Minimizing vector norm I have a vector-valued function $F(t)$, where $t$ is a scalar.  I want to minimize $\left\lVert F(t)\right\rVert$.  (I know already for my case that the only extreme point of this function is a global minimum.)  But I can't simply take the derivative and set it to $0$, since the derivative of $\left\lVert F(t)\right\rVert$ is 
$$ \frac{F(t)}{\left\lVert F(t)\right\rVert} $$
i.e., a vector.  That is, for scalar-valued $f$ we just do
$$f'(t) = 0$$
and solve.  What is the analog to this for $\left\lVert F(t)\right\rVert$?
 A: The chain rule (or at least the version we want to work with here) says the following: if $f:\mathbb{R}^n\rightarrow\mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}^n$ are differentiable, then the function $h:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable and satisfies
$$ h'(t) = (\nabla f)(g(t))\cdot g'(t). $$
In our case, $f(x) = \|x\|$ and so $\nabla f(x)$ is the vector $\frac{x}{\|x\|}$, while $g = F$, so
$$\frac{d}{dt}(\|F(t)\|) = \frac{F(t)}{\|F(t)\|}\cdot F'(t).$$
Thus, to find extrema, you want to find $t$ such that $F(t)\cdot F'(t) = 0$.

An alternate way to find the extrema of $\|F(t)\|$ is to find the extrema of $\|F(t)\|^2$. We can apply the same chain rule to find the derivative of $\|F(t)\|^2$, or simply notice that $\|F(t)\|^2 = F(t)\cdot F(t)$. The dot product ends up satisfying a differentiation rule that mimics the ordinary product rule, i.e. $(f\cdot g)' = f'\cdot g + f\cdot g'$. This implies that
$$\frac{d}{dt}(\|F(t)\|^2) = \frac{d}{dt}(F(t)\cdot F(t)) = 2F(t)\cdot F'(t) $$
and we obtain the same criterion as above.
