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.