Given the vertical cylinder with radius 1 ($x^2 + y^2 = 1$) oriented with an outward pointing normal and its intersection with the plane $z = xtan(\phi)$
Consider the coordinates $\tilde{x} = \sqrt{x^2 + z^2}$ and $\tilde{y} = y$
- Show that this intersection is an ellipse
- Find the curvature, geodesic curvature, and normal curvature of this ellipse (I do not actually need help solving these, but need help to find the parameterization so I can start solving them)
So, I did the following
$\tilde{x} = \sqrt{x^2 + (x tan(\phi))^2} = \sqrt{x^2 + x^2*tan^2(\phi)} = \sqrt{x^2 (1 + tan(\phi))} = xsec(\phi)$
and of course $\tilde{y} = y$
and thus the intersction is $\tilde{x}^2 + \tilde{y}^2 = 1$
and to make it look like an ellipse you can do
$\frac{x^2}{cos^2(\phi)} + \frac{y^2}{1^2} = 1$
where $ a = cos(\phi)$ and $b = 1$
Now that I have found the ellipse, I need to work on finding the curvature, geodesic curvature, and the normal curvature. To do that the hint was given to parameterize the ellipse in terms of the roational angle $\theta$ around the cylinder. So, I looked online and I saw that a general paraneterization for an ellipse is
$x = acos(t)$ and $y = bsin(t)$
However, upon running this information by my instructor I was told that this is only the ellipse, if the ellipse was planar. Therefore my paramterization needs a z-component. I was told that the parameterization is usually something along the lines of
$(x, y, z) = (cos(\theta), sin(\theta), 0)$
But for this I need to rotate through phi in the xz plane to get my parameterization and I'm not exactly sure what to do/what that looks like. Any advice? Thanks!
Best Answer
The surface of the cylinder can be parametrized as ..
$$(x, y, z) = (\cos\theta, \sin\theta, t)$$ intersect with the plane ...
$$ z = x \tan\phi \implies t=\cos\theta \tan \phi $$
so the intersection can be parametrized as ...
$$(x, y, z) = (\cos\theta, \sin\theta, \cos\theta \tan \phi)$$