Showing that a tangent to an involute of a circle is perpendicular to the circle It's been established that the parametric equations of a circle involute are: $$\begin{align} x &= r (\cos\theta + \theta \sin\theta) \\\ y &= r (\sin\theta - \theta \cos\theta) \end{align}$$ where $r$ is the radius of the circle and $\theta$ is angle measured from the horizontal. The derivatives of these are just: $$\begin{align} x' &=r \theta\cos\theta \\\ y' &=r \theta\sin\theta \end{align}$$ > While visually, I can see that the tangent vector of a point on the involute curve is perpendicular to the circle it turns around, how can I prove this?

Consider the parametrization of the circle $(x_c(\theta), y_c(\theta))=(r\cos \theta,r\sin \theta)$. It's derivative $(x_c'(\theta), y_c'(\theta))=(-r\sin \theta,r\cos \theta)$ gives the direction of the tangent line at $\theta$. For every $\theta$, this vector is perpendicular to the tangent of the involute: $$ (x_c'(\theta),y_c'(\theta))\cdot(x'(\theta),y'(\theta)) = (-r\sin\theta)(r\theta \cos\theta) + (r\cos\theta)(r\theta \sin \theta) = 0. $$ This proves the statement.

Alternatively, it suffices to note that for every $\theta$ the vector $(\cos \theta,\sin \theta)$ is normal to the circle. And since $(x'(\theta),y'(\theta))$ is just a multiple of the normal, the statement follows.