I am learning the Hessian comparison theorem on Riemannian manifold. It refers to the gradient of distant function.
Fix $x\in M$. Let $\rho(y)=d(y,x)$, and $r:I\to M$ is a minimal geodesic curve with $r(0)=x$. Namely $d(r(t),x)=|r'(0)|t$.
So can we get the conclusion $$\text{grad}\rho=r'$$
Any advice is helpful. Thank you.
$\nabla \rho$ is the dual of $d\rho$ under the metric $g$. But by Gauss's lemma, we know that
$$\left\langle\frac{\partial}{\partial \rho}, \frac{\partial }{\partial \rho}\right\rangle = 1,\ \ \ \ \left\langle\frac{\partial}{\partial \rho}, \frac{\partial }{\partial \theta_i}\right\rangle = 0,$$ where $\theta_i, i = 1, \cdots, n-1$ are coordinate under the geodesic polar coordinate. Thus
$$\nabla \rho = g^{11} \frac{\partial}{\partial \rho} + \sum_{i=1}^{n-1} g^{1i}\frac{\partial }{\partial \theta_i} = \frac{\partial}{\partial \rho}$$
as $g^{11} = 1$, $g^{1i} = 0$.
Note that $\frac{\partial}{\partial \rho}$ is in general not the $r'$ you wrote down. In particular, you did not tell what $|r'(0)|$ is. If $|r'(0)| = 1$, then they are the same.