Let $S$ be a surface. Let $\alpha:[0,L] \rightarrow S$ be a curve. A family of curves $\{\alpha_t:[0,L]\rightarrow S\}_{t \in (-\epsilon,\epsilon)}$ is said to be a variation of $\alpha$ if $\alpha_0=\alpha$, $\alpha(0)=\alpha_t(0)$ and $\alpha(L)=a_t(L)$ for any $t \in (-\epsilon,\epsilon)$. Moreover, for any fixed $s \in [0,L]$, $\alpha_t(s)$ is a smooth function in $t$. The vector field $V(s)=\frac{d}{dt}|_{t=0}\alpha_t(s)$ is called the variation vector field.
I want to show that $\alpha$ is a geodesic iff for any variation $\{\alpha_t:[0,L]\rightarrow S\}_{t \in (-\epsilon,\epsilon)}$ of $\alpha$, we have $$\frac{d}{dt}|_{t=0} Length(\alpha_t)=0$$ where $Length$ denotes the arc length of the curve.
For the 'if' direction, it is the direct consequence of the first variation formula for arc length.
What I cannot understand it is 'only if' direction. By the assumption and the first variation formula, we have $$\int_{0}^LV(s) \bullet(\alpha''(s))^Tds=\int_{0}^LV(s) \bullet\alpha''(s)ds=0$$ for any variation vector field, where $(\bullet)^T$ denotes the component of the vector on the tangent plane. By my lecture note, it immediately concludes that $(\alpha''(s))^T=0$. I am confused how it can jump that fast to conclusion. Anyone have any idea?