Further Pure | Topic 2: Polar Coordinates

Curriculum Objectives:

• Understand the relations between cartesian and polar coordinates (using the convention $r \ge 0$), and convert equations of curves from cartesian to polar form and vice versa;
• sketch simple polar curves, for 0 < θ < 2π or −π < θ < π or a subset of either of these intervals (detailed plotting of curves will not be required, but sketches will generally be expected to show significant features, such as symmetry, the form of the curve at the pole and least/greatest values of r );
• recall the formula $\frac{1}{2}\int_\alpha ^\beta  {{r^2}d\theta } $ for the area of a sector, and use this formula in simple cases. 

Polar Coordinates:

I can say this, without exaggerating, that the cartesian coordinates are the most used of any coordinate system. However there are curves, the representation whereof in the cartesian system is an onerous task. Fortunately though, we have other systems of coordinates available, which can conveniently simplify the task. One such is the Polar Coordinate System.

Consider the point (3,4) on the cartesian plane. With the origin defined, you can without ambiguity fix this point to a plane. If you were to relay these coordinates to someone else, they too would be able to identify this unique point on the plane. But is that the only way to define that point? No, one alternative would be tell whoever is to plot the point that it is a distance a from some fixed point, let us in this example assume that it is the origin of the cartesian plane considered previously, and furthermore tell him that it makes an angle of say b radians, with some given line, which we may most unimaginatively denote by the initial line. Carrying our analogy of cartesian plane further, let the initial line be the x-axis. In this example a would be 5 and b would be 0.927 rad.

Something to this effect. (Disregard the crudeness of the plot)

This system at the moment does have some problems, that can be adequately overcome, by agreeing on the following conventions:

1) The length from the fixed point, r, shall always be equal to or greater than zero. 

2) The angle made with the initial line, $\theta $, will remain in the confines of the following inequality: $ - \pi  < \theta \le \pi $

3) The angle measured counter-clockwise from the initial line is positive, and that measured clockwise is negative.

If we agree upon these rules, then (r, $\theta $) are the polar coordinates of a point. 

The following graph might help in visualising the system:

Conversion of a curve from Polar to Cartesian form

Using basic trigonometry we can derive the relationship between a polar and cartesian system. Consider the follow the plot:

It is evident from above the sketch, that if we positioned the pole (the fixed point wherefrom the distance is measured) at the origin of a cartesian plane, and take the positive x axis as the initial line, we have the following relationships:

$r\cos \theta  = x$

$r\sin \theta  = y$

${r^2} = {x^2} + {y^2}$

The general method is explained through the following examples:

Ex. Convert the following polar equation into its cartesian form:

${r^2} = 4\cos (2\theta )$

Sol: We consider the following relationships:

${r^2} = {x^2} + {y^2}$

$\cos (2\theta ) = {\cos ^2}\theta  - {\sin ^2}\theta $

$\sin \theta  = \frac{y}{r}$

$\cos \theta  = \frac{x}{r}$

Using these it is not difficult to see that the cartesian equation can be derived thusly:

${x^2} + {y^2} = 4\left( {\frac{{{x^2}}}{{{r^2}}} - \frac{{{y^2}}}{{{r^2}}}} \right)$

Then taking $r^2$ common, and using the suitable relationship, we arrive at the cartesian equation:

$r\left( {{x^2} + {y^2}} \right) = 4\left( {{x^2} - {y^2}} \right)$

${\left( {{x^2} + {y^2}} \right)^2} = 4\left( {{x^2} - {y^2}} \right)$

Ex. Oct/Nov 2012 Paper 12, Question 1:

Find the cartesian equation corresponding to the polar equation 

$r = \sqrt 2 \sec \left( {\theta  - \frac{\pi }{4}} \right)$

Sol: 

We rearrange the equation as follows:

$r\left( {\cos \theta  - \frac{\pi }{4}} \right) = \sqrt 2 $

$r\cos \theta cos\left( {\frac{\pi }{4}} \right) + r\sin \theta sin\left( {\frac{\pi }{4}} \right) = \sqrt 2 $

$\frac{x}{{\sqrt 2 }} + \frac{y}{{\sqrt 2 }} = \sqrt 2 $

$y = 2 - x$

Curve Sketching

The most effective strategy in sketching a curve, $r = f(\theta )$, is to find the polar coordinates of a number of points on the curve. Following are plots of curves of the type expected by your examiners and some points of note on the matter.

Ex. Sketch the curve 

$r = 2a(1 + \cos \theta )$

Before the actual plotting, we find by inspection alone, that the function is an even function, which means that the function yields the same value of $\theta $, as it does for - $\theta $. This is because of the well known identity:

$\cos ( - \theta ) = \cos (\theta )$

This means the function will be symmetrical about the initial line, and you will need only the values in the region $0 \le \theta  \le \pi $

We then put in values of $\theta $ with the increments of $\frac{\pi }{6}$ in the range given above.

Using this information you can get the plot of the following shape.

This plot was made by taking a = 2. You however would plot it by using
values of u, 2u, 3u... so forth on the axes. 

Area of a Sector

For any curve, the area bound by the curve and the lines, $\theta  = \alpha $ and $\theta = \beta $, is given by:

 $\frac{1}{2}\int_\alpha ^\beta  {{r^2}d\theta } $

Perhaps the above graph will hep you visualise the result. The formula will give you the area shaded in the above diagram.

Ex. Find the area of the cardioid  $r = 2(1 + \cos \theta )$

We note that:

${r^2} = {\left( {2(1 + \cos \theta )} \right)^2}$
${r^2} = 4(1 + 2\cos \theta  + {\cos ^2}\theta )$
${r^2} = 4 + 8\cos \theta  + 4{\cos ^2}\theta $

Further considering:

$\cos (2\theta ) = 2{\cos ^2}\theta  - 1$
$2{\cos ^2}\theta  = \cos (2\theta ) + 1$
$4co{s^2}\theta  = 2\cos (2\theta ) + 2$

We have then:

${r^2} = 4 + 8\cos \theta  + 2\cos (2\theta ) + 2 = 2\cos (2\theta ) + 8\cos (\theta ) + 6$
$\frac{1}{2}\int_0^{2\pi } {{r^2}d\theta } $ = $\frac{1}{2}\int_0^{2\pi } {\left( {2\cos (2\theta ) + 8\cos (\theta ) + 6} \right)d\theta } $ $ = 6\pi $

Because the cardioid is symmetrical about the x-axis, you could evaluated the area by using the formula from zero to $\pi $ instead of $2 \pi $ and then multiplying the resulting area by two. But as this makes the evaluation only marginally less tedious, and introduces room for error, it is not recommended.