A unit-speed plane curve $\gamma (s)$ rolls without slipping along a straight line $l$ parallel to a unit vector $a$, and initially touches $l$ at a point $p = \gamma (0)$. Let $q$ be a point fixed relative to $\gamma$. Let $\Gamma (s)$ be the point to which $q$ has moved when $\gamma$ has rolled a distance $s$ along $l$ (note that $\Gamma$ will not usually be unit-speed). Let $\theta (s)$ be the angle between $a$ and the tangent vector $\dot \gamma$ . Show that $$\Gamma (s) = p + sa + \rho_{−\theta (s)}(q − \gamma (s)),$$ where $\rho_{\phi}$ is the rotation about the origin through an angle $\phi$. Show further that $$\dot \Gamma (s) \cdot \rho_{−\theta(s)}(q − \gamma (s)) = 0.$$
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We have that $\Gamma (s)$ is the point to which $q$ has moved when $\gamma$ has rolled a distance $s$ along $l$, why isn't then $\Gamma (s)=q+s$ ?
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EDIT:
What does it mean that "$\gamma (s)$ rolls without slipping along a straight line $l$" ?