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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.

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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.

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  • $\begingroup$ Got it. I was thinking of $(x, y)$ as the matrix with lines $x$ and $y$, instead of columns. Thanks a lot. $\endgroup$ Commented Mar 3, 2020 at 0:15

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