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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 (cartesian) coordinates in two main areas, converting points and converting equations.

### Search 17Calculus

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.

determining θ for $$\tan(\theta)=y/x$$

$$\theta = \arctan(y/x)$$

$$x > 0$$

$$\theta = \arctan(y/x)+\pi$$

$$x < 0, y \geq 0$$

$$\theta = \arctan(y/x)-\pi$$

$$x < 0, y < 0$$

$$\theta = +\pi/2$$

$$x = 0, y > 0$$

$$\theta = -\pi/2$$

$$x = 0, y < 0$$

$$\theta = 0$$

$$x=0, y=0$$

Okay, time for some practice problems.

Practice 1

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

solution

Practice 2

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

solution

Practice 3

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

solution

 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.

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

Practice 4

$$x^2+y^2=1$$ → $$g(r,\theta)$$

solution

Practice 5

$$y=2x+1$$ → $$g(r,\theta)$$

solution

Practice 6

$$y=3/x$$ → $$g(r,\theta)$$

solution

Practice 7

$$y=x^2$$ → $$g(r,\theta)$$

solution

Practice 8

$$xy=1$$ → $$g(r,\theta)$$

solution

Practice 9

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.

solution

 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 $$y = r \sin(\theta)$$.

 Basic Problems

Practice 10

$$(4,\pi/3)$$ → $$(x,y)$$

solution

Practice 11

$$(2,\pi)$$ → $$(x,y)$$

solution

Practice 12

$$(1,\pi/4)$$ → $$(x,y)$$

solution

Practice 13

$$(-2,2\pi/3)$$ → $$(x,y)$$

solution

Practice 14

$$(1,-\pi/3)$$ → $$(x,y)$$

solution

Practice 15

$$(3,3\pi/2)$$ → $$(x,y)$$

solution

Practice 16

$$(2,-\pi/4)$$ → $$(x,y)$$

solution

 Intermediate Problems

Practice 17

$$(-7,-2\pi/3)$$ → $$(x,y)$$

solution

 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 identity $$\cos^2(\theta) + \sin^2(\theta) = 1$$. In terms of the above equations, we have
$$\displaystyle{ \begin{array}{rcl} x^2 + y^2 & = & [ r \cos(\theta) ]^2 + [ r \sin(\theta) ]^2 \\ & = & r^2[ \cos^2(\theta) + r \sin^2(\theta)] \\ & = & r^2 \end{array} }$$

 Basic Problems

Practice 18

$$r=3\cos(\theta)$$ → $$f(x,y)$$

solution

Practice 19

$$r=-5\cos(\theta)$$ → $$f(x,y)$$

solution

 Intermediate Problems

Practice 20

$$r^2=-3\sec(\theta)$$ → $$f(x,y)$$

$$r=3\sin(\theta)\tan(\theta)$$ → $$f(x,y)$$