This is an exercise from Do Carmo's Differential Geometry which I doubt whether or not my working is correct.
Let $\alpha: I\to R^3$ be a regular parametrized curve (not necessarily by arc length) and let $\beta:J\to R^3$ be a reparametrization of $\alpha(I)$ by the arc length $s=s(t)$, measured from $t_0\in I$. Let $t=t(s)$ be the inverse function of $s$ and set $d\alpha/dt=\alpha'$, $d^2\alpha/dt^2=\alpha''$, etc. Prove that the curvature of $\alpha$ at $t\in I$ is
$$k(t)=\frac{|\alpha'\wedge\alpha''|}{{|\alpha'|}^3}$$
My attempt:
We first simplify the RHS: $\frac{|\alpha'\wedge\alpha''|}{{|\alpha'|}^3}=\frac{|\alpha'||\alpha''|}{{|\alpha'|}^3}=\frac{|\alpha''|}{{|\alpha'|}^2}$, where I used the definition $|a\wedge b|=|a||b|\cos\theta$.
Then we find the unit tangent vector of $\alpha$: $T(t)=\frac{\alpha'}{|\alpha'|}$.
So $\frac{dT}{ds}=\frac{dT}{dt}\frac{dt}{ds}=T'(t)\frac{1}{|\alpha'|}$.
Now it is very tempting to write $T'(t)=\frac{\alpha''}{|\alpha'|}$, then we have $\frac{dT}{ds}=\frac{\alpha''}{{|\alpha'|}^2}$ and since $\frac{dT}{ds}=kn$ where $n$ is the unit normal vector, so $k=\frac{|\alpha''|}{{|\alpha'|}^2}$.
I really doubt my correctness, the part that I found it hard is to find $T'(t)$. And I think my working $T'(t)=\frac{\alpha''}{|\alpha'|}$ is wrong since the denominator might also be a function of $t$. But I am not sure how to fix it. Helps are really greatly appreciated! Thanks.
Best Answer
Let $s'(t)=\left|\gamma'(t)\right|$. Then, the second derivative with respect to arclength is $$ \frac1{s'}\left[\frac{\gamma'}{s'}\right]'=\frac{\gamma''s'-\gamma's''}{s'^3}\tag{1} $$ Since $\frac{\gamma'}{s'}$ is a unit vector, the vector in $(1)$ is perpendicular to it. Therefore, it has the same length as $$ \frac{\gamma'}{s'}\times\frac{\gamma''s'-\gamma's''}{s'^3}=\frac{\gamma'\times\gamma''}{\left|\gamma'\right|^3}\tag{2} $$ Therefore, $$ \kappa=\frac{\left|\gamma'\times\gamma''\right|}{\left|\gamma'\right|^3}\tag{3} $$