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I'm having trouble differentiating $(M \circ \alpha)(t)$ where $\alpha:(a,b)\to \mathbb{R}^n$ is a regular curve and $M:\mathbb{R}^n \to \mathbb{R}^n$ is a rigid motion.

For context, I'm reading Do Carmo's Differential Geometry, and in the section on the local theory of curves, he states the following:

Fundamental Theorem of the Local Theory of Curves
If $k(s)>0$ and $\tau(s)$ are differentiable functions on some interval $I=(a,b)\owns s$, then there exists a regular curve $\alpha:I\to\mathbb{R}^3$ parametrized by arc length such that $k(s)=\lvert \alpha''(s)\rvert$ is the curvature and $\tau(s)$ is the torsion of $\alpha$. Moreover, any other such curve $\bar\alpha:I\to\mathbb{R}^3$ differs from $\alpha$ by a rigid motion; that is, there exists some $\rho\in\mathrm{O}(3)$ with $\det\rho>0$ and some $c\in\mathbb{R}^3$ such that $\alpha=\rho\circ \bar{\alpha}+c$.

He forgoes the proof of the existence of $\alpha$, but he does offer a sketch of a proof of the part on uniqueness. I say "sketch" because his proof starts off by claiming

$$\int_{t_0}^{t} \left\lvert \dfrac{d\alpha(t')}{dt'} \right\rvert dt' = \int_{t_0}^{t} \left\lvert \dfrac{d(M\circ\alpha)(t')}{dt'} \right\rvert dt'$$

for $t_0<t$ in $I$. He doesn't actually verify this, however; instead waving his hands about the symmetries of the dot and cross products.

How would I verify this second result?

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By writing out the coordinates of the matrix-vector product $M \circ \alpha$, we see that $$ \frac{d(M \circ \alpha)(t')}{dt'} = M \circ \frac{d\alpha(t')}{dt'}. $$ In fact, this is true for any Matrix $M$, not just for a rigid motion.

Now, if $M$ is a rigid motion, we have $M^TM = E$, where the superscript $T$ denotes transposition, and $E$ is the identity matrix. From this, we get $$ \left| \frac{d(M \circ \alpha)(t')}{dt'} \right| = \left(\left(\frac{d(M \circ \alpha)(t')}{dt'}\right)^T\left(\frac{d(M \circ \alpha)(t')}{dt'}\right)\right)^{1/2} = \left(\left(M \circ \frac{d\alpha(t')}{dt'}\right)^T\left(M \circ \frac{d\alpha(t')}{dt'}\right)\right)^{1/2} = \left(\left(\frac{d\alpha(t')}{dt'}\right)^TM^TM\left(\frac{d\alpha(t')}{dt'}\right)\right)^{1/2} = \left(\left(\frac{d\alpha(t')}{dt'}\right)^T\left(\frac{d\alpha(t')}{dt'}\right)\right)^{1/2} = \left| \frac{d\alpha(t')}{dt'} \right|. $$ Integrating this equality proves the second yellow box in your question.

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