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How can I parametrize Viviani's Curve ?

$\textbf{Viviani’s curve}$ is the intersection of the unit sphere with center $(\frac{-1}{2},0,0)$ and the cylinder with center $(0,0,0)$ and radius $1/2$

Paramatrization of the cylinder is

$(\frac{1}{2}cos(t),\frac{1}{2}sin(t),?)$

and the unit sphere

$(cos(\theta)cos(\phi)-\frac{1}{2},cos(\theta)sin(\phi),sin(\theta))$

I think, i have to use some trigonometric equations but i see only 1 in this case.

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Let $a=1/2$.

The sphere is given by $(x+a)^2+y^2+z^2=1$ and the cylinder is given by $x^2+y^2=a^2$.

As you have noticed, the cylinder gives $x=a\cos t$, $y=a\sin t$. Plug these into the sphere and get: $$ 1 = (x+a)^2+y^2+z^2 = a^2+2a^2\cos t + a^2\cos^2 t + a^2\sin^2 t + z^2 = 2a^2(1+\cos t)+z^2 $$ Now solve for $z$ using that $a=1/2$. You may want to look up the half-angle formula.

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  • $\begingroup$ then, z is equal to $sin(\theta)$ or what ? $\endgroup$ – derivative Feb 7 '14 at 12:56
  • $\begingroup$ @derivative, forget the parametrization of the sphere. $\endgroup$ – lhf Feb 7 '14 at 12:57
  • $\begingroup$ since $1/4=(1/2cost)^2+(1/2sint)^2=(cos\theta cos\phi-1/2)^2+(cos\theta sin\phi)^2$ i get $cos^2\theta-cos\theta cos\phi=0$$\quad$$\Rightarrow$$sin\phi=sin\theta$, $cos\phi=cos\theta$ So, finally i get $(cos^2\theta-1/2,cos\theta sin\theta, sin\theta)$, Ok thanks. $\endgroup$ – derivative Feb 7 '14 at 13:19

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