Chapter 4.5 Polar Coordinates
In this section, we will present a different coordinate system that can be used to describe a point in \(\mathbb{R}^2\). We will discuss the differences between polar coordinates and rectangular coordinates.
Polar Coordinates
In rectangular coordinates, a point \(p\) in \(\mathbb{R}^2\) is represented by an ordered pair \((a,b)\), where \(a\) is the number of units to move in the \(x\)-axis and \(b\) is the number of units to move in the \(y\)-axis.
Instead of specifying \(p\) like this, we can specify \(p\) by an ordered pair \((r\colon \theta)\), where \(r\) is the distance of \(p\) from the origin \((0,0)\) and \(\theta\) is the angle in radian measure that the ray from the origin to \(p\) makes with the \(x\)-axis.
This representation is known as polar coordinates.
Polar Coordinates Definition
A point \(p=(r:\theta)\) is the polar coordinate representation of \(p\), where \(r\geq 0\) is the distance from \(p\) to the origin \((0,0)\) and \(\theta\) in \([0,2\pi)\) is the angle the ray from \((0,0)\) to \(p\) makes from the \(x\)-axis.
Usually, a point is plotted in polar coordinates using a graph like this:
Each ring represents the radius \(r\).
In this example, we graph a point in polar coordinates.
Example 1
Graph the point \(p=(5:\frac{\pi}{6})\) in polar coordinates.
The radius is \(r=5\) and the angle is \(\theta=\frac{\pi}{6}\). So the point looks like this.
Any point in polar coordinates can be converted to rectangular coordinates. If a point \(q\) is on the unit circle, then there is an angle \(\theta\) in \([0,2\pi)\) so that \(q=(\cos(\theta),\sin(\theta))\). If \(q\) is not on the unit circle, then take \(r\) to be its distance from the origin so that \(q=(r\cos(\theta),r\sin(\theta))\).
Therefore, \((r\colon \theta)\) in rectangular coordinates is \((r\cos(\theta),r\sin(\theta)).\)
Example 2
Use rectangular coordinates to represent the point \(p\) in the plane that is given in polar coordinates by \(p=(5:\frac{\pi}{6})\).
The radius is \(r=5\) and the angle is \(\theta=\frac{\pi}{6}\). So the point in rectangular coordinates is
\[\left(5\colon \frac{\pi}{6}\right)\to \left(5\cos\left(\frac{\pi}{6}\right),5\sin\left(\frac{\pi}{6}\right)\right)=\left(\frac{5\sqrt{3}}{2},\frac{5}{2}\right).\]
Now practice with the next example.
Example 3
Use rectangular coordinates to represent the point \(p\) in the plane that is given in polar coordinates by \(p=(2:\frac{\pi}{2})\). Also graph \(p\) on a polar grid.
The radius is \(r=2\) and the angle is \(\theta=\frac{\pi}{2}\). So the point in rectangular coordinates is
\[\left(2\colon \frac{\pi}{2}\right)\to \left(2\cos\left(\frac{\pi}{2}\right),2\sin\left(\frac{\pi}{2}\right)\right)=\left(0,2\right).\]
On a polar grid this is where the point is
To convert a point \(p=(x,y)\) from rectangular to polar, use the definition of \(r\) and \(\theta\). The number \(r\) is the distance from \((0,0)\) to \(p\), so \[\sqrt{x^2+y^2}=r\quad\text{so}\quad x^2+y^2=r^2.\] The angle \(\theta\) in radian measures is so that \[x=r\cos(\theta)\quad\text{and}\quad y=r\sin(\theta).\] Altogether, the conversion is the solution \(r\geq 0\) and \(\theta \in [0,2\pi)\) to this system of equations:
\[ \begin{cases} x^2+y^2=r^2\\ x=r\cos(\theta)\\ y=r\sin(\theta). \end{cases} \]
Now practice with the next example.
Example 4
Convert \((0,-4)\) to polar coordinates.
For this example, \(x=0\) and \(y=-4.\)
The system of equations is
\[ \begin{cases} 0^2+(-4)^2=r^2\\ 0=r\cos(\theta)\\ -4=r\sin(\theta) \end{cases}\quad\text{which is equivalent to}\quad \begin{cases} 16=r^2\\ 0=r\cos(\theta)\\ -4=r\sin(\theta) \end{cases}. \]
The solution to the first equation is \(r=4\). So the remaining two equations can be rewritten like this:
\[ \begin{cases} 0=4\cos(\theta)\\ -4=4\sin(\theta) \end{cases}\quad\text{which is equivalent to}\quad \begin{cases} 0=\cos(\theta)\\ -1=\sin(\theta) \end{cases}. \]
The radian measure solution to both equations is \(\theta=\frac{3\pi}{2}.\)
Therefore, the point \((0,-4)\) in polar coordinates is \((4\colon \frac{3\pi}{2}).\)
Here is an additional example like the one we just completed.
Example 5
Convert \((3,-3)\) to polar coordinates.
For this example, \(x=3\) and \(y=-3.\)
The system of equations is
\[ \begin{cases} 3^2+(-3)^2=r^2\\ 3=r\cos(\theta)\\ -3=r\sin(\theta) \end{cases}\quad\text{which is equivalent to}\quad \begin{cases} 18=r^2\\ 3=r\cos(\theta)\\ -3=r\sin(\theta) \end{cases}. \]
The solution to the first equation is \(r=\sqrt{18}\) or \(3\sqrt{2}\). So the remaining two equations can be rewritten like this:
\[ \begin{cases} 3=3\sqrt{2}\cos(\theta)\\ -3=3\sqrt{2}\sin(\theta) \end{cases}\quad\text{which is equivalent to}\quad \begin{cases} \frac{1}{\sqrt{2}}=\cos(\theta)\\ -\frac{1}{\sqrt{2}}=\sin(\theta) \end{cases}. \]
The radian measure solution to both equations is \(\theta=\frac{7\pi}{4}.\)
Therefore, the point \((3,-3)\) in polar coordinates is \((3\sqrt{2}\colon \frac{7\pi}{4}).\)
Polar and Rectangular Equations
A rectangular equation is an equation of the form \[f(x,y)=C\] where \(C\) is a constant and \(f\) is a function of the variables \(x\) and \(y\).
Here are three examples of rectangular equations: \[x+y=2,\quad x^2+4=0,\quad\text{and}\quad y^2+x=1.\]
A polar equation is an equation of the form \[f(r\colon \theta)=C\] where \(C\) is a constant and \(f\) is a function of the variables \(r\) and \(\theta\).
Here are three examples of rectangular equations: \[r-3=0,\quad \sin(\theta)=0,\quad\text{and}\quad \cos(\theta)-\sin(\theta)=0.\]
In the next examples, we will convert a polar equation to a rectangular equation.
Example 6
Convert the following polar equation to a rectangular equation and sketch the equation: \[r=5\csc(\theta).\]
Rewrite everything in terms of \(x\) and \(y\), where
\[x=r\cos(\theta)\quad\text{and}\quad y=r\sin(\theta).\]
First, \(\csc(\theta)=\frac{1}{\sin(\theta)}\),
so
\[\begin{align*} r&=5\csc(\theta)\\ r&=\frac{5}{\sin(\theta)}\\ r\sin(\theta)&=5\\ y&=5. \end{align*}\]
The equivalent rectangular equation is thus \[y=5.\]
This means the original equation is all points on the horizontal line \(y=5.\)
Here is an additional example like the one we just completed.
Example 7
Convert the following polar equation to a rectangular equation and sketch the equation: \[r=3\sec(\theta).\]
Rewrite everything in terms of \(x\) and \(y\), where
\[x=r\cos(\theta)\quad\text{and}\quad y=r\sin(\theta).\]
First, \(\sec(\theta)=\frac{1}{\cos(\theta)}\),
so
\[\begin{align*} r&=3\sec(\theta)\\ r&=\frac{3}{\cos(\theta)}\\ r\cos(\theta)&=3\\ x&=3. \end{align*}\]
The equivalent rectangular equation is thus \[x=3.\]
This means the original equation is all points on the vertical line \(x=3.\)
In the next example, we go from a rectangular equation to a polar equation.
Example 8
Convert the following rectangular equation to a polar equation and sketch the equation: \[x^2+y^2=16.\]
Rewrite everything in terms of \(r\) and \(\theta\), where
\[x=r\cos(\theta)\quad\text{and}\quad y=r\sin(\theta).\]
The equation looks like this:
\[\begin{align*} x^2+y^2&=16\\ (r\cos(\theta))^2+(r\sin(\theta))^2&=16\\ r^2\cos^2(\theta)+r^2\sin^2(\theta)&=16\\ r^2\left(\cos^2(\theta)+\sin^2(\theta)\right)&=16\\ r^2\cdot 1&=16\\ r^2&=16\\ r&=4. \end{align*}\]
The equivalent polar equation is thus \[r=4.\] There is no restriction on \(\theta\), so any angle in \([0,2\pi)\) satisfies the equation. Therefore, the equation represents a circle with center \((0,0)\) of radius \(4.\)
Here is another example. In this example, we do additional rewriting to determine the shape.
Example 9
Convert the following rectangular equation to a polar equation and sketch the equation: \[x^2+y^2=2x.\]
Rewrite everything in terms of \(r\) and \(\theta\), where
\[x=r\cos(\theta)\quad\text{and}\quad y=r\sin(\theta).\]
The equation looks like this:
\[\begin{align*} x^2+y^2&=2x\\ (r\cos(\theta))^2+(r\sin(\theta))^2&=2r\cos(\theta)\\ r^2\cos^2(\theta)+r^2\sin^2(\theta)&=2r\cos(\theta)\\ r^2\left(\cos^2(\theta)+\sin^2(\theta)\right)&=2r\cos(\theta)\\ r^2\cdot 1&=2r\cos(\theta)\\ r^2&=2r\cos(\theta)\\ r^2-2r\cos(\theta)&=0\\ r(r-2\cos(\theta))&=0, \end{align*}\]
which can be rewritten as \[r=0\quad\text{or}\quad r=2\cos(\theta).\]
The equation \(r=0\) is called a degenerate equation because the only solution is \((0:0).\)
To understand what its sketch looks like, look at the original rectangular equation and rewrite it.
\[\begin{align*} x^2+y^2&=2x\\ x^2-2x+y^2&=0. \end{align*}\]
The term \(x^2-2x\) can be rewritten as \((x-1)^2-1\).
So,
\[\begin{align*} x^2+y^2&=2x\\ x^2-2x+y^2&=0\\ (x-1)^2-1+y^2&=0\\ (x-1)^2+y^2&=1. \end{align*}\]
This a circle with center \((1,0)\) and radius \(r=1.\)
Here is another example.
Example 10
Convert the following rectangular equation to a polar equation and sketch the equation: \[x^2+y^2=4y.\]
Rewrite everything in terms of \(r\) and \(\theta\), where
\[x=r\cos(\theta)\quad\text{and}\quad y=r\sin(\theta).\]
The equation looks like this:
\[\begin{align*} x^2+y^2&=4y\\ (r\cos(\theta))^2+(r\sin(\theta))^2&=4r\sin(\theta)\\ r^2\cos^2(\theta)+r^2\sin^2(\theta)&=4r\sin(\theta)\\ r^2\left(\cos^2(\theta)+\sin^2(\theta)\right)&=4r\sin(\theta)\\ r^2\cdot 1&=4r\sin(\theta)\\ r^2&=4r\sin(\theta)\\ r^2-4r\sin(\theta)&=0\\ r(r-4\sin(\theta))&=0, \end{align*}\]
which can be rewritten as \[r=0\quad\text{or}\quad r=4\sin(\theta).\]
The equation \(r=0\) is called a degenerate equation because the only solution is \((0:0).\)
To understand what its sketch looks like, look at the original rectangular equation and rewrite it.
\[\begin{align*} x^2+y^2&=4y\\ x^2+y^2-4y&=0. \end{align*}\]
The term \(y^2-4y\) can be rewritten as \((y-2)^2-4\).
So,
\[\begin{align*} x^2+y^2&=4y\\ x^2+y^2-4y&=0\\ x^2+(y-2)^2-4&=0\\ x^2+(y-2)^2&=4. \end{align*}\]
This a circle with center \((0,2)\) and radius \(r=2.\)
