An involute of a shortest path in a strictly convex surface in $\mathbb{R}^3$. Consider a smooth closed surface $\Sigma$ in $\mathbb{R}^3$ of positive Gaussian curvature, homeomorphic to a sphere. Here $U$ is the interior of the convex hull of $\Sigma$. When $c:[0,l]\rightarrow \Sigma$ is a minimizing geodesic of unit speed in it, then we defines $$a :[0,l]\rightarrow \mathbb{R}^3,\ a(t)=c(t)-t c'(t)$$ and $$f(t)=d(x,a(t))$$ for any fixed $x\in \Sigma$. Here $d$ is the intrinsic distance on $ \mathbb{R}^3- U$.

Problem : Then prove that $f(t)$ is increasing.

Thank you in advance.
 A: Edit: the current answer only shows that the distance function on $\mathbb{R}^3$ is increasing. The OP wants to show that the distance on $\mathbb{R}^3 \setminus U$ is increasing.
Let $g(t)$ be the squared distance $\langle x - a(t), x-a(t) \rangle$. Its derivative is
$$
\begin{align*}
 g'(t) &= 2 \langle x - c(t) + tc'(t), - c'(t) + c'(t) + tc''(t) \rangle \\
       &= 2 t\langle x - c(t) + tc'(t), c''(t) \rangle \\
       &= 2 t \langle x - c(t), c''(t) \rangle.
\end{align*}
$$
In the last step we used that $\langle c'(t), c''(t) \rangle = 0$ since $c$ has unit speed.
Since $c(t)$ is a geodesic, the acceleration vector $c''(t)$ is parallel with the surface normal of $\Sigma$ at $c(t)$. So the tangent plane at $c(t)$ is given by $\langle p - c(t), c''(t)\rangle = 0$. Also note that $c''(t)$ is an inward pointing normal vector at $c(t)$.
Finally, by the convexity of $\Sigma$, all points of $\Sigma$ lie on one part the tangent plane. Since $c''(t)$ is inward pointing, $\langle x - c(t), c''(t)\rangle\geq 0 $ and consequently
$$
  g'(t) = 2 t \langle x - c(t), c''(t) \rangle \geq 0.
$$
This proves that the Euclidean distance function is increasing.
A: Fix a point $x$ And if $c$ is a minimizing geodesic in $\Sigma$, we define $a(t)=c(t)-tc'(t)$ and $f(t)=d(x,a(t))$. We define $b$ s.t. $$\bigg\langle
 a(t+\varepsilon ),\frac{-c''(t)
}{|c''(t) |}  \bigg\rangle \frac{-c''(t) }{|c''(t) | } + a(t)=
b(\varepsilon )$$
If $\gamma=\gamma_\varepsilon$ is a shortest path from $x$ to
$b(\varepsilon)$, then assume that $$ \gamma(0)=x,\ \gamma(s)=y $$
and $$ y\in T_{c(t)}\Sigma \bigcap \gamma $$
Hence by Earnie060's answer, we have $$ f(t) = {\rm length}\ \gamma_0\leq {\rm length} \bigg(
\gamma|[0,s]\cap [y a(t)] \bigg)\leq {\rm length } \gamma $$
Hence $$\frac{d}{d\varepsilon }{\rm length}\ \gamma\geq 0 $$
Further note that
$$|f(t+\varepsilon )- {\rm length}\ \gamma |<C\varepsilon^2$$ for some $C>0$ and some $0<
\varepsilon <\delta$.
Hence $$ \frac{d}{d \varepsilon } f(t+\varepsilon ) = \frac{d}{dt}
{\rm length}\ \gamma $$
which completes the proof.
