91Ó°ÊÓ

A cardioid is the path traced by a point on the perimeter of a circle rolling around another fixed circle of the same radius. Its parametric equations can be given as: $$ x = a(1 + \cos \phi)\cos \phi $$ and $$ y = a(1 + \cos \phi)\sin \phi $$ where \(a\) is the radius of the circles, \(\phi\) is the rolling angle, and \((x, y)\) are the coordinates of the point on the cardioid.

To prove that \(\psi = \frac{\phi}{2}\), we can differentiate the parametric equations above with respect to the parameter \(\phi\). By doing so, we can find out the relationship between the angles \(\psi\) and \(\phi\) through their corresponding differentials. Differentiate \(x\) with respect to \(\phi\): $$ \frac{dx}{d\phi} = -2a\cos^2(\frac{\phi}{2})\sin(\phi) $$ Differentiate \(y\) with respect to \(\phi\): $$ \frac{dy}{d\phi} = 2a\sin^2(\frac{\phi}{2})\cos(\phi) $$ Now divide \(\frac{dy}{d\phi}\) by \(\frac{dx}{d\phi}\) to find \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = \frac{2a\sin^2(\frac{\phi}{2})\cos(\phi)}{-2a\cos^2(\frac{\phi}{2})\sin(\phi)} = -\frac{\sin^2(\frac{\phi}{2})}{\cos^2(\frac{\phi}{2})} $$ Now, we can recognize that \(\frac{dy}{dx}\) is the tangent of the angle \(\psi\). Therefore: $$ \tan(\psi) = -\frac{\sin^2(\frac{\phi}{2})}{\cos^2(\frac{\phi}{2})} = -\tan^2(\frac{\phi}{2}) $$ Using the inverse tangent function, we find: $$ \psi = \arctan(-\tan^2(\frac{\phi}{2})) $$

Finally, we need to show that \(\psi = \frac{\phi}{2}\). Since the arctangent function \(\arctan(x)\) is an odd function, we have \(\arctan(-x) = -\arctan(x)\). Therefore: $$ \psi = \arctan(-\tan^2(\frac{\phi}{2})) = -\arctan(\tan^2(\frac{\phi}{2})) $$ Since \(\tan(\arctan(x)) = x\), we can write: $$ \psi = -\arctan(\tan^2(\frac{\phi}{2})) = -\frac{\phi}{2} $$ Given that \(\psi\) and \(\phi\) are measured in the same direction, their signs are the same. Thus: $$ \psi = \frac{\phi}{2} $$ We have proven that for the cardioid \(\psi = \frac{\phi}{2}\).