Prove that $\|c\|$ is constant

Question

I've been trying to prove the following Lemma:

Let $c: I\to \mathbb{R}^n$ be a normal (edit: normal) parameterized curve. Then, $\|c(t)\|$ is a non-zero constant $\forall t\in I$ iff $c(t) \perp c'(t),\,\forall t\in\mathbb{R}$.

Attempt of proof:

$(\Rightarrow)$ Suppose that $\|c(t)\|=r\neq0,\,\forall t\in I.$ Consider the function $f:I\to \mathbb{R}$ with $f(t):=\langle c(t), c(t)\rangle$, where$\langle \cdot, \cdot \rangle$ denotes the Eucledean inner product. Then $f(t)=\|c(t)\|^2=r^2,\,\forall t\in I$ and thus $f'(t)=0,\,\forall t\in I$, i.e. (using the bi-linearity of the inner product and elementary properties of the function $f$) $2\langle c(t), c'(t)\rangle=0,\,\forall t\in I$, i.e. $\langle c(t), c'(t)\rangle=0,\,\forall t\in I$, i.e. . $c(t) \perp c'(t),\,\forall t\in\mathbb{R}$.

The opposite direction gives me some trouble. I've reached up to this point:

$(\Leftarrow)$ Suppose that $c(t) \perp c'(t),\,\forall t\in\mathbb{R}$. Then $f'(t)=0,\,\forall t\in I$ and thus there exists a constant $c\in\mathbb{R}$ such that $f(t)=c,\,\forall t\in I$, i.e. $\langle c(t), c(t)\rangle=\|c(t)\|^2=c\geq0,\,\forall t\in I$.

My question is how to justify that $c$ cannot be zero.

Note: I do realize that (in the case $n=2$) ' $c(t) \perp c'(t),\,\forall t\in\mathbb{R}$' is equivalent that the curve $c$ lies on an cyclical arc.

Answer 1

You have had correctly that the lemma is basically proved by observing $$ \frac{d}{dt}\|c\|^2=2\langle c,c'\rangle. $$

My question is how to justify that cannot be zero.

The nonzero part follows from the assumption that $c$ is a "normal parameterized curve": if $c$ is constantly zero, then its derivative will be zero as well.