Introduction

Welcome to this comprehensive guide on polar coordinates and differentiation in polar form! This topic introduces a powerful alternative coordinate system that naturally describes curves with rotational symmetry, such as circles, spirals, and rose curves. While rectangular coordinates use horizontal and vertical distances, polar coordinates use distance from the origin and angle of rotation, often simplifying the equations of complex curves.

Differentiation in polar coordinates extends familiar derivative techniques to this new coordinate system, allowing us to analyze slopes, rates of change, and curve properties. You'll learn to convert between coordinate systems, find derivatives with respect to both the angle parameter and the rectangular coordinates, and use these derivatives to understand curve behavior in ways that would be much more complex using rectangular coordinates alone.

Understanding Polar Coordinates

The polar coordinate system uses two values to locate any point in the plane:

A point is written as $(r,\theta )$ in polar coordinates, where $r\ge 0$ and $\theta$ can be any real number.

Conversion Between Coordinate Systems

The relationships between rectangular coordinates $(x,y)$ and polar coordinates $(r,\theta )$ are:

Polar to Rectangular:

$x=r\cos (\theta )$

$y=r\sin (\theta )$

Rectangular to Polar:

$r=\sqrt{x^2+y^2}$

$\theta =\arctan \left(\frac{y}{x}\right)\text{ (with proper quadrant adjustment)}$

Example 1: Coordinate Conversion

Convert the point $(3,4)$ from rectangular to polar coordinates.

$r=\sqrt{3^2+4^2}=\sqrt{9+16}=5$

$\theta =\arctan \left(\frac{4}{3}\right)\approx 0.927\text{ radians}$

The polar coordinates are $(5,\arctan (\frac{4}{3}))$.

Convert the point $(2,\frac{\pi }{3})$ from polar to rectangular coordinates.

$x=2\cos \left(\frac{\pi }{3}\right)=2\cdot \frac{1}{2}=1$

$y=2\sin \left(\frac{\pi }{3}\right)=2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}$

The rectangular coordinates are $(1,\sqrt{3})$.

Polar Equations and Common Curves

Polar equations have the form $r=f(\theta )$, where the radius depends on the angle. Common polar curves include:

Example 2: Graphing Polar Equations

The equation $r=2+2\cos (\theta )$ represents a cardioid. Key points:

Differentiation in Polar Coordinates

The essential knowledge tells us that methods for calculating derivatives of real-valued functions can be extended to functions in polar coordinates. There are several types of derivatives we can calculate.

Finding dr/dθ

For a polar equation $r=f(\theta )$, we can find $\frac{dr}{d\theta }$ using standard differentiation rules.

Example 3: Basic Polar Derivatives

Find $\frac{dr}{d\theta }$ for $r=3{\theta }^2+2\cos (\theta )$.

$\frac{dr}{d\theta }=6\theta -2\sin (\theta )$

Finding dx/dθ and dy/dθ

Since $x=r\cos (\theta )$ and $y=r\sin (\theta )$, we can find these derivatives using the product rule:

$\frac{dx}{d\theta }=\frac{d}{d\theta }[r\cos (\theta )]=\frac{dr}{d\theta }\cos (\theta )-r\sin (\theta )$

$\frac{dy}{d\theta }=\frac{d}{d\theta }[r\sin (\theta )]=\frac{dr}{d\theta }\sin (\theta )+r\cos (\theta )$

Example 4: Parametric Derivatives in Polar Form

For $r=1+\cos (\theta )$, find $\frac{dx}{d\theta }$ and $\frac{dy}{d\theta }$.

First, find $\frac{dr}{d\theta }=-\sin (\theta )$.

$\frac{dx}{d\theta }=(-\sin (\theta ))\cos (\theta )-(1+\cos (\theta ))\sin (\theta )$

$=-\sin (\theta )\cos (\theta )-\sin (\theta )-\cos (\theta )\sin (\theta )$

$=-\sin (\theta )(1+2\cos (\theta ))$

$\frac{dy}{d\theta }=(-\sin (\theta ))\sin (\theta )+(1+\cos (\theta )(\cos (\theta ))$

$=-{\sin }^2(\theta )+\cos (\theta )+{\cos }^2(\theta )$

$=\cos (\theta )+\cos (2\theta )$

Finding the Slope dy/dx

The slope of the tangent line to a polar curve is found using the chain rule:

$\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{\frac{dr}{d\theta }\sin (\theta )+r\cos (\theta )}{\frac{dr}{d\theta }\cos (\theta )-r\sin (\theta )}$

This formula is crucial for finding tangent lines and analyzing curve behavior.

Example 5: Finding Slope of Polar Curves

Find the slope of $r=2\cos (\theta )$ at $\theta =\frac{\pi }{4}$.

First, $\frac{dr}{d\theta }=-2\sin (\theta )$.

At $\theta =\frac{\pi }{4}$:

$\frac{dy}{dx}=\frac{(-\sqrt{2})\sin (\frac{\pi }{4})+\sqrt{2}\cos (\frac{\pi }{4})}{(-\sqrt{2})\cos (\frac{\pi }{4})-\sqrt{2}\sin (\frac{\pi }{4})}$

$=\frac{(-\sqrt{2})(\frac{\sqrt{2}}{2})+\sqrt{2}(\frac{\sqrt{2}}{2})}{(-\sqrt{2})(\frac{\sqrt{2}}{2})-\sqrt{2}(\frac{\sqrt{2}}{2})}=\frac{-1+1}{-1-1}=\frac{0}{-2}=0$

The slope is $0$, indicating a horizontal tangent line.

Second Derivatives and Concavity

To find the second derivative $\frac{d^{2}y}{d{x}^2}$, we use:

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left[\frac{dy}{dx}\right]=\frac{d}{d\theta }\left[\frac{dy}{dx}\right]\cdot \frac{d\theta }{dx}=\frac{d}{d\theta }\left[\frac{dy}{dx}\right]\cdot \frac{1}{dx/d\theta }$

This allows us to determine concavity and inflection points of polar curves.

Example 6: Second Derivative Analysis

For $r=1+\cos (\theta )$, find $\frac{d^{2}y}{d{x}^2}$ at $\theta =0$.

Refer to Example 4, where we calculated $\frac{dy}{d\theta }=\cos (\theta )+\cos (2\theta )$ and $\frac{dx}{d\theta }=-\sin (\theta )(1+2\cos (\theta ))$

From previous work: $\frac{dy}{dx}=\frac{\cos (\theta )+\cos (2\theta )}{-\sin (\theta )(1+2\cos (\theta ))}$

Using the formula: $\frac{d^{2}y}{d{x}^2}=\frac{-\sin (\theta )-2\sin (2\theta )}{-\sin (\theta )(1+2\cos (\theta ))}$

At $\theta =0$: $\frac{dy}{dx}=\frac{1+1}{-0}=\frac{2}{0}$ (undefined, indicating a vertical tangent)

At $\theta =0$: $\frac{d^{2}y}{d{x}^2}=\frac{0}{0}$ (indeterminate)

This requires careful analysis near $\theta =0$ using limits. This can be used to determine of $\theta =0$ is a point of inflection, and this would occur the second derivative's limit approaches 0 and has a confirmed sign shift.

Let's set up the limit and perform algebraic manipulation.

$\underset{\theta \to 0}{\lim }\frac{-\sin (\theta )-2\sin (2\theta )}{-\sin (\theta )(1+2\cos (\theta ))}$

$\underset{\theta \to 0}{\lim }\frac{\sin (\theta )+2\sin (2\theta )}{\sin (\theta )(1+2\cos (\theta ))}$

$\underset{\theta \to 0}{\lim }\frac{\sin (\theta )+2\sin (2\theta )}{\sin (\theta )+2\sin (\theta )\cos (\theta ))}$

$\underset{\theta \to 0}{\lim }\frac{\sin (\theta )+2\sin (2\theta )}{\sin (\theta )+\sin (2\theta )}$

Then, we can apply L'Hopital's Rule as the limit yields the $\frac{0}{0}$ indeterminate form.

$\underset{\theta \to 0}{\lim }\frac{\cos (\theta )+4\cos (2\theta )}{\cos (\theta )+2\cos (2\theta )}=\frac{5}{3}$

Thus, the polar coordinate $(0,4)$ is NOT a point of inflection and a point of vertical tangency on the polar function.

Horizontal and Vertical Tangents

Horizontal tangents occur when $\frac{dy}{d\theta }=0$ and $\frac{dx}{d\theta }\ne 0$.

Vertical tangents occur when $\frac{dx}{d\theta }=0$ and $\frac{dy}{d\theta }\ne 0$.

Example 7: Finding Tangent Lines

For $r=2+2\cos (\theta )$, find points with horizontal and vertical tangents.

$\frac{dr}{d\theta }=-2\sin (\theta )$

$\frac{dx}{d\theta }=-2\sin (\theta )\cos (\theta )-(2+2\cos (\theta ))\sin (\theta )=-2\sin (\theta )(1+2\cos (\theta ))$

$\frac{dy}{d\theta }=-2\sin (\theta )\sin (\theta )+(2+2\cos (\theta ))\cos (\theta )=2\cos (\theta )(1+\cos (\theta ))-2{\sin }^2(\theta )$

Horizontal tangents: $\frac{dy}{d\theta }=0$

$2\cos (\theta )(1+\cos (\theta ))-2{\sin }^2(\theta )=0$

$2\cos (\theta )+2{\cos }^2(\theta )-2(1-{\cos }^2(\theta ))=0$

$4{\cos }^2(\theta )+2\cos (\theta )-2=0$

$2(2\cos (\theta )-1)(\cos (\theta )+1)=0$

This gives $\cos (\theta )=\frac{1}{2}$ or $\cos (\theta )=-1$

Resolving these, we see points of horizontal tangencies occur when $\theta =\frac{\pi }{3}+2\pi n,-\frac{\pi }{3}+2\pi n,\pi +2\pi n$, where $n\in \mathbb{Z}$

Vertical tangents: $\frac{dx}{d\theta }=0$

$-2\sin (\theta )(1+2\cos (\theta ))=0$

This gives $\sin (\theta )=0$ or $\cos (\theta )=-1/2$.

Resolving these, we see points of vertical tangencies occur when $\theta =\pi n,\frac{2\pi }{3}+2\pi n,-\frac{2\pi }{3}+2\pi n$, where $n\in \mathbb{Z}$

Applications of Polar Differentiation

Example 8: Rate of Change Problems

A point moves along the spiral $r=\theta$ with $\frac{d\theta }{dt}=2$ rad/s. Find $\frac{dr}{dt}$ when $\theta =\pi$.

$\frac{dr}{dt}=\frac{dr}{d\theta }\cdot \frac{d\theta }{dt}=1\cdot 2=2$

The radius is increasing at $2$ units/s when $\theta =\pi$.

Complex Polar Curves

Example 9: Rose Curve Analysis

For the rose curve $r=3\cos (2\theta )$, analyze its properties.

$\frac{dr}{d\theta }=-6\sin (2\theta )$

The curve has four petals, with maximum $r=3$ at $\theta =0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}$ and $r=0$ at $\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$.

The slope at any point is: $\frac{dy}{dx}=\frac{-6\sin (2\theta )\sin (\theta )+3\cos (2\theta )\cos (\theta )}{-6\sin (2\theta )\cos (\theta )-3\cos (2\theta )\sin (\theta )}$

Practice Section

Test your understanding with these multiple-choice questions: