Derivatives of Polar Functions

To find the derivative of a polar function r = f(θ), treat θ as a parameter: x = r cos θ and y = r sin θ. Then the slope is dy/dx = (dy/dθ)/(dx/dθ), where dy/dθ = r′ sin θ + r cos θ and dx/dθ = r′ cos θ − r sin θ. The derivative dr/dθ alone is not the slope.

Before this lesson: Graphing Polar Equations, Derivatives of Parametric Equations and Product Rule

The idea: polar curves are parametric curves

A polar curve r = f(θ) gives a distance r for each angle θ. Converting to x and y, as in polar coordinates, gives x = f(θ) cos θ and y = f(θ) sin θ.

Those are parametric equations with θ as the parameter. So the slope of the curve is the same rise-rate-over-run-rate idea from parametric curves. Each of x and y is a product, so each derivative needs the product rule.

The derivative dr/dθ answers a different question. It tells whether the point is moving away from the origin (positive) or toward it (negative) as θ increases.

Polar derivative formula

\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}, \qquad r' = \dfrac{dr}{d\theta}

• Horizontal tangent: dy/dθ = 0 and dx/dθ ≠ 0.
• Vertical tangent: dx/dθ = 0 and dy/dθ ≠ 0.
• If both are 0, the formula doesn’t decide; look for a cusp, often at the pole (the origin).
• Tangent at the pole: if r = 0 at θ = α and r′(α) ≠ 0, the tangent line there is the line θ = α, with slope tan α.
• For a tangent line, convert the point to (x, y) = (r cos θ, r sin θ) first.

Derivatives of polar functions examples

Example 1: A circle

Find dy/dx for r = 3 sin θ, and its value at θ = π/3.

r' = 3\cos\theta

\dfrac{dy}{d\theta} = 3\cos\theta\sin\theta + 3\sin\theta\cos\theta = 3\sin 2\theta

\dfrac{dx}{d\theta} = 3\cos^2\theta - 3\sin^2\theta = 3\cos 2\theta

\dfrac{dy}{dx} = \dfrac{3\sin 2\theta}{3\cos 2\theta} = \tan 2\theta

\tan\dfrac{2\pi}{3} = -\sqrt{3}

Answer: dy/dx = tan 2θ, which is −√3 at θ = π/3. The point is (r cos θ, r sin θ) = (3√3/4, 9/4).

Example 2: Tangent line to a cardioid

Find the tangent line to the cardioid r = 1 + cos θ at θ = π/2.

r = 1 + \cos\dfrac{\pi}{2} = 1, \quad r' = -\sin\dfrac{\pi}{2} = -1

\dfrac{dy}{d\theta} = (-1)(1) + (1)(0) = -1

\dfrac{dx}{d\theta} = (-1)(0) - (1)(1) = -1

\dfrac{dy}{dx} = \dfrac{-1}{-1} = 1

(x, y) = (1 \cdot 0,\; 1 \cdot 1) = (0, 1)

y - 1 = 1(x - 0)

y = x + 1

Answer: y = x + 1.

Example 3: Horizontal tangents

Find the points where r = 1 + cos θ has a horizontal tangent, for θ in [0, 2π).

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

-(1 - \cos^2\theta) + \cos\theta + \cos^2\theta

2\cos^2\theta + \cos\theta - 1

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

\cos\theta = \tfrac{1}{2} \Rightarrow \theta = \tfrac{\pi}{3}, \tfrac{5\pi}{3}

\cos\theta = -1 \Rightarrow \theta = \pi

\dfrac{dx}{d\theta}\Big|_{\pi/3} = -\sqrt{3}

\dfrac{dx}{d\theta}\Big|_{\pi} = 0

Answer: horizontal tangents at θ = π/3 and 5π/3, where r = 3/2, at the points (3/4, 3√3/4) and (3/4, −3√3/4). At θ = π the curve has a cusp at the origin.

Example 4: dr/dθ versus dy/dx

For r = 3 − 2 cos θ, at θ = π/2, is the curve moving toward or away from the origin? What is the slope dy/dx?

r' = 2\sin\theta

r'\left(\tfrac{\pi}{2}\right) = 2

r\left(\tfrac{\pi}{2}\right) = 3 - 0 = 3

\dfrac{dy}{d\theta} = 2(1) + 3(0) = 2

\dfrac{dx}{d\theta} = 2(0) - 3(1) = -3

\dfrac{dy}{dx} = -\dfrac{2}{3}

Answer: moving away from the origin (dr/dθ = 2), while the curve’s slope is −2/3. The two derivatives answer different questions.

Example 5: A sprinkler’s spray edge (word problem)

The edge of the area a lawn sprinkler reaches is r = 4 + 2 sin θ, in meters. Find the slope of the edge at θ = 0, and the angles where the spray reaches farthest and least far.

r' = 2\cos\theta

r(0) = 4, \quad r'(0) = 2

\dfrac{dy}{d\theta} = 2(0) + 4(1) = 4

\dfrac{dx}{d\theta} = 2(1) - 4(0) = 2

\dfrac{dy}{dx} = \dfrac{4}{2} = 2

2\cos\theta = 0 \Rightarrow \theta = \tfrac{\pi}{2}, \tfrac{3\pi}{2}

r\left(\tfrac{\pi}{2}\right) = 6, \quad r\left(\tfrac{3\pi}{2}\right) = 2

Answer: the edge has slope 2 at the point (4, 0). The spray reaches farthest, 6 m, at θ = π/2 and least far, 2 m, at θ = 3π/2.

Common mistakes

Related methods

Parametric derivatives

This lesson is derivatives of parametric equations with x = r cos θ and y = r sin θ. Second derivatives and concavity work the same way, with θ in place of t.

Graphing polar curves

Knowing where tangents are horizontal or vertical, and where r is largest, helps when graphing polar equations by hand.

Area and length of polar curves

The next lessons find the area inside a polar curve and the length of a polar curve. Both use the same conversion to x and y in the background.

Practice problems

1. Find dy/dx for r = 4 cos θ and its value at θ = π/6.
2. Find dy/dx for the spiral r = θ at θ = π/2.
3. Find dy/dx for the circle r = 2 at θ = π/4.
4. Find the tangent line to r = 1 + sin θ at θ = 0.
5. For r = 2 + cos θ, is the curve moving toward or away from the origin at θ = π/2?
6. Find the vertical tangents of r = 3 sin θ for θ in [0, π).
7. Find the tangent lines at the pole for r = 2 sin 3θ, θ in [0, π).
8. Find dy/dx for r = e^θ at θ = 0.
9. Find dy/dx for r = 1 − 2 sin θ at θ = 0.
10. A garden light’s beam edge is r = 5 − 3 cos θ meters. Where is the edge closest to and farthest from the light, and what is its slope at θ = π/2?

\dfrac{dy}{d\theta} = -4\sin^2\theta + 4\cos^2\theta = 4\cos 2\theta

\dfrac{dx}{d\theta} = -8\sin\theta\cos\theta = -4\sin 2\theta

\dfrac{dy}{dx} = -\cot 2\theta

-\cot\dfrac{\pi}{3} = -\dfrac{\sqrt{3}}{3}

\dfrac{dy}{d\theta} = 1(1) + \tfrac{\pi}{2}(0) = 1

\dfrac{dx}{d\theta} = 1(0) - \tfrac{\pi}{2}(1) = -\tfrac{\pi}{2}

\dfrac{dy}{dx} = \dfrac{1}{-\pi/2} = -\dfrac{2}{\pi}

\dfrac{dy}{d\theta} = 0 + 2\cos\tfrac{\pi}{4} = \sqrt{2}

\dfrac{dx}{d\theta} = 0 - 2\sin\tfrac{\pi}{4} = -\sqrt{2}

r(0) = 1, \quad r'(0) = \cos 0 = 1

\dfrac{dy}{d\theta} = 1(0) + 1(1) = 1, \quad \dfrac{dx}{d\theta} = 1(1) - 1(0) = 1

y - 0 = 1(x - 1)

r' = -\sin\theta

r'\left(\tfrac{\pi}{2}\right) = -1

\dfrac{dx}{d\theta} = 3\cos 2\theta = 0

\theta = \tfrac{\pi}{4}, \tfrac{3\pi}{4}

r = \tfrac{3\sqrt{2}}{2} \Rightarrow \left(\tfrac{3}{2}, \tfrac{3}{2}\right), \left(-\tfrac{3}{2}, \tfrac{3}{2}\right)

2\sin 3\theta = 0 \Rightarrow \theta = 0, \tfrac{\pi}{3}, \tfrac{2\pi}{3}

r' = 6\cos 3\theta = \pm 6

y = 0,\; y = \sqrt{3}\,x,\; y = -\sqrt{3}\,x

r(0) = 1, \quad r'(0) = 1

\dfrac{dy}{d\theta} = 1(0) + 1(1) = 1, \quad \dfrac{dx}{d\theta} = 1(1) - 1(0) = 1

r(0) = 1, \quad r'(0) = -2\cos 0 = -2

\dfrac{dy}{d\theta} = -2(0) + 1(1) = 1, \quad \dfrac{dx}{d\theta} = -2(1) - 1(0) = -2

r' = 3\sin\theta = 0 \Rightarrow \theta = 0, \pi

r(0) = 2, \quad r(\pi) = 8

\dfrac{dy}{d\theta} = 3(1) + 5(0) = 3, \quad \dfrac{dx}{d\theta} = 3(0) - 5(1) = -5

Polar derivatives FAQ

How do you find dy/dx for a polar curve?

Write x = r cos θ and y = r sin θ, differentiate both with respect to θ using the product rule, and divide: dy/dx = (r′ sin θ + r cos θ)/(r′ cos θ − r sin θ).

Is dr/dθ the slope of a polar curve?

No. dr/dθ tells how fast the distance from the origin changes. The slope of the tangent line is dy/dx.

How do you find horizontal tangents of a polar curve?

Set dy/dθ = r′ sin θ + r cos θ equal to 0 and solve. Keep the angles where dx/dθ is not 0.

What is the tangent line at the pole?

If r = 0 at θ = α and r′(α) is not 0, the curve passes through the origin heading along the line θ = α, so that line is the tangent.

Summary

• Treat θ as a parameter: x = r cos θ, y = r sin θ.
• dy/dθ = r′ sin θ + r cos θ and dx/dθ = r′ cos θ − r sin θ.
• dy/dx = (dy/dθ)/(dx/dθ); dr/dθ is not the slope.
• Horizontal: dy/dθ = 0. Vertical: dx/dθ = 0. Both 0: check for a cusp.
• Convert the point to (x, y) before writing a tangent line.

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