I have been reading about Jacobi fields in do Carmo's book "Riemannian geometry", and I have a little question.

Let $(M,g)$ be a riemannian manifold and let $p\in M$ and $v\in T_p M$ such that $\exp_p v$ is defined.

It begins with a parametrized surface $$f(t,s)=\exp_{p}tv(s)\qquad 0\leq t\leq 1,\, -\varepsilon\leq s\leq\varepsilon$$ where $v(s)$ is a curve in $T_p M$ such that $v(0)=v$ and $v'(0)=w\in T_v(T_p M)$.
By the Gauss lemma we have $$(d\exp_p)_v w=\frac{\partial f}{\partial s}(1,0)$$ So far so good. Then it goes on by saying

"It is convenient to extend our objective slightly and study the field $$(d\exp_p)_{tv}(tw)=\frac{\partial f}{\partial s}(t,0)$$ along the geodesic $\gamma(t)=\exp_p(tv)$."

Since $\gamma$ is a geodesic, we have $\nabla_{\dot\gamma}\dot\gamma=0$. But in the book they claim that we also have $$\nabla_{\dot\gamma(t)}\frac{\partial f}{\partial t}(t,s)=0$$ for all $(t,s)$. Why is that?
I see why this holds for $s=0$, since $v(0)=v$.
But why would that be for $s\neq 0$?

Any help would be very much appreciated!

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    $\begingroup$ It is not true that $\dot{\gamma}(t)=v$ for all $t$. In fact, the above phrase does not even make any sense, as $v$ is a tangent vector at $p$ whereas $\gamma(t)\ne p$ for $t\ne0$. $\endgroup$ – Amitai Yuval Mar 16 at 14:03
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    $\begingroup$ On the other hand, letting $\gamma_s(t):=f(t,s)$, the path $\gamma_s$ is a geodesic for any $s$. By construction, we have $\dot{\gamma}_s(t)=\partial f/\partial t$, and the claim follows. $\endgroup$ – Amitai Yuval Mar 16 at 14:07
  • $\begingroup$ @AmitaiYuval thank you, i have edited my question. So we have $\nabla_{\dot\gamma}=\nabla_{\dot\gamma_0}$. Why would that imply $\nabla_{\dot\gamma_0}\dot\gamma_s=0$ for all $s$? $\endgroup$ – Pink Panther Mar 16 at 14:30
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    $\begingroup$ No, this is not what do Carmo says. The direction of derivative, $\dot{\gamma}(t)$ is taken with respect to the varying value of $s$. Otherwise it makes no sense. $\endgroup$ – Amitai Yuval Mar 16 at 15:12
  • $\begingroup$ @AmitaiYuval I think I got it now: In the statement (this is a quote from the book) "In fact, since $\gamma$ is a geodesic, we have for all $(t,s), \frac{D}{dt}\frac{\partial f}{\partial t}=0$", the 'since' seems kinda redundant then, right? Because it seems clear that $f(t,s)=\gamma_s(t)$ is a geodesic for fixed $s$. Also the $\frac{D}{dt}\dot\gamma_s$ then becomes $\nabla_{\dot\gamma_s}\dot\gamma_s$, right? $\endgroup$ – Pink Panther Mar 17 at 0:26

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