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A rigid motion of $\mathbb {R^{n}}$ is a map

$M = T ◦ R : \mathbb {R^{n}} \to \mathbb {R^{n}}$

where $ \mathbb {R^{n}} \to \mathbb {R^{n}} $ is linear and satisfies $|R(v)| = |v| \space \space \space \space \space \space ∀v \in \mathbb {R^{n}}$

And T : $ \mathbb {R^{n}} \to \mathbb {R^{n}} $ is translation, i.e. there exists $ w \in \mathbb {R^{n}}$ such that $T (v) = v + w$ $\space \space \space \space \space∀v \in \mathbb {R^{n}} $

(Q) Let $α : (a, b) \to \mathbb {R^{n}} $ be a curve and let $M :\mathbb {R^{n}} \to \mathbb {R^{n}}$ be a rigid motion.

Show that $(M ◦ α)' (t)$ = $R ◦ α'(t)$ for all $t \in (a, b)$.

i claim that $(M ◦ α)' (t) = (T◦R◦α)'(t) = (T◦(A(α)))'(t)= (A(α)+w)'(t) = ((A(α))'+(w)')(t)= (A(α))' +0= A'α'(t)$

i also claim that $A'α'(t)= R_1 ◦α' t$

Q1) is this correct?

Q2) how do i show that $R_1=R$?

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    An $A$ suddenly appeared above... – 2017-01-25
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    A is a rotational matrix its how im used to it existing its got sin n cos on diagonals to rotate by theta. i think i understand what u did below but im not sure. why i cant treat a composition as a a rotational matrix – 2017-01-25

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You have $DM(x) = DT(R(x)) D R(x)$, and $DT(x) = I$, hence $DM(x) = DR(x)$.

Since $R$ is linear, $DR(x) = R$.

Then $D (M \circ \alpha) (t) = DM(\alpha(t)) D \alpha(t) = DR(\alpha(t)) D \alpha(t) = R \alpha'(t)$.