Convert between polar and Cartesian coordinates

One point, two names

The orange point P is 4 units from the pole, the center of the grid. The angle \(\theta\) turns counterclockwise from the positive x-axis to the ray, the half-line from the pole through P.

P has polar coordinates \((4, \theta)\), a distance and an angle. Its Cartesian coordinates \((x, y)\) give how far it is to the right of the y-axis and above the x-axis.

Set \(\theta = \frac{\pi}{3}\). The dashed legs and the orange line form a right triangle with long side 4. The horizontal leg is \(4\cos\frac{\pi}{3} = 2\) and the vertical leg is \(4\sin\frac{\pi}{3} = 2\sqrt{3}\), so P is at \((2, 2\sqrt{3})\).

Drag \(\theta\) past \(\frac{\pi}{2}\). The horizontal leg turns red because it points left, so it counts as negative. A green leg counts as positive.

Any distance, any angle

P is at distance \(r\) from the pole, in the direction \(\theta\).

Drag \(\theta\) through all four quadrants. The legs are always \(r\cos\theta\) and \(r\sin\theta\), because cosine and sine have the same signs as the legs in every quadrant.

Press the first button to change \(r\) to \(-r\). P moves through the pole to the opposite side, because a negative radius points away from the faint ray at angle \(\theta\).

This means that \((-r, \theta)\) names the same point as \((r, \theta + \pi)\). One point has many polar names, and the formulas work for all of them.

Notation

1. The legs of the right triangle are \(x\) and \(y\), with their signs. Multiply \(\cos\theta = \frac{x}{r}\) and \(\sin\theta = \frac{y}{r}\) by \(r\).

   \[x = \cy{r}\cos\theta, \qquad y = \cy{r}\sin\theta\]

   The orange \(r\) is the distance from the pole to P.

2. The Pythagorean theorem on the same triangle gives \(r\) from \(x\) and \(y\).

   \[\cy{r}^2 = x^2 + y^2\]

   If \(r \ge 0\), then \(r = \sqrt{x^2 + y^2}\).

3. The tangent of \(\theta\) is the vertical leg over the horizontal leg. This needs \(x \ne 0\).

   \[\tan\theta = \dfrac{y}{x}\]

   If \(x = 0\), then \(\theta\) is \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\), according to the sign of \(y\).

4. The tangent repeats every \(\pi\), so two points on opposite sides of the pole give the same ratio. Choose \(\theta\) in the quadrant of the point.

5. Adding \(2\pi\) to \(\theta\) names the same point. Adding \(\pi\) names it too if the sign of \(r\) changes.

   \[(r, \theta) = (r, \theta + 2\pi) = (-r, \theta + \pi)\]

   The pole is \((0, \theta)\) for every \(\theta\).

When it fails

Press the first button to choose the point \((-1, \sqrt{3})\), which the white ring marks. There \(r = 2\) and \(\frac{y}{x} = -\sqrt{3}\). The second button uses the arctangent of \(-\sqrt{3}\), which is \(-\frac{\pi}{3}\). That ray lies in the fourth quadrant, so it names \((1, -\sqrt{3})\), the red ring on the opposite side of the pole.

If \(x < 0\), add \(\pi\) to the arctangent angle, as the third button does. The angle \(-\frac{\pi}{3} + \pi = \frac{2\pi}{3}\) returns P to the white ring.

Example 1: polar to Cartesian

Find the Cartesian coordinates of the point with polar coordinates \((6, \frac{5\pi}{6})\).

1. Put \(r = 6\) and \(\theta = \frac{5\pi}{6}\) into the two formulas.

   \[x = 6\cos\tfrac{5\pi}{6}, \qquad y = 6\sin\tfrac{5\pi}{6}\]

2. The angle \(\frac{5\pi}{6}\) is in the second quadrant, so cosine is negative and sine is positive.

   \[\cos\tfrac{5\pi}{6} = -\tfrac{\sqrt{3}}{2}, \qquad \sin\tfrac{5\pi}{6} = \tfrac12\]

3. Multiply. The red leg shows that \(x\) is negative, and the green leg shows that \(y\) is positive.

   \[x = \cr{-3\sqrt{3}}, \qquad y = \cg{3}\]

4. As a check, P is \(\sqrt{27 + 9} = 6\) from the pole.

Example 2: Cartesian to polar

Write the point \((-3, -3)\) in polar coordinates with \(r > 0\) and \(0 \le \theta < 2\pi\).

1. Find \(r > 0\) with the Pythagorean theorem. Both legs are 3 long and red, because \(x\) and \(y\) are negative.

   \[\begin{aligned} r^2 &= (-3)^2 + (-3)^2 = 18 \\ r &= 3\sqrt{2} \end{aligned}\]

2. A tempting wrong move: The ratio \(\frac{y}{x}\) is 1, so \(\theta = \frac{\pi}{4}\).

   \[\theta = \arctan 1 = \cr{\tfrac{\pi}{4}}\]

   The ray at \(\frac{\pi}{4}\) points up and to the right, but \((-3, -3)\), the white ring, is down and to the left. Going \(3\sqrt{2}\) along that ray reaches \((3, 3)\), the red ring on the opposite side of the pole.

3. The tangent is also 1 at \(\frac{\pi}{4} + \pi\). The point is in the third quadrant, so use that angle.

   \[\theta = \tfrac{\pi}{4} + \pi = \tfrac{5\pi}{4}\]

4. The point is \((3\sqrt{2}, \frac{5\pi}{4})\), which \((-3\sqrt{2}, \frac{\pi}{4})\) also names.

Practice

The point P has polar coordinates \((-4, \frac{7\pi}{6})\). Find its Cartesian coordinates. Then give a polar name for P with \(r > 0\).

A negative radius puts P on the side of the pole opposite the ray at \(\frac{7\pi}{6}\).

Questions

The circles are 1 unit apart, and the rays are \(\frac{\pi}{6}\) apart. The orange point is P in question 1 and Q in question 2. The white ring marks Q.

1. What are the Cartesian coordinates of the orange point P?

   • \((\tfrac32, -\tfrac{3\sqrt{3}}{2})\)
   • \((\tfrac{3\sqrt{3}}{2}, -\tfrac32)\)
   • \((\tfrac32, \tfrac{3\sqrt{3}}{2})\)
   Answer

   \((\tfrac32, -\tfrac{3\sqrt{3}}{2})\)

2. Which pair also names the point Q, marked by the white ring?

   • \((-2, \frac{7\pi}{4})\)
   • \((-2, \frac{3\pi}{4})\)
   • \((2, \frac{7\pi}{4})\)
   Answer

   \((-2, \frac{7\pi}{4})\)

3. Which pair names the point \((-2\sqrt{3}, -2)\) with \(r > 0\) and \(0 \le \theta < 2\pi\)?

   • \((4, \frac{7\pi}{6})\)
   • \((4, \frac{\pi}{6})\)
   • \((16, \frac{7\pi}{6})\)
   Answer

   \((4, \frac{7\pi}{6})\)

Summary

You convert with \(x = r\cos\theta\) and \(y = r\sin\theta\), or with \(r^2 = x^2 + y^2\) and \(\tan\theta = \frac{y}{x}\), and you choose \(\theta\) in the quadrant of the point.

This works because \(x\) and \(y\) are the legs of a right triangle whose long side is \(r\).

You use it when a polar curve needs x and y, or when a point needs a distance and an angle.

This is one of four skills in topic 9.7 of Which & Why: AP Calculus.