3rd: A

1st: C   2nd: B   3rd: A

1st: AB (tie)   2nd: C

1st: AC (tie)   2nd: B

1st: BC (tie)    2nd: A

1st: A   2nd: BC (tie)

1st: B   2nd: AC (tie)

1st: C   2nd: BC (tie)

1st: ABC (tie)

That comes out to 13 different ways these horses unicorns can finish the race.

That’s the answer for 3 unicorns. What’s the answer for 8 unicorns?

(FYI: If you want to know if you’re on the right track… I have 541 for 5 unicorns…)

# Implicit Differentiation

Normally, I don’t have trouble teaching implicit differentiation. However, I’m never satisfied with what I do. I’m fairly certain that I have taught it four different ways in the past four years. But what’s common is that we do a lot of algebra. By the end, they can find $\frac{dy}{dx}$ for a relation like $\sin(xy)+y^3=2x+y$. Or something like that. But we lose the meaning of what we’re doing.

I realized we can do all this algebra, but it’s all procedure. And so there’s no real depth.

So today, after introducing implicit differentiation (including some visual motivation), I assigned 5 basic problems from the textbook. Each of the problems has an equation like $3y^3+x^2=5$ and students are asked to find $\frac{dy}{dx}$. My kids are going to go home today and struggle with it. We’ll spend about 20 or 25 minutes in our next class going over their solutions, talking about things, whatever.

And then… then… I’m going to hand out this sheet I wrote today.

[.doc, .pdf]
[if you're wondering, the graphs were made by the fabulous winplot which I adore... it can do implicit plotting!]

My kids found $\frac{dy}{dx}$ for homework. Now in class, my kids are going to interrogate what that means.

I am not sure yet how I’m going to structure the class. I think I might have us all work together on the first problem (#9), and then assign pairs to work on two of the remaining problems. And then I’ll pick one problem for each pair to present to the class. But what I’m truly happy about is that each problem gets kids to relate implicit differentiation to a graphical understanding of the derivative. It forces my kids to look at the derivative equation, and make connections to the original graph.

Although I’m proud of it, I’m honestly just not sure if this investigation is beyond the scope of my kids’s abilities. It pulls together a lot of concepts. I think it’ll work for them. This year I have a really really strong crew so I have faith. However, it’s an activity I’m going to have to give my kids time to do, and room to struggle. I know me, and I’m going to want to rush it, and I’m going to want to help them in ways that aren’t good for them. The struggle is where they’re going to learn in this, so I have to give it time and stay out.

I am in the middle of a hellish week, but if I have time, I’ll try to report back how it goes after we do it in class.

# Taking a Moment… in Calculus

In calculus, I’ve historically asked kids to take the derivative of:

$f(x)=\frac{2x^2+\sqrt{x}}{\sqrt{x}}$

and students will immediately go to the quotient rule. OBVIOUSLY! There’s a numerator and denominator. Duh. So go at it!

Unfortunately, this is VERY UNWISE because it leads to a lot more work. And I was sick of my kids not taking a moment to think: what are my options, and what might be the best option available? Also, kids generally found it hard to deal when we started mixing the derivative rules up!

So I came up with a sheet to address this and paired kids to work on it.

(I’ve also had kids think they can do some crazy algebra with $g(x)=\frac{x^2+1}{x+1}$. This sheet also helped me talk with kids individually about that.)

For a little context, my kids have only learned the power rule, the product rule, the quotient rule, and that the derivative of $e^x$ is $e^x$. They have not yet been formally exposed to the chain rule.

[.pdf, .doc]

# Believe it or not… a log question which was briefly stumping us

Hi all,

A teacher approached me with the following question.

The function $\ln(x^{-2})$ has a graph that looks like:

It makes sense that the function exists for all negative x values, because when you raise a negative number to the -2 power, you’re going to get a positive number. And you can take the natural log of a positive number.

Then the teacher said to consider the following function: $-2 \ln(x)$, and the graph looks like:

Notice that you can’t input negative x values, because the domain of natural log doesn’t allow for it.

Here’s the question.

According to the log rules/properties, we know that:

$\ln(a^b)=b\ln(a)$ (obviously).

So $\ln(x^{-2})=-2\ln(x)$. But the graphs are different.

We went a little crazy trying to figure out what’s going on… For about 3 minutes, we were having a great conversation. But we quickly converged on the little text that accompanies the log rules in any textbook… and this text says that these rules work but are only valid for $a>0$.

I kinda love this as an in-class exercise (I’ll probably forget this when I get to logarithms, but maybe posting it here will prevent me from forgetting it). Because it will force kids to (a) be confuzzled, (b) talk through ideas, (c) go back to the definition and qualifications for the log rules, and (d) see that these rules are indeed valid (we didn’t break math), but they are a bit more restrictive that we might have thought.

What I love is that $\ln(x^{-2})=-2\ln(x)$ isn’t actually an identity. But we are so used to using the rules blindly, robotically, that we never think about it. But for it to be a good mathematical statement, you need to qualify it! You need to say this is only an equivalence for $x>0$. This was a good reminder for us.

# My Favorite Test Question of All Time

In Calculus, we just finished our limits unit. I gave a test. It had a great question on it, inspired by Bowman and his limit activity.

Then I ask part (b)…

Which reads: “Scratch off the missing data. With the new information, now answer the question: What do you think the limit as x approaches 2 of the function is (and say “d.n.e.” if it does not exist)? Explain why (talk about what a limit is!).

So then they get this…

This is what I predicted. (And this was conjecture.) Almost all my kids are going to get part (a) right. I’ve done them well by that. However, with part (b), there are going to be two types of thinkers…

one kind of thinker, where they think “Mr. Shah wouldn’t give us this scratch off and this new data if the answer doesn’t change. So it has to change. What can it be? Clearly it has to be 2.5, because that’s the new information given to us. So I’m going to put 2.5 for the answer and then come up with some way to explain it, like saying since the function has a height of 2.5 when $x$ is 2, clearly it means the limit is 2.5.” (WRONG.)

the other kind of thinker, who will get the problem right and for the right reasons.

What’s the difference between the two kinds of thinkers? My guess: confidence. More than anything, this is a question that really gets at how confident kids are with the knowledge they have. You have to be pretty sure of yourself to come up with the right response, methinks.

My conjecture was pretty spot on. Let me tell you the responses are fascinating. So far my conjecture seems to be holding water. And it’s just the most intriguing thing to read the responses from students who got it wrong. The phrase that springs to mind is cognitive dissonance. There are a number of kids who are saying two totally contradictory things in their explanation, even from sentence to sentence, but they don’t recognize the contradiction. They’ll say “a limit is what y is approaching as x is approaching a number, and doesn’t have anything to do with the value of the function at the point” and then in the next sentence say “since the value of the function at x=2 is 2.5, we know the limit of the function as x approaches 2 must be 2.5.”

It’s a great question… my favorite test question of all time I think… but I wonder if that’s because of the scratch off.

I know we don’t tend to share student work often on blogs, but I asked and my kids were okay with me anonymously sharing their responses.

I don’t know exactly why I wanted to post student work. I don’t have anything specific I wanted to get out of it right now. But I know I was fascinated by it, and I figured y’all would be too.

But for me, it’d be interesting at the most basic level to just see the different ways our kids respond to questions in other classes… Even regular, basic non-writing problems! Just to see if anyone has ways to get kids to organize their work? Or if we could find a way to examine one student response to a question and throw around ideas about how to best proceed with the kid? Or talk about how we actually write feedback and what kind of feedback we give (and why)? Just a thought… Not for now, but something to mull over…

# When you get too lost in the algebra…

I was hunting for a book on my bookshelves when I got distracted and started browsing. In one book, I came across this great idea that I didn’t want to lose. So I thought I’d type it here in an attempt to remember.

One of the hard things about working with derivatives, for me, is that I can easily get caught up in the wonderful (to me, annoying to my kids) algebra. We have the chain rule, the product rule, the quotient rule, and strange and funky derivatives like the derivatives of the inverse trig functions. And I admit it. I love going overboard with these sorts of questions. There’s something really cool about being able to have an answer to a problem take up the length of a page. It looks cool, darnit! And when we get to this point in the curriculum, I often lose sight of the meaning of the derivative. The process takes precedence. And for weeks, we’re swimming (drowning?) in a sea of equations.

When I get to that point, I hope to remember to give my kids this problem:

Find the derivative of $\log(\log(\sin(x)))$. I’m confident that by the time I’m done with them, my kids will get $\frac{\cos(x)}{\sin(x)\log(\sin(x))}$.

But then I have to ask them to sketch a graph of $\log(\log(\sin(x)))$.

This great setup is on pages 64 and 65 of Ian Stewart’s Concepts of Modern Mathematics. He continues, describing what happened when he gave this problem to his class:

This caused great consternation, because it revealed that the formula didn’t make any sense. For any value of $x$, $\sin(x)$ is at most equal to 1, so $\log(\sin(x)) \leq 0$. Since logarithms of negative numbers cannot be defined, the value $\log(\log(\sin(x)))$ does not exist; the formula is a fraud.

On the other hand, the ‘derivative’ … does make sense for certain values of $x$

Some people might enjoy living in a world where one can take a function which does not exist, differentiate it, and end up with one that does exist. I am not one of them.

There’s a great moral here, about remembering that taking the derivative of a function means something. Yes, you can talk about composition of functions and domains and ranges and all that stuff, but that’s not the enduring understanding I would pull from this. It is: divorcing calculus from meaning and focusing on routine procedures is a dangerous road to travel — so one must always be vigilant.

It actually reminds me of one of my most favorite calculus problems, which to solve it needs one to stop focusing on procedure and start thinking. I would never give this to my calculus kids, but for the very high achieving AP Calculus BC kid, this might throw them for a loop (in a good way):

$\int_{0}^{\pi/2} \frac{dx}{1+(\tan x)^{\sqrt{2}}}$

I first saw this problem in Loren C. Larson’s Problem-Solving Through Problems (pages 32-33). I don’t quite want to share the solution in case you want to try it yourself. After the jump, I’ll throw down the answer (but not solution) so you can see if you got it right.