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I can't understand how parametrics equations are found. For example, I realize that the parametrization of the curve given by the intersection of the plane $\ 2x+2y+z=2$ and $z=x^2+y^2$ is:

  • $x=-1+\cos(t)$
  • $y=-1+\sin(t)$
  • $z=6-2\cos(t)-2\sin(t)$
  • $0\leqslant t\leqslant2\pi$

Or that the surface of $x^2+y^2=2$ delimited by $x^2+y^2+z^2=4$ is:

  • $x=\sqrt2\cos(u)$
  • $y=\sqrt2\sin(u)$
  • $z=v$
  • $0\leqslant u\leqslant\pi/2$ and $0\leqslant v\leqslant\sqrt2$

But what is the step by step to find those equations?

Tianlalu
  • 5,177
J.Doe
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  • Are you sure you got these right? For the first problem $x^2+y^2=2-2x-2y\implies (x+1)^2+(y+1)^2=4$, which is inconsistent with your first two bullet points. – J.G. Dec 06 '18 at 06:36
  • Now I see it.I got them from a list and didn't check. What would be a correct parametrization then? – J.Doe Dec 06 '18 at 21:36
  • You need to double the trigonometric functions' coefficients. – J.G. Dec 06 '18 at 23:30

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