Precalculus

Converting an equation between the two systems

A curve is described by an equation in x and y, but the problem is about distances and directions from one center, or the other way round. How do you rewrite the whole curve in the other system, and how do you recognize it afterward?

You need

Converting between polar and rectangularPrecalculus

Where you are 3 in Polar coordinates

Trigonometry7

Polar coordinates and vectors10

Polar coordinates3

Converting between polar and rectangular Plotting polar points

Converting an equation between the two systems

When you reach for it

The relations from Converting between polar and rectangular,

\begin{aligned} x &= r\cos\theta, & y &= r\sin\theta, \\ x^2 + y^2 &= r^2, & \tan\theta &= \frac{y}{x} \end{aligned}

move one point at a time between the two systems. Used on a whole equation, they move a whole curve. The circle x^2 + y^2 = 49 needs two variables in rectangular form but is simply r = 7 in polar form: every point at distance 7. The other way, an unfamiliar polar equation such as r = 6\cos\theta - 8\sin\theta turns out, once converted, to be a circle whose center and radius can be read off. Rewrite in polar form when the curve is about distance from one center or about direction; rewrite in rectangular form to recognize a curve you have only seen in polar form.

Throughout, the pole sits at the origin and the polar axis runs along the positive x-axis.

The procedure

What a converted equation has to say

A point lies on the graph of a polar equation when at least one of its polar names satisfies the equation. A point has many names, (r, \theta), (r, \theta + 2\pi), (-r, \theta + \pi) and so on, and the pole is (0, \theta) for every \theta; only one name needs to work.

The relations x = r\cos\theta and y = r\sin\theta hold for every name of a point, negative r included. So if a rectangular equation in x and y holds at a point, it holds after x and y are replaced by r\cos\theta and r\sin\theta, for every name of that point; and if the substituted equation holds for one name, the point’s x and y satisfy the original. The two equations describe the same set of points. Nothing else is needed to justify the substitution.

From rectangular to polar

Replace x by r\cos\theta and y by r\sin\theta, and wherever x^2 + y^2 appears, replace it by r^2, which is shorter and is the same thing by the Pythagorean identity. Then, when it can be done, solve for r, so the equation reads as a distance given a direction. A factor \cos\theta or \sin\theta divided out must be nonzero, and the directions where it is zero have to be looked at separately.

From polar to rectangular

Here the aim is to build the three combinations that have rectangular names: r\cos\theta, r\sin\theta and r^2. Three moves do most of the work.

Write \displaystyle \sec\theta = \frac{1}{\cos\theta} and \displaystyle \csc\theta = \frac{1}{\sin\theta} and clear the fraction, which brings r\cos\theta or r\sin\theta together.
Multiply both sides by r. A bare \cos\theta becomes r\cos\theta = x, and r becomes r^2 = x^2 + y^2.
Square both sides, to turn a lone r into r^2.

An equation \theta = c needs none of this. With r free to be any real number, (r, c) runs over the ray at angle c (positive r), the opposite ray (negative r) and the pole (r = 0): the whole line through the origin at angle c. When \cos c \ne 0 that line is y = (\tan c)\,x, and when \cos c = 0 it is the y-axis, x = 0.

Multiplying by r, dividing by r, and the pole

Multiplying both sides by r adds solutions: the new equation also holds when r = 0, so it contains the pole, whatever the old one did. Dividing both sides by r does the reverse; it can throw the pole away. Either way, only one point is at stake, and checking it is quick. The pole is on a polar graph exactly when some name (0, \theta) satisfies the equation; for an equation r = f(\theta) that means f(\theta) = 0 for some \theta. After multiplying by r, test the equation you started from: if it reaches r = 0, the pole was already on the curve and nothing changed; if it never does, the converted equation has one extra point, the origin, and it has to be removed in words. After dividing by r, test the equation you ended with: if it reaches r = 0, nothing was lost.

Squaring can also add points, those where r equals the negative of what the equation asked for. For the curves in this lesson those points turn out to be on the curve already, and each example says why.

Recognizing the result

After converting, put the rectangular equation in a form you know. A linear equation is a line. An equation x^2 + y^2 + Dx + Ey = 0, with D and E not both 0, is a circle through the origin, and completing the square in x and in y gives its center and radius. An equation with x^2 and y to the first power only is a parabola (The parabola as a locus).

Worked examples

Example 1

Lines that are easy in rectangular form

Write x = -4, y = 7 and \displaystyle y = -\sqrt{3}x in polar form.

Solution

x = -4 becomes r\cos\theta = -4. A point with \cos\theta = 0 would have x = 0, which is not on this line, so dividing by \cos\theta loses nothing: r = -4\sec\theta. Check: at \theta = \pi, r = -4 \cdot (-1) = 4, the point 4 units out along the negative x-axis, (-4, 0).

y = 7 becomes r\sin\theta = 7, and in the same way r = 7\csc\theta.

\displaystyle y = -\sqrt{3}x becomes \displaystyle r\sin\theta = -\sqrt{3}\,r\cos\theta. At the pole this holds with r = 0, and the pole is on the line. Away from it r \ne 0, and \cos\theta \ne 0 (otherwise \sin\theta would be 0 too), so dividing by r\cos\theta gives \displaystyle \tan\theta = -\sqrt{3}. One angle with that tangent is \displaystyle \frac{2\pi}{3}, so the line is

\theta = \frac{2\pi}{3}

With r > 0 that is the half in quadrant II; with r < 0 it is the opposite half, in quadrant IV. \displaystyle \theta = -\frac{\pi}{3} is an equally good answer, since it is the same line.

Example 2

A general line

Write 2x - 7y = 14 in polar form, and explain the directions where the polar equation gives no r.

Solution

Substitute and factor out r:

\begin{aligned} 2r\cos\theta - 7r\sin\theta &= 14 \\ r(2\cos\theta - 7\sin\theta) &= 14 \end{aligned}

When 2\cos\theta - 7\sin\theta \ne 0, divide:

r = \frac{14}{2\cos\theta - 7\sin\theta}

Check two points: \theta = 0 gives r = 7, the x-intercept (7, 0), and \displaystyle \theta = -\frac{\pi}{2} gives \displaystyle r = \frac{14}{7} = 2 in the downward direction, the y-intercept (0, -2).

The denominator is 0 when \displaystyle \tan\theta = \frac{2}{7}. That is exactly the Slope of the line, \displaystyle y = \frac{2}{7}x - 2, so the ray in that direction runs parallel to the line and never reaches it: no distance, no r. The formula is not failing; it is reporting the geometry.

Figure 1Example 2. Each ray from the origin meets the line once, at distance \displaystyle r = \frac{14}{2\cos\theta - 7\sin\theta}, except the direction \displaystyle \tan\theta = \frac{2}{7}, which is parallel to the line.

Example 3

Circles centered at the origin and through it

Write x^2 + y^2 = 49 and x^2 + y^2 = -8x in polar form.

Solution

x^2 + y^2 = 49 becomes r^2 = 49, so r = 7 or r = -7. These are not two curves. The point (-7, \theta) is the point (7, \theta + \pi), so r = -7 traces the same circle; r = 7 alone is the answer.

x^2 + y^2 = -8x becomes r^2 = -8r\cos\theta, so r(r + 8\cos\theta) = 0:

r = 0 \quad\text{or}\quad r = -8\cos\theta

The first is only the pole. The second already contains it, since \displaystyle \theta = \frac{\pi}{2} gives \displaystyle r = -8\cos\frac{\pi}{2} = 0. Dividing by r lost nothing, and the answer is r = -8\cos\theta. To recognize the curve, complete the square in the original: x^2 + 8x + 16 + y^2 = 16, or (x + 4)^2 + y^2 = 16, the circle with center (-4, 0) and radius 4, which does pass through the origin.

Example 4

Polar lines and a polar circle, back to rectangular

Rewrite r = 4\csc\theta, \displaystyle \theta = \frac{5\pi}{6} and r = -5 in rectangular form, and name each curve.

Solution

r = 4\csc\theta means \displaystyle r = \frac{4}{\sin\theta}, so r\sin\theta = 4: the horizontal line y = 4.

\displaystyle \theta = \frac{5\pi}{6} is the whole line through the origin at that angle. Its points satisfy \displaystyle y = x\tan\frac{5\pi}{6}, and \displaystyle \tan\frac{5\pi}{6} = -\frac{1}{\sqrt{3}}, so the line is \displaystyle y = -\frac{\sqrt{3}}{3}x.

r = -5: squaring gives r^2 = 25, so x^2 + y^2 = 25. Squaring also lets in r = 5, but that is the same circle, so nothing was added. The graph is the circle of radius 5 centered at the origin; the minus sign only changes which name each point is reached by.

Example 5

An unfamiliar polar equation

Riikka meets r = 6\cos\theta - 8\sin\theta. Rewrite it in rectangular form and identify the curve.

Solution

A bare \cos\theta and \sin\theta suggest multiplying by r:

\begin{aligned} r^2 &= 6r\cos\theta - 8r\sin\theta \\ x^2 + y^2 &= 6x - 8y \end{aligned}

Multiplying by r put the pole into the new equation. Was it on the curve already? r = 0 when 6\cos\theta = 8\sin\theta, that is \displaystyle \tan\theta = \frac{3}{4}, which has solutions; so the pole was on the curve and nothing changed. Now complete the square in each variable, adding 9 and 16 to both sides:

\begin{aligned} x^2 - 6x + 9 + y^2 + 8y + 16 &= 25 \\ (x - 3)^2 + (y + 4)^2 &= 25 \end{aligned}

The curve is the circle with center (3, -4) and radius 5. It passes through the origin, as 3^2 + 4^2 = 25 confirms. Read in polar terms, r is the length of the chord from the origin in direction \theta; at \displaystyle \theta = \frac{3\pi}{2}, for example, r = 8, and the point is (0, -8), where the circle crosses the y-axis a second time.

Figure 2Example 5. The chord from the origin in direction \theta has length r = 6\cos\theta - 8\sin\theta. Where the ray only touches the circle, the length is 0 and the point is the pole.

Polar circles of this kind, and the symmetry tests that sketch them without converting, are the subject of Polar circles and symmetry tests.

Example 6

When multiplying by r adds the pole

Branko converts r = 2 + \cos\theta to rectangular form. Find a rectangular equation, and decide whether it describes exactly the same points.

Solution

Multiply by r, then replace what has a rectangular name:

\begin{aligned} r^2 &= 2r + r\cos\theta \\ x^2 + y^2 - x &= 2r \end{aligned}

A lone r is left, so square both sides and use r^2 = x^2 + y^2 again:

(x^2 + y^2 - x)^2 = 4(x^2 + y^2)

Check a point: \theta = 0 gives r = 3, the point (3, 0), and (9 - 3)^2 = 36 = 4 \cdot 9.

Now the two steps that can add points. Squaring also admits points with x^2 + y^2 - x = -2r, that is, r = \cos\theta - 2 after dividing by r. Those are not new: if (r, \theta) satisfies r = \cos\theta - 2, the same point’s other name (-r, \theta + \pi) has -r = 2 - \cos\theta = 2 + \cos(\theta + \pi), so it is on Branko’s curve.

The pole is a different matter. It satisfies the final equation, 0 = 0. But \cos\theta \ge -1, so r = 2 + \cos\theta \ge 1 for every \theta: the curve never reaches r = 0, and the pole is not on it. The rectangular description is therefore

\begin{aligned} &(x^2 + y^2 - x)^2 = 4(x^2 + y^2), \\ &(x, y) \ne (0, 0) \end{aligned}

The curve itself is a limaçon, one of the shapes in Limacons and cardioids.

Boundaries

Dividing by a trigonometric factor needs it nonzero. r\cos\theta = -4 may be divided by \cos\theta only because \cos\theta = 0 gives no point of the line; in general the excluded directions must be checked, as in Example 2.

Multiplying or dividing by r moves only the pole. The pole is on a polar graph exactly when r = 0 for some \theta. Check that once, and say “except the origin” when the converted equation has gained it.

Squaring can let in negative r. The extra points are those where r is the negative of what the equation requires; for each curve, check whether they are already on it.

Not every polar equation has a tidy rectangular form, or the reverse. Spirals such as r = \theta and most roses (Rose curves) become long and unhelpful in x and y; a curve like y = x^3 becomes awkward in polar form. Convert when the result is simpler.

Where this goes next

Polar equations and their graphs (Polar equations and their graphs) are sketched directly in the polar plane; conversion is how a sketch is confirmed, and it identifies every polar circle in Polar circles and symmetry tests. The same idea, removing one set of variables by substitution, returns in Eliminating the parameter, where a curve given by x and y in terms of a parameter is rewritten as one equation in x and y. Polar equations of the form \displaystyle r = \frac{\ell}{1 + e\cos\theta}, with \ell > 0 and e > 0, convert to parabolas, ellipses and hyperbolas; they are the polar form of the Conic sections when a focus sits at the pole.

Exercises

Rectangular to polar

Write each equation in polar form, solved for r (or for r^2) where possible.

x = -6
y = 9
y = x
5x + 3y = 15
y = 3x - 5
x^2 + y^2 = 18
x^2 + y^2 = 10x
x^2 + y^2 + 12y = 0
xy = 8
x^2 - y^2 = 1

Polar to rectangular

Write each equation in rectangular form and name the curve; for a circle, give its center and radius.

r = 7\sec\theta
r = -3\csc\theta
\displaystyle \theta = \frac{\pi}{3}
r = 11
r = 8\cos\theta
r = 2\cos\theta + 4\sin\theta
\displaystyle r = \frac{8}{\cos\theta - 4\sin\theta}
r = \tan\theta\sec\theta
\displaystyle r = \frac{2}{1 + \sin\theta}

Reasoning

Branko says r = 3 and r = -3 are different curves. Explain why they have the same graph.
Sampo rewrites x^2 + y^2 = 14y as r^2 = 14r\sin\theta, divides by r and gets r = 14\sin\theta. He worries the origin was lost. Was it?
Riikka converts r = 3 + \sin\theta and gets (x^2 + y^2 - y)^2 = 9(x^2 + y^2). Show that the origin satisfies her equation, and explain why it is not on the polar curve.
Branko says \displaystyle \theta = \frac{\pi}{4} is only a ray into quadrant I. Explain why its graph is the whole line y = x.
For the polar form of 5x + 3y = 15 in exercise 4, find the directions where r does not exist, and explain them geometrically.
Show that a circle with center (a, b) that passes through the origin has polar equation r = 2a\cos\theta + 2b\sin\theta.

Answers

r = -6\sec\theta. r\cos\theta = -6, and \cos\theta = 0 gives no point of the line.
r = 9\csc\theta. r\sin\theta = 9.
\displaystyle \theta = \frac{\pi}{4}. Away from the pole, \tan\theta = 1; negative r gives quadrant III.
\displaystyle r = \frac{15}{5\cos\theta + 3\sin\theta}. r(5\cos\theta + 3\sin\theta) = 15.
\displaystyle r = \frac{5}{3\cos\theta - \sin\theta}. r\sin\theta = 3r\cos\theta - 5, so r(3\cos\theta - \sin\theta) = 5.
\displaystyle r = 3\sqrt{2}. r^2 = 18, and \displaystyle r = -3\sqrt{2} is the same circle.
r = 10\cos\theta. r^2 = 10r\cos\theta; the pole is kept, at \displaystyle \theta = \frac{\pi}{2}.
r = -12\sin\theta. r^2 + 12r\sin\theta = 0; the pole is kept, at \theta = 0.
r^2 = 16\csc 2\theta. r^2\cos\theta\sin\theta = 8, and 2\sin\theta\cos\theta = \sin 2\theta (Double angle formulas), so r^2\sin 2\theta = 16.
r^2 = \sec 2\theta. r^2(\cos^2\theta - \sin^2\theta) = 1, and \cos^2\theta - \sin^2\theta = \cos 2\theta.
x = 7, a vertical line. r\cos\theta = 7.
y = -3, a horizontal line. r\sin\theta = -3.
\displaystyle y = \sqrt{3}x, the line through the origin at \displaystyle \frac{\pi}{3}. \displaystyle \tan\frac{\pi}{3} = \sqrt{3}.
x^2 + y^2 = 121, the circle with center (0, 0) and radius 11.
(x - 4)^2 + y^2 = 16, center (4, 0), radius 4. x^2 + y^2 = 8x; r = 0 at \displaystyle \theta = \frac{\pi}{2}, so the pole was already on it.
(x - 1)^2 + (y - 2)^2 = 5, center (1, 2), radius \displaystyle \sqrt{5}. x^2 + y^2 = 2x + 4y; r = 0 where \displaystyle \tan\theta = -\frac{1}{2}.
x - 4y = 8, a line. r\cos\theta - 4r\sin\theta = 8.
y = x^2, a parabola. r\cos^2\theta = \sin\theta; times r, r^2\cos^2\theta = r\sin\theta. The pole was already on the curve (r = 0 at \theta = 0).
x^2 = 4 - 4y, a parabola opening down with vertex (0, 1). r + r\sin\theta = 2 gives r = 2 - y; squaring, x^2 + y^2 = 4 - 4y + y^2. On this parabola y \le 1, so 2 - y > 0 and squaring added no points.
(-3, \theta) is the point (3, \theta + \pi), so every point of r = -3 is at distance 3 from the pole, and each point at distance 3 is (-3, \theta) for a suitable \theta. Both are x^2 + y^2 = 9.
No. r = 14\sin\theta gives r = 0 at \theta = 0, so the origin is on it.
At (0, 0) both sides are 0. On the curve, \sin\theta \ge -1 makes r = 3 + \sin\theta \ge 2, so r is never 0 and the pole is not on it. The extra point came from multiplying by r (r^2 = 3r + r\sin\theta, so x^2 + y^2 - y = 3r) before squaring.
r may be any real number: r > 0 gives the ray into quadrant I, r < 0 the opposite ray into quadrant III, and r = 0 the origin. Each point (x, y) on the line is \displaystyle \left(r\frac{\sqrt{2}}{2}, r\frac{\sqrt{2}}{2}\right) with \displaystyle r = \sqrt{2}x.
Where 5\cos\theta + 3\sin\theta = 0, that is \displaystyle \tan\theta = -\frac{5}{3}. The line \displaystyle y = -\frac{5}{3}x + 5 has that slope, so rays in those two directions run parallel to it and never meet it.
The radius is the distance to the origin, \displaystyle \sqrt{a^2 + b^2}, so the circle is (x - a)^2 + (y - b)^2 = a^2 + b^2, which expands to x^2 + y^2 = 2ax + 2by. In polar form, r^2 = 2ar\cos\theta + 2br\sin\theta, so r = 0 or r = 2a\cos\theta + 2b\sin\theta. The second already contains the pole, at a \theta with a\cos\theta + b\sin\theta = 0.