The polar coordinates of a point in the plane are determined by first choosing a point as the 'pole', and a polar axis indicating a direction away from the pole.

• The polar axis is often drawn in the horizontal direction and identified with the positive x axis; the pole is often identified with the origin of an x-y coordinate system.
• The polar coordinates of a point in the plane consists of r, the displacement between the pole and the point, and `theta, which is the angle made by the displacement with the polar axis.
• The polar coordinates are indicated by the ordered pair (r, `theta).

Before you continue you should attempt to sketch the points with polar coordinates (3, `pi/6) and (5, 2 `pi/3)

The point (3, `pi/6) is located at distance 3 from the pole and along the line making angle `pi/6 with the polar axis, as indicated below.

The point (-3, `pi/4) is located opposite to the point (3, `pi/4).

• This point could also be represented by the polar coordinates (3, 5 `pi / 4).

You should now attempt to find the rectangular coordinates of the three points indicated in the figure below.

• The circular model of the sine and cosine functions should be helpful to you.

The rectangular coordinates of the point lying at displacement r from the pole and at angle `theta are the x and y coordinates, which by the circular definitions of the sine and cosine functions are x = r cos(`theta) and y = r sin(`theta).

• The rectangular coordinates of (3, `pi/6) are therefore x = 3 cos(`pi/6) and y = 3 sin(`pi/6).
• Similarly the rectangular coordinates of (5, 2 `pi / 3) are x = 5 cos(2 `pi / 3) and y = 5 sin(2 `pi / 3).
• The rectangular coordinates of (-3, `pi/4) are x = -3 cos(`pi/4) and y = -3 sin(`pi/4).
• You can easily determine the numerical values of these coordinates and match them to the graphs; you should do so (be sure you are in radian mode...).

From the x and y coordinates of a point we can easily determine its distance from the origin, which if the origin is identified with the pole will match the polar coordinate r.

From the x and y coordinates of a point we can easily determine its angle with the positive x axis, which is the x axis is identified with the polar axis will be tan^-1(y / x) or tan^-1(y / x) + `pi.

• For the point (3,4), we easily find from the Pythagorean Theorem that r = `sqrt(x^2 + y^2), and that the first-quadrant angle is `theta = tan^-1(4/3).
• A similar set of calculations for the point (-5, 7) tells us that r = `sqrt(5^2 + 7^2) and that `theta = tan^-1(-5 / 7) + `pi.

   

Attempt to sketch the thee sets of points specified in the figure below.

The set of points from which r = 7 is the set of points of which lie at distance 7 from the pole.

• This is the set of points lie on the circle of radius 7 and centered at the pole.

The set of points for which `theta = `pi/4 consists of all points which are displaced from the origin at an angle of `pi/4 from the polar axis.

• Since the displacement can be in the positive or negative direction, the set of points is identical to the line `theta = `pi/4 indicated in the figure below; the line makes angle `pi/4 with the polar axis.

To get the locus of points for which r = `theta, we can first make a table of r vs. `theta for this relationship.

• The table is as indicated.
• We see that as the angle increases, the distance from the pole increases by the same amount.
• We might plot the points indicated by the table and then construct a smooth curve which moves further and further from the pole as we move away from the polar axis in the counterclockwise direction.
• The result is a spiral much like the one depicted below.

A table for r = cos(`theta) is depicted below.

• For this table we chose to let `theta change by `pi/4.
• The points corresponding to the table are as indicated in the figure below.
• The shape of the figure is not really as drawn, but without knowing the shape it would be easy to conjecture that the shape might well be like the one drawn.

It turns out that the actual shape of the graph of r = cos(`theta) is a circle.

We show below how we can verify this.

• We know the equation of a circle in rectangular coordinates.
• Therefore we first express r and cos(`theta) and rectangular coordinates.
• The expressions for r and cos(`theta) are as indicated.
• We thus translate r = cos(`theta) into rectangular coordinate says indicated in the fourth line.
• In the fifth line we have multiplied by the denominator `sqrt(x^2 + y^2).

To obtain the standard form of this circle, we complete the square as indicated below.

• We thus see that the graph should be a circle with center (1/2, 0) and radius 1/2.
• This agrees well with the graph we sketched above for r = cos(`theta).

As a final example we sketch r = sin(2 `theta).

• Since we are taking the sine of 2 `theta, we choose to let `theta change by `pi / 8 so that the argument of the sine function will change by `pi/4.
• For `theta from 0 to `pi, we obtain the table indicated below.
• Plotting the points and following the curve as `theta changes from 0 to `pi, we obtain the indicated graph.
• Note that the second-quadrant angles `theta give us fourth-quadrant points because of the corresponding negative values of r.
• This graph consists of 2 `petals' of a 'rose'; the petals should be congruent in shape and size, though they aren't quite sketched that way.
• You should as an exercise complete the table all the way to `theta = 2 `pi to see the completed figure.