Since $\beta=f^{-1}df$,
$$
\frac d{dt}f(\gamma(t))
=f(\gamma(t))\,\beta_{\gamma(t)}(\dot\gamma(t)).
$$
The <fundamental theorem of calculus> and the <chain rule> then give
$$
F'(t)
=e^{-\int_0^t\gamma^*\beta}
\left(\frac d{dt}f(\gamma(t))-f(\gamma(t))\beta(\dot\gamma(t))\right)=0.
$$
Thus $F$ is constant.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 115 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-115/2/d/solution