Arc Length of Lissajous Curves, or Pretty, Pretty Pictures!

In MV Calc today, we were learning about arc length. In 3D, if you have parametric equations defining a curve, you can find the arc length by calculating:

L=\int_a^b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt

I asked them to calculate the length of this curve, which will repeat itself, over and over and over:

x(t)=\cos(3t), y(t)=\sin(5t), z(t)=\cos(2t)

If you graph it, it looks like this (it’s a 3D Lissajous curve):

Vodpod videos no longer available.
more about “2008-10-27_1849“, posted with vodpod

Which is awesome! They had to figure out the limits of integration (the function will go back to it’s original starting point when t=2\pi, so the limits of integration are from 0 to 2\pi. And we had to use our fnInt function on our graphing calculators to actually calculate the length. But it was cool.

During the class, I starting thinking of all the extensions and projects that could come out of this. For example, we could have students study x(t)=\cos(at), y(t)=\sin(bt), z(t)=\cos(ct). If a, b, and c are all rational numbers, we can prove that the curve will repeat itself. However, as soon as we make one of them irrational, we can prove the curve will not repeat itself. Look at this video to see how cool it looks!

Vodpod videos no longer available.
more about “2008-10-27_1852“, posted with vodpod

What if all three (a, b, and c) are irrational? What constraints do we need if we want the function to repeat? (The answer won’t be tough, I imagine, but worth exploring.)

Also, on the subway ride home, I wondered if we could come up with an explicit formula (rather than parametric) for a surface containing every point on the curve. (A harder question, for sure.)

And another: what is the smallest volume you can design to enclose a curve which does repeat, like x(t)=\cos(3t), y(t)=\sin(5t), z(t)=\cos(2t). Since sine and cosine bounce between 1 and -1, I will say that your volume had better be better than 2^3.



Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s