Converting Between Rectangular and Polar Coordinates
You CAN Ace Calculus |
---|
17trek > calculus > polar coordinates > converting |
Topics You Need To Understand For This Page
Page Tools, Related Topics and Links
sections on this page |
---|
external links you may find helpful |
Wikipedia - Converting Between Polar and Cartesian Coordinates |
Search 17calculus
17calculus is a 17trek subject |
---|
quick links |
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 | |
---|---|---|
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
\(\displaystyle{
\begin{array}{rcl}
x^2 + y^2 & = & [ r \cos(\theta) ]^2 + [ r \sin(\theta) ]^2 \\
& = & r^2[ \cos^2(\theta) + r \sin^2(\theta)] \\
& = & r^2
\end{array}
}\)