Cartesian to Polar Conversionr

By using the theorem of Pythagoras, we can convert between Polar Coordinates and Cartesian Coordinates. Going from Cartesian to Polar is really quite simple, provided we do not need or desire algebraically simplified equations. All we have to do is take the Cartesian equation in $x$ and $y$ and substitute $r\cdot cos(\theta)$ and $r\cdot sin(\theta)$ for every instance of $x$ and $y$ respectively. Look at the conversion formula in figure 1. There, a single point, $P$, has been graphed with coordinates given in both systems. By constructing the Pythagorean triangle, one can understand how to convert from Cartesian to Polar. Converting in the other direction, from Polar to Cartesian may require solving for $r$ and $\theta$, although in many instances, the $r\cdot cos(\theta)$ or $r^{2}$ terms may just appear in the equation and substitutions can be made directly.

PolarCartesianConversion.png
Figure 1: Cartesian to Polar Conversion. The graph represents the single point, $P$, on both the Polar and Cartesian coordinate system simultaneously. In the polar system, the point has coordinates $(r,\theta)$ where $\theta$ is the angle from the horizontal axis counter-clockwise to a line between the origin and point $P$. In the Cartesian system, point $P$ has coordinates, $(x,y)$. This sets up a Pythagorean relationship between the Polar and Cartesian coordinates. Using the three trigonometric definitions and the circle formula, it is possible to convert from one system to the other. As shown, the relationships are easiest to go from Cartesian to polar. However, one can solve for $r$ and $\theta$ and thus convert in the other direction.
Example: A circle centered at the origin is given by $x^{2}+y^{2}=r^{2}$. Convert it to polar, where $0\le\theta\le2\pi$.
Answer: Substitute $x=r\cos\theta$ and $y=r\sin\theta$. We will see that $\theta$ is about to cancel out!$$ r^{2}\cos^{2}\theta+r^{2}\sin^{2}\theta=r^{2}$$ $$r^{2}(\cos^{2}\theta+\sin^{2}\theta)=r^{2}$$ $$r^{2}\cdot1=r^{2}$$ $$r=r$$ So for a circle centered at the origin, when plotting in pure polar coordinates, there is no dependence on $\theta$. At any $\theta$ (and for all $\theta$), each point on the circle is at distance $r$. If the radius were $2$, then the equation would be $r=2$.
{Aside: A very basic trig identity is $\cos^{2}\theta+\sin^{2}\theta=1$}
Example: " Given the equation for an ellipse, convert it to polar coordinates. The Cartesian equation for an ellipse with its center point at the origin is $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1. \tag{EQ1} \label{EQ1}$$ Answer: " Make the substitutions $x=r\cdot cos(\theta)$ and $y=r\cdot sin(\theta)$. $$\frac{r^{2}cos^{2}\theta}{a^{2}}+\frac{r^{2}sin^{2}\theta}{b^{2}}=1$$ $$r=\frac{a^{2}b^{2}}{\pm\sqrt{b^{2}cos^{2}\theta+a^{2}sin^{2}\theta}}$$ $$b^{2}r^{2}cos^{2}\theta+a^{2}r^{2}sin^{2}\theta=a^{2}b^{2} \tag{EQ2} \label{EQ2}$$ Very clearly, equation $\eqref{EQ2}$ is not as simple as equation $\eqref{EQ1}$. Also, if we had included an $(h,k)$ center point that was not at the origin, the result would have been still more complicated. There is a much easier way to parametrize an ellipse.
Polar Offset Circle.png
Figure 2: Polar Equation of Circle that is not at the origin. This is the polar graph of $r=2\cos\theta+3\sin\theta\pm\sqrt{4\cos^{2}\theta+12\cos\theta\sin\theta+9\sin^{2}\theta-4}$.
Example: What is the polar equation of the offset Cartesian Circle, $(x-2)^{2}+(y-3)^{2}=9$.
Answer: Do the substitution.$$\left(r\cos\theta-2\right)^{2}+\left(r\sin\theta-3\right)^{2}=9.$$ Clearly this is going to be algebraically messy, but it can be solved for $r$ in terms of $\theta$ to get: $$r=2\cos\theta+3\sin\theta\pm\sqrt{4\cos^{2}\theta+12\cos\theta\sin\theta+9\sin^{2}\theta-4}$$ The plus or minus is not relevant, as it just determines the starting point going all the way around the circle, so we can use either plus or minus. We can see this graphed in polar coordinates in figure 2.