Show that the curvature of simple arc C is given by $\kappa(q)=\frac{\Vert f'(a)\times f''(a)\Vert}{\Vert f''(a)\Vert^3}$

Question

I need help with this problem:

Let $C$ be a simple arc smoothly parametrized by $f:D\subset\mathbb{R}\rightarrow\mathbb{R}^3$. Show that at the point $\mathbf q\in C$, where $\mathbf q=f(a), a\in D$, the curvature is given by $$\kappa(\mathbf q)=\frac{\Vert f'(a)\times f''(a)\Vert}{\Vert f'(a)\Vert^3}$$ Hint: apply 2.8.20 and 2.8.21.

2.8.20 $f'(t)=v(t)T_f(t)$

2.8.21 $f''(t)=v'(t)T_f(t)+(v(t))^2\kappa_f(t)N_f(t)$

I don't know how to begin, I find differential geometry confusing. Can you pleas guide me on how to begin showing this? Thanks.

Answer 1

Just cross-multiply $f'(t)$ and $f''(t)$ using your two formulas. You get $T\times T=0$ and $T\times N=B$ (the unit binormal vector). That is, $$ f'(t)\times f''(t)=(v(t))^3\kappa(t)B(t) $$ Taking the norm, $$ \|f'(t)\times f''(t)\|=(v(t))^3\kappa(t)\|B(t)\|=(v(t))^3\kappa(t). $$ Now just divide by $v(t)^3$ and use the fact that $v(t)=\|f'(t)\|$ ('speed').