tangent to a level surface 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.
 A: 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)
A: 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!
