We introduce a new coordinate system called the polar coordinate system which is useful in certain situations like dealing with circles or circular regions. The polar coordinate system has a fixed point called the pole (or the origin) denoted by \(O\) and a fixed horizontal line from the pole called the polar axis (or the positive \(x\)-axis). The polar coordinates of a point \(P\) in the plane are written as \((r,\theta)\) where \(r\geq 0\) is the distance between \(P\) and the pole (or the origin) and \(\theta\) is the angle in \([0,2\pi)\) that is between the polar axis (or the positive \(x\)-axis) and the line joining \(P\) and the pole (or the origin). Here an angle is positive when it is measured counterclockwise. Note that \(r=0\) or \((0,\theta)\) correspond to the pole or the origin for any value of \(\theta\).

Note the following convention for polar coordinates not in the above unique representation of \((r,\theta)\):

1. \((r,\theta)=(r,\theta+2n\pi)\) for all integers \(n\).
   
   
2. \((-r,\theta)\) is \((r,\theta+\pi)\) when \(0\leq \theta < \pi\) and \((r,\theta-\pi)\) when \(\pi\leq \theta < 2\pi\).
   
   

Example.

1. \(\left( 2,\frac{\pi}{4} \right)=\left( 2,\frac{9\pi}{4} \right)\) as \(\frac{9\pi}{4}=2\pi+\frac{\pi}{4}\). Similarly \(\left( 2,\frac{\pi}{4} \right)=\left( 2,-2\pi+\frac{\pi}{4} \right)=\left( 2,-\frac{7\pi}{4} \right)\).
   
   
2. \(\left( -2,\frac{\pi}{4} \right)=\left( 2,\frac{\pi}{4}+\pi \right)=\left( 2,\frac{5\pi}{4}\right)\).
   
   

Conversion between Polar and Cartesian Coordinates:
The Cartesian coordinates \((x,y)\) of the polar coordinates \((r,\theta)\) are given by \[x=r\cos \theta,\; y=r\sin \theta.\] Note from the preceding equations that \[x^2+y^2=r^2\cos^2 \theta+r^2\sin^2 \theta=r^2(\cos^2 \theta+\sin^2\theta)=r^2\] and \[\displaystyle\frac{y}{x}=\frac{r\sin \theta}{r\cos \theta}=\tan \theta.\] The polar coordinates \((r,\theta)\) of the Cartesian coordinates \((x,y)\) are give by \[r=\sqrt{x^2+y^2},\; \theta=\tan^{-1}\left( \frac{y}{x} \right),\] where \(\theta\) is the angle in \([0,2\pi)\) that is between the positive \(x\)-axis and the line joining \((x,y)\) and the origin. Note that \(-\frac{\pi}{2} <\tan^{-1}\left( \frac{y}{x} \right) < \frac{\pi}{2}\) and there are two angles in \([0,2\pi)\) that satisfy \(\tan \theta=\displaystyle\frac{y}{x}\). Therefore, when \(x < 0\), the correct \(\theta\) is \(\tan^{-1}\left( \frac{y}{x} \right) +\pi\).

Example.

1. Find the Cartesian coordinates of the following points in polar coordinates: (a) \((2,\pi/4)\), (b) \((4,-11\pi/6)\).
   
   
2. Find the polar coordinates \((r,\theta)\) (with \(r\geq 0,\: 0\leq \theta < 2\pi\)) of the following points: (a) \((\sqrt{3},1)\), (b) \((-2,2)\).
   
   

Solution.

1. (a) Here \(r=2\) and \(\theta=\frac{\pi}{4}\). \[x=2\cos\left( \frac{\pi}{4} \right)=2\frac{\sqrt{2}}{2}=\sqrt{2},\; y=2\sin \left( \frac{\pi}{4} \right)=2\frac{\sqrt{2}}{2}=\sqrt{2}.\] So the point \((2,\pi/4)\) in polar coordinates is \((\sqrt{2},\sqrt{2})\) in Cartesian coordinates.
   
   (b) Here \(r=4\) and \(\theta=-\frac{11\pi}{6}\). \[\begin{align*} x &=4\cos\left( -\frac{11\pi}{6} \right)=4\cos\left( \frac{\pi}{6}-2\pi \right)=4\cos\left( \frac{\pi}{6} \right)=4 \frac{\sqrt{3}}{2}=2\sqrt{3},\\ y &=4\sin\left( -\frac{11\pi}{6} \right)=4\sin\left( \frac{\pi}{6}-2\pi \right)=4\sin\left( \frac{\pi}{6} \right)=4 \frac{1}{2}=2. \end{align*}\] So the point \((4,-11\pi/6)\) in polar coordinates is \((2\sqrt{3},2)\) in Cartesian coordinates.
   
   
2. (a) Here \(x=\sqrt{3}\) and \(y=1\). \[ r=\sqrt{(\sqrt{3})^2+1^2}=2,\; \theta=\tan^{-1}\left( \frac{1}{\sqrt{3}} \right)=\frac{\pi}{6}. \] So the point \((\sqrt{3},1)\) in Cartesian coordinates is \((2,\frac{\pi}{6})\) in polar coordinates.
   
   (b) Here \(x=-2\) and \(y=2\). \[ r=\sqrt{(-2)^2+2^2}=2\sqrt{2},\; \theta=\tan^{-1}\left( \frac{2}{-2} \right)=\tan^{-1}(-1)=-\frac{\pi}{4}. \] Note that \(\theta=-\frac{\pi}{4}\) is not correct as it corresponds to the fourth quadrant where the point \((-2,2)\) lies in the second quadrant. So the correct \(\theta\) is \(\theta=-\frac{\pi}{4}+\pi=\frac{3\pi}{4}\). Thus the point \((-2,2)\) in Cartesian coordinates is \((2\sqrt{2},\frac{3\pi}{4})\) in polar coordinates.
   
   

The Polar Equation of a Curve:
The equation of a curve in rectangular coordinates can be converted to a polar equation by substituting \(x=r\cos \theta,\; y=r\sin \theta\). A polar equation is written in the form \(r=f(\theta)\) or \(F(r,\theta)=0\).

Example.

1. The polar equation of the circle \(x^2+y^2=a^2\), \(a > 0\) is \(r=a\).
   Since \(x^2+y^2=r^2\), \(r^2=a^2 \implies r=a\).
   
   
2. The polar equation of the circle \((x-a)^2+(y-b)^2=a^2+b^2\) is \(r=2a\cos\theta+2b\sin\theta\). \[\begin{align*} (x-a)^2+(y-b)^2=a^2+b^2 &\implies (r\cos \theta-a)^2+(r\sin \theta-b)^2=a^2+b^2\\ &\implies r^2-2r(a\cos \theta+b\sin \theta) +a^2+b^2=a^2+b^2\\ &\implies r(r-2a\cos \theta-2b\sin \theta) =0\\ &\implies r=2a\cos\theta+2b\sin\theta. \end{align*}\]
   
   
3. The polar equation of the line \(y=mx\) is \(\theta=\tan^{-1}m\) (where \(-\infty < r < \infty\)).
   \[y=mx \implies r\sin \theta=mr\cos \theta \implies\tan \theta=m.\] Note that \(\theta=\tan^{-1}m,\; r\geq 0\) and \(\theta=\pi+\tan^{-1}m,\; r\geq 0\) represent two lines starting from the origin whose union is the line \(y=mx\). If we allow \(r\) to be negative, \(\theta=\pi+\tan^{-1}m,\; r\geq 0\) can be written as \(\theta=\tan^{-1}m,\; r\leq 0\). Therefore, the equation simply becomes \(\theta=\tan^{-1}m\), \(-\infty < r < \infty\).
   
   
4. The polar equation of the line whose closest point from the origin is \((d,\alpha)\) in polar coordinates is \(r=d\sec (\theta-\alpha)\).
   
   Consider an arbitrary point \(P(r,\theta)\) on the line. From the right triangle formed by \(P\), \((d,\alpha)\), and the origin, we have \[\frac{d}{r}=\cos (\theta-\alpha) \implies r=d\sec (\theta-\alpha).\]
5. There are famous curves with simple polar equations such as spiral (e.g., \(r=\theta\)), cardioid (e.g., \(r=1-\cos \theta\)), lemniscate (e.g., \(r^2=\cos(2\theta)\)), and polar rose (e.g., \(r=\cos(2\theta)\)).
   
   

Example. Write a Cartesian equation for the cardioid \(r=1-\cos \theta\).

Solution. Substituting \(r=\sqrt{x^2+y^2}\) and \(\cos \theta=\frac{x}{r}=\frac{x}{\sqrt{x^2+y^2}}\) in \(r=1-\cos \theta\), we have \[\begin{align*} \sqrt{x^2+y^2}=1-\frac{x}{\sqrt{x^2+y^2}} & \implies x^2+y^2= \sqrt{x^2+y^2} -x\\ & \implies x^2+y^2+x= \sqrt{x^2+y^2}\\ & \implies (x^2+y^2+x)^2= x^2+y^2. \end{align*}\]

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