Let $r(t)$ be a unit speed curve on a sphere $x^2+y^2+z^2=R^2$. Show that the curve $c(t)=\int^t_0 r(u) du$ has a constant curvature $\frac{1}{R^2}$
I am still a little shaky with this stuff, so I don't know if I'm going about it the right way.
Curvature is given $k=\frac{|c''(t)\times c'(t)|}{|c'(t)|^3}$. So,
$c'=r(t)-r(0)$
$c''=r'(t)$
$\implies \frac{|r'(t) \times (r(t)-r(0))|}{|r(t)|^3}$
Am I going about this the right way? I don't know where else to go. I assumed I couldn't just let $r(t)=(R\cos^2(t),R\cos(t)\sin(t),R\sin(t))$ since we don't know more about the given curve. Any help in getting further would be much appreciated.
Edit: I am using the local formula for curvature and not that of unit speed, since I don't know if we can make any assumptions about $c$ based on $r$ having unit speed.
Best Answer
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!!!