Proposition 3.6 of do Carmo's Riemannian Geometry, why $r(1)=l(\gamma)$?

Question

Picture below is from the do Carmo's Riemannian geometry. I don't know why $r(1)=l(\gamma)$ ?

Answer 1

Fix $p \in M$ and take some normal geodesic ball $B_R(p) \subset M$ with polar coordinates \begin{align} E \colon (0,R)\times S &\to B_R(p) \\ (r,v) & \mapsto \exp_p(rv) \end{align}

Fix $q \in B_R(p)$, which has polar coordinates $(r(q),v(q))$. Any curve $c \colon [0,1] \to B_R(p)$ joining $p$ to $q$ has some polar coordinates $(r(t),v(t))$. The fact that $c$ joins $p$ and $q$ is equivalent to saying that $r(0) = 0$ and $r(1) = r(q)$, $v(1) = v(q)$. Now, as the ball is supposed normal, the only geodesic joining $p$ and $q$ is, in coordinates, $\gamma(t) = (t\cdot r(q), v(q))$. If follows that $r(q) = d(p,q) = l(\gamma)$ and thus, $r(1) = r(q) = l(\gamma)$.