Tangent Lines in Polar
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.4: “Arc Length of a Curve and Surface Area” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-4-arc-length-of-a-curve-and-surface-area |
| Textbook used in class | Stewart, Calculus, Section 10.4: “Areas and Lengths in Polar Coordinates” |
Opening Scenario
A polar curve $r = f(\theta)$ is, at its core, a parametric curve with parameter $\theta$: $x = r\cos\theta = f(\theta)\cos\theta$ and $y = r\sin\theta = f(\theta)\sin\theta$. The slope of the tangent to the curve at a point is therefore $dy/dx$, computed using the parametric derivative formula: $dy/dx = (dy/d\theta)/(dx/d\theta)$.
Quick Reference
For $r = f(\theta)$, write $x = r\cos\theta$, $y = r\sin\theta$. Then:
$$\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta, \qquad \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta.$$
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}.$$
Horizontal tangents: $dy/d\theta = 0$ and $dx/d\theta \neq 0$.
Vertical tangents: $dx/d\theta = 0$ and $dy/d\theta \neq 0$.
Key Concepts
1. Convert to Parametric, Then Differentiate
A polar curve $r = f(\theta)$ in Cartesian coordinates is $x = f(\theta)\cos\theta$, $y = f(\theta)\sin\theta$. Differentiate both with respect to $\theta$ using the product rule:
$$\frac{dx}{d\theta} = f'(\theta)\cos\theta - f(\theta)\sin\theta = r'\cos\theta - r\sin\theta.$$
$$\frac{dy}{d\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta = r'\sin\theta + r\cos\theta.$$
The slope is $dy/dx = (dy/d\theta)/(dx/d\theta)$.
2. Horizontal and Vertical Tangents
Horizontal tangent at angle $\theta_0$: numerator $dy/d\theta = 0$, denominator $dx/d\theta \neq 0$.
Vertical tangent at $\theta_0$: denominator $dx/d\theta = 0$, numerator $dy/d\theta \neq 0$.
When both are zero, the tangent direction requires more analysis.
3. Tangent at the Origin
When a curve passes through the origin ($r = 0$ at $\theta = \theta_0$), the tangent line to the curve at the origin has slope: $$\frac{dy}{dx}\bigg|_{r=0} = \frac{r'(\theta_0)\sin\theta_0 + 0}{r'(\theta_0)\cos\theta_0 - 0} = \tan\theta_0.$$
This means the tangent line at the origin is in the direction $\theta_0$ -- the direction in which the curve departs from the origin.
Worked Example
Find the horizontal and vertical tangent lines to the cardioid $r = 1 + \cos\theta$.
$r' = -\sin\theta$.
$$\frac{dy}{d\theta} = (-\sin\theta)\sin\theta + (1+\cos\theta)\cos\theta = -\sin^2\theta + \cos\theta + \cos^2\theta = \cos^2\theta - \sin^2\theta + \cos\theta = \cos 2\theta + \cos\theta.$$
$$\frac{dx}{d\theta} = (-\sin\theta)\cos\theta - (1+\cos\theta)\sin\theta = -\sin\theta\cos\theta - \sin\theta - \sin\theta\cos\theta = -2\sin\theta\cos\theta - \sin\theta = -\sin(2\theta) - \sin\theta.$$
Horizontal tangents ($dy/d\theta = 0$): $\cos 2\theta + \cos\theta = 0 \Rightarrow 2\cos^2\theta - 1 + \cos\theta = 0 \Rightarrow (2\cos\theta-1)(\cos\theta+1) = 0$.
$\cos\theta = 1/2 \Rightarrow \theta = \pi/3, 5\pi/3$; $\cos\theta = -1 \Rightarrow \theta = \pi$ (check: $dx/d\theta$ at $\pi$: $-\sin(2\pi)-\sin\pi = 0$, so both zero; skip $\theta = \pi$ for now).
At $\theta = \pi/3$: $r = 1+1/2 = 3/2$. Point: $(r\cos(\pi/3), r\sin(\pi/3)) = (3/4, 3\sqrt{3}/4)$.
At $\theta = 5\pi/3$: $r = 3/2$. Point: $(3/4, -3\sqrt{3}/4)$ (symmetric by $x$-axis symmetry).
Vertical tangents ($dx/d\theta = 0$): $-\sin 2\theta - \sin\theta = 0 \Rightarrow -2\sin\theta\cos\theta - \sin\theta = 0 \Rightarrow -\sin\theta(2\cos\theta+1) = 0$.
$\sin\theta = 0 \Rightarrow \theta = 0, \pi$; $\cos\theta = -1/2 \Rightarrow \theta = 2\pi/3, 4\pi/3$.
At $\theta = 0$: $r = 2$. Point: $(2, 0)$. $dy/d\theta = \cos 0 + \cos 0 = 2 \neq 0$. Vertical tangent at $(2,0)$.
At $\theta = 2\pi/3$: $r = 1/2$. Point: $(-1/4, \sqrt{3}/4)$. Vertical tangent there.
Boxed answers: Horizontal tangents at $(3/4, \pm 3\sqrt{3}/4)$; vertical tangents at $(2,0)$ and $(-1/4, \pm\sqrt{3}/4)$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Treating $dr/d\theta$ as $dy/dx$ | Using $r' = -\sin\theta$ directly as the slope | The slope in Cartesian coordinates is $dy/dx = (r'\sin\theta + r\cos\theta)/(r'\cos\theta - r\sin\theta)$ |
| Forgetting the product rule for $dy/d\theta$ | Writing $dy/d\theta = r'\sin\theta$ (missing the $r\cos\theta$ term) | $y = r\sin\theta$ is a product; $dy/d\theta = r'\sin\theta + r\cos\theta$ by the product rule |
| Declaring a horizontal tangent without checking $dx/d\theta$ | Reporting $\theta = \pi$ as a horizontal tangent of the cardioid | When both $dy/d\theta = 0$ and $dx/d\theta = 0$, the tangent type is indeterminate; additional analysis is needed |
Common Misconceptions
the slope of a polar curve $r = f(\theta)$ at a point is $dr/d\theta$.
This is the height-vs-slope error in polar form. The quantity $dr/d\theta$ measures how the radius changes with angle, not the slope of the tangent in the Cartesian plane. The Cartesian slope is $dy/dx = (dy/d\theta)/(dx/d\theta)$, computed by first writing $x = r\cos\theta$ and $y = r\sin\theta$ and then differentiating with respect to $\theta$ using the product rule. These are genuinely different calculations.
$dy/d\theta = r'\sin\theta$ (forgetting the $r\cos\theta$ term).
This is a composition error in applying the product rule. Because $y = r(\theta)\sin\theta$ is a product of two functions of $\theta$, the product rule gives $dy/d\theta = r'(\theta)\sin\theta + r(\theta)\cos\theta$. Omitting the second term $r\cos\theta$ is the single most common computational mistake on polar tangent problems.
Leveled Practice
Level 1 -- Compute the Slope
Problem 1. For $r = 2\cos\theta$, find $dy/dx$ at $\theta = \pi/4$.
Show answer
$r = 2\cos\theta$, $r' = -2\sin\theta$.
$dy/d\theta = r'\sin\theta + r\cos\theta = -2\sin^2\theta + 2\cos^2\theta = 2\cos 2\theta$.
$dx/d\theta = r'\cos\theta - r\sin\theta = -2\sin\theta\cos\theta - 2\sin\theta\cos\theta = -2\sin 2\theta$.
At $\theta = \pi/4$: $dy/d\theta = 2\cos(\pi/2) = 0$; $dx/d\theta = -2\sin(\pi/2) = -2$.
$dy/dx = 0/(-2) = 0$. Horizontal tangent at $\theta = \pi/4$.
Level 2 -- Tangent at the Origin
Problem 2. The rose $r = \cos(2\theta)$ passes through the origin at several $\theta$-values. Find the direction of the tangent at the origin for $\theta = \pi/4$.
Show answer
When $r = 0$ (at $\theta = \pi/4$, since $\cos(2\cdot\pi/4) = \cos(\pi/2) = 0$), the tangent line at the origin is in the direction $\theta_0 = \pi/4$. Its slope is $\tan(\pi/4) = 1$. The tangent line is $y = x$.
Mastery Checklist
Mental Model
A polar curve is secretly a parametric curve with $\theta$ as the parameter. Everything from parametric calculus (slope = $dy/d\theta$ divided by $dx/d\theta$, horizontal and vertical tangents) applies directly. The only new step is differentiating $x = r\cos\theta$ and $y = r\sin\theta$ with respect to $\theta$, which requires the product rule because $r$ itself depends on $\theta$.
Connections
Looking back
- Parametric derivatives (Section 10.2): The slope formula $dy/dx = (dy/dt)/(dx/dt)$ applies with $t = \theta$.
- Polar curves (Section 10.3): Must know the curve shape to interpret where horizontal/vertical tangents occur.