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.

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Maybe I want to apply the Schwartz inequality? – Arkamis Nov 3 '12 at 16:21
It might be helpful to notice that $|f(t)|=1$ is equivalent to $f(t)\cdot f(t)=1$. Do you know the analog of the product rule in this situation? – Benji Nov 3 '12 at 16:23
@bobobinks I did note that in my statement of the problem. I also noted that I don't see how that can help. – Arkamis Nov 3 '12 at 16:23
@bobobinks re: product rule. Sort of. I need to establish it, but I know it. I see where this is going. I've added an answer to my own question; mind reviewing to see if I've made a mistake? – Arkamis Nov 3 '12 at 16:26

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.

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Your proof is fine. – Benji Nov 3 '12 at 16:31
@bobobinks Much obliged for the push in the right direction. – Arkamis Nov 3 '12 at 16:34
...but it should have dots in the second display. – GEdgar Nov 3 '12 at 16:41
@Gedgar And now it does :) – Arkamis Nov 3 '12 at 16:58

$$\left|{\bf f}\left(t\right)\right| = \sqrt{{\bf f}\left(t\right) \bullet {\bf f} \left(t\right)} = 1 \Rightarrow {\bf f}\left(t\right) \bullet {\bf f} \left(t\right) = 1 \Rightarrow \frac{d}{dt} \left[{\bf f}\left(t\right) \bullet {\bf f} \left(t\right) \right] = 2 {\bf f}\left(t\right) \bullet \frac{d}{dt}{\bf f} \left(t\right) = 0$$

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Here is a geometric way to see it. If $f$ is a unit-norm curve, it is is confined to a sphere. Any tangent to a sphere is perpendicular to its radius. The tangent vector to such a curve must live inside a tangent plane, so it is perpenducular to the radius vector f.

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I mentioned exactly this in my post :) – Arkamis Nov 4 '12 at 1:30