Prove for the time derivative of a vector with constant magnitude

Question

First of all, sorry for my bad English, I have several doubts about the geometric demonstration made by Kleppner for the derivative of a vector that has a constant magnitude, page 25, which is attached in this photo:

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Is this proof rigorously validated? I understand that $\sin (x) \approx x$ for small $x$, by dividing both sides by $\Delta t$, and take the limit for the equality. I understand that equality is correct, but is the argument correct? In the book, i understand the intuitive part, but for example what would have happened if I had taken an approximation of higher order, this would not have had any contribution in the limit? How are these terms depreciated rigorously, since I think that depends on how the time angle $θ (t)$ depends, for example, you can say that the vector varies with $Δθ (t)$, very slow or very fast so that the terms of higher order have a contribution in the limit, and therefore the equality raised is not correct, I apologize in case the question is a triviality.

Answer 1

The argument can be made more rigorous by using the well-known $\displaystyle\lim_{x\to0}{\sin x\over x}=1$: $$ \left|{\Delta A\over\Delta t}\right|={2A\sin({\Delta\theta/2})\over\Delta t} =2A{\sin({\Delta\theta/2})\over{\Delta\theta/2}}{{\Delta\theta/2}\over\Delta t} =A{\sin({\Delta\theta/2})\over{\Delta\theta/2}}{{\Delta\theta}\over\Delta t}, $$ and $\displaystyle{\sin({\Delta\theta/2})\over{\Delta\theta/2}}\to1$, because $\Delta\theta\to0$ as $\Delta t\to0$.

Answer 2

If you are looking for an intuitive way to justify your book, I repeat my comment above:

Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $\Delta \theta$ is the difference between the vectors $A(t+Δt)$ and $A(t)$, and as $Δt→0$, $Δθ→0$. The rate at which $Δθ→0$ is linearly related to the rate at which $Δt→0$ by some basic geometry about circles

I also provide the following if you want an alternative, perhaps more mathematically satisfying approach:

Consider the image at the bottom of this post (from Wikipedia), where we can denote the vector in the diagram with radius $r$ by $\bf{A}$$(t)$ as your textbook does. Then we have $\bf{A}$$(\theta)= \left[r\cos\theta,r\sin\theta\right]$ so $\bf{v}$$(t) = \frac{d{\bf A}(t)}{dt} = \frac{d{\bf A}(\theta)}{d\theta} \frac{d\theta}{dt} =\left[-r\sin\theta,r\cos\theta\right]\frac{d\theta}{dt}$ (where we used the chain rule) so that $|{\bf{v}}(t)| = \sqrt{(-r\sin\theta)^2+(r\cos\theta)^2}\frac{d\theta}{dt} = \sqrt{r^2(\sin^2\theta+\cos^2\theta)}\frac{d\theta}{dt} = r\frac{d\theta}{dt}$

Which is precisely what your textbook claims.