= Solution
Convert the <cardioid> from <polar coordinates> to <Cartesian coordinates>:
$$
x=(1+\cos\theta)\cos\theta,\qquad y=(1+\cos\theta)\sin\theta.
$$
The <curve> is symmetric about the horizontal axis. It has a <cusp> at the origin at $\theta=\pi$, passes through $(0,\pm1)$, and reaches $(2,0)$ on the right. Its leftmost horizontal coordinate is $-1/4$, since $x=(\cos\theta+1/2)^2-1/4$.
For the largest $y$, differentiate the parametrization:
$$
\frac{dy}{d\theta}=\cos\theta+\cos2\theta
=(2\cos\theta-1)(\cos\theta+1).
$$
The candidates have $\cos\theta=1/2$ or $-1$. The positive maximum occurs at $\theta=\pi/3$, where $r=3/2$. Comparing with the negative extremum and the zero <cusp> value proves
$$
\boxed{(x_M,y_M)=\left(\frac34,\frac{3\sqrt3}{4}\right).}
$$
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1-cardioid.png]
{title=The cardioid with its cusp, axis crossings and point of greatest vertical coordinate}
Back to article page
