Calculus with Polar Curves: Slopes and Tangents

Q: How do I find dy/dx for a polar curve r = f(θ)?

Write x = f(θ) cos θ and y = f(θ) sin θ, differentiate both with respect to θ, and divide: dy/dx = (dy/dθ)/(dx/dθ).

Q: Is dr/dθ the slope of a polar curve?

No. dr/dθ is how fast the distance from the origin changes as θ increases. The slope is dy/dx, which needs both x and y.

Q: What does a positive dr/dθ mean?

The curve is moving away from the origin as θ increases (when r > 0). A negative dr/dθ means it is moving toward the origin.

Quick answer

A polar curve r = f(θ) is the parametric curve x = f(θ) cos θ, y = f(θ) sin θ, with θ as the parameter. Its slope is therefore dy/dx = (dy/dθ)/(dx/dθ), which works out to (f′ sin θ + f cos θ)/(f′ cos θ − f sin θ). Horizontal tangents occur where dy/dθ = 0 and vertical ones where dx/dθ = 0. The derivative dr/dθ answers a different question: whether the curve is moving away from the origin or toward it as θ increases.

What you'll learn

Write a polar curve as a parametric curve in θ
Find dy/dx and tangent lines for polar curves
Locate horizontal and vertical tangents of polar curves
Interpret dr/dθ as motion toward or away from the origin

A polar curve is a parametric curve

The polar curve $r = f(\theta)$ puts the point at distance $f(\theta)$ in the direction $\theta$ . In rectangular coordinates, that is

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

a parametric curve with $\theta$ as the parameter. Everything from parametric calculus applies, including the slope:

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

The numerator and denominator come from the product rule. There is no need to memorize the fraction: write $x$ and $y$ and differentiate.

Why dr/dθ is not the slope

The derivative $\tfrac{dr}{d\theta}$ measures how fast the distance from the origin changes as the direction turns. The slope measures how the curve tilts in the $xy$ -plane. On the circle $r = 2$ , $\tfrac{dr}{d\theta} = 0$ everywhere, yet the slope takes every value as the circle goes around. $\tfrac{dr}{d\theta}$ says whether the curve moves away from the origin or toward it; $\tfrac{dy}{dx}$ says which way it tilts. When $r > 0$ , a positive $\tfrac{dr}{d\theta}$ means moving away, and a negative one means moving closer.

r = 1 + cos θ
tangent lines

The cardioid r = 1 + cos θ with three tangent lines

Example 1 — A tangent line to the cardioid

Find the slope of $r = 1 + \cos\theta$ at $\theta = \tfrac{\pi}{2}$ .

Here $x = (1 + \cos\theta)\cos\theta$ and $y = (1 + \cos\theta)\sin\theta$ . With $f = 1 + \cos\theta$ and $f' = -\sin\theta$ :

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

At $\theta = \tfrac{\pi}{2}$ : $\tfrac{dx}{d\theta} = -1$ and $\tfrac{dy}{d\theta} = -1$ , so the slope is $1$ . The point is $(0, 1)$ , so the tangent line is $y = x + 1$ .

Horizontal and vertical tangents

As for any parametric curve, the tangent is horizontal where $\tfrac{dy}{d\theta} = 0$ with $\tfrac{dx}{d\theta} \ne 0$ , and vertical where $\tfrac{dx}{d\theta} = 0$ with $\tfrac{dy}{d\theta} \ne 0$ . Where the curve passes through the origin, with $r = 0$ at $\theta = \alpha$ and $\tfrac{dr}{d\theta} \ne 0$ , it leaves along the ray at angle $\alpha$ .

Worked examples

Example 2 — The highest point of the cardioid

Find the horizontal tangents of $r = 1 + \cos\theta$ .

Write $y = \sin\theta + \sin\theta\cos\theta = \sin\theta + \tfrac{1}{2}\sin 2\theta$ , so

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

It is $0$ where $\cos\theta = \tfrac{1}{2}$ , at $\theta = \tfrac{\pi}{3}$ and $\tfrac{5\pi}{3}$ , and where $\cos\theta = -1$ , at $\theta = \pi$ . At $\theta = \pi$ the curve is at the origin and $\tfrac{dx}{d\theta} = 0$ too: that is the cusp. The horizontal tangents are at $\theta = \tfrac{\pi}{3}$ and $\tfrac{5\pi}{3}$ , where $r = 1.5$ , at the points $\left(0.75, \pm\tfrac{3\sqrt{3}}{4}\right) \approx (0.75, \pm 1.30)$ .

Common mistakes

Practice problems

Write $r = 3\sin\theta$ as parametric equations in $\theta$ .

Answer

$x = 3\sin\theta\cos\theta$ , $y = 3\sin^2\theta$

Full solution

Multiply $r$ by $\cos\theta$ for $x$ and by $\sin\theta$ for $y$ .
Find the slope of $r = 2$ at $\theta = \tfrac{\pi}{6}$ .

Answer

$-\sqrt{3}$

Full solution

$x = 2\cos\theta$ and $y = 2\sin\theta$ , so $\tfrac{dy}{dx} = \tfrac{2\cos\theta}{-2\sin\theta} = -\cot\theta = -\sqrt{3}$ at $\tfrac{\pi}{6}$ .
Find the slope of $r = \theta$ at $\theta = \tfrac{\pi}{2}$ .

Answer

$-\tfrac{2}{\pi}$

Full solution

$x = \theta\cos\theta$ , $y = \theta\sin\theta$ . $\tfrac{dx}{d\theta} = \cos\theta - \theta\sin\theta = -\tfrac{\pi}{2}$ and $\tfrac{dy}{d\theta} = \sin\theta + \theta\cos\theta = 1$ at $\tfrac{\pi}{2}$ . The slope is $\tfrac{1}{-\pi/2}$ .
Find the tangent line to $r = 1 + \cos\theta$ at $\theta = \tfrac{\pi}{2}$ .

Answer

$y = x + 1$

Full solution

Example 1 gives slope $1$ at the point $(0, 1)$ .
For $r = 2 + \sin\theta$ , is the curve moving toward the origin or away from it at $\theta = \pi$ ?

Answer

Toward it

Full solution

$\tfrac{dr}{d\theta} = \cos\theta = -1 < 0$ at $\theta = \pi$ , and $r = 2 > 0$ , so the distance from the origin is shrinking.
Find the vertical tangents of $r = 2\cos\theta$ for $0 \le \theta < \pi$ .

Answer

At $(0, 0)$ and $(2, 0)$

Full solution

$x = 1 + \cos 2\theta$ , so $\tfrac{dx}{d\theta} = -2\sin 2\theta = 0$ at $\theta = 0$ and $\tfrac{\pi}{2}$ . There $\tfrac{dy}{d\theta} = 2\cos 2\theta = \pm 2 \ne 0$ . $\theta = 0$ gives $(2, 0)$ and $\theta = \tfrac{\pi}{2}$ gives the origin.
At which angles does the rose $r = \cos 2\theta$ pass through the origin, and along which lines does it leave?

Answer

At $\theta = \tfrac{\pi}{4}$ , $\tfrac{3\pi}{4}$ , $\tfrac{5\pi}{4}$ , $\tfrac{7\pi}{4}$ , along the lines $y = x$ and $y = -x$

Full solution

$r = 0$ when $2\theta$ is an odd multiple of $\tfrac{\pi}{2}$ . There $\tfrac{dr}{d\theta} = -2\sin 2\theta = \pm 2 \ne 0$ , so the curve leaves along the rays at those angles, which lie on $y = \pm x$ .
Find $\tfrac{dy}{dx}$ for $r = e^{\theta}$ in general, and at $\theta = 0$ .

Answer

$\tfrac{\sin\theta + \cos\theta}{\cos\theta - \sin\theta}$ ; $1$ at $\theta = 0$

Full solution

With $f = f' = e^{\theta}$ , the factor $e^{\theta}$ cancels from top and bottom of the slope formula. At $\theta = 0$ the slope is $\tfrac{0 + 1}{1 - 0} = 1$ .
Where on $0 \le \theta \le \pi$ does $r = 1 + \cos\theta$ have a vertical tangent?

Answer

At $\theta = 0$ , the point $(2, 0)$ , and at $\theta = \tfrac{2\pi}{3}$ , the point $\left(-\tfrac{1}{4}, \tfrac{\sqrt{3}}{4}\right)$

Full solution

$x = \cos\theta + \cos^2\theta$ , so $\tfrac{dx}{d\theta} = -\sin\theta(1 + 2\cos\theta)$ , which is $0$ at $\theta = 0$ , $\pi$ and $\tfrac{2\pi}{3}$ . At $\theta = \pi$ , $\tfrac{dy}{d\theta} = 0$ too: the cusp. At the other two, $\tfrac{dy}{d\theta} = \cos\theta + \cos 2\theta$ is $2$ and $-1$ , not $0$ .
A student says $r = 3$ has slope $0$ everywhere, because $\tfrac{dr}{d\theta} = 0$ . What went wrong?

Hint

What shape is $r = 3$ ?

Answer

$\tfrac{dr}{d\theta}$ is not the slope. $r = 3$ is a circle, whose slope is $-\cot\theta$ .

Full solution

$\tfrac{dr}{d\theta} = 0$ says the distance from the origin never changes, which is exactly what a circle centered at the origin does.

The slope comes from $x = 3\cos\theta$ and $y = 3\sin\theta$ : $\tfrac{dy}{dx} = \tfrac{3\cos\theta}{-3\sin\theta} = -\cot\theta$ .

Frequently asked questions

How do I find dy/dx for a polar curve r = f(θ)?

Write x = f(θ) cos θ and y = f(θ) sin θ, differentiate both with respect to θ, and divide: dy/dx = (dy/dθ)/(dx/dθ).

Is dr/dθ the slope of a polar curve?

No. dr/dθ is how fast the distance from the origin changes as θ increases. The slope is dy/dx, which needs both x and y.

What does a positive dr/dθ mean?

The curve is moving away from the origin as θ increases (when r > 0). A negative dr/dθ means it is moving toward the origin.

Where does a polar curve have a horizontal tangent?

Where dy/dθ = 0 and dx/dθ ≠ 0. Where dx/dθ = 0 and dy/dθ ≠ 0, the tangent is vertical.

What is the slope of a polar curve where it passes through the origin?

If r = 0 at θ = α and dr/dθ ≠ 0 there, the curve leaves the origin along the line at angle α, so the slope is tan α.

What to learn next

Area in Polar CoordinatesFind areas bounded by polar curves with A = ½∫r² dθ: one curve, the region between two curves, and finding the limits where curves meet.

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