Differentiable $\mathbf{f}:\Bbb R \to \Bbb R^3$, $|\mathbf{f}(t)| = 1$ implies $\mathbf{f}(t)\cdot\mathbf{f}'(t) = 0$ This question comes from Rudin's PMA problem 9.13.
For a differentiable function $\mathbf{f} : \Bbb R \to \Bbb R^3$ and $|\mathbf{f}(t)|=1$ for all $t$, I want to prove that $\mathbf{f}(t)\cdot\mathbf{f}'(t) = 0$.
Notionally, this makes sense to me. $|\mathbf{f}(t)|=1$ means that $\mathbf{f}(t)$ describes a curve on the unit sphere in $\Bbb R^3$. Of course, the tangent to this curve at any point will lie in the tangent plane to that point on the unit sphere, which is normal to the vector from the origin to that point. Therefore, by orthogonality, the dot product will be zero.
However, I am struggling to prove it rigorously using the definition of derivatives.
I have
$$\mathbf{f}(t)\cdot\mathbf{f}'(t) = \lim_{h\to 0} \frac{1}{h} \left(\mathbf{f}(t) \cdot \left(\mathbf{f}(t+h)-\mathbf{f}(t)\right)\right)$$
by applying properties of limits, etc., but this seems to bring me right back in a circle (no pun intended). I also know that $|\mathbf{f}(t)| = 1$ means that $\mathbf{f}(t)\cdot\mathbf{f}(t) = 1$. I'm just not sure how to proceed from here. I want to argue that the numerator of the limit must be zero, but I can't; evaluation of the limit just brings me back to $\mathbf{f}\cdot\mathbf{f}'$.
This is a homework problem, so please just provide a nudge in the right direction.
Edit: I also don't "know" that $\mathbf{a \cdot b} = |\mathbf{a}||\mathbf{b}|\cos \theta$. I suppose I could prove it, but I feel as if that's avoiding the intent of the question.

Based on @bobobinks' comment, I can employ
$$ |\mathbf{f}(t)| = 1 \implies \mathbf{f}(t)\cdot \mathbf{f}(t) = 1$$
Differentiating this, applying the product rule, I would have
$$\mathbf{f}(t)\cdot\mathbf{f}'(t)+ \mathbf{f}'(t)\cdot\mathbf{f}(t) = 2\mathbf{f}(t)\cdot\mathbf{f}'(t) = 0$$
and this would finish the problem.
