Math Core

Lesson 7.2 · Polar and Parametric Equations

Polar equations and graphs

A polar equation relates rr and θ\theta, usually in the form r=f(θ)r = f(\theta). Curves that are messy in xx and yy, like flowers, hearts and spirals, often have short, elegant polar equations. In this lesson you will graph the main families and convert equations between the two systems.

Graphing by plotting points

The graph of r=f(θ)r = f(\theta) is every point (r,θ)(r, \theta) that satisfies the equation. The basic method is the same as for y=f(x)y = f(x): make a table, plot, and connect. Think of sweeping a ray counterclockwise from θ=0\theta = 0 and watching how far out the point sits as the ray turns.

Take r=2+2cos⁡θr = 2 + 2\cos\theta.

θ\theta00π3\frac{\pi}{3}π2\frac{\pi}{2}2π3\frac{2\pi}{3}π\pi4π3\frac{4\pi}{3}3π2\frac{3\pi}{2}5π3\frac{5\pi}{3}
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The point starts 4 units out on the positive xx-axis, shrinks to the pole at θ=π\theta = \pi, then grows back. The result is a heart shape called a cardioid.

The cardioid r = 2 + 2cos θ touches the pole at θ = π.Open in grapher →

The standard families

You don't need a table every time. A few families come up again and again, and recognizing the form tells you the shape. In each family, aa and bb are positive constants.

EquationGraph
r=ar = acircle of radius aa centered at the pole
θ=α\theta = \alphaline through the pole at angle α\alpha
r=acos⁡θr = a\cos\theta or r=asin⁡θr = a\sin\thetacircle of diameter aa passing through the pole
r=a±bcos⁡θr = a \pm b\cos\theta or r=a±bsin⁡θr = a \pm b\sin\thetalimaçon (cardioid when a=ba = b)
r=acos⁡(nθ)r = a\cos(n\theta) or r=asin⁡(nθ)r = a\sin(n\theta)rose with petals of length aa

Circles through the pole

The graph of r=4sin⁡θr = 4\sin\theta is a circle of diameter 4 sitting on top of the pole. The graph of r=4cos⁡θr = 4\cos\theta is the same circle turned to the right.

r = 4 sin θ (above the pole) and r = 4 cos θ (right of the pole)Open in grapher →

Limaçons

For r=a+bcos⁡θr = a + b\cos\theta (or with sin⁡\sin), the ratio ab\tfrac{a}{b} decides the shape:

  • a<ba < b: the curve has an inner loop, because rr becomes negative for some angles.
  • a=ba = b: a cardioid, touching the pole at one point.
  • a>ba > b: a dimpled or convex limaçon that never reaches the pole.
r = 1 + 2 sin θ has an inner loop; r = 3 + 2 sin θ does not reach the pole.Open in grapher →

Roses

Petals of a rose

The rose r=acos⁡(nθ)r = a\cos(n\theta) or r=asin⁡(nθ)r = a\sin(n\theta), with nn a positive integer, has petals of length aa and

  • nn petals when nn is odd,
  • 2n2n petals when nn is even.

Why the difference? When nn is odd, the negative values of rr retrace petals that were already drawn. When nn is even, the negative values draw brand-new petals in the gaps.

r = 3 sin(2θ) has 4 petals; r = 3 cos(3θ) has 3 petals.Open in grapher →

Common mistake

The number of petals is not always nn. A common error is to say r=3sin⁡(2θ)r = 3\sin(2\theta) has 2 petals. Because 22 is even, it has 2⋅2=42 \cdot 2 = 4 petals.

Symmetry tests

Symmetry halves the plotting work. For a polar equation:

  • Replacing θ\theta with −θ-\theta gives an equivalent equation ⇒\Rightarrow symmetric about the polar axis (xx-axis). Every equation in cos⁡θ\cos\theta alone passes, since cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta.
  • Replacing θ\theta with π−θ\pi - \theta gives an equivalent equation ⇒\Rightarrow symmetric about the line θ=π2\theta = \tfrac{\pi}{2} (yy-axis). Equations in sin⁡θ\sin\theta alone pass, since sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta.
  • Replacing rr with −r-r gives an equivalent equation ⇒\Rightarrow symmetric about the pole.

These tests are sufficient but not necessary: a curve can be symmetric even when a test fails.

Converting between polar and rectangular equations

Use the same tools as for points: x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, r2=x2+y2r^2 = x^2 + y^2. The trick is often to multiply by rr so those combinations appear.

Worked example: Polar to rectangular

Convert r=6cos⁡θr = 6\cos\theta to rectangular form and describe the graph.

Solution. Multiply both sides by rr: r2=6rcos⁡θr^2 = 6r\cos\theta. Substitute: x2+y2=6xx^2 + y^2 = 6x.

Complete the square: x2−6x+9+y2=9x^2 - 6x + 9 + y^2 = 9, so

(x−3)2+y2=9.(x - 3)^2 + y^2 = 9.

This is a circle with center (3,0)(3, 0) and radius 3. It passes through the pole, as the family table promised.

Worked example: Rectangular to polar

Convert the line y=2y = 2 and the circle x2+y2=25x^2 + y^2 = 25 to polar form.

Solution. For the line, substitute y=rsin⁡θy = r\sin\theta: rsin⁡θ=2r\sin\theta = 2, so r=2sin⁡θ=2csc⁡θr = \dfrac{2}{\sin\theta} = 2\csc\theta.

For the circle, x2+y2=r2x^2 + y^2 = r^2, so r2=25r^2 = 25 and r=5r = 5. (The equation r=−5r = -5 traces the same circle, so you only need one.)

Worked example: A line hiding in polar form

Convert r=42cos⁡θ−sin⁡θr = \dfrac{4}{2\cos\theta - \sin\theta} to rectangular form.

Solution. Clear the fraction: 2rcos⁡θ−rsin⁡θ=42r\cos\theta - r\sin\theta = 4. Substitute: 2x−y=42x - y = 4, which is the line y=2x−4y = 2x - 4.

Intersections of polar curves

To find where two polar curves meet, set the rr-expressions equal and solve for θ\theta. Then also check the pole separately: two curves can both pass through the pole at different angles, and setting them equal will miss it.

Tip

Graph both curves before you solve. The picture tells you how many intersection points to expect, so you know when you've found them all.

Practice

Practice 1

What is the graph of r=5r = 5?

Practice 2

How many petals does the rose r=2sin⁡(4θ)r = 2\sin(4\theta) have?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

The polar equation r=6sin⁡θr = 6\sin\theta is a circle. Convert it to rectangular form and give the center as (x,y)(x, y).

Enter a point like (2, -3)

Practice 4

Which polar equation describes the vertical line x=4x = 4?

Practice 5

The limaçon r=1+2sin⁡θr = 1 + 2\sin\theta passes through the pole. Find every θ\theta in [0,2π)[0, 2\pi) where r=0r = 0.

Separate answers with commas, e.g. 2, -5

Practice 6

Convert r=3cos⁡θ+sin⁡θr = \dfrac{3}{\cos\theta + \sin\theta} to a rectangular equation. Solve for yy.

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 7

The circles r=2r = 2 and r=4cos⁡θr = 4\cos\theta are graphed below. Find every θ\theta in [0,2π)[0, 2\pi) where they meet, that is, where 4cos⁡θ=24\cos\theta = 2.

r = 2r = 4cos(θ)Open in grapher →

Separate answers with commas, e.g. 2, -5

Practice 8

Which statement about r=3−3cos⁡θr = 3 - 3\cos\theta is true?