Parametrization of a circle on a sphere

[TimNguyen]

Homework Statement

Parametrize a circle of radius r on a sphere of radius R>r by arclength.

Homework Equations

Circle Equation: (cos [theta], sin[theta], 0)

The Attempt at a Solution

I don't know if the professor is tricking us, but isn't the parametrization just

Circle: (r*sin[theta]/s, r*cos[theta]/s, s), where s is the arc length?

 

[hunt_mat]

The circumference of a circle is 2\pi, so when you find the length of your circle, you should obtain length=1.

Mat

 

[HallsofIvy]

You can always set up your coordinate system so that the circle is at a fixed z.

In spherical coordinates we have x= Rcos(\theta)sin(\phi), y= R sin(\theta)sin(\phi) and z= R cos(\phi). At fixed z, \phi= cos^{-1}(z/R) and it is easy to show that
sin(cos^{-1}(z/R))= \frac{\sqrt{R^2- z^2}}{R^}

It is also true that with radius r, we have r^2+ z^2= R^2 so that R^2- z^2= r^2 and so
sin(\phi)= sin(cos^{-1}(z/R))= \frac{r}{R}

That is, x= r cos(\theta), y= r sin(\theta), and z= \sqrt{R^2- r^2}, a constant.

But, since \theta is in radians, the arclength of a segment over angle \theta is s= r\theta so that
\theta= \frac{s}{r}

 

[TimNguyen]

Oh, there should not be a [theta] in the trigonometric functions, but rather the value (s/r).

Hence, the parametrization of a circle would be:

(r*cos(s/r), r*sin(s/r), sqrt[R^2 - r^2])

Thanks for all the help!