15. Polar Coordinates

a. Polar Coordinate System

2. Converting Between Rectangular and Polar Coordinates

If you need to review the definitions of the trig functions, you can use 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

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

Most often, \(r\) is the positive square root: \(r=\sqrt{x^2+y^2}\).
On rare occations, as in plotting, \(r\) is 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. Why? The correct formula is:

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

However, in practice, we usually use the equation \(\tan\theta=\dfrac{y}{x}\) and adjust \(\theta\) to give the correct quadrant. On the rare occasions when we take \(r\) to be negative, we must add \(\pi\) to these values of \(\theta\).

For the point \(P\) with rectangular coordinates \((-\sqrt{3},1)\), find polar coordinates for \(P\) with a positive \(r\).

Algebraically, \[ r^2=(-\sqrt{3})^2+(1)^2=4 \qquad \text{and} \qquad \tan\theta=\dfrac{1}{-\sqrt{3}} \] 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 a pair of polar coordinates is \(\left(2,\dfrac{5\pi}{6}\right)\).

A point \(Q\) has polar coordinates \((-2,\dfrac{3\pi}{4})\). Obtain a pair of rectangular coordinates for \(Q\).

When converting from polar to rectangular coordinates, it does not matter if the radius is positive or negative. The formula will automatically put the point in the correct quadrant. We compute: \[ x=r\cos\theta =-2\cos\dfrac{3\pi}{4} =-2\left(-\,\dfrac{1}{\sqrt{2}}\right) =\sqrt{2} \] \[ y=r\sin\theta =-2\sin\dfrac{3\pi}{4} =-2\left(\dfrac{1}{\sqrt{2}}\right) =-\sqrt{2} \] Therefore, the rectangular coordinates are \((x,y)=(\sqrt{2},-\sqrt{2})\). We notice this point is in the quadrant IV, which is correct.