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You CAN Ace Calculus

17calculus > polar coordinates > converting

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Converting Between Rectangular and Polar Coordinates

On this page, we discuss converting between polar and rectangular coordinates in two main areas, converting points and converting equations.

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.

 PatrickJMT - using the Pythagorean Theorem
 Rectangular → Polar Points

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.

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

 Rectangular → Polar Equations

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.

 $$x = r \cos(\theta)$$ $$y = r \sin(\theta)$$
 Polar → Rectangular Points

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.

 Polar → Rectangular Equations

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
$$\displaystyle{ \begin{array}{rcl} x^2 + y^2 & = & [ r \cos(\theta) ]^2 + [ r \sin(\theta) ]^2 \\ & = & r^2[ \cos^2(\theta) + r \sin^2(\theta)] \\ & = & r^2 \end{array} }$$

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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.

 Level A - Basic

Practice A01

given $$(x,y)=(0,2)$$
calculate $$(r,\theta)$$

solution

Practice A02

given $$(x,y)=(-3,4)$$
calculate $$(r,\theta)$$

solution

Practice A03

given $$(x,y)=(1,-5)$$
calculate $$(r,\theta)$$

solution

Practice A04

calculate $$(x,y)$$
given $$(r,\theta)=(4,\pi/3)$$

solution

Practice A05

calculate $$(x,y)$$
given $$(r,\theta)=(2,\pi)$$

solution

Practice A06

calculate $$f(x,y)=0$$
given $$r=3\cos(\theta)$$

solution

Practice A07

given $$x^2+y^2=1$$
calculate $$g(r,\theta)=0$$

solution

Practice A08

given $$y=2x+1$$
calculate $$g(r,\theta)=0$$

solution

Practice A09

given $$y=3/x$$
calculate $$g(r,\theta)=0$$

solution

Practice A10

calculate $$(x,y)$$
given $$(r,\theta)=(1,\pi/4)$$

solution

Practice A11

calculate $$(x,y)$$
given $$(r,\theta)=(-2,2\pi/3)$$

solution

Practice A12

calculate $$(x,y)$$
given $$(r,\theta)=(1,-\pi/3)$$

solution

Practice A13

calculate $$(x,y)$$
given $$(r,\theta)=(3,3\pi/2)$$

solution

Practice A14

calculate $$(x,y)$$
given $$(r,\theta)=(2,-\pi/4)$$

solution

Practice A15

calculate $$f(x,y)=0$$
given $$r=-5\cos(\theta)$$

solution

Practice A16

given $$y=x^2$$
calculate $$g(r,\theta)=0$$

solution

Practice A17

given $$xy=1$$
calculate $$g(r,\theta)=0$$

solution

 Level B - Intermediate

Practice B01

calculate $$(x,y)$$
given $$(r,\theta)=(-7,-2\pi/3)$$

solution

Practice B02

calculate $$f(x,y)=0$$
given $$r^2=-3\sec(\theta)$$

solution

Practice B03

calculate $$f(x,y)=0$$
given $$r=3\sin(\theta)\tan(\theta)$$

For the two 3-dim curves, $$2z=x^2-y^2+2x$$ and $$3z=4x^2+y^2-2x$$, determine the projection of the intersection of these curves in the xy-plane and describe the projection in polar coordinates.