Help! Parametric equation and distance...

[world of math]

HI everyone,
I got a problem I'm stuck on. It says:

A circle of radius 1 rolls around the circumference of a larger circle of radius 4. The epicycloid traced by the point on the circumference of the smaller circle is given by x=5cost-cos5t and y=5sint-sin5t. Find the distance travedled by the point on one complete trip about the larger circle.

So I graphed that parametric equation and it looks like a four-leafed clover. I'm not sure how exactly to start. Am I supposed integrate from a to b sqrt(1+(dx/dy)^2), like finding the length of the graph? How should I start this?

Thank you for any help!

 

[galactus]

parametric arc length is given by:

\(\displaystyle \L\\\int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt\)


There's something about the epicycloid that may prove easier

The resulting integral is a monster. So, just use

\(\displaystyle \L\\8(4+1)=40\) instead. It'll give you the same answer.

 

[world of math]

Thanks! I knew something was wrong with the formula I was using. However, I just noticed something strange. dx=-5sint+5sint, which cancels out to be 0. Same with dy, which is 5cost-5cost, canceling out to be 0 again. I think I did something wrong somewhere else again...any ideas? Thanks so much!

 

[galactus]

I just gave you a shortcut. I edited my post.

And yes, you did something wrong with the differentiation.

\(\displaystyle \L\\\frac{d}{dt}[5cos(t)-cos(5t)]=5sin(5t)-5sin(t)\)

\(\displaystyle \L\\\frac{d}{dt}[5sin(t)-sin(5t)]=5cos(t)-5cos(5t)\)

 

[world of math]

Cool, I learned something new about epicycloids (my teacher never taught me about them before). I see now, thanks for all the help!

 

[galactus]

Actually, I was looking at your integral and it ain't bad.

At first glance I thought it was a monster but it simplifies nicely.

Let's expand the derivatives:

\(\displaystyle \L\\(5sin(5t)-5sin(t))^{2}+(5cos(t)-5cos(5t))^{2}\\=25sin^{2}(5t)-50sin(5t)sin(t)+25sin^{2}(t)+25cos^{2}(t)-50cos(t)cos(5t)+25cos^{2}(5t)\)

The \(\displaystyle sin^{2}+cos^{2}\) terms result to 1 and we have:

\(\displaystyle \L\\50-50(\underbrace{sin(5t)sin(t)+cos(t)cos(5t)}_{\text{cos(4t)}})\\=50(\underbrace{1-cos(4t)}_{\text{8sin^2tcos^2t}})\\=400sin^{2}(t)cos^{2}(t)\)

\(\displaystyle \L\\\sqrt{400sin^{2}(t)cos^{2}(t)}\\=20sin(t)cos(t)\\=10sin(2t)\)

So, we have:

\(\displaystyle \L\\40\int_{0}^{\frac{\pi}{2}}sin(2t)dt=\fbox{40}\)

 

[galactus]

You had better go with the integral. I am not 100% sure that easy formula will work all the way around. I worked that out rather quickly. It may not be a general case formula. I should amend my post.

I have been playing around with it and I believe a general rule appears to be \(\displaystyle \L\\8(a+b)\). Where a and b are the radii of the respective circles.