Prove that $\langle\, f,f'\,\rangle=0$

Question

Given $f:\mathbb{R}\rightarrow\mathbb{R}^n$ differential and for every $x\in\mathbb{R}$ $\|f\|=1$.

Prove $\langle\,f(x),f'(x)\,\rangle=0$ for every $x\in\mathbb{R}$.

I've got the idea that the function is on the unit circle and the tangent space is vertical but I've got stuck after trying to assume that the inner product isn't $0$.

Answer 1

Note that\begin{align}\lVert f\rVert=1&\iff\lVert f\rVert^2=1\\&\iff\langle f,f\rangle=1.\end{align}So, if you differentiate $\langle f,f\rangle$, you get $0$. But\begin{align}\langle f,f\rangle'&=\langle f,f'\rangle+\langle f',f\rangle\\&=2\langle f,f'\rangle.\end{align}

Answer 2

Let $f(x)=[f_1(x), ..., f_n(x)]$, for all $x \in \mathbb R$ for some unique (scalar) functions $f_i: \mathbb R \to \mathbb R$, $i=1,...,n$

We're given that $f_1^2(x) + ... + f_n^2(x) = 1$ for all $x \in \mathbb R$. We want to show that $f_1(x)f_1'(x) + ... + f_n(x)f_n'(x) = 0$ for all $x \in \mathbb R$

Differentiate both sides of $f_1^2 + ... + f_n^2 = 1$ to get

$$2f_1f_1' + ... + 2f_nf_n' = 0$$

Equivalently,

$$f_1f_1' + ... + f_nf_n' = 0$$