Polar Functions Polar Functions

Understanding Polar Coordinates

Cartesian coordinates use

(x, y)
to represent points on a plane, where
x
is the horizontal distance from the origin and
y
is the vertical distance. Explicit functions using cartesian coordinates are usually of the form:
y = f(x)

Polar coordinates use
(r, \theta)
to represent those points, where
r
is the distance from the origin and
θ
is the angle of the point from the positive x-axis. Polar functions are usually of the form:
r = f(\theta) \\

Examples

r = 1
r = \theta
r = \sin(6\theta)
r = 1-\sin(\theta)

Converting From Cartesian to Polar

To convert from cartesian coordinates to polar, we use the following equations:

r = \sqrt{x^2 + y^2} \\ \theta = \arctan\left(\frac{y}{x}\right)
Since
x
and
y
are the horizontal and vertical distances from the origin, we use euclidean distance to find the distance,
r
, between the point and the origin.

The equation for
\theta
(theta) comes from the following:
\tan(\theta)=y/x
.

Converting From Polar to Cartesian

To convert from polar coordinates to cartesian, we use the following equations:

x = r\cos(\theta) \\ y = r\sin(\theta)
For a right triangle with a hypotenuse of length
1
:
The length of the side adjacent to a corner with angle,
\theta
, is
\cos(\theta)
.
The length of the side opposite to a corner with angle,
\theta
, is
\sin(\theta)
.


By multiplying these two equations by
r
, we find the lengths of the sides of a right triangle with a hypotenuse of length
r
and an angle of
\theta
.

Differentiation in Polar Form

Polar functions are secretly parametric functions. For the function,

f(\theta)
x = r\cos(\theta) = f(\theta)\cos(\theta) \\ y = r\sin(\theta) = f(\theta)\sin(\theta)
Using parametric differentiation:
\begin{align*} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} \\ \\ &= \frac{f'(\theta)\sin(\theta) + f(\theta)\cos(\theta)}{f'(\theta)\cos(\theta) - f(\theta)\sin(\theta)} \end{align*}

Examples

Example 1: Find

dy/dx
for
r = 2\sin(\theta)
\begin{align*} f(\theta) &= 2\sin(\theta) \quad f'(\theta) = 2\cos(\theta) \\\\ \frac{dy}{dx} &= \frac{2\cos(\theta)\sin(\theta) + 2\sin(\theta)\cos(\theta)}{2\cos(\theta)\cos(\theta) - 2\sin(\theta)\sin(\theta)} \\\\ &= \frac{4\sin(\theta)\cos(\theta)}{2(\cos^2\theta - \sin^2\theta)} \\\\ &= \frac{\sin(2\theta)}{\cos(2\theta)} \\\\ &= \tan(2\theta) \\\\ \end{align*}
Example 2: Find
dy/dx
for
r = 1 + \cos(\theta)
(a cardioid)
\begin{align*} f(\theta) &= 1 + \cos(\theta) \quad f'(\theta) = -\sin(\theta) \\\\ \frac{dy}{dx} &= \frac{(-\sin(\theta))\sin(\theta) + (1 + \cos(\theta))\cos(\theta)}{(-\sin(\theta))\cos(\theta) - (1 + \cos(\theta))\sin(\theta)} \\\\ &= \frac{-\sin^2\theta + \cos(\theta) + \cos^2\theta}{-2\sin(\theta)\cos(\theta) - \sin(\theta)} \\\\ &= \frac{\cos(\theta) + (\cos^2\theta - \sin^2\theta)}{-\sin(\theta)(2\cos(\theta) + 1)} \\\\ &= \frac{\cos(\theta) + \cos(2\theta)}{-\sin(\theta)(2\cos(\theta) + 1)} \end{align*}
Example 3: Find
dy/dx
for
r = e^\theta
(a logarithmic spiral)
\begin{align*} f(\theta) &= e^\theta \quad f'(\theta) = e^\theta \\\\ \frac{dy}{dx} &= \frac{e^\theta\sin(\theta) + e^\theta\cos(\theta)}{e^\theta\cos(\theta) - e^\theta\sin(\theta)} \\\\ &= \frac{\sin(\theta) + \cos(\theta)}{\cos(\theta) - \sin(\theta)} \\\\ &= \frac{1 + \tan\theta}{1 - \tan\theta} \\\\ &= \tan\left(\theta + \frac{\pi}{4}\right) \end{align*}