Day: April 22, 2010

MEAN (grrr) value of a function!

In calculus today I was talking about how to find the average height of a function. Some kids just have a hard time understanding the concept. I always show them a few functions on certain intervals and I ask them what they think the average height would be. Just to initially test their intuition on the concept.

Some see it, and understand it; some don’t. All certainly have trouble articulating why they chose that value.

So I have two things that work for me, when explaining this. There’s some handwaving, but the focus is on the idea, and building intuition.

The first thing is we talk about how we would approximate the average temperature somewhere:

We take a bunch of temperature readings, and we add them together and divide by the number of readings.

How do you make it more accurate?

MORE READINGS!

How do you make it more accurate?

INFINITY OF READINGS!

What helps us deal with infinities and infinitessimals?

CALCULUS!

So that’s how we get started.

Then when I want them to understand the formula — f_{avg}=\frac{\int_a^b f(x)dx}{b-a} — I give them a little dumb, cute story.

So an EVIL mathematician has an almost 2 dimensional fish tank. Really thin. Sad for the fish. Which are almost 2 D. And the mathematician likes to lay a strip of plastic on top of the water,  and constrain the fish in these weird shapes.

(In this case, the mathematician is constraining the fish in an x^2 from [0,1].)

You come along and want to GIVE THE FISH WHAT THEY WANT: a normal rectangular water to swim in.

So you yank the plastic strip away, and what happens to the water?

IT ALL LEVELS OUT!

What shape does it make?

A RECTANGLE!

Does the amount of water change?

NO!

What’s the height of the rectangle?

THE AVERAGE HEIGHT OF THE FUNCTION!

So by then, we have on the board:

And since the amount of water didn’t change, they know that the area of the red rectangle and the area of the blue rectangle are the same.

That makes sense to them.

I then threw this up and almost all of ’em got it!

So that’s my way of building their intuition when it comes to average height of a function. It’s not like it’s hard for them to apply the formula, but I think this little thing makes it more conceptually manageable. And if they forget the formula, they can just do the “fish tank problem.”

Solutions to parabola problem

I’m going to post the way I worked out the two recent problems that I posted. Today I’m going to focus on the second problem first.  Other people had different solutions they threw in the comments to the original posts, so you can look there too.

FIRST PROBLEM:

Statement: A particle is moving along the curve y=x^2-x at a constant speed of 2\sqrt{10}. When it reaches the point (2,2), you know \frac{dx}{dt}>0. Find the value of \frac{dy}{dt} at that point.

Solution: I imagined the particle moving along the curve, and it being played on a film. The particle follows this path and is going at a constant speed:

So then I said: we only care about the particle around (2,2), so I mentally zoomed in near that point:

So we don’t care about the rest of the picture. The particle is actually moving in a straight line in the area we care about, and this line is y=3x-4 (we found the equation of the line tangent to the original curve at the point (2,2)). So this greatly simplifies how I had to think about the problem. Where we care about things, the particle is moving in a straight line at a rate of 2\sqrt{10}.

So then I thought about the velocity vector for the particle — moving in the direction of the line at a rate of 2\sqrt{10}. And this vector is composed of the velocity in the x-direction and the velocity in the y-direction: \frac{dx}{dt} and \frac{dy}{dt}.

And we just have to use the last piece of information that we haven’t used… That the ratio of height/length of this triangle is 3 (the slope of the line — the direction of velocity — is 3). So we can solve this with a bunch of different ways, but I found the easiest to just make similar triangles and solve:

(I calculated that the hypotenuse of the second triangle was \sqrt{10}.) Clearly we can deduce that dy/dt=6.

In essence, this is the exact same method that other people used to solve it, but it took me to actually zoom in and picture what was going on with the particle to figure this problem out so that I conceptually had mastered it.

The way I approached the second problem comes later.

Parametrization, Parabolas, Calculus, OH MY!

Okay, so a second problem in a row! This one is a straight up calculus one, from the 2008 AP Calculus BC exam — multiple choice section. The teacher of that class asked me if I could work this problem — and I admit I struggled. She showed me her solution, and then I left thinking “it couldn’t be that hard…”

When trying to fall asleep today, I started thinking of it and I was able to solve it in a different way.

Without any more preamble, if you care to try your hand at this:

A particle is moving along the curve y=x^2-x at a constant speed of 2\sqrt{10}. When it reaches the point (2,2), you know \frac{dx}{dt}>0. Find the value of \frac{dy}{dt} at that point.

As usual, feel free to throw your thoughts, solutions, etc. in the comments below, if you want. I bet for many of you this will be super easy, but for the few of you who struggle through it (sigh) like me, you might find it actually frustratingly enjoyable.

Oh, and also throw down there if you get stuck and care to see my solution… It’ll motivate me to actually type it up in a timely fashion.