Polar Coordinates

Definition of Polar Coordinates

Recall that we define a point (x,y) on the plane as x units to the right of the origin and y units to the left of the origin.  This works great for lines and parabolas, but circles have somewhat messy equations.  As an alternative we define a new coordinate system where the first coordinate r is the distance from the origin to the point and the second coordinate q is the angle that the ray from the origin to the point makes with the positive x-axis.  From trigonometry, we have

 x = rcosq     y = rsinq

Your calculator has a special mode for polar coordinates.  We use the calculator to graph

r = 5cosq

and

r = cos( ) A circle centered at the origin has equation x2 + y2 = R2

In polar form we have

r = R

For example the circle of radius 3 centered at (0,0) has polar equation

r = 3

If

y = mx + b

we can write

r sin(q) = m r cos q + b

or

b
r  =
sin q  -  m cos q

Conic Sections

Recall that a conic section is defined as follows:
Let f ( called the focus) be a fixed point in the plane, m (called the directrix) be a fixed line, and e (called the eccentricity) a positive constant.  Then the set of points P in the plane with

|PF|
=   e
|Pm|

is a conic section.  If e < 1 then the section is an ellipse, if e = 1, then its a parabola,
and if e = 0 then it is a hyperbola.

Note:   If F is the origin m is x = d then

|PF| = r,     |Pm| = d - r cos q

so that the equation becomes

r = e(d - rcos(q)) = ed - recos(q)

or

 ed   r   =                                      1 + e cosq

Derivatives in Polar Coordinates

Theorem

Let r = r(q) represent a polar curve, then

dy              dy/dq                  r' sinq + r cosq
=                      =
dx              dx/dq                  r' cosq - r sinq

 dy             r' sinq + r cosq                         =                                                                dx                r' cosq - r sinq

Proof:

Since

x = r cosq,     and     y = r sinq,

we can substitute in r = r(q) to get

x = r cosq

Taking a derivative,

x' = r' cosq - r sinq

and

y' = r' sinq + r cosq.

Since

dy              dy/dq
=
dx              dx/dq

dividing gives the result.

Example

Let

r = q sinq

Then

dy              (sinq + q cosq) sinq  +  q sinq cosq
=
dx               (sinq + qcosq) cosq  - q sinq sinq

sin2q + 2qsinqcosq
=
sinq cosq + qcos2q - qsin2q

sinq cosq + qcos(2q)
=
sin2q + qsin(2q)

e-mail Questions and Suggestions