# Differential Geometry Question

1. Feb 10, 2008

### mXCSNT

1. The problem statement, all variables and given/known data
Assume that $$\tau(s) \neq 0$$ and $$k'(x) \neq 0$$ for all $$s \in I$$. Show that a necessary and sufficient condition for $$\alpha(I)$$ to lie on a sphere is that $$R^2 + (R')^2T^2 = const$$ where $$R = 1/k$$, $$T = 1/\tau$$, and $$R' = \frac{dr}{ds}$$

2. Relevant equations
$$\alpha(s)$$ is a curve in R3 parametrized by arc length
$$k = curvature = |\alpha''|$$
$$\tau = torsion = -\frac{\alpha' \times \alpha'' \cdot \alpha'''}{k^2}$$ (note sign; this is opposite of some conventions)

3. The attempt at a solution
I've approached this from 2 directions, but I haven't gotten them to meet. First, a necessary and sufficient condition is that $$|\alpha - P|$$ is constant, where P is the center of the circle. Alternatively, $$(\alpha - P) \cdot \alpha' = 0$$.
And I've expanded out $$R^2 + (R')^2T^2 = const$$ to get
$$\frac{(\alpha' \times \alpha'' \cdot \alpha''')^2 + (\alpha'' \cdot \alpha''')^2}{(\alpha'' \cdot \alpha'')(\alpha' \times \alpha'' \cdot \alpha''')^2} = const$$
Also, I'm going to guess that the const on the right hand side is some function of the radius of the sphere, maybe the square of it (which would be $$(\alpha - P) \cdot (\alpha - P)$$), because what else is constant in a sphere?

But I don't know where to go from here. I'm just looking for a hint at an avenue of approach, please nothing specific.

2. Feb 10, 2008

### mXCSNT

So if I assume that the "const" in question is the square of the radius, that could make R and (R')T the lengths of respective legs of a right triangle whose hypotenuse is the radius. I know that R is the radius of curvature, so drawing a diagram I'm guessing that the center P is
$$P = Rn - (R')Tb$$
(where n is the normal vector $$\alpha''/|\alpha''|$$ and b = $$t \times n$$ is the binormal vector).
So then the sphere radius I want to test, $$(\alpha-P) \cdot (\alpha-P)$$, becomes $$(\alpha - Rn + (R')Tb) \cdot (\alpha - Rn + (R')Tb)$$
$$= \alpha \cdot \alpha - 2 R \alpha \cdot n + 2 R' T \alpha \cdot b + R^2 + (R')^2T^2$$
Now if I assume that $$R^2 + (R')^2T^2$$ is constant I need to show that $$\alpha \cdot \alpha - 2 R \alpha \cdot n + 2 R' T \alpha \cdot b$$ is constant to show the sphere radius is constant. Am I on the right track?

3. Feb 10, 2008

### mXCSNT

Revised guess for center P:
$$P = \alpha + Rn - (R')Tb$$
so now the constant radius follows immediately and I simply have to show that P itself is constant.