1. Coordinate Systems

c. Polar Coordinates - 2D

For a review of Trigonometry, including the definitions of the trig functions and all the trig identities, see the chapter on Trigonometry.

2. 2D Converting between Rectangular and Polar

It is easy to convert between rectangular coordinates and polar coordinates because of the Pythagorean Theorem and the definitions of sine and cosine: \[ x^2+y^2=r^2 \] \[ \cos{\theta}=\dfrac{x}{r} \qquad \sin{\theta}=\dfrac{y}{r} \] Consequently:

If you know the polar coordinates \((r,\theta)\), then the rectangular coordinates \((x,y)\) are \[ x=r\cos\theta \qquad \text{and} \qquad y=r\sin\theta \qquad \qquad (1) \] Conversely, if you know the rectangular coordinates \((x,y)\), then the polar coordinates \((r,\theta)\) are found from \[ r^2=x^2+y^2 \qquad \text{and} \qquad \tan\theta=\dfrac{y}{x} \qquad \qquad (2) \]

Most often, \(r\) is the positive square root: \(r=\sqrt{x^2+y^2}\).
On rare occations, as in plotting, \(r\) may be the negative square root: \(r=-\sqrt{x^2+y^2}\).
It is tempting to also solve for \(\theta\) and write \(\theta=\arctan{\dfrac{y}{x}}\), but this is not quite right. The correct formula is:

Why is the formula \(\theta=\arctan{\dfrac{y}{x}}\) not always correct?

Looking at the graph of \(\theta=\arctan{z}\) we see that \(\arctan\) always produces an angle between \(\dfrac{-\pi}{2}\) and \(\dfrac{\pi}{2}\), which is in quadrant I or IV.

When the point, \((x,y)\), is in quadrant II or III (where x is negative), the formula \[ \theta=\arctan{\dfrac{y}{x}}=\arctan{\dfrac{-y}{-x}} \] gives an angle in quadrant IV or I respectively. We fix this by adding \(\pi\) or \({180^\circ}\) to the \(\arctan\): \[ \theta=\arctan{\dfrac{y}{x}}+\pi \]

\[ \theta=\arctan{\dfrac{y}{x}}+ \begin{cases} 0 \quad \text{if} \quad (x,y) \quad \text{is in quadrant I or IV} \\ \pi \quad \text{if} \quad (x,y) \quad \text{is in quadrant II or III} \end{cases} \]

In practice, we compute \(\theta=\arctan\dfrac{y}{x}\) and add \(\pi\) if \((x,y)\) is in the \(2^\text{nd}\) or \(3^\text{rd}\) quadrant. On the rare occasions when we take \(r\) to be negative, we must add \(\pi\) to all of these values of \(\theta\).

A point \(P\) has rectangular coordinates \((-\sqrt{3},1)\). Find two sets of polar coordinates for \(P\), one with \(r>0\) and one with \(r<0\).

Use the formulas for the first pair of coordinates.
Then, add or subtract \(\pi\) from the angle and reverse the sign of the radius.

\(\left(2,\dfrac{5\pi}{6}\right)\)
\(\left(-2,\dfrac{11\pi}{6}\right)\) or \(\left(-2,-\,\dfrac{\pi}{6}\right)\)
Alternatively, you can add any multiple of \(2\pi\) to the angles.

Algebraically, \[ r^2=\left(-\sqrt{3}\right)^2+(1)^2=4 \qquad \text{and} \qquad \tan\theta=\dfrac{1}{-\sqrt{3}} \] The positive solution for the radius is \(r=2\). Next, \(\arctan\dfrac{1}{-\sqrt{3}}=-\,\dfrac{\pi}{6}\) which is in quadrant IV. However, \((-\sqrt{3},1)\) is quadrant II. So we add \(\pi\) and conclude: \[ \theta=\arctan\dfrac{1}{-\sqrt{3}}+\pi =-\,\dfrac{\pi}{6}+\pi=\dfrac{5\pi}{6} \] So one pair of polar coordinates is \(\left(2,\dfrac{5\pi}{6}\right)\). For the second pair of coordinates, we reverse the sign of the radius and add or subtract \(\pi\) from the angle. So the alternate coordinates are: \[ \left(-2,\dfrac{11\pi}{6}\right) \qquad \text{or} \qquad \left(-2,-\,\dfrac{\pi}{6}\right) \]
In addition, you can add any multiple of \(2\pi\) to the angles.

Find the rectangular coordinates for the point \(R\) whose polar coordinates are \(\left(6,-\,\dfrac{3\pi}{4}\right)\).

\(\left(-3\sqrt{2},-3\sqrt{2}\right)\)

We compute: \[ x=r\cos\theta =6\cos\left(-\,\dfrac{3\pi}{4}\right) =6\left(-\,\dfrac{1}{\sqrt{2}}\right) =-3\sqrt{2} \] \[ y=r\sin\theta =6\sin\left(-\,\dfrac{3\pi}{4}\right) =6\left(-\,\dfrac{1}{\sqrt{2}}\right) =-3\sqrt{2} \] Therefore, the rectangular coordinates are \((x,y)=\left(-3\sqrt{2},-3\sqrt{2}\right)\). We notice this point is in the quadrant III, from both the rectangular and polar coordinates.

You can also review the definitions of the trig functions, using the following Maplets (requires Maple on the computer where this is executed):

Triangle Definitions of Trigonometric Functions Rate It

Circle Definitions of Trigonometric Functions Rate It

Visualizing Circle Definitions of Trig Functions Rate It

1. Coordinate Systems

c. Polar Coordinates - 2D

2. 2D Converting between Rectangular and Polar

Why is the formula \(\theta=\arctan{\dfrac{y}{x}}\) not always correct?

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