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# 6.4: Polar Coordinates

• • Contributed by Michael Corral
• Professor (Mathematics) at Schoolcraft College

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ Figure 6.4.2. We call the pair $$(r,\theta)$$ the polar coordinates of $$P$$, and the positive $$x$$-axis is called the polar axis of this coordinate system. Note that $$(r,\theta) = (r,\theta + 360^\circ k)$$ for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$, so (unlike for Cartesian coordinates) the polar coordinates of a point are not unique. FIGURE 6.4.2 Polar coordinates $$(r,θ)$$ FIGURE 6.4.3 Negative $$r:\, (−r,θ)$$

In polar coordinates we adopt the convention that $$r$$ can be negative, by defining $$(-r,\theta) = (r,\theta + 180^\circ)$$ for any angle $$\theta$$. That is, the ray $$\overrightarrow{OP}$$ is drawn in the opposite direction from the angle $$\theta$$, as in Figure 6.4.3. When $$r=0$$, the point $$(r,\theta) = (0,\theta)$$ is the origin $$O$$, regardless of the value of $$\theta$$.

You may be familiar with graphing paper, for plotting points or functions given in Cartesian coordinates (sometimes also called rectangular coordinates). Such paper consists of a rectangular grid. Similar graphing paper exists for plotting points and functions in polar coordinates, similar to Figure 6.4.4. FIGURE 6.4.4 Polar Coordinate Graph

The angle $$\theta$$ can be given in either degrees or radians, whichever is more convenient. Radians are often preferred when graphing functions in polar coordinates. The reason is that, unlike degrees, radians can be considered "unitless'' (as we mentioned in Chapter 4). This is desirable when a function given in polar coordinates is expressed as $$r$$ as a function of $$\theta$$ (similar to how, in Cartesian coordinates $$(x,y)$$, functions are usually expressed as $$y$$ as a function of $$x$$). For example, if a
function in polar coordinates is written as $$r = 2\,\theta$$, then $$r$$ would have the same units as $$\theta$$. But $$r$$ should be a unitless quantity, hence using radians for $$\theta$$ makes more sense in this case.

Example 6.14

Express the spiral from Figure 6.4.1 in polar coordinates.

Solution:
We will use radians for $$\theta$$. The goal is to find some equation involving $$r$$ and $$\theta$$ that describes the spiral. We see that

\nonumber \begin{align*} \theta ~=~ 0 \quad&\Rightarrow\quad r ~=~ 1\\ \nonumber \theta ~=~ 2\pi \quad&\Rightarrow\quad r ~=~ 2\\ \nonumber \theta ~=~ 4\pi \quad&\Rightarrow\quad r ~=~ 3\\ \nonumber &\vdots\\ \nonumber \theta ~=~ 2\pi\,k \quad&\Rightarrow\quad r ~=~ 1+k\\ \nonumber \end{align*}

for $$k=0,1,2,\ldots$$. In fact, that last relation holds for any nonnegative real number $$k$$ (why?). So for any $$\theta \ge 0$$,

$\theta ~=~ 2\pi\,k \quad\Rightarrow\quad k ~=~ \frac{\theta}{2\pi} \quad\Rightarrow\quad r ~=~ 1 + k ~=~ 1 + \frac{\theta}{2\pi} ~.\nonumber$

Hence, the spiral can be written as $$\fbox{\(r ~=~ 1 + \frac{\theta}{2\pi}$$} \) for $$\theta \ge 0$$. The graph is shown in Figure 6.4.5, along with the Gnuplot commands to create the graph. FIGURE 6.4.5 $$r=1+\frac{\theta}{2\pi}$$

Note that when using the $$\texttt{set polar}$$ command, Gnuplot will assume that the function being plotted is $$r$$ as a function of $$\theta$$ (represented by the variable $$\texttt{t}$$ in Gnuplot).

Figure 6.4.6 shows how to convert between polar coordinates and Cartesian coordinates. For a point with polar coordinates $$(r,\theta)$$ and Cartesian coordinates $$(x,y)$$: FIGURE 6.4.6

Polar to Cartesian:
$\label{eqn:polartorect} \boxed{ x ~=~ r\,\cos\;\theta \qquad y ~=~ r\,\sin\;\theta }$

Cartesian to Polar:
$\label{eqn:recttopolar} \boxed{ r ~=~ \pm\;\sqrt{x^2 ~+~ y^2} \qquad \tan\;\theta ~=~ \frac{y}{x} ~~\text{if $$x \ne 0$$} }$

Note that in Equation \ref{eqn:recttopolar}, if $$x = 0$$ then $$\theta = \pi/2$$ or $$\theta = 3\pi/2$$. Also, if $$x \ne 0$$ and $$y \ne 0$$ then the two possible solutions for $$\theta$$ in the equation $$\tan\;\theta ~=~ \frac{y}{x}$$ are in opposite quadrants (for $$0 \le \theta < 2\pi$$). If the angle $$\theta$$ is in the same quadrant as the point $$(x,y)$$, then $$r = \sqrt{x^2 ~+~ y^2}$$ (i.e. $$r$$ is positive); otherwise $$r = -\sqrt{x^2 ~+~ y^2}$$ (i.e. $$r$$ is negative).

Example 6.15

Convert the following points from polar coordinates to Cartesian coordinates:
(a) $$(2,30^\circ)$$; (b) $$(3,3\pi/4)$$; (c) $$(-1,5\pi/3)$$

Solution:
(a)
Using Equation \ref{eqn:polartorect} with $$r=2$$ and $$\theta = 30^\circ$$, we get:
$\nonumber (x,y) ~=~ ( r\,\cos\;\theta, r\,\sin\;\theta) ~=~ (2\,\cos\;30^\circ,2\,\sin\;30^\circ) ~=~ \left(2 \;\cdot\; \tfrac{\sqrt{3}}{2}, 2 \;\cdot\; \tfrac{1}{2} \right) \quad\Rightarrow\quad \boxed{(x,y) ~=~ \left( \sqrt{3},1 \right)}$

(b) Using Equation \ref{eqn:polartorect} with $$r=3$$ and $$\theta = 3\pi/4$$, we get:
$\nonumber (x,y) ~=~ ( r\,\cos\;\theta, r\,\sin\;\theta) ~=~ \left( 3\,\cos\;\tfrac{3\pi}{4},3\,\sin\;\tfrac{3\pi}{4} \right) ~=~ \left(3 \;\cdot\; \tfrac{-1}{\sqrt{2}}, 3 \;\cdot\; \tfrac{1}{\sqrt{2}} \right) \quad\Rightarrow\quad \boxed{(x,y) ~=~ \left( \tfrac{-3}{\sqrt{2}},\tfrac{3}{\sqrt{2}} \right)}$

(c) Using Equation \ref{eqn:polartorect} with $$r=-1$$ and $$\theta = 5\pi/3$$, we get:
$\nonumber (x,y) ~=~ ( r\,\cos\;\theta, r\,\sin\;\theta) ~=~ \left( -1\,\cos\;\tfrac{5\pi}{3},-1\,\sin\;\tfrac{5\pi}{3} \right) ~=~ \left(-1 \;\cdot\; \tfrac{1}{2},-1 \;\cdot\; \tfrac{-\sqrt{3}}{2} \right) \quad\Rightarrow\quad \boxed{(x,y) ~=~ \left( -\tfrac{1}{2},\tfrac{\sqrt{3}}{2} \right)}$

Example 6.16

Convert the following points from Cartesian coordinates to polar coordinates:
(a) $$(3,4)$$; (b) $$(-5,-5)$$

Solution:
(a) Using formula \ref{eqn:recttopolar} with $$x=3$$ and $$y=4$$, we get:
$\nonumber \tan\;\theta ~=~ \frac{y}{x} ~=~ \frac{4}{3} \quad\Rightarrow\quad \theta ~=~ 53.13^\circ \quad \text{or}\quad \theta ~=~ 233.13^\circ$
Since $$\theta = 53.13^\circ$$ is in the same quadrant (QI) as the point $$(x,y) = (3,4)$$, we can take
$$r ~=~ \sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = 5$$. Thus, $$\boxed{(r,\theta) = (5,53.13^\circ)}$$.

Note that if we had used $$\theta = 233.13^\circ$$, then we would have $$(r,\theta) =(-5,233.13^\circ)$$.

(b) Using Equation \ref{eqn:recttopolar} with $$x=-5$$ and $$y=-5$$, we get:
$\nonumber \tan\;\theta ~=~ \frac{y}{x} ~=~ \frac{-5}{-5} ~=~ 1 \quad\Rightarrow\quad \theta ~=~ 45^\circ \quad \text{or}\quad \theta ~=~ 225^\circ$
Since $$\theta = 225^\circ$$ is in the same quadrant (QIII) as the point $$(x,y) = (-5,-5)$$, we can take
$$r ~=~ \sqrt{x^2 + y^2} = \sqrt{(-5)^2 + (-5)^2} = 5\,\sqrt{2}$$. Thus, $$\boxed{(r,\theta) =(5\,\sqrt{2},225^\circ)}$$.

\noindent Note that if we had used $$\theta = 45^\circ$$, then we would have $$(r,\theta) = (-5\,\sqrt{2},45^\circ)$$.
\end{exmp}

Example 6.17

Write the equation $$x^2 + y^2 = 9$$ in polar coordinates.

Solution:
This is just the equation of a circle of radius $$3$$ centered at the origin. Since $$r = \pm\sqrt{x^2 + y^2} = \pm\sqrt{9}$$, in polar coordinates the equation can be written as simply $$\boxed{r = 3}$$.

Example 6.18

Write the equation $$x^2 + (y-4)^2 = 16$$ in polar coordinates.

Solution: This is the equation of a circle of radius $$4$$ centered at the point $$(0,4)$$. Expanding the equation, we get:
\nonumber \begin{align*} x^2 ~+~ (y-4)^2 ~&=~ 16\\ \nonumber x^2 ~+~ y^2 ~-~ 8y ~+~ 16 ~&=~ 16\\ \nonumber x^2 ~+~ y^2 ~&=~ 8y\\ \nonumber r^2 ~&=~ 8\,r\sin\;\theta\\ \nonumber r ~&=~ 8\,\sin\;\theta \end{align*}

Why could we cancel $$r$$ from both sides in the last step? Because the point $$(0,0)$$ is on the circle, canceling $$r$$ does not eliminate $$r=0$$ as a potential solution of the equation (since $$\theta = 0^\circ$$ would make $$r = 8\,\sin\;\theta = 8\,\sin\;0^\circ = 0$$). Thus, the equation is $$\boxed{r = 8\,\sin\;\theta}$$.

Example 6.19

Write the equation $$y = x$$ in polar coordinates.

Solution:
This is the equation of a line through the origin. So when $$x=0$$, we know that $$y=0$$. When $$x \ne 0$$, we get:
\nonumber \begin{align*} y ~&=~ x\\ \nonumber \frac{y}{x} ~&=~ 1\\ \nonumber \tan\;\theta ~&=~ 1\\ \nonumber \theta ~&=~ 45^\circ \end{align*}

Since there is no restriction on $$r$$, we could have $$r=0$$ and $$\theta = 45^\circ$$, which would take care of the case $$x = 0$$ (since then $$(x,y) = (0,0)$$, which is the same as $$(r,\theta) = (0,45^\circ))$$. Thus, the equation is $$\boxed{\theta = 45^\circ}$$.

Example 6.20

Prove that the distance $$d$$ between two points $$(r_1 , \theta_1)$$ and $$(r_2 , \theta_2)$$ in polar coordinates is
$\nonumber d ~=~ \sqrt{r_1^2 ~+~ r_2^2 ~-~ 2r_1r_2\,\cos\;(\theta_1 - \theta_2)} ~~.$

Solution:
The idea here is to use the distance formula in Cartesian coordinates, then convert that to polar coordinates. So write
\nonumber \begin{alignat*}{4} x_1 ~&=~ r_1 \,\cos\;\theta_1 \qquad& y_1 ~&=~ r_1 \,\sin\;\theta_1\\ \nonumber x_2 ~&=~ r_2 \,\cos\;\theta_2 \qquad& y_2 ~&=~ r_2 \,\sin\;\theta_2 ~. \end{alignat*}

Then $$(x_1,y_1)$$ and $$(x_2,y_2)$$ are the Cartesian equivalents of $$(r_1 , \theta_1)$$ and $$(r_2 , \theta_2)$$, respectively. Thus, by the Cartesian coordinate distance formula,
\nonumber \begin{align*} d^2 ~&=~ (x_1 - x_2)^2 ~+~ (y_1 - y_2)^2\\ \nonumber &=~ (r_1 \,\cos\;\theta_1 - r_2 \,\cos\;\theta_2)^2 ~+~ (r_1 \,\sin\;\theta_1 - r_2 \,\sin\;\theta_2)^2\\ \nonumber &=~ r_1^2 \cos^2\;\theta_1 ~-~ 2r_1 r_2 \cos\;\theta_1~\cos\;\theta_2 ~+~ r_2^2 \cos^2\;\theta_2 ~+~ r_1^2 \sin^2\;\theta_1 ~-~ 2r_1 r_2 \sin\;\theta_1~\sin\;\theta_2 ~+~ r_2^2 \sin^2\;\theta_2\\ \nonumber &=~ r_1^2 (\cos^2\;\theta_1 ~+~ \sin^2\;\theta_1) ~+~ r_2^2 (\cos^2\;\theta_2 ~+~ \sin^2\;\theta_2) ~-~ 2r_1 r_2 (\cos\;\theta_1~\cos\;\theta_2 ~+~ \sin\;\theta_1~\sin\;\theta_2)\\ \nonumber d^2 ~&=~ r_1^2 ~+~ r_2^2 ~-~ 2r_1r_2\,\cos\;(\theta_1 - \theta_2) ~, \end{align*}
so the result follows by taking square roots of both sides.

In Example 6.17 we saw that the equation $$x^2 + y^2 = 9$$ in Cartesian coordinates could be expressed as $$r = 3$$ in polar coordinates. This equation describes a circle centered at the origin, so the circle is symmetric about the origin. In general, polar coordinates are useful in situations when there is symmetry about the origin (though there are other situations), which arise in many physical applications.