Coordinate systems are tools that let us use algebraic methods to
understand geometry. While the
**rectangular**
(also called
**Cartesian**)
coordinates that we
have been using are the most common, some problems are easier to
analyze in alternate coordinate systems.

A coordinate system is a scheme that allows us to identify any point
in the plane or in three-dimensional space by a set of numbers. In
rectangular coordinates these numbers are interpreted, roughly
speaking, as the lengths of the sides of a rectangle.
In **polar
coordinates** a
point in the plane is identified by a pair of numbers $(r,\theta)$.
The number $\theta$ measures the angle between the positive
$x$-axis and a ray that goes through the point,
as shown in figure 12.1.1; the number
$r$ measures the distance from the origin to the
point. Figure 12.1.1 shows the point with
rectangular coordinates $\ds (1,\sqrt3)$ and polar coordinates
$(2,\pi/3)$, 2 units from the origin and $\pi/3$ radians from the
positive $x$-axis.

Just as we describe curves in the plane using equations involving $x$ and $y$, so can we describe curves using equations involving $r$ and $\theta$. Most common are equations of the form $r=f(\theta)$.

Example 12.1.1 Graph the curve given by $r=2$. All points with $r=2$ are at distance 2 from the origin, so $r=2$ describes the circle of radius 2 with center at the origin. $\square$

Example 12.1.2 Graph the curve given by $r=1+\cos\theta$. We first consider $y=1+\cos x$, as in figure 12.1.2. As $\theta$ goes through the values in $[0,2\pi]$, the value of $r$ tracks the value of $y$, forming the "cardioid'' shape of figure 12.1.2. For example, when $\theta=\pi/2$, $r=1+\cos(\pi/2)=1$, so we graph the point at distance 1 from the origin along the positive $y$-axis, which is at an angle of $\pi/2$ from the positive $x$-axis. When $\theta=7\pi/4$, $\ds r=1+\cos(7\pi/4)=1+\sqrt2/2\approx 1.71$, and the corresponding point appears in the fourth quadrant. This illustrates one of the potential benefits of using polar coordinates: the equation for this curve in rectangular coordinates would be quite complicated. $\square$

Each point in the plane is associated with exactly one pair of numbers in the rectangular coordinate system; each point is associated with an infinite number of pairs in polar coordinates. In the cardioid example, we considered only the range $0\le \theta\le2\pi$, and already there was a duplicate: $(2,0)$ and $(2,2\pi)$ are the same point. Indeed, every value of $\theta$ outside the interval $[0,2\pi)$ duplicates a point on the curve $r=1+\cos\theta$ when $0\le\theta< 2\pi$. We can even make sense of polar coordinates like $(-2,\pi/4)$: go to the direction $\pi/4$ and then move a distance 2 in the opposite direction; see figure 12.1.3. As usual, a negative angle $\theta$ means an angle measured clockwise from the positive $x$-axis. The point in figure 12.1.3 also has coordinates $(2,5\pi/4)$ and $(2,-3\pi/4)$.

The relationship between rectangular and polar coordinates is quite easy to understand. The point with polar coordinates $(r,\theta)$ has rectangular coordinates $x=r\cos\theta$ and $y=r\sin\theta$; this follows immediately from the definition of the sine and cosine functions. Using figure 12.1.3 as an example, the point shown has rectangular coordinates $\ds x=(-2)\cos(\pi/4)=-\sqrt2\approx 1.4142$ and $\ds y=(-2)\sin(\pi/4)=-\sqrt2$. This makes it very easy to convert equations from rectangular to polar coordinates.

Example 12.1.3 Find the equation of the line $y=3x+2$ in polar coordinates. We merely substitute: $r\sin\theta=3r\cos\theta+2$, or $\ds r= {2\over \sin\theta-3\cos\theta}$. $\square$

Example 12.1.4 Find the equation of the circle $\ds (x-1/2)^2+y^2=1/4$ in polar coordinates. Again substituting: $\ds (r\cos\theta-1/2)^2+r^2\sin^2\theta=1/4$. A bit of algebra turns this into $r=\cos(t)$. You should try plotting a few $(r,\theta)$ values to convince yourself that this makes sense. $\square$

Example 12.1.5 Graph the polar equation $r=\theta$. Here the distance from the origin exactly matches the angle, so a bit of thought makes it clear that when $\theta\ge0$ we get the spiral of Archimedes in figure 12.1.4. When $\theta< 0$, $r$ is also negative, and so the full graph is the right hand picture in the figure. $\square$

Converting polar equations to rectangular equations can be somewhat trickier, and graphing polar equations directly is also not always easy.

Example 12.1.6 Graph $r=2\sin\theta$. Because the sine is periodic, we know that we will get the entire curve for values of $\theta$ in $[0,2\pi)$. As $\theta$ runs from 0 to $\pi/2$, $r$ increases from 0 to 2. Then as $\theta$ continues to $\pi$, $r$ decreases again to 0. When $\theta$ runs from $\pi$ to $2\pi$, $r$ is negative, and it is not hard to see that the first part of the curve is simply traced out again, so in fact we get the whole curve for values of $\theta$ in $[0,\pi)$. Thus, the curve looks something like figure 12.1.5. Now, this suggests that the curve could possibly be a circle, and if it is, it would have to be the circle $\ds x^2+(y-1)^2=1$. Having made this guess, we can easily check it. First we substitute for $x$ and $y$ to get $\ds (r\cos\theta)^2+(r\sin\theta-1)^2=1$; expanding and simplifying does indeed turn this into $r=2\sin\theta$. $\square$

## Exercises 12.1

**Ex 12.1.1**
Plot these polar coordinate points on one graph:
$(2,\pi/3)$, $(-3,\pi/2)$, $(-2,-\pi/4)$, $(1/2,\pi)$, $(1,4\pi/3)$,
$(0,3\pi/2)$.

Find an equation in polar coordinates that has the same graph as the given equation in rectangular coordinates.

**Ex 12.1.2**
$\ds y=3x$
(answer)

**Ex 12.1.3**
$\ds y=-4$
(answer)

**Ex 12.1.4**
$\ds xy^2=1$
(answer)

**Ex 12.1.5**
$\ds x^2+y^2=5$
(answer)

**Ex 12.1.6**
$\ds y=x^3$
(answer)

**Ex 12.1.7**
$\ds y=\sin x$
(answer)

**Ex 12.1.8**
$\ds y=5x+2$
(answer)

**Ex 12.1.9**
$\ds x=2$
(answer)

**Ex 12.1.10**
$\ds y=x^2+1$
(answer)

**Ex 12.1.11**
$\ds y=3x^2-2x$
(answer)

**Ex 12.1.12**
$\ds y=x^2+y^2$
(answer)

Sketch the curve. You can check your work with Sage.

**Ex 12.1.13**
$\ds r=\cos\theta$

**Ex 12.1.14**
$\ds r=\sin(\theta+\pi/4)$

**Ex 12.1.15**
$\ds r=-\sec\theta$

**Ex 12.1.16**
$\ds r=\theta/2$, $\theta\ge0$

**Ex 12.1.17**
$\ds r=1+\theta/\pi^2$

**Ex 12.1.18**
$\ds r=\cot\theta\csc\theta$

**Ex 12.1.19**
$\ds r={1\over\sin\theta+\cos\theta}$

**Ex 12.1.20**
$\ds r^2=-2\sec\theta\csc\theta$

In the exercises below, find an equation in rectangular coordinates that has the same graph as the given equation in polar coordinates.

**Ex 12.1.21**
$\ds r=\sin(3\theta)$
(answer)

**Ex 12.1.22**
$\ds r=\sin^2\theta$
(answer)

**Ex 12.1.23**
$\ds r=\sec\theta\csc\theta$
(answer)

**Ex 12.1.24**
$\ds r=\tan\theta$
(answer)