Example

(a) Graph the polar equation \(r=1-\sin \theta,\;0\leq \theta\leq 2\pi.\)

(b) Find parametric equations for \(r=1-\sin \theta.\)

Solution (a) The polar equation \(r=1-\sin \theta\) contains \(\sin \theta\), which has the period \(2\pi .\) We construct Table 2 using common values of \(\theta\) that range from \(0\) to \(2\pi ,\) plot the points \(( r,\theta )\), and trace out the graph, beginning at the point \((1,0)\) and ending at the point \(( 1, \ 2\pi )\), as shown in Figure 43(a). Figure 43(b) shows the graph using technology.

TABLE 2

\(\theta\)	\(r=1-\sin \theta\)	\(( r,\theta )\)
\(0\)	\(1-0=1\)	\(( 1,0)\)
\(\dfrac{\pi }{6}\)	\(1-\dfrac{1}{2}=\dfrac{1}{2}\)	\(\left(\dfrac{1}{2},\dfrac{\pi }{6}\right)\)
\(\dfrac{\pi }{2}\)	\(1-1=0\)	\(\left( 0,\dfrac{\pi }{2}\right)\)
\(\dfrac{5\pi }{6}\)	\(1-\dfrac{1}{2}=\dfrac{1}{2}\)	\(\left( \dfrac{1}{2},\dfrac{5\pi }{6}\right)\)
\(\pi\)	\(1-0=1\)	\(\left( 1, \ \pi \right)\)
\(\dfrac{7\pi }{6}\)	\(1-\left( -\dfrac{1}{2}\right) =\dfrac{3}{2}\)	\(\left( \dfrac{3}{2},\dfrac{7\pi }{6}\right)\)
\(\dfrac{3\pi }{2}\)	\(1-\left( -1\right) =2\)	\(\left( 2,\dfrac{3\pi }{2}\right)\)
\(\dfrac{11\pi }{6}\)	\(1-\left( -\dfrac{1}{2}\right) =\dfrac{3}{2}\)	\(\left( \dfrac{3}{2},\dfrac{11\pi }{6}\right)\)
\(2\pi\)	\(1-0=1\)	\(( 1, \ 2\pi )\)

(b) We obtain parametric equations for \(r=1-\sin \theta\) by using the conversion formulas \(x=r \ \cos \theta\) and \(y=r \ \sin\theta\): \[ x=r \ \cos \theta =( 1-\sin \theta ) \ \cos \theta \qquad y=r \ \sin \theta =( 1-\sin \theta ) \ \sin \theta \]

Here, \(\theta\) is the parameter, and if \(0\leq \theta \leq 2\pi ,\) then the graph is traced out exactly once in the counterclockwise direction.

Graphs of polar equations of the form \[ \boxed{\bbox[#FAF8ED,5pt]{ \begin{array}{rcl@{\qquad}crcl} r&=&a(1+\cos \theta ) & r=a(1+\sin \theta )\\ r&=&a(1-\cos \theta ) & r=a(1-\sin \theta ) \end{array} }} \]

where \(a>0,\) are called cardioids. A cardioid contains the pole and is heart-shaped (giving the curve its name).

Problem 5.