Polar Coordinates

Section 2.2 Polar Coordinates

The standard way of representing point in

R^{2}

is using rectangular coordinates,

(x, y) .

However, sometimes is more convenient to instead use polar coordinates

(r, θ) .

The polar coordinates and the rectangular coordinates are related to each other by the following relations.

From cartesian to polar.
🔗
$r = \sqrt{x^{2} + y^{2}}, and θ = \tan^{- 1} \frac{y}{x} .$
From polar to cartesian.
🔗
$x = r \cos θ, and y = r \sin θ .$

Find the distance between the two points:

p_{1} = (1, \frac{π}{6})

and

p_{2} = (3, - \frac{π}{2})

The distance between two points in cartesian coordinates,

(x_{1}, y_{1})

and

(x_{2}, y_{2}),

is defined as

distance = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} .

To find the distance between two polar points, we first convert each point to cartesian coordinates and then apply the formula for distance.

\begin{array}{ccl} (x_{1}, y_{1}) & = & (1 \cos (\frac{π}{6}), 1 \sin (\frac{π}{6})) \\ = & (\frac{\sqrt{3}}{2}, \frac{1}{2}) . \\ (x_{2}, y_{2}) & = & (3 \cos (- \frac{π}{2}), 3 \sin (- \frac{π}{2})) \\ = & (0, - 3) . \end{array}

\begin{array}{ccl} distance & = & \sqrt{{(\frac{\sqrt{3}}{2} - 0)}^{2} + {(\frac{1}{2} - (- 3))}^{2}} \\ = & \sqrt{\frac{3}{4} + \frac{49}{4}} = \sqrt{\frac{52}{4}} \\ = & \sqrt{13} \approx 3.61 . \end{array}

For each set of polar coordinates

(r, θ),

match the equivalent cartesian coordinates

(x, y) .

\begin{array}{cl} A & (2.93893, - 4.04508) \\ B & (- 2.93893, - 4.04508) \\ C & (- 1.61803, - 1.17557) \\ D & (2.42705, 1.76336) \end{array}

C

B

A

D

For each set of cartesian coordinates

(x, y),

match the equivalent polar coordinates

(r, θ) .

\begin{array}{cl} A & (A, 0.6 π) \\ B & (B, 0.8 π) \\ C & (C, - 0.5 π) \\ D & (D, 0.1 π) \end{array}

B

D

C

A