Curvature of a curve lying on a sphere?

I'm going to use the "dot" notation for derivatives with respect to $s$, so that $\dot{\mathbf r}(s) = \mathbf r'(s)$, etc. Then, assuming that the "natural parameter" $s$ is the arc-length along $\mathbf r(s)$, we have:

Since $\mathbf r(s)$ is contained in the unit sphere,

$\mathbf r(s) \cdot \mathbf r(s) = 1, \tag{1}$

from which it readily follows upon differentiation with respect to the arc-length parameter $s$, that

$\mathbf r(s) \cdot \dot{\mathbf r}(s) = 0; \tag{2}$

furthermore we see that

$\dot{\mathbf r}(s) \cdot \dot{\mathbf r}(s) = 1 \tag{3}$

by virtue of the fact that $s$ is the arc-length along $\mathbf r(s)$; $\dot {\mathbf r}(s) = \mathbf T(s)$ is the unit tangent to $\mathbf r(s)$, and a member of the Frenet-Serret apparatus (or frame) of $\mathbf r(s)$. It then follows from (1)-(3) and the elementary properties of the vector cross product that $\mathbf r(s) \times \dot{\mathbf r}(s)$ is itself a unit vector and is orthogonal to both the unit vectors $\mathbf r(s)$ and $\dot{\mathbf r}(s)$, i.e.,

$(\mathbf r(s) \times \dot{\mathbf r}(s)) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)) = 1, \tag{4}$

$(\mathbf r(s) \times \dot{\mathbf r}(s)) \cdot \mathbf r(s) = (\mathbf r(s) \times \dot{\mathbf r}(s)) \cdot \dot{\mathbf r}(s) = 0. \tag{5}$

(1)-(5) show that $\mathbf r(s)$, $\dot{\mathbf r}(s)$, and $\mathbf r(s) \times \dot{\mathbf r}(s)$ themselves form an orthonormal frame at each point of $\mathbf r(s)$; we expand the Frenet-Serret normal

$\mathbf N(s) = (1 / \kappa)\dot{\mathbf T}(s) = (1 / \kappa) \ddot {\mathbf r}(s) \tag{6}$

in terms of this frame, starting by noting that differentiation of (2) yields

$\dot{\mathbf r}(s) \cdot \dot{\mathbf r}(s) + \mathbf r(s) \cdot \ddot {\mathbf r}(s) = 0, \tag{7}$

which by virtue of (3) and (6) reads

$1 + \kappa \mathbf N(s) \cdot \mathbf r(s) = 0. \tag{8}$

(7) also has the advantage of showing that $\ddot {\mathbf r}(s) \ne 0$, so that $\kappa = \Vert \ddot {\mathbf r}(s) \Vert \ne 0$ and $\mathbf N(s)$ is well-defined at all points of $\mathbf r(s)$. (8) shows that the component of $\mathbf N(s)$ along the unit vector $\mathbf r(s)$ is in fact

$\mathbf N(s) \cdot \mathbf r(s) = -\dfrac{1}{\kappa}. \tag{9}$

The $\dot{\mathbf r}(s)$ component of $\mathbf N(s)$ vanishes for all $s$, since

$\mathbf N(s) \cdot \dot{\mathbf r}(s) = \mathbf N(s) \cdot \mathbf T(s) = 0, \tag{10}$

and finally the component of $\mathbf N(s)$ along $\mathbf r(s) \times \dot{\mathbf r}(s)$ is

$\mathbf N(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)) = \dfrac{1}{\kappa}\ddot{\mathbf r}(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)), \tag{11}$

so that $\mathbf N(s)$ may in fact be written

$\mathbf N(s) = -\dfrac{1}{\kappa} \mathbf r(s) + \dfrac{1}{\kappa}\ddot{\mathbf r}(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s))(\mathbf r(s) \times \dot{\mathbf r}(s)), \tag{12}$

and using (12) to compute $\mathbf N(s) \cdot \mathbf N(s) = 1$ yields

$\dfrac{1}{\kappa^2}(1 + (\ddot{\mathbf r}(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)))^2 = 1, \tag{13}$

which is easily re-arranged:

$\kappa = \sqrt{1 + (\ddot{\mathbf r}(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)))^2}, \tag{14}$

the requisite result. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!