curvesdifferential-geometry
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.
I think this is a lot easier than what you think: Take (1) above: $$\frac{\tau }{\kappa }=\frac{d}{ds}\left (\frac{\dot \kappa}{\tau \kappa^2}\right ) \tag 1$$
Rewrite this in terms of $\sigma$ and $\rho$ $$\frac{\rho}{\sigma }=\frac{d}{ds}\left (\sigma (-\dot \rho)\right ) = -\frac{d}{ds}\left (\dot{\rho} \sigma \right ) \tag 2$$
Now you're almost done.
Take the derivative of the LHS of your expression:
$$\frac{d}{ds}\left(\rho^2 + (\dot{\rho} \sigma)^2 \right) = 2\rho \dot{\rho} + 2(\dot{\rho} \sigma)\frac{d}{ds}(\dot{\rho} \sigma)$$ But we just figured out what that last term is in terms of $\rho$ so substitute: $$\frac{d}{ds}\left(\rho^2 + (\dot{\rho} \sigma)^2 \right) = 2\rho \dot{\rho} + 2(\dot{\rho} \sigma)\left(\frac{-\rho}{\sigma}\right)$$ You can now see that $$\frac{d}{ds}\left(\rho^2 + (\dot{\rho} \sigma)^2 \right) = 0$$
Or in other words $\rho^2 + (\dot{\rho} \sigma)^2$ is a constant (it must be positive, do you see why?). And if you call that constant $r^2$ you are done.
Put $a = \rho N + \sigma \dot{\rho} B$
We have just shown that this vector has a constant length of $r$. You can use the Frenet equations to show that $\dot{a} = -T = -\dot{\gamma}$
Then it is clear that $\gamma + a$ is a constant vector (the center of the circle) and the path $\gamma$ is at a constant distance $r$ from it.
For Viviani's curve put $$ \gamma(t) = \left(\frac{r}{2}(1 + \cos t), \frac{r}{2} \sin(t), r \sin(t/2)\right)$$ Which you can get from Wikipedia or Mathworld (remember we have a constant $r$ for our radius). then $$ \gamma'(t) = \frac{r}{2}\left(-\sin (t), \cos(t), \cos(t/2)\right)$$ $$ \gamma''(t) = -\frac{r}{2}\left(\cos (t), \sin(t), \sin(t/2)/2\right)$$ This is different from what you have. You need to calculate this out to finish.
For example I get $||\gamma'' \times \gamma'|| = \frac{r^2}{8\sqrt{2}}\sqrt{13 + \cos (t)}$
Can you take it from there?
A curve lies on a sphere if its distance from a fixed point remains constant. If we optimistically assume that the centre of the sphere is the origin, we can just look at $\lVert \alpha(t) \rVert^2$, but it turns out if you do this calculation, it doesn't come out right. Instead, we don't like the look of the $1/2$ in the first coordinate, so try $\lVert \alpha(t)-(-1/2,0,0) \rVert^2$: $$ \lVert \alpha(t)+(1/2,0,0) \rVert^2 = (\cos^2{t})^2+\sin^2{t}\cos^2{t}+\sin^2{t} = \cos^4{t} +(1-\cos^2{t})(1+\cos^2{t}) = 1$$ using difference of two squares. Thus $\alpha(t)$ lies on the sphere of radius $1$ with centre $(-1/2,0,0)$ If we hadn't been able to guess, we would have had to try subtracting a general point $(a,b,c)$ and trying to make the answer constant.
The curve will lie on a cylinder with centre on the $z$-axis if its distance from the $z$-axis is constant. The square of this is given by $$ (\cos^2{t}-1/2)^2 + \sin^2{t}\cos^2{t} = \cos^4{t}-\cos^2{t}+ \frac{1}{4} + \cos^2{t}-\cos^4{t} = \frac{1}{4}, $$ and as the question said, the curve lies on the cylinder centred on the $z$-axis with radius $\sqrt{1/4}=1/2$. And the intersection of the sphere and the cylinder turn out to be this curve: here's a Mathematica plot.
As you noticed, the curvature doesn't tell you much about a curve: this is because a curve that lies on a surface has both curvature in the surface ("geodesic curvature") and curvature that is the extrinsic curvature of the surface ("normal curvature"). When all you have is the curve, you can't separate the two (since it lies on plenty of different surfaces).
Best Answer
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.