Page 72 exercise 4.5, there is the following situation:
There is a curve $\underline{x}(t)$ with $t$ not a natural parameter. I have to find the curvature vector $\underline{k}$ and the curvature $k$ in the corresponding point at $t=1$ (details of the problem below).
I've found the tangent unit vector $t=\frac{\underline{x'}}{|\underline{x'}|}$
I write the derivative of a vector $\underline{v}$ with respect to a natural parameter $s$ as $\underline{v}^{*}$
Then I know that $\underline{k}= \underline{t}^{*}$ Then on the book this continues as $\underline{k}= \underline{t}^{*} = \frac{\underline{t}'}{|\underline{x}'|}$
Why the last equality holds? It holds in general? I can't see why it holds, but I think it's quite easy.
Details of the exercise
The curve is $\underline{x}=t \underline{e_1}+ \frac{1}{2} t^2 \underline{e_2}+ \frac{1}{3} t^3 \underline{e_3}$
That's the chain rule: If $u$ denotes the curve parameter and $s$ the arc length parameter, then $ds/du = |x'(u)|$ is the speed, so $$ t^{*} = \frac{dt}{ds} = \frac{dt}{du} \cdot \frac{du}{ds} = \frac{dt/du}{ds/du} = \frac{t'}{|x'|}. $$