## Converting Between Rectangular and Polar Coordinates

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On this page, we discuss converting between polar and rectangular coordinates in two main areas, converting points and converting equations. Here is a table showing the main topics with links to jump to the topic.

Converting Points In The Plane Converting Equations rectangular → polar rectangular → polar polar → rectangular polar → rectangular

As discussed on the main polar coordinates page, the two main equations we use to convert between polar and rectangular (cartesian) coordinates are

 $$x = r \cos(\theta)$$ $$y = r \sin(\theta)$$

All other equations can be derived from these two equations.

 This video clip (less than a minute long) explains really well how to use trigonometry and the Pythagorean Theorem to get these equations.

In this case, we are given a specific point $$(x,y)$$ and we want to find $$r$$ and $$\theta$$ that describes this point. This direction is a little more complicated because we need to pay special attention to the signs of $$x$$ and y.

The equations we use are $$r = \sqrt{x^2+y^2}$$ and $$\theta= \arctan(y/x)$$. Let's discuss these equations in more detail. This discussion is critical for you to understand in order to correctly determine the polar coordinates.

The first equation looks easy but there is a hidden assumption that you need to be aware of. Looking at the graph on the right, you know from the Pythagorean Theorem that $$r^2 = x^2 + y^2$$. So far, so good. Now, to solve for $$r$$ we take the square root of both sides.

$$\begin{array}{rcl} r^2 & = & x^2 + y^2 \\ \sqrt{r^2} & = & \pm \sqrt{x^2 + y^2} \\ r & = & \pm \sqrt{x^2 + y^2} \end{array}$$

In the second line above, we could have put the $$\pm$$ sign on either side, but to be strictly true, we needed it. (Do you see why?)

Now, in the equation $$r=\sqrt{x^2+y^2}$$ we dropped the $$\pm$$ sign. What does that mean? That means we are assuming that $$r$$ is always positive. So, when we convert from rectangular to polar coordinates, we will take $$r$$ to be positive. This is a subtle point but you need to keep that in mind. (As a teacher, one of my favorite questions on homework or exams will be to ask what happens when $$r$$ is negative. See the practice problems below for examples of this case.)

Okay, now for the equation $$\theta= \arctan(y/x)$$. This takes special care. You cannot just divide y by x and plug that value into your calculator. You need to know the sign of both x and y, which will determine the quadrant of your answer and thus the angle $$\theta$$. For discussion here, we want the angle to be in the interval $$(-\pi, \pi]$$. Notice this is a half-open (or half-closed) interval and includes $$\pi$$ but excludes $$-\pi$$. Of course, you need to check with your instructor to see what they require.

The best way that we've found to determine this angle, can be found using these equations. [ source ]

$$\theta = \left\{ \begin{array}{lll} \arctan(y/x) & x > 0 & \text{quadrant 1 or 4} \\ \arctan(y/x)+\pi & x < 0, y \geq 0 & \text{quadrant 2} \\ \arctan(y/x)-\pi & x < 0, y < 0 & \text{quadrant 3} \\ +\pi/2 & x = 0, y > 0 & \\ -\pi/2 & x = 0, y < 0 & \\ 0 & x=0, y=0 & \end{array} \right.$$

In this case, we are given an equation $$f(x,y)=0$$ and we need to find an equation $$g(r, \theta)=0$$. This is pretty easy. We just substitute for x and y using these equations and simplify.

 $$x = r \cos(\theta)$$ $$y = r \sin(\theta)$$

In this case, we are given specific $$r$$ and $$\theta$$ values and we want to find the point $$(x,y)$$. This is the easiest direction because we already have the equations in the form we need, i.e. $$x = r \cos(\theta)$$ and $$y = r \sin(\theta)$$. For examples, check out the practice problems.

Equations - - Here, we are given a function $$g(r,\theta)=0$$ and we need to find an equation $$f(x,y)=0$$. One of the keys to solving these problems is to use the fact $$\cos^2(\theta) + \sin^2(\theta) = 1$$. In terms of the above equations, we have
$$\displaystyle{ \begin{array}{rcl} x^2 + y^2 & = & [ r \cos(\theta) ]^2 + [ r \sin(\theta) ]^2 \\ & = & r^2[ \cos^2(\theta) + r \sin^2(\theta)] \\ & = & r^2 \end{array} }$$

### Polar Converting Resources

filters

rect → polar

polar → rect

points

equations

Practice Problems

Instructions - - Unless otherwise instructed, convert the following points or equations accordingly, giving angles in radians and all other answers in exact terms.
Note: When you are asked to calculate $$g(r,\theta) = 0$$ this means to find the equation in polar form and move all terms to one side of the equal sign for your final answer. Similarly for equations in rectangular form.
[ Click on the practice problem to reveal/hide the solution. ]

 Level A - Basic

 Level B - Intermediate