Riemannian geometry: ...Why is it called 'Exponential' map?

The exponential map $exp_{p}:T_{p}M \to M$ given a suitable $v \in T_{p}M$, returns $p$, displaced along the geodesic uniquely determined by $(p,v) \in TM$ for unit "time".

So, what does the above have to do with the familiar concept/s of exponentiation?
Why does this map carry this name?

• In the case of matrix groups, it really is the matrix exponential.
– user27126
Jan 10, 2014 at 8:39

The function $f(t)=\mathrm{e}^{at}$ can be viewed as the solution of the initial value problem: $$x'=ax, \quad x(0)=1.$$ More generally, if $A\in\mathbb R^{n\times n}$ and $u_0\in\mathbb R^n$, then the solution of the initial value problem: $$x'=Ax, \quad x(0)=u_0,$$ is $u(t)=\mathrm{e}^{tA}u_0$, where $\mathrm{e}^{tA}$ is the exponential of the matrix $A$. It $A$ possessed imaginary eigenvalues, then $u(t)$ returns to a multiple of itself, for suitable initial $u_0$.