# Chain Rule and Integration in Matrix Calculus

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 $$t_0 in $$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?

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.