Suppose $f$ is a 3-dimensional differential mapping of $\mathbb{R}$ into $\mathbb{R}^3$ such that $||f(t)||_2 = 1$ for all $t \in \mathbb{R}$.
Prove that $f'(t) \cdot f(t) = 0$.
My thought's to that are, that $||f(t)||_2 = 1$ is equivalent to $(f_1(t))^2 + (f_2(t))^2 + (f_3(t))^2 = 1$.
Furthermore, it is $f'(t) \cdot f(t) = f_1'(t) \cdot f_1(t) + f_2'(t) \cdot f_2(t) + f_3'(t) \cdot f_3(t)$.
I tried squaring $f'(t) \cdot f(t)$ to hopefully implementing the condition $||f(t)||_2 = 1$
$[f'(t) \cdot f(t)]^2 = (f_1'(t))^2(f_1(t))^2 + (f_2'(t))^2(f_2(t))^2 + (f_3'(t))^2(f_3(t))^2$
$+ 2\; [\; f_1'(t)f_1(t)f_2'(t)f_2(t) + f_1'(t)f_1(t)f_3'(t)f_3(t) + f_2'(t)f_2(t)f_3'(t)f_3(t) \; ]$,
however, I didn't manage to use $(f_1(t))^2 + (f_2(t))^2 + (f_3(t))^2 = 1$ anywhere except of maybe substituting e.g. $(f_1(t))^2 = 1 - (f_2(t))^2 - (f_3(t))^2$, but this just makes the term much more complicated than it already is.
How could this be done? Do you have any hints?
Since $f(t) \cdot f(t) = f_1(t) \cdot f_1(t) + f_2(t) \cdot f_2(t) + f_3(t) \cdot f_3(t)=1$ then $(f(t) \cdot f(t))'=0$ and this is $$f(t)' \cdot f(t))+f(t) \cdot f(t)'=0,$$ by the Leibniz property of derivatives.
But the inner product is symmetric, so $2[f'(t) \cdot f(t)] = 2[f_1'(t) \cdot f_1(t) + f_2'(t) \cdot f_2(t) + f_3'(t) \cdot f_3(t)]=0$ too, hence
$$f_1'(t) \cdot f_1(t) + f_2'(t) \cdot f_2(t) + f_3'(t) \cdot f_3(t)=0.$$