I'd like some hint for this problem:
Show that every unit-speed curve $ \alpha : I \subset \mathbb{R} \longrightarrow \mathbb{R}^3$, whose image is in the sphere of radius $R$ , has curvature $\displaystyle k_{\alpha}(s)\geq \frac{1}{R}$ .
Differential Geometry – Curvature of a Curve on a Sphere
curvaturecurvesdifferential-geometry
Related Solutions
I would do it like this:
We have
$c(t) = \int_0^t r(u)du, \tag{1}$
whence
$\dot c = \dfrac{dc}{dt} = r(t), \tag{2}$
and since $r(t)$ lies in the sphere of radius $R$, we have
$\Vert r(t) \Vert = R \tag{3}$
for all $t$. This shows that the tangent vector $\dot c$ has constant magnitude $R$:
$\Vert \dot c(t) \Vert = \Vert r(t) \Vert = R, \tag{4}$
holding for all $t$. But $\Vert \dot c \Vert$ is the rate of change of arc-length $s$ along $c(t)$ with respect to the parameter $t$:
$\dfrac{ds}{dt} = \Vert \dot c(t) \Vert = R, \tag{5}$
and (5) implies
$\dfrac{dt}{ds} = \dfrac{1}{R} \tag{6}$
along $c(t)$ as well. By (2) and (3) we see that the unit tangent field $\mathbf t$ to $c(t)$ is
$\mathbf t = \dfrac{\dot c}{R} = \dfrac{r(t)}{R}; \tag{7}$
the curvature $\kappa$ of $c(t)$ is thus given by the Frenet-Serret equation
$\dfrac{d\mathbf t}{ds} = \kappa \mathbf n, \tag{8}$
where $\Vert \mathbf n \Vert = 1$. From (6)-(8), using the chain rule for derivatives,
$\kappa \mathbf n = \dfrac{d\mathbf t}{ds} = \dfrac{dt}{ds} \dfrac{d\mathbf t}{dt} = \dfrac{1}{R}\dfrac{\dot r}{R} = \dfrac{\dot r}{R^2}. \tag{9}$
Since $r(t)$ is a unit speed curve, $\Vert \dot r \Vert = 1$, so taking the norm of each side of (9) yields
$\kappa = \dfrac{1}{R^2}, \tag{10}$
that is, the curvature of $c(t)$ is $R^{-2}$. QED!!!
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
You want to show that the torsion is zero. Note that If $\mathbf{r}$ is the unit radius vector to the curve then $$\mathbf{t}=\frac{d \mathbf{r}}{ds}$$ Now $\mathbf{t} \cdot \mathbf{r}=0$ so on differentiating we get
$$\frac{d \mathbf{t}}{ds}\mathbf{r}+\mathbf{t}\frac{d \mathbf{r}}{ds}=0$$ or $$\kappa \mathbf{n}\cdot \mathbf{r}+ \mathbf{t}\cdot \mathbf{t}=0$$ (Note that $\kappa\neq 0$, since otherwise we have a straight line.) So $$\mathbf{n}\cdot \mathbf{r}=-\frac{1}{\kappa}$$ differentiating this and using that $\kappa$ is constant we get
$$(-\kappa \mathbf{t}+\tau \mathbf{b})\cdot \mathbf{r}+\mathbf{n}\cdot \mathbf{t}=0$$ which simplifies to $$\tau \mathbf{b}\cdot \mathbf{r}=0$$ If $\tau=0$ we are done so assume
$$\mathbf{b}\cdot \mathbf{r}=0$$
and differentiate which gives
$$-\tau \mathbf{n}\cdot \mathbf{r}+\mathbf{b}\cdot \mathbf{t}=0$$
$$-\tau \mathbf{n}\cdot \mathbf{r}=0$$ But $\mathbf{n}\cdot \mathbf{r}=-\frac{1}{\kappa}$ so again $\tau=0$.
Best Answer
Regard $\alpha\colon I \to \mathbb{S}^2_R$ as a map from the interval $I$ to the sphere of radius $R$. Let $\langle \cdot, \cdot \rangle$ denote the dot product. Here are two hints:
(1) Consider the quantity $\frac{d}{ds}\langle \alpha(s), \alpha'(s)\rangle$. What do you know about the quantity $\langle \alpha(s), \alpha'(s)\rangle$ based on the geometry of the sphere?
(2) Since you're trying to prove an inequality, perhaps you know of some inequality involving the inner product $\langle v, w\rangle$ of two vectors.