Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 15

Convert the cardioid from polar coordinates to Cartesian coordinates:The curve is symmetric about the horizontal axis. It has a cusp at the origin at $\theta =\pi$, passes through $(0,\pm 1)$, 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: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

Each radial segment from $0$ to $1+\cos \theta$ lies inside the cardioid. Integrating the polar area element givesThe three integrals are respectively $2\pi$, $0$ and $\pi$. Thusin square coordinate units, agreeing with the scale-one case of the area and perimeter of a cardioid.

Differentiate $x=r\cos \theta$, $y=r\sin \theta$:Squaring and adding cancels the mixed terms and uses ${\cos }^2\theta +{\sin }^2\theta =1$, givingFor this cardioid, the polar arc length integrand isThe cosine inside the absolute value changes sign at $\theta =\pi$. Therefore the perimeter isin coordinate length units. The isolated cusp does not obstruct this piecewise smooth arc length integral.