# Curvature of the composition between a rigid motion and a curve

Consider the following problem (Exercise 1.18 in the book Curves and Surfaces, by Montiel and Ros)

Let $$\alpha : I \longrightarrow \mathbb{R}^2$$ be a regular curve parametrized by the arc length. Let $$M$$ be a rigid motion. Let $$\beta = M \circ \alpha$$. Show that $$k_\beta(s) = \begin{cases} k_\alpha(s) \quad \text{ if } M \text{ is direct } \\ - k_\alpha(s) \quad \text{ if } M \text{ is inverse } \end{cases},$$ where $$k$$ denotes the curvature.

Now, $$M$$ being a rigid motion is of the form $$Mx = Ax + b$$. Then $$k_\beta(s) = \det(\beta'(s), \beta''(s)) = \det((M \circ \alpha)'(s), (M \circ \alpha)''(s)) = \det(A\alpha'(s), A\alpha''(s)).$$ The result suggests that we should be able to write $$k_\beta(s) = \det A \det(\alpha'(s), \alpha''(s))$$, but I cannot see why is this true.

Any hints will be the most appreciated.

You are almost there! Just note that if $$x,y\in\mathbb{R}^2$$ and $$A\in\mathbb{R}^{2\times 2}$$ then $$(Ax,Ay) = A(x,y).$$ Indeed, write $$A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}, \quad x=\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} ,\qquad y= \begin{pmatrix} y_1 \\ y_2 \end{pmatrix},$$ then $$(Ax,Ay) = \begin{pmatrix} a_{11}x_1+a_{12}x_2 & a_{11}y_1 + a_{12}y_2\\ a_{21}x_1+a_{22}x_2 & a_{21}y_1+a_{22}y_2 \end{pmatrix} = A=\begin{pmatrix} x_1 & y_1\\ x_2 & y_2 \end{pmatrix} = A(x,y).$$ So, we have that $$\det(Ax,Ay) = \det(A(x,y)) = \det(A)\det(x,y).$$ With this, $$\det(A(\alpha'(s),A\alpha''(s))) = \det(A)\det(\alpha'(s),\alpha''(s)),$$ as desired.
• Got it. I was thinking of $(x, y)$ as the matrix with lines $x$ and $y$, instead of columns. Thanks a lot. Commented Mar 3, 2020 at 0:15