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I am trying to understand why the hessian of a function from a surface in $\mathbb{R}$, $\left. \frac{d^2}{dt^2} \right\rvert_{0} (f \circ \alpha)(t)$ does not depend of $\alpha$.

In my book you can see: ($\alpha(t)=X(u(t),v(t)), X$ is a parametrization of the surface ) $$\left. \frac{d^2}{dt^2} \right\rvert_{0} (f \circ \alpha)(t)= \left. \frac{d^2}{dt^2} \right\rvert_{0} (f \circ X)(u(t), v(t))=\left. \frac{d}{dt} \right\rvert_{0} ((f \circ X)_u \dot\ u'(t)+(f \circ X)_v \dot\ v'(t))=$$ $$[(f \circ X)_{uu} \dot\ u'(0)+(f \circ X)_{uv} \dot\ v'(0)]\dot\ u'(0)+(f \circ X)_{u} \dot\ u''(0) + [(f \circ X)_{vu} \dot\ u'(0)+(f \circ X)_{vv} \dot\ v'(0)]\dot\ v'(0)+(f \circ X)_{v} \dot\ v''(0)$$ And now it says that $(f \circ X)_{u} \dot\ u''(0)+(f \circ X)_{v} \dot\ v''(0)=0 $ since the point we are working on is a critical point. I do not understant this last equality. Any suggestion?

davidivadful
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    It seems the statement under consideration is made under the additional assumption that $f$ has a critical point at $X(u(0), v(0))$.. If thats true then by elementary reasoning $(f\circ X)_u= 0 = (f\circ X)_v$ in that point. – Thomas Jul 26 '17 at 18:41
  • Yes. As you say we have that critical point. Could you please explain how to reach that equality? – davidivadful Jul 26 '17 at 21:04
  • And of course it depends on $\alpha$. But if you fix $\alpha(0)$ and $\alpha'(0)$, then it does not depend on $\alpha$. But yes, you need to assume you're at a point where the first derivatives of $f$ are $0$. – Ted Shifrin Jul 26 '17 at 21:40

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