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17calculus > polar coordinates > converting

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.

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.

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 \(\tan(\theta)=y/x\)

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

\(x > 0\)

quadrant 1 or 4

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

\(x < 0, y \geq 0\)

quadrant 2

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

\(x < 0, y < 0\)

quadrant 3

\(\theta = +\pi/2\)

\(x = 0, y > 0\)

\(\theta = -\pi/2\)

\(x = 0, y < 0\)

\( \theta = 0 \)

\(x=0, y=0\)

[ source: wikipedia ]

Okay, time for some practice problems.

Practice 1

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

answer

solution

Practice 2

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

answer

solution

Practice 3

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

answer

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

\(x^2+y^2=1\) → \(g(r,\theta) \)

answer

solution

Practice 5

\(y=2x+1\) → \(g(r,\theta) \)

answer

solution

Practice 6

\(y=3/x\) → \(g(r,\theta) \)

answer

solution

Practice 7

\( y=x^2 \) → \( g(r,\theta) \)

answer

solution

Practice 8

\(xy=1\) → \(g(r,\theta)\)

answer

solution

Advanced Problems

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

\((4,\pi/3)\) → \((x,y)\)

answer

solution

Practice 11

\((2,\pi)\) → \((x,y)\)

answer

solution

Practice 12

\((1,\pi/4)\) → \((x,y)\)

answer

solution

Practice 13

\((-2,2\pi/3)\) → \((x,y)\)

answer

solution

Practice 14

\((1,-\pi/3)\) → \((x,y)\)

answer

solution

Practice 15

\((3,3\pi/2)\) → \((x,y)\)

answer

solution

Practice 16

\((2,-\pi/4)\) → \((x,y)\)

answer

solution

Intermediate Problems

Practice 17

\((-7,-2\pi/3)\) → \((x,y)\)

answer

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

\(r=3\cos(\theta)\) → \(f(x,y)\)

answer

solution

Practice 19

\(r=-5\cos(\theta)\) → \(f(x,y)\)

answer

solution

Intermediate Problems

Practice 20

\(r^2=-3\sec(\theta)\) → \(f(x,y)\)

answer

solution

Practice 21

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

answer

solution

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