# Arc length parameterization lying on a sphere

Show that if $\alpha$ is an arc length parameterization of a curve $C$
which lies on a sphere of radius $R$ about the origin then $$R^2 = (\frac{1}{\kappa(s)})^2+((\frac{1}{\kappa(s)})’\frac{1}{\tau(s)})^2.$$

I know I can write the unit tangent $T$, normal $N$, and the binormal $B$ as a linear combination $\alpha(s) = a(s)T(s)+b(s)N(s)+c(s)B(s)$ and try to determine what $a, b, c$ are, but how can I continue solving this?

#### Solutions Collecting From Web of "Arc length parameterization lying on a sphere"

First, we need that $\kappa$ and $\tau$ vanish nowhere. Clearly we have $\langle \alpha, T\rangle=0$, thus $\langle \alpha, T’\rangle+\langle T’,T’\rangle=0$, so we derive
$$\frac{1}{\kappa}=-\langle\alpha,N\rangle.$$
Now show that
$$\left(\frac{1}{\kappa}\right)’=\langle\alpha,\tau B\rangle, \quad\text{hence}\quad\left(\frac{1}{\kappa}\right)’\frac{1}{\tau}=-\langle\alpha,B\rangle.$$

Finale:
$$\alpha=-\frac{1}{\kappa}N-\left(\frac{1}{\kappa}\right)’\frac{1}{\tau}B,$$
now evaluate $\|\alpha\|^2$.

Michael

The key is to construct a vector $\alpha$ as @Michael Hoppe did.:

If $R$ is on a sphere. Then think of the vector $\alpha$ as exactly the vector that points from the centre of sphere to the curve $R$. So $\alpha$ is different from the original curve $R$ by just by a translation, hence they have the same tangent, curvature and torsion.

We can always express $\alpha$ as a linear combination (as suggested by Shifrin) of:

$$\alpha(s) = \lambda(s)T(s) + \mu(s)N(s)+\nu(s)B(s)$$

for some functions $\lambda$, $\mu$, $\nu$

(Note that $\alpha$ itself is not necessarily arc length parametrised.)

Then since the $R$ is on the sphere, and that $\alpha$ is from the centre to the sphere, we require that $\alpha$ is tangent to its tangent vector (notice that if $\alpha$ does not point from the centre to the sphere, it will not be tangent to its tangent vector $T$), that is,

$$<\alpha, T>=0$$

By differentiating above, we will find the expression for $\alpha$ as shown in @Michael Hoppe’s post:

$$\alpha=-\frac{1}{\kappa}N-\left(\frac{1}{\kappa}\right)’\frac{1}{\tau}B,$$

$\alpha$ must be of constant length. This implies hat the relationship in the problem holds.