(a) Graph the polar equation \(r=1-\sin \theta,\;0\leq \theta\leq 2\pi.\)
(b) Find parametric equations for \(r=1-\sin \theta.\)
\(\theta\) | \(r=1-\sin \theta\) | \(( r,\theta )\) |
---|---|---|
\(0\) | \(1-0=1\) | \(( 1,0)\) |
\(\dfrac{\pi }{6}\) | \(1-\dfrac{1}{2}=\dfrac{1}{2}\) | \(\left(\dfrac{1}{2},\dfrac{\pi }{6}\right)\) |
\(\dfrac{\pi }{2}\) | \(1-1=0\) | \(\left( 0,\dfrac{\pi }{2}\right)\) |
\(\dfrac{5\pi }{6}\) | \(1-\dfrac{1}{2}=\dfrac{1}{2}\) | \(\left( \dfrac{1}{2},\dfrac{5\pi }{6}\right)\) |
\(\pi\) | \(1-0=1\) | \(\left( 1, \ \pi \right)\) |
\(\dfrac{7\pi }{6}\) | \(1-\left( -\dfrac{1}{2}\right) =\dfrac{3}{2}\) | \(\left( \dfrac{3}{2},\dfrac{7\pi }{6}\right)\) |
\(\dfrac{3\pi }{2}\) | \(1-\left( -1\right) =2\) | \(\left( 2,\dfrac{3\pi }{2}\right)\) |
\(\dfrac{11\pi }{6}\) | \(1-\left( -\dfrac{1}{2}\right) =\dfrac{3}{2}\) | \(\left( \dfrac{3}{2},\dfrac{11\pi }{6}\right)\) |
\(2\pi\) | \(1-0=1\) | \(( 1, \ 2\pi )\) |
Graphs of polar equations of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{array}{rcl@{\qquad}crcl} r&=&a(1+\cos \theta ) & r=a(1+\sin \theta )\\ r&=&a(1-\cos \theta ) & r=a(1-\sin \theta ) \end{array} }} \]
where \(a>0,\) are called cardioids. A cardioid contains the pole and is heart-shaped (giving the curve its name).
(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.