There are many ways to mark points in the plane or in 3-dim space for purposes of navigation. In the familiar rectangular coordinate system, a point is chosen as the origin and a perpendicular set of lines is drawn through that point, one horizontal and one vertical. A unit of Length is chosen, and every point is given a pair of coordinates (x, y) indicating its distance, horizontally and vertically from the origin.

The choice of origin, axes, and length is completely arbitrary.

Polar Coordinates

(0,0)1

1

2 3-1-2-3

-1

-2

-3

2

3 (2, 3)

An alternative method of assigning coordinates is shown below.

Here we draw a straight line from the origin to the point. We then assign to the point the angle θ that the line makes with the positive x axis, and the distance r from the origin to the point.

Some examples are:

θ

r(r,θ )

(2, π/4)

(1, π/4)

(2, 0)

(1, π/2)

(3, π/6)(1, 3π/4)

(2, −π/3)(2, −π/2)

The relationship between rectangular and polar coordinates is shown in the diagram below.

It is summarized by the equations:

x = rcos(θ)

y = rsin(θ)

2 2r x y= +

1arctan tany yx x

θ −= =

The angle θ is taken to be between −π and π.

θr y

x

Problem. The polar coordinates of several points are given below. Find the rectangular coordinates of each of those points.

(a) (3, π/4) (b) (1, −π/3) (c) (4, 3π/4) (d) (1, −π/6)

Solution.(a)

(b)

(c)

(d)

3 3cos( /4) ;2

x π= = 3 3sin( /4) ;2

y π= = 3 3,2 2

1 cos( /3) ;2

x π= − = 3 sin( /3) ;2

y π= − =−1 3,2 2

−

4 4cos(3 /4) ;2

x π= =− 4 4sin(3 /4) ;2

y π= =4 4,2 2

−

3 cos( /6) ;2

x π= − =1 sin( /6) ;2

y π= − =−3 1,

2 2

−

Problem. The rectangular coordinates of several points are given below. Find the polar coordinates of each of those points.

(a) (5, 5) (b) (1, −√3) (c) (−3, 3√3) (d) (−2, −2)

Solution.

(a) 2 2 50 5 2;r x y= + = = 51 tan ;5 4

πθ −= = 5 2,4π

(5, 5)

Solution.

(b) 2 2 2;r x y= + = ( )1 tan 3 ;3πθ −= − =− 2,

3π −

(1, −√3)

Problem. The rectangular coordinates of several points are given below. Find the polar coordinates of each of those points.

(a) (5, 5) (b) (1, −√3) (c) (−3, 3√3) (d) (−2, −2)

2

Solution.

(c) 2 2 6;r x y= + = ( ) 21 tan 3 ;3πθ −= − =

26,3π

(− 3, 3√3)

Problem. The rectangular coordinates of several points are given below. Find the polar coordinates of each of those points.

(a) (5, 5) (b) (1, −√3) (c) (−3, 3√3) (d) (−2, −2)

6

Solution.

(d) 2 2 8 2 2;r x y= + = = ( ) 31 tan 1 ;4πθ −= − =−

32 2,4π −

(−2, −2)

Problem. The rectangular coordinates of several points are given below. Find the polar coordinates of each of those points.

(a) (5, 5) (b) (1, −√3) (c) (−3, 3√3) (d) (−2, −2)

We consider the problem of graphing functions of the form r = f(θ) in polar coordinates, where the angle is measured in radians. First we look at two simple cases

Plot the function r = c in polar coordinates, where c is a positive constant.

This is the case where c = 2.

Plot the function θ = c in polar coordinates, where c is a constant.

θ = c radians

In trigonometric calculations, we usually require r to be positive. However, when graphing curves r = f(θ) in polar coordinates, we allow r to be negative (since f(θ) often is) and interpret this to mean measuring “backward” from the origin along the ray.

positive r

negative r

We now consider the problem of graphing more complicated functions.

Plot the function r = sin(θ) in polar coordinates.

There are two ways to see how this plot looks. One is to change from polar to rectangular coordinates. The equation r = sin(θ) can also be written as r2 = rsin(θ) , or x2 + y2 = y. By completing the square, we can write this as

1 12 24 4

x y y+ − + = or2 21 12

2 2x y + − =

A second way is to note that as θ increases, the line connecting the origin to the corresponding point on the curve sweeps around counterclockwise like the hand of a clock. At each value of θ the curve is plotted on that line a distance f(θ) from the origin. Thus as the line revolves around the origin, the point on the curve slides up and down the line. Proceed as follows:

Locate the value of r forθ = 0. Then as θ increases to 90 degrees, the point slides in or out until it reaches the correctlocation of r corresponding to θ = π/2. This gives an idea of how the curve looks in the first quadrant.

Continue this process in each quadrant.

For the function r = sin(θ) , the initial position of the point isr = 0 when θ = 0. At θ = π/2, the point is at a distance sin(π/2) = 1 along the “clock hand”. Thus as the hand sweeps out that 90 degree angle, the point moves up the hand from 0 to 1. The result is clearly the picture shown below, at least approximately.

As the clock hand moves another 90 degrees to π, the point must move back along the clock hand to sin(π) = 0. Thus the picture continues as shown below.

At this point we have the entire circle. In the next two quadrants, r is negative. Thus the point moves to the negative part of the hand. The circle is painted out again in this way, and the process then repeats forever.

Now let us look at the similar curve r = 2cos(θ). When θ = 0, r = 2. At θ = π/2, r = 2cos(π/2) = 0. Therefore, as the angle of the hand moves from 0 to π/2, the point on the curve moves down the hand from 2 to 0. This is shown below.

As the angle of the hand moves from π/2 to π, the cosine becomes negative, so the point is plotted back through the origin, a negative distance. Thus r continues down the hand from 0 to −2.

This figure appears to be a circle as well. To verify this we can translate into rectangular coordinates. If r = 2cos(θ), then r2 = 2rcos(θ), or x2 + y2 = 2x. As before, we can complete the square to obtain the equation

2 22 1 1x x y− + + =or

2 2( 1) 1x y− + =

This is clearly the equation of a circle with radius 1 and center (1, 0).

Graph the curve r = 1 + cos(θ) in polar coordinates.

In this case, the change to rectangular coordinates would not yield a familiar curve. Thus we proceed in the straightforward way. At θ = 0, r is 2 and at θ = π/2, r is 1. This is illustrated below.

Finally, as the angle of the hand moves from 3π/2 to 2π , the cosine goes from 0 to 1, and so r moves up the hand from 1 to 2.

The result is called a cardioid because of its heart shape. Here are some other examples.

r = 1 − cos(θ) r = 1 + sin(θ)

r = 1 − sin(θ)