second fundamental form: unknown derivation

Question

I'm unable to compute the second line in the proof double dot $r=r_{uu}\dot{u}^2+2r_{uv}...$

I think that I should use the chain rule together with the Leibnitz rule, but I do not know how.

Answer 1

Yes, we must use (multivariable) chain rule here. Let $r = r(u(t), v(t))$. Then the chain rule says $$\frac{dr}{dt} = \frac{\partial r}{\partial u}(u(t),v(t))\frac{du}{dt}(t) + \frac{\partial r}{\partial v}(u(t),v(t))\frac{dv}{dt}(t),$$ or, more succinctly, $$\dot{r} = r_u\dot{u}+r_v\dot{v}.$$ Now, the time derivative of $r_u\dot{u}$ is $$\frac{dr_u}{dt}\,\dot{u}+r_u\ddot{u} = (r_{uu}\dot{u}+r_{uv}\dot{v})\dot{u}+r_u\ddot{u} = r_{uu}\dot{u}^2+r_{uv}\dot{u}\dot{v}+r_u\ddot{u}$$ where we have used product rule and chain rule again. Similarly, the time derivative of $r_v\dot{v}$ is $$\frac{dr_v}{dt}\dot{v}+r_v\ddot{v} = (r_{vu}\dot{u}+r_{vv}\dot{v})\dot{v}+r_v\ddot{v} = r_{vv}\dot{v}^2+ r_{vu}\dot{u}\dot{v}+r_v\ddot{v}.$$ Combining the two and using equality of mixed partials ($r_{uv} = r_{vu}$) we find that $$\ddot{r} = r_{uu}\dot{u}^2+2r_{uv}\dot{u}\dot{v}+r_{vv}\dot{v}^2+r_u\ddot{u}+r_v\ddot{v},$$ exactly as claimed.