Mathematics Asked by Nono4271 on January 16, 2021
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 $tilde{y} = y$
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!
The surface of the cylinder can be parametrized as ..
$$(x, y, z) = (costheta, sintheta, t)$$ intersect with the plane ...
$$ z = x tanphi implies t=costheta tan phi $$
so the intersection can be parametrized as ...
$$(x, y, z) = (costheta, sintheta, costheta tan phi)$$
Correct answer by WW1 on January 16, 2021
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