Introduction to polar coordinates

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Polar coordinates are a special type of parametric curves.

Polar coordinates

Now we focus on a special type of parametric equations, those of the form:

$\begin{align*} x(\theta) &= r(\theta) \cdot \cos(\theta)\\ y(\theta) &= r(\theta) \cdot \sin(\theta) \end{align*}$ where $r(\theta )$ is a function of $\theta$ . When working with parametric equations of this form, it is common to notate $(r \cdot \cos (\theta ), r\cdot \sin (\theta )) \text { as } (r,\theta )$ and state that we are working in polar coordinates.

An ordered pair consisting of a radius and an angle $(r,\theta )$ can be graphed as $\begin{align*} x &= r\cdot \cos(\theta)\\ y &= r\cdot \sin(\theta) \end{align*}$ meaning:

Coordinates of this type are called polar coordinates.

Polar coordinates are great for certain situations. However, there is a price to pay. Every point in the plane has more than one of description in polar coordinates.

Which of the following represent the origin, $(0,0)$ , in $(x,y)$ -coordinates?

$(0,0)$ $(0,\pi )$ $(0,-\pi )$

All of these represent the origin, since $(0,\theta )$ represents the origin for any angle $\theta$ .

Plot the following points in polar coordinates: $A =(1,\pi /4)\quad B=(1.5,\pi )\quad C = (2,-\pi /3)\quad D = (-1,\pi /4)$

It helps to use a “polar grid” to plot these points:

To place the point $A$ , go out $1$ unit along the horizontal axis (putting you on the inner circle shown on the grid), then rotate clockwisecounterclockwise $\pi /4$ radians (or $45^\circ$ ).

To plot $B$ , go out $1.5$ units along the horizontal axis and rotate $\pi$ radians ( $180^\circ$ ).

To plot $C$ , go out 2 units along the initial ray then rotate $\pi /3$ radians, as the angle given is negative.

To plot $D$ , move along the initial ray “ $-1$ ” units, in other words, “back up” $1$ unit, then rotate by $\pi /4$ .

It is useful to recognize both the rectangular (or, Cartesian) coordinates of a point in the plane and its polar coordinates.

Given a point $P=(r,\theta )$ in polar coordinates, rectangular coordinates are given by $x=r\cos \theta \qquad y=r\sin \theta .$ Given a point $Q=(x,y)$ in rectangular coordinates, polar coordinates are given by $r^2=x^2+y^2\qquad \tan \theta = \frac yx.$

Let $P=(2,2\pi /3)$ be a point in polar coordinates. Describe $P$ in rectangular coordinates. $P = (\answer {2\cos (2\pi /3)}, \answer {2\sin (2\pi /3)})$

Let $Q=(-1,5\pi /4)$ be a point in polar coordinates. Describe $Q$ in rectangular coordinates. $Q = (\answer {-1\cos (5\pi /4)}, \answer {-1\sin (5\pi /4)})$

Let $P=(1,2)$ be a point in rectangular coordinates. Describe $P$ in polar coordinates. $P = (\answer {\sqrt {5}}, \answer {\arctan (2)})$

Let $Q=(-1,1)$ be a point in rectangular coordinates. Describe $Q$ in polar coordinates. $Q = (\answer {-\sqrt {2}}, \arctan (-1))$

We’ll tell you the angle, you think about the radius.

Polar graphs

Let’s talk about how to plot polar functions. A polar function $r(\theta )$ corresponds to the parametric function:

$\begin{align*} x(\theta) &= r(\theta) \cdot \cos(\theta)\\ y(\theta) &= r(\theta) \cdot \sin(\theta) \end{align*}$ However, if you are sketching a polar function by hand, there are some tricks that can help. If you want to sketch $r(\theta )$ , it is often useful to first set $\theta = x$ , and plot $y=r(x)$ in rectangular coordinates. Let’s just work examples. It is my belief that “doing things” is better than “describing.”

Sketch the polar function $r=1+\cos \theta$ on $[0,2\pi ]$ .

While one could make a table of values and plot them in polar coordinates, it is often more useful to first set $\theta =x$ and then plot $y=r(x)$ . This is what we’ll do, starting with $x = \pi /4$ :

Sketch the polar function $r=\cos (2 \theta )$ on $[0,2\pi ]$ .

Converting to and from polar coordinates

It is sometimes desirable to refer to a graph via a polar equation, and other times by a rectangular equation. Therefore it is necessary to be able to convert between polar and rectangular functions. Here is the basic idea:

Given a function $y=f(x)$ in rectangular coordinates, polar coordinates are given by setting $x=r\cos (\theta )\qquad y=r\sin (\theta ).$ and solving for $r$ .

Given a function $r(\theta )$ in polar coordinates, rectangular coordinates harder to find. The basic idea is to “find” $r\cdot \cos (\theta )$ and $r\cdot \sin (\theta )$ and write: $r\cos (\theta ) = x\qquad r\sin (\theta ) = y.$ Sometimes it is useful to remember that: $r^2=x^2+y^2\qquad \tan \theta = \frac yx.$

Convert $y=x^2$ from rectangular coordinates to polar coordinates.

Replace $y$ with $r\sin (\theta )$ and replace $x$ with $r\cos (\theta )$ , giving: $\begin{align*} y &=x^2\\ r\sin(\theta) &= \answer[given]{r^2\cos^2(\theta)}\\ \answer[given]{\frac{\sin(\theta)}{\cos^2(\theta)}} &= r. \end{align*}$ We have found that $r=\sin \theta /\cos ^2\theta = \tan \theta \sec \theta$ . The domain of this polar function is $[\answer [given]{-\pi /2},\answer [given]{\pi /2}]$ . Plot a few points to see how the familiar parabola is traced out by the polar equation.

Convert $xy = 1$ from rectangular coordinates to polar coordinates.

We again replace $x$ and $y$ using the standard identities and work to solve for $r$ : $\begin{align*} xy &= 1 \\ \answer[given]{r\cos\theta\cdot r\sin\theta} & = 1\\ r^2 & = \frac{1}{\cos\theta\sin\theta}\\ r & = \frac{1}{\sqrt{\cos\theta\sin\theta}}\\ \end{align*}$ This function is valid only when the product of $\cos \theta \sin \theta$ is positive. This occurs in the first and third quadrants, meaning the domain of this polar function is $(0,\pi /2)$ with $(\pi ,3\pi /2)$ .

We can rewrite the original rectangular equation $xy=1$ as $y=1/x$ . Note it only exists in the first and third quadrants.

Convert $r=\frac {2}{\sin \theta -\cos \theta }$ from polar coordinates to rectangular coordinates.

There is no set way to convert from polar to rectangular; in general, we look to form the products $r\cos \theta$ and $r\sin \theta$ , and then replace these with $x$ and $y$ , respectively. We start in this problem by multiplying both sides by $\sin \theta -\cos \theta$ : $\begin{align*} r &= \frac{2}{\sin\theta-\cos\theta} \\ r(\sin\theta-\cos\theta) &= 2\\ \answer[given]{r\sin\theta}-r\cos\theta &= 2. \qquad \text{Now replace with $y$ and $x$:}\\ y-x &= 2\\ y &= x+2. \end{align*}$ The original polar equation, $r=2/(\sin \theta -\cos \theta )$ does not easily reveal that its graph is simply a line. However, our conversion shows that it is.

Convert $r =2\cos \theta$ from polar coordinates to rectangular coordinates.

By multiplying both sides by $r$ , we obtain both an $r^2$ term and an $r\cos \theta$ term, which we replace with $x^2+y^2$ and $x$ , respectively. $\begin{align*} r &=2\cos\theta \\ r^2 &= 2r\cos\theta \\ \answer[given]{x^2+y^2} &= \answer[given]{2x}. \end{align*}$ We recognize this as a circle. By completing the square we can find its radius and center. $\begin{align*} x^2-2x+y^2 &= 0 \\ \answer[given]{(x-1)^2 + y^2} &=1. \end{align*}$ The circle is centered at $(1,0)$ and has radius $1$ .