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Wikipedia  Converting Between Polar and Cartesian Coordinates 
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.
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 halfopen (or halfclosed) 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 
\(\theta = +\pi/2\) 
\(x = 0, y > 0\) 

\(\theta = \pi/2\) 
\(x = 0, y < 0\) 

\( \theta = 0 \) 
\(x=0, y=0\) 

Okay, time for some practice problems.
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 
Advanced Problems 
Practice 9 

For the two 3dim curves, \(2z=x^2y^2+2x\) and \(3z=4x^2+y^22x\), determine the projection of the intersection of these curves in the xyplane 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 \( y = r \sin(\theta)\).
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
\(\displaystyle{
\begin{array}{rcl}
x^2 + y^2 & = & [ r \cos(\theta) ]^2 + [ r \sin(\theta) ]^2 \\
& = & r^2[ \cos^2(\theta) + r \sin^2(\theta)] \\
& = & r^2
\end{array}
}\)