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This should be easy, but I cannot figure out what I'm doing wrong, and it's killing me.

Let $\mathbf{f}(t)$ be a function from $\mathbb{R}$ to $\mathbb{R}^3$. I want to find the "rate of change of the angle between $\mathbf{f}(t)$ and nearby vectors (from $\mathbf{f}$)".

I'm going to assume $|\mathbf{f}(t)|=1$, because it makes the writing easier, and doesn't change the answer I get.

So we want to look at nearby vectors $\mathbf{f}(t+h)$, and the angle between that and $\mathbf{f}(t)$ is just $\mathbf{f}(t+h)\cdot\mathbf{f}(t)$. The rate of change is then

$$ \lim_{h\to0}\dfrac{\mathbf{f}(t+h)\cdot\mathbf{f}(t)-1}{h} $$

And this is just $\mathbf{f}'(t)\cdot\mathbf{f}(t)$, which is $0$ (because of the assumption above about the norm of $\mathbf{f}$).

But the right answer is (I think) $|\mathbf{f}'(t)|$. What am I doing wrong?

For context, this comes up in do Carmo's book on Differential Geometry, where it is claimed that for an arc-length parameterized curve, $|\mathbf{\alpha}''(t)|$ measures the rate of change of direction for the tangent curves. Think of $\mathbf{f}$ above as $\mathbf{\alpha}'$.

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    explain why you think that $|f'(t)|$ is the answer – 2017-01-02
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    "the angle between that and $\mathbf{f}(t)$ is just $\mathbf{f}(t+h)\cdot\mathbf{f}(t)$." no it's not. the dot product gives the *cosine* of the angle. – 2017-01-02
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    @symplectomorphic: Yes, you're right! I thought it wouldn't matter much, but using $\arccos$ above in the numerator of the limit, and after some manipulation, I get $\sqrt{-\mathbf{f}''(t)\cdot\mathbf{f}(t)}$, which is indeed $|\mathbf{f}'(t)|$! Do you know of a geometric/cleaner proof of this, without going through the tricky limit derivation? – 2017-01-02

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Here are two ways to think about this: the first uses differentiation to get the result, the second is perhaps more what you want, when we look at small variations.

We have $\cos{\theta(t)}=\mathbf{f}(t) \cdot \mathbf{f}(a) $. Putting $t=a$ gives the consistent $\cos{\theta(a)}=1$, so $\theta(a)=0$. Differentiating both sides once and putting $t=a$, $$ \theta'(a) \sin{\theta(a)} = \mathbf{f}'(a) \cdot \mathbf{f}(a) = 0, $$ again consistent, but of no use, since it does not tell us $\theta'(a)$. But differentiating again, $$ \theta''(a) \sin{\theta(a)} - \theta'(a)^2 \cos{\theta(a)} = \mathbf{f}''(a) \cdot \mathbf{f}(a), $$ and simplifying this gives $$ \theta'(a)^2 = -\mathbf{f}''(a) \cdot \mathbf{f}(a) $$. But, as you note in your comment, differentiating $ 0 = \mathbf{f}'(t) \cdot \mathbf{f}(t) $ gives $ 0 = \mathbf{f}''(a) \cdot \mathbf{f}(a) + \mathbf{f}'(a) \cdot \mathbf{f}'(a) $, so we find $\theta'(a)^2 = \lvert \mathbf{f}'(a) \rvert^2$.

Now let's get the same result by looking at $\mathbf{f}(a+\delta t)$, and see what the variation in the angle, from $\theta$ to $\theta+\delta \theta$, is. Of course we know that $\theta=0$. We have

$$\cos{\delta\theta}=1-\frac{1}{2}(\delta\theta)^2 + o((\delta \theta)^2), $$ while $$ \begin{align} \mathbf{f}(a+\delta t) \cdot \mathbf{f}(a) &= \left( \mathbf{f}(a) + \mathbf{f}'(a) \delta t + \frac{1}{2} \mathbf{f}''(a) (\delta t)^2 + o((\delta t)^2) \right) \cdot \mathbf{f}(a) \\ &= 1 + 0 + \frac{1}{2} \mathbf{f}''(a) \cdot \mathbf{f}(a) (\delta t)^2 + o((\delta t)^2), \end{align}$$ and again, we find that $$ \left( \frac{\delta \theta}{\delta t} \right)^2 = -\mathbf{f}''(a) \cdot \mathbf{f}(a) + o(\delta t) = \lvert \mathbf{f}(a) \rvert^2 + o(\delta t). $$

Note that these approaches are really rather similar in the actual calculation, although the notation looks different. We need $\mathbf{f}$ to be twice-differentiable: since the first derivative tells us nothing by itself, we have to go to second-order.