9.7 - Defining Polar Coordinates and Differentiating in Polar Form

Polar Coordinate System Basics

Rectangular vs. Polar Coordinates

Rectangular (Cartesian) coordinates: points are written as

(x, y)

Polar coordinates: points are written as

(r, \theta)

, where: -

r

= the directed distance from the origin to the point -

\theta

= the directed angle from the positive

x

-axis

9.7 - Defining Polar Coordinates and Differentiating in Polar Form

Important: - Positive

r

extends in the direction of the angle

\theta

- Negative

r

extends in the opposite direction - Positive

\theta

is measured counterclockwise - Negative

\theta

is measured clockwise

Multiple Representations

The same point can have multiple polar coordinate representations.

Example: these all represent the same point: -

\left(2, \frac{\pi}{3}\right)

\left(-2, \frac{4\pi}{3}\right)

\left(2, -\frac{5\pi}{3}\right)

Converting Between Coordinate Systems

From Polar to Rectangular

x = r\cos(\theta) \qquad y = r\sin(\theta)

From Rectangular to Polar

r^2 = x^2 + y^2 \qquad \theta = \arctan\left(\frac{y}{x}\right)

Also useful:

\sin(\theta) = \frac{y}{r} \qquad \cos(\theta) = \frac{x}{r}

Careful with

\arctan

: it only gives angles between

-\frac{\pi}{2}

and

\frac{\pi}{2}

, which point into Quadrant I or IV. If the point is in Quadrant II or III, add

\pi

. For example,

(-1, 1)

gives

\arctan(-1) = -\frac{\pi}{4}

, but the point is in Quadrant II, so

\theta = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}

, with

r = \sqrt{2}

Converting Polar Equations to Rectangular Form

Example 1: Vertical Line

Convert

r\cos(\theta) = -5

Since

r\cos(\theta) = x

x = -5

This is a vertical line at

x = -5

Example 2: Circle

Convert

6x = r^2

Since

r\cos(\theta) = x

and

r^2 = x^2 + y^2

6x = x^2 + y^2

0 = x^2 - 6x + y^2

Complete the square on

x

0 + 9 = x^2 - 6x + 9 + y^2

9 = (x - 3)^2 + y^2

This is a circle with center

(3, 0)

and radius

3

Example 3: Linear Equation

Convert

r(3\cos(\theta) - \sin(\theta)) = 5

Distribute the

r

3r\cos(\theta) - r\sin(\theta) = 5

Substitute

x

and

y

3x - y = 5 \quad \Rightarrow \quad y = 3x - 5

This is a line with slope

3

and

y

-intercept

-5

Graphing Polar Equations with Technology

Calculator setup:

Set the mode to POLAR, not function or parametric

In the equation editor, you’ll see $r_1 =$ , $r_2 =$ , and so on

Enter your equations in the form $r = f(\theta)$

Adjust the window settings:

$\theta_{\text{min}} = 0$ , $\theta_{\text{max}} = 2\pi$

$\theta_{\text{step}}$ controls smoothness, and smaller means smoother

Tracing: - You can trace along the curve or jump to specific

\theta

values - Enter a specific angle, like

\frac{\pi}{4}

, to see the coordinates at that point

Finding the Slope of Polar Curves

Conversion to Parametric Form

A polar equation

r = f(\theta)

can be written parametrically:

x = f(\theta)\cos(\theta) \qquad y = f(\theta)\sin(\theta)

Derivative Formula

The slope of a polar curve is:

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{y'(\theta)}{x'(\theta)}

Expanded form:

\frac{dy}{dx} = \frac{f'(\theta)\sin(\theta) + f(\theta)\cos(\theta)}{f'(\theta)\cos(\theta) - f(\theta)\sin(\theta)}

where

f(\theta) = r

, and you use the product rule on both the numerator and the denominator.

Example 4: Slope at a Point

Find the slope of

r = 2 + 3\sin(\theta)

\theta = 0

First, express it in parametric form:

y = (2 + 3\sin\theta)\sin\theta = 2\sin\theta + 3\sin^2\theta

x = (2 + 3\sin\theta)\cos\theta = 2\cos\theta + 3\sin\theta\cos\theta

Find the derivatives:

\frac{dy}{d\theta} = 2\cos\theta + 6\sin\theta\cos\theta

\frac{dx}{d\theta} = -2\sin\theta + 3\cos^2\theta - 3\sin^2\theta = -2\sin\theta + 3(\cos^2\theta - \sin^2\theta)

Evaluate at

\theta = 0

\frac{dy}{d\theta}\bigg|_{\theta=0} = 2\cos(0) + 6\sin(0)\cos(0) = 2(1) + 0 = 2

\frac{dx}{d\theta}\bigg|_{\theta=0} = -2\sin(0) + 3\cos^2(0) - 3\sin^2(0) = 0 + 3(1) - 0 = 3

Therefore:

\frac{dy}{dx}\bigg|_{\theta=0} = \frac{2}{3}

Horizontal and Vertical Tangent Lines

Finding Horizontal Tangents

Horizontal tangent lines happen when the slope equals zero:

\frac{dy}{dx} = 0 \implies \frac{dy}{d\theta} = 0 \text{ (and } \frac{dx}{d\theta} \neq 0\text{)}

Steps: 1. Find

y = r\sin(\theta)

2. Compute

\frac{dy}{d\theta}

3. Solve

\frac{dy}{d\theta} = 0

Finding Vertical Tangents

Vertical tangent lines happen when the slope is undefined:

\frac{dy}{dx} \text{ undefined} \implies \frac{dx}{d\theta} = 0 \text{ (and } \frac{dy}{d\theta} \neq 0\text{)}

Steps: 1. Find

x = r\cos(\theta)

2. Compute

\frac{dx}{d\theta}

3. Solve

\frac{dx}{d\theta} = 0

Example 5: Finding Tangent Lines

Find the horizontal and vertical tangents for

r = 2 - \sin(\theta)

[0, 2\pi)

For horizontal tangents:

y = (2 - \sin\theta)\sin\theta = 2\sin\theta - \sin^2\theta

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

Set it equal to zero:

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

Solutions:

\cos\theta = 0 \implies \theta = \frac{\pi}{2}, \frac{3\pi}{2}

1 - \sin\theta = 0 \implies \sin\theta = 1 \implies \theta = \frac{\pi}{2}

We’ll check which ones are valid after finding the vertical tangents.

For vertical tangents:

x = (2 - \sin\theta)\cos\theta = 2\cos\theta - \sin\theta\cos\theta

\frac{dx}{d\theta} = -2\sin\theta - \cos^2\theta + \sin^2\theta

= -2\sin\theta - (1 - \sin^2\theta) + \sin^2\theta

= -2\sin\theta - 1 + \sin^2\theta + \sin^2\theta

= 2\sin^2\theta - 2\sin\theta - 1

Set it equal to zero, treating it as a quadratic in

\sin\theta

2\sin^2\theta - 2\sin\theta - 1 = 0

Using the quadratic formula with

u = \sin\theta

\sin\theta = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}

\sin\theta = \frac{1 + \sqrt{3}}{2} \approx 1.366 \quad \text{(impossible, since } |\sin\theta| \leq 1\text{)}

\sin\theta = \frac{1 - \sqrt{3}}{2} \approx -0.366

Find

\theta

where

\sin\theta \approx -0.366

\theta \approx 3.516, \, 5.908 \text{ radians}

When both are zero: if a value of

\theta

makes both

\frac{dy}{d\theta} = 0

and

\frac{dx}{d\theta} = 0

, the slope formula gives

\frac{0}{0}

, so you can’t call it a horizontal or vertical tangent yet. For a polar curve this only happens at the origin, and it’s often a cusp (a sharp point), like the tip of the cardioid

r = 1 + \cos\theta

\theta = \pi

. It takes more work, like a limit, to know for sure.

In this example,

\theta = \frac{\pi}{2}

shows up twice in the horizontal list, but not in the vertical list: at that angle dx/dθ = 2 - 2 - 1 = -1, not 0. So it isn’t a cusp. It’s a horizontal tangent at the point (0, 1).

Final answers: - Horizontal tangents:

\theta = \frac{\pi}{2}, \, \frac{3\pi}{2}

- Vertical tangents:

\theta \approx 3.516, \, 5.908

Advanced Factoring Techniques

When you’re solving trig equations that come from derivatives, keep these in mind.

Pythagorean Identities

\sin^2(\theta) + \cos^2(\theta) = 1

1 - \sin^2(\theta) = \cos^2(\theta)

1 - \cos^2(\theta) = \sin^2(\theta)

Factoring Trigonometric Polynomials

Treat

\sin(\theta)

\cos(\theta)

as a variable, like

x

, and factor accordingly.

Example:

2\sin^2\theta - 3\sin\theta + 1 = 0

Factor:

(2\sin\theta - 1)(\sin\theta - 1) = 0

Solutions:

\sin\theta = \frac{1}{2}

\sin\theta = 1

What the Derivatives Tell You About a Polar Curve

A polar curve

r = f(\theta)

comes with several derivatives, and each one answers a different question about the curve.

Derivative	What it tells you
$\frac{dr}{d\theta}$	How the distance from the origin is changing as $\theta$ increases (sign rule below)
$\frac{dx}{d\theta}$	Positive: the point moves right as $\theta$ increases. Negative: it moves left
$\frac{dy}{d\theta}$	Positive: the point moves up as $\theta$ increases. Negative: it moves down
$\frac{dy}{dx}$	The slope of the tangent line
$\frac{d^2y}{dx^2}$	Concavity: positive means concave up, negative means concave down

Moving Toward or Away from the Origin

The distance from the origin is

|r|

, not

r

, because

r

can be negative. So compare the signs of

r

and

\frac{dr}{d\theta}

Same sign (both positive or both negative): $|r|$ is increasing, so the point is moving away from the origin

Opposite signs: $|r|$ is decreasing, so the point is moving toward the origin

When

r > 0

, this is just “

\frac{dr}{d\theta} > 0

means moving away.” The sign check matters when

r

is negative.

Example 6: Distance from the Origin

For

r = 1 + 2\sin\theta

, is the point moving toward or away from the origin at

\theta = \frac{\pi}{6}

? At

\theta = \frac{4\pi}{3}

\frac{dr}{d\theta} = 2\cos\theta

\theta = \frac{\pi}{6}

r = 1 + 2\left(\frac{1}{2}\right) = 2

and

\frac{dr}{d\theta} = 2\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}

. Both are positive, so the point is moving away from the origin.

\theta = \frac{4\pi}{3}

r = 1 + 2\left(-\frac{\sqrt{3}}{2}\right) = 1 - \sqrt{3} \approx -0.732

and

\frac{dr}{d\theta} = 2\left(-\frac{1}{2}\right) = -1

. Both are negative, so the point is moving away from the origin here too.

r

is decreasing, but it’s getting more negative, so the distance

|r|

is growing.

Since

r

is negative at

\theta = \frac{4\pi}{3}

, the point is plotted in the opposite direction. It lands on the inner loop in Quadrant I, not in Quadrant III.

Second Derivative in Polar Form

Since

x

and

y

are both functions of

\theta

, the second derivative works just like Topic 9.2, with

\theta

in place of

t

\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left[\frac{dy}{dx}\right]}{\frac{dx}{d\theta}}

Example 7: Concavity of a Polar Curve

Is the curve

r = 4\cos\theta

concave up or concave down at

\theta = \frac{\pi}{8}

Step 1: Write

x

and

y

in terms of

\theta

. Use the double angle identities

\cos^2\theta = \frac{1 + \cos(2\theta)}{2}

and

2\sin\theta\cos\theta = \sin(2\theta)

x = 4\cos^2\theta = 2 + 2\cos(2\theta) \qquad y = 4\sin\theta\cos\theta = 2\sin(2\theta)

Step 2: Find

\frac{dy}{dx}

\frac{dx}{d\theta} = -4\sin(2\theta) \qquad \frac{dy}{d\theta} = 4\cos(2\theta) \qquad \frac{dy}{dx} = \frac{4\cos(2\theta)}{-4\sin(2\theta)} = -\cot(2\theta)

Step 3: Find

\frac{d^2y}{dx^2}

. The derivative of

-\cot(2\theta)

2\csc^2(2\theta)

(chain rule), and the denominator is

\frac{dx}{d\theta}

\frac{d^2y}{dx^2} = \frac{2\csc^2(2\theta)}{-4\sin(2\theta)} = -\frac{1}{2}\csc^3(2\theta)

Step 4: Evaluate at

\theta = \frac{\pi}{8}

. Here

2\theta = \frac{\pi}{4}

, and

\csc\left(\frac{\pi}{4}\right) = \sqrt{2}

\frac{d^2y}{dx^2}\bigg|_{\theta=\pi/8} = -\frac{1}{2}\left(\sqrt{2}\right)^3 = -\frac{1}{2}\left(2\sqrt{2}\right) = -\sqrt{2} \approx -1.414

Answer: negative, so the curve is concave down at

\theta = \frac{\pi}{8}

. That makes sense:

r = 4\cos\theta

is the circle

(x - 2)^2 + y^2 = 4

, and

\theta = \frac{\pi}{8}

lands on its top half, at

\left(2 + \sqrt{2}, \sqrt{2}\right)

Key Takeaways

Polar coordinates: $(r, \theta)$ , where $r$ is the distance and $\theta$ is the angle

Multiple representations are possible for the same point

Conversion formulas: $x = r\cos\theta$ , $y = r\sin\theta$ , $r^2 = x^2 + y^2$

Slope of a polar curve: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ , so treat it like a parametric curve

The product rule is required for both $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$

Horizontal tangents: $\frac{dy}{d\theta} = 0$ , and $\frac{dx}{d\theta} \neq 0$

Vertical tangents: $\frac{dx}{d\theta} = 0$ , and $\frac{dy}{d\theta} \neq 0$

Both derivatives zero at the same $\theta$ gives $\frac{0}{0}$ , so there’s no conclusion yet (it’s often a cusp)

Pythagorean identities are often needed to simplify expressions

Distance from the origin: moving away when $r$ and $\frac{dr}{d\theta}$ have the same sign, toward when they have opposite signs

Second derivative: $\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left[\frac{dy}{dx}\right]}{\frac{dx}{d\theta}}$ , the same as for parametric equations

Graphing technology helps for visualizing polar curves and checking tangent lines

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