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Let be $M$ a Riemannian manifold and $X,Y$ vector fields over $M.$ Now take $p\in M$ arbitrarily, my question is, how construc a variation $f:U\to M,$ $$U\subset \mathbb{R}^2,~~U=\{-\epsilon<t<1+\epsilon,~~-\epsilon<s<1+\epsilon, ~~\epsilon>0\}$$ satisfying $f(s,0)=f(0,0)$ for all $s, $ such that

$$\dfrac{\partial f}{\partial t}(0,1)=X(p),~~~\dfrac{\partial f}{\partial s}(0,1)= Y(p)~~~~~~ ??$$

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Take a totally convex neighborhood $U$ of $p$ in $M$ and choose $q\in U$ such that the unique geodesic $\gamma:[0,1]\rightarrow M$ joinning $q$ to $p$ satisfies $$\gamma(1)=p$$ $$\gamma'(1)=X(p).$$ Moreover, there exists $v\in T_qM$ such that $\exp_q(v)=\gamma(1)$. We have that $\exp_q$ is a diffeomorphism of some neighborhood of $0\in T_qM$ onto $U$. Then, there exists $Y\in T_v(T_qM)$ such that $$d(\exp_q)_v(Y)=Y(p)$$ Consider the curve $c(t)=v+tY$ in $T_qM$ and the variation $$f(s,t)=\exp_q(t\cdot c(s))$$ We have by definition that $\partial_tf(0,1)=\gamma'(1)=X(p)$. We have, at the same time, \begin{eqnarray} \partial_sf(0,1)&=&\frac{d}{ds}f(s,1)_{s=0}\\ &=&\frac{d}{ds}\exp_q(c(s))_{s=0}\\ &=&d(\exp_q)_{c(0)}\cdot c'(0)\\ &=&d(\exp_q)_v\cdot Y\\ &=&Y(p) \end{eqnarray}

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