Polar Coordinates and their Conversion

Polar coordinates

Polar coordinate system is a plane with Pole(point O) and polar axis which is horizontal axis from point O. Any point P in this plane is assigned polar coordinates represented as P(r,θ ) as shown below.

we measure as positive when moving counterclockwise and negative when moving clockwise.

If r > 0, then P is on the terminal side of θ. If r < 0, then P is on the terminal side of π+θ.

Plotting Points in the Polar Coordinate System

Example1.Plot the following points with given polar coordinates

a) P(2, π/3) b) Q(-1, 3π/4 ) c) R(3,-45° )

Finding all Polar Coordinates of a point

Let point P have polar coordinates (r,θ). Any other polar coordinates of P must be of the form

(r, θ+2nπ ) Or

(-r, θ+(2n+1)π ) Where n is any integer.

The coordinates (r,θ ) , (r, θ+2π), and (-r, θ+π) all name the same point.

Example2.If the point P has Polar coordinates (1.5,-20° ), then find all other polar coordinates for P

Solution: using the formula given above

(1.5, -20+2nπ )

(-1.5, -20+(2n+1)π )

These are all the coordinates of point P. Using these formulas we can find first two coordinates using n=0,1

(1.5, -20+2π ) = (1.5 , 340°)

(-1.5,-20+π ) = (-1.5 , 160°)

Coordinate conversion:

Using the following equations Polar coordinates can be converted to rectangular coordinates and vice versa .

Let P have polar coordinates (r,θ ) and rectangular coordinates (x,y), then

x = r cosθ y = r sinθ

$\dpi{120} \mathbf{r^{2}=x^{2}+y^{2}}$ $\dpi{120} \mathbf{\theta =tan^{-1}\left ( \frac{y}{x} \right )}$

Example3Use an algebraic method to find the rectangular coordinates of the points with given polar coordinates.

(2, 270° )

Solution: we have r=2 and θ=270

x = r cosθ y =r sinθ

x = 2cos(270) y =2sin(270)

x = 2(0) = 0 y = 2(-1) = -2

So the rectangular coordinates are (0,-2)

Example4.Convert the rectangular coordinates (-1,1) to two polar coordinates algebraically.

Solution: Given that x=-1 and y=1

$\dpi{120} \mathbf{r=\sqrt{x^{2}+y^{2}}}$ $\dpi{120} \mathbf{\theta =tan^{-1}\left ( \frac{y}{x} \right )}$

$\dpi{120} \mathbf{r=\sqrt{(-1)^{2}+(1)^{2}}}$ $\dpi{120} \theta =tan^{-1}\left ( \frac{1}{-1} \right )$

$\dpi{120} r=\sqrt{2}$ $\dpi{120} \theta =tan^{-1}(-1)= \frac{3\pi }{4}$

So the first polar coordinates are (√2 , 3π/4 )

Other polar coordinate would be (-√2 , 3π/4 + π) = (-√2 , 7π/4)

which can also be written as (-√2 , -π/4 )

Converting equations:

Example 5. Convert the following polar equations to rectangular form and identify the graph.

a) r secθ = 3

r/cosθ = 3

r = 3cosθ

r*r= 3*r*cosθ

$\dpi{120} r^{2}=3x$

$\dpi{120} x^{2}+y^{2}=3x$

$\dpi{120} x^{2}-3x +y^{2}=0$

By completing squares,

$\dpi{120} x^{2}-3x + \frac{9}{4}+y^{2}=0 +\frac{9}{4}$

$\dpi{120} \left ( x-\frac{3}{2} \right )^{2}+\left ( y-0 \right )^{2}=\left ( \frac{3}{2} \right )^{2}$
Which is equation of a circle.

Center = (3/2 ,0)

Radius(r) = 3/2

b) r=2 sinθ-4cosθ

r*r = r(2 sinθ-4cosθ)

$\dpi{120} r^{2}= 2rsin\theta -4 rcos\theta$

$\dpi{120} x^{2}+y^{2}= 2y-4x$

$\dpi{120} x^{2}+4x+y^{2}-2y = 0$

$\dpi{120} x^{2}+4x+4+y^{2}-2y+1= 0+4+1$

$\dpi{120} \left ( x+2 \right )^{2}+\left ( y-1 \right )^{2}= 5$

$\dpi{120} \left ( x+2 \right )^{2}+\left ( y-1 \right )^{2}= \left (\sqrt{5} \right )^{2}$

Which is the equation of circle.

center = (-2,1)

radius(r)= √5

Example6: Convert the following rectangular equations to polar form.

a) 2x-3y =5

2*rcosθ -3*rsinθ = 5

r(2cos -3sinθ ) = 5

$\dpi{120} r = \frac{5}{2cos\theta -3sin\theta }$

b) $\dpi{120} \left ( x+3 \right )^{2}+\left ( y+3 \right )^{2}=18$

x^2+6x+9 +y^2 +6y+9 = 18

$\dpi{120} x^{2}+y^{2}+6x+6y+18=18$

$\dpi{120} x^{2}+y^{2}+6x+6y=0$

$\dpi{120} r^{2}+6rcos\theta +6rsin\theta =0$

r^2 = -r(6cosθ +6sinθ )

r = -(6cosθ +6sinθ )

Finding Distance using Polar Coordinates

Example7. The location, given in polar coordinates, of two planes approaching the Vicksburg airport are (4mi, 12° ) and (2mi,72° ) . Find the distance between the airplanes.

Solution: let the two planes be A and B as shown in figure given below.

To find the distance between A and B we use cosine rule. Angle between OA and OB is given as,

θ=72-12=60°

$\dpi{120} AB^{2}=OA^{2}+OB^{2}-2OA*OB cos(60)$

$\dpi{120} AB^{2}=4^{2}+2^{2}-2*4*2 cos(60)$

$\dpi{120} AB^{2}=20-16\left ( \frac{1}{2} \right ) =12$

AB = √12 = 3.46 miles

Distance between two planes is 3.46 miles.

Example8. A square with sides of length a and center at the origin has two sides parallel to the x-axis. Find polar coordinates of the vertices.

Solution: A square with sides of length a and center at origin would be drawn as below.

Rectangular coordinates of vertex A are (a/2 , a/2). We find its equivalent polar coordinates.

x = a/2 , y = a/2

$\dpi{120} r=\sqrt{x^{2}+y^{2}}$

$\dpi{120} r=\sqrt{\left ( \frac{a}{2} \right )^{2}+\left ( \frac{a}{2} \right )^{2}}=\sqrt{\frac{2a^{2}}{4}}= \frac{a}{\sqrt{2}}$

$\dpi{120} \theta =tan^{-1}\left ( \frac{\frac{a}{\sqrt{2}}}{\frac{a}{\sqrt{2}}} \right )= tan^{-1}(1) = \frac{\pi }{4}$

Polar coordinates of vertex

$\dpi{120} A = \left ( \frac{a}{\sqrt{2}},\frac{\pi }{4} \right )$

For all other vertices value of r remain same as sides of squares are equal just θ changes in each quadrant. In each quadrant it get increased by π/2

$\dpi{120} B= \left ( \frac{a}{\sqrt{2}},\frac{3\pi }{4} \right )$

$\dpi{120} C = \left ( \frac{a}{\sqrt{2}},\frac{5\pi }{4} \right )$

$\dpi{120} D = \left ( \frac{a}{\sqrt{2}},\frac{7\pi }{4} \right )$

Practice problems:

Find all the polar coordinates for given point P

(2, π/6 )

Convert the following rectangular equation to polar form.

$\dpi{120} \left ( x-3 \right )^{2}+y^{2}=9$

Convert the given polar equation to rectangular form r cscθ = 1

Radar detects two airplanes at the same altitude. Their polar coordinates are (8 mi,110°) and (5 mi, 15°). How far apart are the airplanes?

Answers :

$\dpi{120} \left ( 2, \frac{\pi }{6}+2n\pi \right ), \left ( -2,\frac{\pi }{6} +(2n+1)\pi \right )$
r=6cosθ
$\dpi{120} x^{2}+\left ( y-\frac{1}{2} \right )^{2}=\frac{1}{4}$
9.80 miles