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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?

Lays
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  • Hint: since $|c|=r$, $\langle c,\dot c\rangle=0$ and $\langle \ddot c, \dot c\rangle=-1$. – Michael Hoppe Sep 30 '13 at 05:54
  • @Michael Hoppe: I'll buy $\langle c, \dot c \rangle = 0$, and I'll buy $\langle \ddot c, c \rangle = -1$, but the market for $\langle \dot c, \ddot c \rangle = -1$ seems pretty bearish right now! ;) – Robert Lewis Sep 30 '13 at 06:39

2 Answers2

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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.

taper
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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

Michael Hoppe
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  • Thanks a bunch, Michael! So to finish I just have to compute $|\alpha|^2 = \alpha \dot\ \alpha$ and differentiate it? – Lays Oct 01 '13 at 06:44
  • We know that $|\alpha|^2=R^2$, now determine $$|-\frac{1}{\kappa}N-\left(\frac{1}{\kappa}\right)'\frac{1}{\tau}B|^2$$ using $\langle N,N\rangle=\langle B,B\rangle=1$ and $\langle N,B\rangle=0$. – Michael Hoppe Oct 01 '13 at 07:52