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Polar Coordinates
and Differentiation

Master the polar coordinate system and learn to find derivatives in polar form using \(\frac{dy}{dx}\) with parametric techniques.

AP Calculus BC

What Are Polar Coordinates?

Instead of locating a point by its horizontal and vertical distances \((x, y)\), polar coordinates use a distance from the origin and an angle from the positive \(x\)-axis.

\(r\) : Radial Distance
The distance from the pole (origin) to the point. Can be negative.
\(\theta\) : Angle
The angle measured counterclockwise from the positive \(x\)-axis.

Visualizing Polar vs Cartesian

\(\theta\) \(r\) \((r, \theta)\) \(x\) \(y\) O

Conversion Formulas

Polar → Cartesian
\(x = r\cos\theta\)
\(y = r\sin\theta\)
Cartesian → Polar
\(r = \sqrt{x^2 + y^2}\)
\(\theta = \arctan\!\left(\frac{y}{x}\right)\)

Why Polar Curves Matter

Many beautiful and important curves are naturally described in polar form. Trying to express them in Cartesian coordinates would be extremely complicated.

Cardioids
\(r = 1 + \cos\theta\)
Rose Curves
\(r = \cos(3\theta)\)
Spirals
\(r = \theta\)

The Core Formula: \(\frac{dy}{dx}\) in Polar

Since \(x = r\cos\theta\) and \(y = r\sin\theta\), and \(r = f(\theta)\), both \(x\) and \(y\) are functions of \(\theta\). We use the parametric derivative:

Deriving the Numerator and Denominator

1
Start with \(y = r\sin\theta\) where \(r = f(\theta)\).
2
Apply the product rule: \(\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta\).
3
Similarly for \(x = r\cos\theta\): \(\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta\).
4
Divide: \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\). This works whenever \(\frac{dx}{d\theta} \neq 0\).

Polar vs Parametric Derivatives

Polar Parametric
Parameter \(\theta\) (angle) \(t\) (any parameter)
Given \(r = f(\theta)\) \(x = g(t),\; y = h(t)\)
\(\frac{dy}{dx}\) \(\frac{dy/d\theta}{dx/d\theta}\) \(\frac{dy/dt}{dx/dt}\)
Key Step Product rule on \(r\sin\theta\), \(r\cos\theta\) Differentiate \(g(t)\), \(h(t)\) directly

Tangent Lines and Special Cases

Horizontal Tangent
Occurs when \(\frac{dy}{d\theta} = 0\) and \(\frac{dx}{d\theta} \neq 0\).
Vertical Tangent
Occurs when \(\frac{dx}{d\theta} = 0\) and \(\frac{dy}{d\theta} \neq 0\).

How to Find \(\frac{dy}{dx}\) in Polar

1
Identify \(r = f(\theta)\) and compute \(\frac{dr}{d\theta}\).
2
Compute the numerator: \(\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta\).
3
Compute the denominator: \(\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta\).
4
Divide: \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\).
5
Evaluate at the given \(\theta\) value and simplify.

Example 1: \(r = 2\cos\theta\)

Find \(\frac{dy}{dx}\) for \(r = 2\cos\theta\) at \(\theta = \frac{\pi}{3}\).

1
\(\frac{dr}{d\theta} = -2\sin\theta\). At \(\theta = \frac{\pi}{3}\): \(\frac{dr}{d\theta} = -\sqrt{3}\), \(r = 1\).
2
\(\frac{dy}{d\theta} = (-\sqrt{3})\!\left(\frac{\sqrt{3}}{2}\right) + (1)\!\left(\frac{1}{2}\right) = -\frac{3}{2} + \frac{1}{2} = -1\).
3
\(\frac{dx}{d\theta} = (-\sqrt{3})\!\left(\frac{1}{2}\right) - (1)\!\left(\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\sqrt{3}\).

Visualizing \(r = 2\cos\theta\)

Example 2: \(r = 1 + \sin\theta\)

Find the slope of the tangent to the cardioid \(r = 1 + \sin\theta\) at \(\theta = \frac{\pi}{6}\).

1
\(\frac{dr}{d\theta} = \cos\theta\). At \(\theta = \frac{\pi}{6}\): \(\frac{dr}{d\theta} = \frac{\sqrt{3}}{2}\), \(r = \frac{3}{2}\).
2
\(\frac{dy}{d\theta} = \frac{\sqrt{3}}{2} \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} + \frac{3\sqrt{3}}{4} = \sqrt{3}\).
3
\(\frac{dx}{d\theta} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4} - \frac{3}{4} = 0\).

Example 3: \(r = \sin(2\theta)\)

Find \(\frac{dy}{dx}\) for the rose curve \(r = \sin(2\theta)\) at \(\theta = \frac{\pi}{4}\).

1
\(\frac{dr}{d\theta} = 2\cos(2\theta)\). At \(\theta = \frac{\pi}{4}\): \(\frac{dr}{d\theta} = 0\), \(r = \sin(\frac{\pi}{2}) = 1\).
2
\(\frac{dy}{d\theta} = 0 \cdot \frac{\sqrt{2}}{2} + 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\).
3
\(\frac{dx}{d\theta} = 0 \cdot \frac{\sqrt{2}}{2} - 1 \cdot \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}\).

Common Mistakes to Avoid

✕ Using \(\frac{dr}{d\theta}\) as the slope
\(\frac{dr}{d\theta}\) is the rate of change of the radius, not the slope of the tangent. Always compute \(\frac{dy}{dx}\).
✕ Forgetting the product rule
Both \(x = r\cos\theta\) and \(y = r\sin\theta\) are products of two \(\theta\)-dependent functions. You need the product rule for each.
✕ Swapping numerator and denominator
\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\). Putting \(dx/d\theta\) on top gives you \(\frac{dx}{dy}\), the reciprocal of the slope.

Summary of Key Formulas

Formula Expression
Conversion: \(x\) \(x = r\cos\theta\)
Conversion: \(y\) \(y = r\sin\theta\)
\(\frac{dy}{d\theta}\) \(\frac{dr}{d\theta}\sin\theta + r\cos\theta\)
\(\frac{dx}{d\theta}\) \(\frac{dr}{d\theta}\cos\theta - r\sin\theta\)
Slope \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\)
Horizontal tangent \(\frac{dy}{d\theta} = 0\), \(\frac{dx}{d\theta} \neq 0\)
Vertical tangent \(\frac{dx}{d\theta} = 0\), \(\frac{dy}{d\theta} \neq 0\)

Question 1 of 5

For a polar curve \(r = f(\theta)\), the slope of the tangent line is given by:

Question 2 of 5

For \(r = 3\sin\theta\), what is \(\frac{dy}{d\theta}\)?

Question 3 of 5

A polar curve has \(\frac{dx}{d\theta} = 0\) and \(\frac{dy}{d\theta} = 5\) at a point. What can you conclude?

Question 4 of 5

For \(r = 1 + \cos\theta\), find \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{2}\).

Question 5 of 5

Which expression correctly gives \(\frac{dx}{d\theta}\) for a polar curve \(r = f(\theta)\)?