10.3 Polar Coordinates

Cartesian vs Polar Coordinates

Cartesian, or rectangular, coordinates are not the only way to describe a point in the 2D plane. The rectangular way is to describe a point as an x and a y, the horizontal and vertical distance from the origin. In many cases, it is more natural to describe a coordinate as an angle and a distance. The angle is usually labeled theta and is measured in standard position while the distance is marked as r as if it were the radius of a circle.

Example: Plotting polar coordinates

Plot the following polar coordinates:

1. $(2,210^\circ)$
2. $(2,-150^\circ)$
3. $(4,60^\circ)$
4. $(4,420^\circ)$

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Negative r and $\theta$

Both r and $\theta$ can be negative in polar coordinates. A negative angle should be intuitive by this point; one only needs to measure clockwise from the positive x-axis. A negative r value is similar in that we're going backwards. We measure the angle as usual, then instead of moving from the origin into the quadrant that the angle is pointing to, we move away from that quadrant exactly $180^\circ$ from the measured $\theta$.

Example: Plot polar coordinates

Plot the following polar coordinates.

1. $(-3,\frac{\pi}{2})$
2. $(-5,\frac{5\pi}{3})$
3. $(-2,-390^\circ)$

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Example: Identify point from graph

For the following graphs, write the point shown in polar coordinates with $r>0$, and $0\leq\theta<360^\circ$ or $0\leq\theta<2\pi$

Show Solution

1. $(1,240^\circ)$ or $(1,\frac{4\pi}{3})$
2. $(5,180^\circ)$ or $(5,\pi)$
3. $(2,330^\circ)$ or $(2,\frac{11\pi}{6})$

Converting Coordinates

Now we'll work on converting between our two coordinate systems. Luckily for us, if you draw a line from the origin to any point in the Cartesian plane, we get a nice right triangle, which means everything we need comes from good ol' Pythagoras.

::solution

1. Use $x=r\cos(\theta)$ and $y=r\sin(\theta)$.
   \[ \begin{align*} x&=6\cos\left(\frac{\pi}{6}\right)=6\cdot\frac{\sqrt{3}}{2}=3\sqrt{3}\\ y&=6\sin\left(\frac{\pi}{6}\right)=6\cdot\frac{1}{2}=3 \end{align*} \]

   So the rectangular coordinates are $(3\sqrt{3},3)$.

2. Use $x=r\cos(\theta)$ and $y=r\sin(\theta)$.
   \[ \begin{align*} x&=-4\cos\left(-\frac{\pi}{4}\right)=-4\cdot\frac{\sqrt{2}}{2}=-2\sqrt{2}\\ y&=-4\sin\left(-\frac{\pi}{4}\right)=-4\cdot\left(-\frac{\sqrt{2}}{2}\right)=2\sqrt{2} \end{align*} \]

   So the rectangular coordinates are $(-2\sqrt{2},2\sqrt{2})$.

3. Use $x=r\cos(\theta)$ and $y=r\sin(\theta)$.
   \[ \begin{align*} x&=3\cos\left(\frac{5\pi}{3}\right)=3\cdot\frac{1}{2}=\frac{3}{2}\\ y&=3\sin\left(\frac{5\pi}{3}\right)=3\cdot\left(-\frac{\sqrt{3}}{2}\right)=-\frac{3\sqrt{3}}{2} \end{align*} \]

   So the rectangular coordinates are $\left(\frac{3}{2},-\frac{3\sqrt{3}}{2}\right)$.

4. Use $x=r\cos(\theta)$ and $y=r\sin(\theta)$.
   \[ \begin{align*} x&=-5\cos\left(\frac{7\pi}{6}\right)=-5\cdot\left(-\frac{\sqrt{3}}{2}\right)=\frac{5\sqrt{3}}{2}\\ y&=-5\sin\left(\frac{7\pi}{6}\right)=-5\cdot\left(-\frac{1}{2}\right)=\frac{5}{2} \end{align*} \]

   So the rectangular coordinates are $\left(\frac{5\sqrt{3}}{2},\frac{5}{2}\right)$.

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::solution

1. First find $r$.
   \[ \begin{align*} r&=\sqrt{x^2+y^2}\\ r&=\sqrt{1^2+(-\sqrt{3})^2}\\ r&=\sqrt{1+3}=2 \end{align*} \]
   \[ \theta=\tan^{-1}\left(\frac{-\sqrt{3}}{1}\right)=\tan^{-1}(-\sqrt{3})=-\frac{\pi}{3} \]

   Since the point $(1,-\sqrt{3})$ is in quadrant IV, $\theta=\frac{5\pi}{3}$.

   So the polar coordinates are $\left(2,\frac{5\pi}{3}\right)$.

2. First find $r$.
   \[ \begin{align*} r&=\sqrt{x^2+y^2}\\ r&=\sqrt{(-3)^2+3^2}\\ r&=\sqrt{9+9}=\sqrt{18}=3\sqrt{2} \end{align*} \]
   \[ \theta_{ref}=\tan^{-1}\left(\left|\frac{3}{-3}\right|\right)=\tan^{-1}(1)=\frac{\pi}{4} \]

   Since the point $(-3,3)$ is in quadrant II, we get $\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$.

   So the polar coordinates are $\left(3\sqrt{2},\frac{3\pi}{4}\right)$.

3. First find $r$.
   \[ \begin{align*} r&=\sqrt{x^2+y^2}\\ r&=\sqrt{(2\sqrt{3})^2+(-2)^2}\\ r&=\sqrt{12+4}=4 \end{align*} \]
   \[ \theta=\tan^{-1}\left(\frac{-2}{2\sqrt{3}}\right)=\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6} \]

   Since the point $(2\sqrt{3},-2)$ is in quadrant IV, we get $\theta=\frac{11\pi}{6}$.

   So the polar coordinates are $\left(4,\frac{11\pi}{6}\right)$.

4. First find $r$.
   \[ \begin{align*} r&=\sqrt{x^2+y^2}\\ r&=\sqrt{0^2+(-9)^2}\\ r&=9 \end{align*} \]

   Since $x=0$, the ratio $\frac{y}{x}$ is undefined, so inverse tangent is not the right tool here. The point $(0,-9)$ is on the negative y-axis, so $\theta=\frac{3\pi}{2}$.

   So the polar coordinates are $\left(9,\frac{3\pi}{2}\right)$.

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Converting Equations to Polar Coordinates

If we can describe points in polar coordinates, then we can describe entire equations in polar coordinates. And in some cases, it is easier to do so. Take for example the equation of a circle.

$$x^2+y^2=1$$

This is a circle centered at the origin with a radius of 1. This isn't a function of either $x$ or $y$, because it would fail the vertical or horizontal line test. If you wanted to graph it in an older TI-83/84 calculator then you would have to create two separate equations when solving for $y$.

This means we would have to plot the top half of the circle, $f(x)=\sqrt{1-x^2}$ and then the bottom half $f(x)=-\sqrt{1-x^2}$.

Y= Screen	Window Settings	Graph

However, a circle can be written as a function of its radius when using polar coordinates, $f(\theta)=1$. When $\theta$ is at $0$, our value is one and we slowly maintain that distance of 1 from the origin as $\theta$ wraps around to $2\pi$. Most modern calculators have a polar option for graphing. For the TI-84 used in this example, the setting can be found under MODE. You can see here that after the setting is changed, the Y= menu asks for $r$ instead of $y$, because now instead of $y=f(x)$ we're working with a radius described as a function of its angle, $r=f(\theta)$. In our case, $r=1$

MODE Screen	r= Screen	Graph

::solution

1. Let's start with the friendliest one: $x^2+y^2=25$.

   In polar form, $x^2+y^2=r^2$, so we can substitute immediately.

   \[ \begin{align*} x^2+y^2&=25\\ r^2&=25 \end{align*} \]

   Now solve for $r$. For radius form, we usually take the positive radius description:

   \[ r=5 \]

   So a clean polar equation is $r=5$.

2. Next, convert the vertical line $x=9$.

   Use $x=r\cos\theta$:

   \[ \begin{align*} x&=9\\ r\cos\theta&=9 \end{align*} \]

   Now isolate $r$ by dividing by $\cos\theta$:

   \[ r=\frac{9}{\cos\theta}=9\sec\theta \]

   So the polar form is $r=9\sec\theta$.

3. Now for $y=3x+2$.

   Replace $y$ with $r\sin\theta$ and $x$ with $r\cos\theta$:

   \[ \begin{align*} y&=3x+2\\ r\sin\theta&=3r\cos\theta+2 \end{align*} \]

   Let's gather the $r$ terms on one side:

   \[ \begin{align*} r\sin\theta-3r\cos\theta&=2\\ r(\sin\theta-3\cos\theta)&=2 \end{align*} \]

   Now isolate $r$:

   \[ r=\frac{2}{\sin\theta-3\cos\theta} \]

   So one good polar form is $r=\dfrac{2}{\sin\theta-3\cos\theta}$.

4. Last one: $x^2+y^2=4x$.

   Use $x^2+y^2=r^2$ and $x=r\cos\theta$:

   \[ \begin{align*} x^2+y^2&=4x\\ r^2&=4r\cos\theta \end{align*} \]

   If we divide by $r$ (we're not concerned with $r$ being zero, a circle with a radius of zero is just a point at the origin), we get:

   \[ r=4\cos\theta \]

   So the standard polar form is $r=4\cos\theta$.

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::solution

1. Convert $r=8\cos\theta$ to rectangular.

   A great move here is to multiply by $r$ so we can use both identities $r^2=x^2+y^2$ and $r\cos\theta=x$:

   \[ \begin{align*} r&=8\cos\theta\\ r^2&=8r\cos\theta \end{align*} \]

   Now substitute:

   \[ \begin{align*} x^2+y^2&=8x \end{align*} \]

   That is already rectangular form. If you want the circle form, complete the square in $x$:

   \[ \begin{align*} x^2-8x+y^2&=0\\ (x-4)^2+y^2&=16 \end{align*} \]
2. Convert $r=10$.

   Square both sides so we can use $r^2=x^2+y^2$:

   \[ \begin{align*} r&=10\\ r^2&=100\\ x^2+y^2&=100 \end{align*} \]

   So the rectangular equation is $x^2+y^2=100$.

3. Convert $r=2\sin\theta$.

   Again, multiply both sides by $r$:

   \[ \begin{align*} r&=2\sin\theta\\ r^2&=2r\sin\theta \end{align*} \]

   Now substitute $r^2=x^2+y^2$ and $r\sin\theta=y$:

   \[ \begin{align*} x^2+y^2&=2y \end{align*} \]

   To see the graph more clearly, complete the square in $y$:

   \[ \begin{align*} x^2+y^2-2y&=0\\ x^2+(y-1)^2&=1 \end{align*} \]

   So a rectangular form is $x^2+y^2=2y$ (equivalently $x^2+(y-1)^2=1$).

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