# Show that $F\circ c$ has the same curvature as $c$. How does a Euclidean motion $F$ that is not orientation-preserving affect the curvature of $c$?

Let $$c$$ be a unit speed curve. Let $$F ∈ E(2)$$ be an orientation preserving Euclidean motion, $$F(x) = Ax + b$$ with $$A ∈ SO(2)$$ and $$b ∈ \mathbb R^2.$$ Show that $$F \circ c$$ has the same curvature as $$c$$. Also, How does a Euclidean motion $$F$$ that is not orientation-preserving affect the curvature of $$c$$?

My attempt. $$c$$ be a unit speed curve. So, $$||\dot c||=1$$. By chain rule $$(F\circ c)'(t)=F'(c(t)).\dot c(t)\implies (F\circ c)''(t)=F''(c(t)).(\dot c(t))^2+F'(c(t)).\ddot c(t)=A.(\dot c(t))^2+A.\ddot c(t)$$. By definition 2.2.2,$$A.(\dot c(t))^2+A.\ddot c(t)=A.(\dot c(t))^2+A.\kappa(t).n(t)$$ How do I complete the solution?

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The curvature $$\kappa$$ of plane curves has a sign. This sign is tied to the conventions that counterclockwise rotation is considered positive, and that the normal $$n(t)$$ at a curve point $$c(t)$$ is obtained by rotating the tangent vector $$\dot c(t)$$ counterclockwise by the amount $${\pi\over2}$$. In this way the vectors $$f_1:=\dot c(t)$$ and $$f_2:=n(t)$$ form a positively oriented orthonormal frame at the point $$c(t)$$.
The unit vector $$\dot c(t)$$ has an argument (polar angle) $$\theta(t):=\arg\bigl(c(t)\bigr)$$ with respect to the given $$(x_1,x_2)$$ coordinate axes. The curvature $$\kappa(t)$$ can then be viewed as rotation speed of $$c(t)$$, i.e., $$\kappa(t)={d\over dt}\theta(t)\ .\tag{1}$$ Assume now that a motion $$F$$ is given, which transforms $$c$$ into the new curve $$c_*:=F\circ c$$. Let $$\theta_*(t)$$ be the argument of $$\dot c_*(t)$$. If $$F$$ is orientation preserving then $$\theta_*(t)\equiv\theta(t)+\alpha$$ for some constant angle $$\alpha$$. From $$(1)$$ it then follows that $$\kappa_*(t)\equiv \kappa(t)$$. If $$F$$ is not orientation preserving then $$\theta_*(t)=\alpha-\theta(t)$$ for some constant angle $$\alpha$$. From $$(1)$$ it then follows that $$\kappa_*(t)\equiv -\kappa(t)$$.
In terms of linear algebra: If $$F(x)=Ax+b$$ is not orientation preserving we still have $$\dot c_*(t)=A.\dot c(t)$$, but $$n_*(t)=-A.n(t)$$. The last minus sign changes the sign of $$\kappa$$ when you compute $$\kappa_*$$ using definition 2.2.2.
• E.g., the application of the chain rule went wrong. In particular, $\bigl(\dot c(t)\bigr)^2$ is not a vector; hence you cannot apply $A$ to it. – Christian Blatter Jul 8 at 8:57