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PostPosted: Wed Oct 12, 2016 5:13 pm 

Joined: Sat Nov 14, 2015 6:32 am
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Location: Melbourne, Australia
Let $\mathbb{F}_1: x^2+y^2+z^2=16$ and $\mathbb{F}_2: (x-2)^2 + y^2 =4$. Consider the curve $c: \mathbb{F}_1 \cap \mathbb{F}_2$. You are asked to parametrize the vertical projection $\gamma$ of $c$ onto the $xy$ plane.

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PostPosted: Sun Oct 16, 2016 8:43 am 
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The surface $F_2=\big\{(x,y,z)\in\mathbb{R}^3\; |\; (x-2)^2 + y^2 =4,\; z\in\mathbb{R}\big\}$ is a cylinder with base the circle $p=\big\{(x,y,0)\in\mathbb{R}^3\; |\; (x-2)^2 + y^2 =4\big\}$ and axis perpedicular to $xy$-plane. The curve obtained by the intersection of the cylinder $F_2$ by any surface which passes throughout the cylinder, has as projection onto the $xy$- plane the circle $p$. It easy to check that the sphere $F_1=\big\{(x,y,z)\in\mathbb{R}^3\; |\; x^2 + y^2+z^2 =4^2\big\}$ passes throughout the cylinder $F_2$.(*) So, a parametrization of the requested projection is \[\overrightarrow{p}(t)=(2+2\cos{t},\,2\sin{t},\,0)\,,\quad t\in[0,2\pi]\,.\]
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proj_cylinder.png [ 51.41 KiB | Viewed 640 times ] proj_cylinder.png [ 51.41 KiB | Viewed 640 times ]

(*) I'll give a more strict (mathematically) proof for this, after the answer of the question below.
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Riemann, allow me to continue with this: Find a parametrization of the curve $c=F_1 \cap F_2$.

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