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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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  • $\begingroup$ As a side, have you left the chatroom for some reason? It has two months since you last saw. Do visit once $\endgroup$ – vidyarthi Sep 29 at 18:21
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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.

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  • $\begingroup$ Sir, Can you help me to find where is my mistake in the differentiation? $\endgroup$ – Unknown x Jul 7 at 15:52
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    $\begingroup$ 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. $\endgroup$ – Christian Blatter Jul 8 at 8:57
  • $\begingroup$ As a side, do visit the chat room once $\endgroup$ – vidyarthi Aug 2 at 11:14

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