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Polar Derivative Formulas

Polar Derivative Formulas

The formula for the first derivative dy/dx of a polar curve is given below.

 

Slope of tangent line = dy/dx = (dr/dθ · sinθ + r·cosθ) / (dr/dθ · cosθ − r·sinθ)

 

 

See also

Slope of a curve, tangent line, parametric derivative formulas

Key Formula

dydx=drdθsin⁡θ+rcos⁡θdrdθcos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{\dfrac{dr}{d\theta}\sin\theta + r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta - r\sin\theta}
Where:
  • rr = The polar function r = f(θ), giving the distance from the origin to the curve
  • θ\theta = The polar angle, measured counterclockwise from the positive x-axis
  • drdθ\frac{dr}{d\theta} = The derivative of the polar function with respect to θ
  • dydx\frac{dy}{dx} = The slope of the polar curve in Cartesian coordinates

Worked Example

Problem: Find the slope of the cardioid r = 1 + cos θ at θ = π/2.
Step 1: Identify r and compute dr/dθ.
r=1+cos⁡θdrdθ=−sin⁡θr = 1 + \cos\theta \qquad \frac{dr}{d\theta} = -\sin\theta
Step 2: Write the numerator of the polar derivative formula: (dr/dθ) sin θ + r cos θ. Evaluate at θ = π/2.
Numerator=(−sin⁡π2)sin⁡π2+(1+cos⁡π2)cos⁡π2=(−1)(1)+(1)(0)=−1\text{Numerator} = (-\sin\tfrac{\pi}{2})\sin\tfrac{\pi}{2} + (1+\cos\tfrac{\pi}{2})\cos\tfrac{\pi}{2} = (-1)(1) + (1)(0) = -1
Step 3: Write the denominator: (dr/dθ) cos θ − r sin θ. Evaluate at θ = π/2.
Denominator=(−sin⁡π2)cos⁡π2−(1+cos⁡π2)sin⁡π2=(−1)(0)−(1)(1)=−1\text{Denominator} = (-\sin\tfrac{\pi}{2})\cos\tfrac{\pi}{2} - (1+\cos\tfrac{\pi}{2})\sin\tfrac{\pi}{2} = (-1)(0) - (1)(1) = -1
Step 4: Divide to find dy/dx.
dydx=−1−1=1\frac{dy}{dx} = \frac{-1}{-1} = 1
Answer: The slope of r = 1 + cos θ at θ = π/2 is 1.

Another Example

This example uses a different polar curve (a circle rather than a cardioid) and involves irrational values, showing that the formula works the same way regardless of the type of polar curve.

Problem: Find the slope of the circle r = 4 sin θ at θ = π/3.
Step 1: Identify r and compute dr/dθ.
r=4sin⁡θdrdθ=4cos⁡θr = 4\sin\theta \qquad \frac{dr}{d\theta} = 4\cos\theta
Step 2: Evaluate r and dr/dθ at θ = π/3.
r=4sin⁡π3=4⋅32=23,drdθ=4cos⁡π3=4⋅12=2r = 4\sin\tfrac{\pi}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}, \qquad \frac{dr}{d\theta} = 4\cos\tfrac{\pi}{3} = 4 \cdot \frac{1}{2} = 2
Step 3: Compute the numerator: (dr/dθ) sin θ + r cos θ.
Numerator=2⋅32+23⋅12=3+3=23\text{Numerator} = 2 \cdot \frac{\sqrt{3}}{2} + 2\sqrt{3} \cdot \frac{1}{2} = \sqrt{3} + \sqrt{3} = 2\sqrt{3}
Step 4: Compute the denominator: (dr/dθ) cos θ − r sin θ.
Denominator=2⋅12−23⋅32=1−3=−2\text{Denominator} = 2 \cdot \frac{1}{2} - 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 1 - 3 = -2
Step 5: Divide to find the slope.
dydx=23−2=−3\frac{dy}{dx} = \frac{2\sqrt{3}}{-2} = -\sqrt{3}
Answer: The slope of r = 4 sin θ at θ = π/3 is −√3.

Frequently Asked Questions

How do you derive the polar derivative formula?
Start with the parametric relationships x = r cos θ and y = r sin θ. Differentiate both with respect to θ using the product rule: dx/dθ = (dr/dθ) cos θ − r sin θ and dy/dθ = (dr/dθ) sin θ + r cos θ. Then apply the parametric derivative rule dy/dx = (dy/dθ) / (dx/dθ) to obtain the formula.
When is the tangent line to a polar curve horizontal or vertical?
A horizontal tangent occurs when the numerator (dy/dθ) equals zero while the denominator (dx/dθ) is nonzero. A vertical tangent occurs when the denominator (dx/dθ) equals zero while the numerator is nonzero. If both are zero simultaneously, further analysis (such as L'Hôpital's rule) is needed.
What is the difference between polar derivative formulas and parametric derivative formulas?
The polar derivative formula is actually a special case of the parametric derivative formula. In parametric form, dy/dx = (dy/dt) / (dx/dt) for any parameter t. For polar curves, the parameter is θ, and the specific expressions for dx/dθ and dy/dθ come from applying the product rule to x = r cos θ and y = r sin θ.

Polar Derivative Formulas vs. Parametric Derivative Formulas

Polar Derivative FormulasParametric Derivative Formulas
General formdy/dx = [(dr/dθ) sin θ + r cos θ] / [(dr/dθ) cos θ − r sin θ]dy/dx = (dy/dt) / (dx/dt)
Parameterθ (the polar angle)t (any parameter)
When to useWhen the curve is given as r = f(θ)When the curve is given as x = x(t), y = y(t)
RelationshipA special case of the parametric formulaThe general framework that includes the polar case

Why It Matters

Polar derivative formulas appear frequently in AP Calculus BC and college-level Calculus II courses. You need them whenever a problem asks for the slope, tangent line, or angle of inclination of a polar curve. They also serve as the foundation for finding horizontal and vertical tangents on curves like cardioids, roses, and limaçons, which are standard exam topics.

Common Mistakes

Mistake: Forgetting to apply the product rule when differentiating x = r cos θ and y = r sin θ, and instead writing dy/dx = (dr/dθ)(sin θ / cos θ).
Correction: Both x and y are products of r(θ) and a trigonometric function of θ. You must use the product rule on each, giving dx/dθ = (dr/dθ) cos θ − r sin θ and dy/dθ = (dr/dθ) sin θ + r cos θ.
Mistake: Swapping the numerator and denominator, or mixing up the signs (e.g., writing + r sin θ in the denominator instead of − r sin θ).
Correction: Remember: the numerator comes from dy/dθ (the y-component), and the denominator comes from dx/dθ (the x-component). A helpful mnemonic: the denominator has a minus sign because differentiating cos θ produces −sin θ.

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