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Let $F:\mathbb{R}\to \mathbb{R}^n$ be differentiable. Let $f:\mathbb{R}^n \to \mathbb{R}$ be continuously differentiable and such that the composition $g(t)=f(F(t))$ exists. If $F'(t_0)$ is tangent to a level surface of $f$ at $F(t_0)$, show that $g'(t_0)=0$

My initial thought was just chain rule and the tangent statement so that $g'(t_0) = \nabla f(F(t_0))\cdot F'(t_0)=0$. But this just doesn't feel right. The book also says that the tangent plane to a level set $S$ at $x_0$ is the set of all points $x$ satisfying $$\nabla f(x_0)\cdot (x-x_0) = 0$$

So my thought is that

$F'(t_0)$ is tangent to a level surface of $f$ at $F(t_0)$

translates to mean that $F'(t_0)\in \{ x | \nabla f(F(t_0))\cdot (x-F(t_0)) = 0 \}$.

From here I have that $\nabla f(F(t_0)) \cdot (F'(t_0) - F(t_0)) = 0$ equivalently that $$g'(t_0) = \nabla f(F(t_0)) \cdot F(t_0)$$

But I'm not sure how to make this last bit go to zero.

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As Thomas pointed out just remember that the tangent plane at the point needs to be shifted to the point first because the gradient (as it is a vector) begins at the origin. Therefore, what you want is

$$\nabla f(F(t_0))\cdot( (F(t_0) + F'(t_0)) - F(t_0)) = \nabla f(F(t_0))\cdot F'(t_0) = 0$$

You essentially had the solution!

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On a level surface $\{f=c\}$ the function $f$ is, by definition, constant. So, whenever $\gamma:(-a,a ) \rightarrow \mathbb{R}^n$ is a curve contained in the surface $\{f=c\}$, $f\circ\gamma$ is a constant function from $(-a,a)$ to $\mathbb{R}$ and, consequently, has derivative $(f\circ\gamma)^\prime = \langle \nabla f,\gamma^\prime\rangle = 0$ (everywhere).

The difference between my $\gamma$ and your $F$ is that your $F$ is assumed to be tangent to $\{f=c\}$ only in a single point $t_0$.

On the other hand, in that point, $F^\prime(t_0)$ coincides with the tangent vector to any curve through that point with initial direction $F^\prime(t_0)$. So if you choose any curve $\gamma \subset \{f=c\}$ such that $\gamma(0)=F(t_0) $ and $\gamma^\prime(0) = F^\prime(t_0)$ the conclusion you are after follows, since $\langle \nabla f,\gamma^\prime\rangle = 0$ does not depend on the curve $\gamma$, but only on the value of the derivative in the point you are interested in.

It only remains to show that such curves exist, but that's not too difficult (and I leave it to you to show)

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  • $\begingroup$ Thanks but could you address the question in terms as I posed it? I don't disagree with anything you said; I'm confused with how I approached the problem and I want to answer it in terms of the set that I defined. $\endgroup$ – Clclstdnt Jul 3 '16 at 21:23
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    $\begingroup$ @Clclstdnt Your reasoning contains an error. The fact that $F^\prime(t_0)$ is tangent to the level surface of $f$ does not translate into the inclusion statement you wrote down. You are mixing up the set of tangent vectors with the tangent plane itself. The set you wrote down is the tangent plane to $f=c$. This is a hyperplane which passes through $x_0$, but not necessarily through $0$. It's the set of all tangent vectors translated by $x_0$. A vector $v$ is tangent to the level surface of $f$ if $v - x_0$ is in the tangent plane, not if $v$ is (unless the tangent plane passes through $0$). $\endgroup$ – Thomas Jul 4 '16 at 4:38

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