Simple curves in polar coordinates

2 Simple curves in polar coordinates

We are used to describing the equations of curves in Cartesian variables $x, y$ . Thus $x^{2} + y^{2} = 1$ represents a circle, centre the origin, and of radius 1, and $y = 2 x^{2}$ is the equation of a parabola whose axis is the $y$ -axis and with vertex located at the origin. (In colloquial terms the vertex is the ‘sharp end’ of a conic.) We can convert these equations into polar form by using the relations $x = r cos θ$ , $y = r sin θ$ .

Example 5

Find the polar coordinate form of

the circle $x^{2} + y^{2} = 1$
the parabola $y = 2 x^{2}$ .

Solution

Using $x = r cos θ$ , $y = r sin θ$ in the expression $x^{2} + y^{2} = 1$ we have
${(r cos θ)}^{2} + {(r sin θ)}^{2} = 1 or r^{2} ({cos}^{2} θ + {sin}^{2} θ) = 1$

giving $r^{2} = 1$ . We simplify this to $r = 1$ (since $r = - 1$ is invalid being a negative distance). Of course we might have guessed this answer since the relation $r = 1$ states that every point on the curve is a constant distance 1 away from the origin.
Repeating the approach used in (1) for $y = 2 x^{2}$ we obtain:
$r sin θ = 2 {(r cos θ)}^{2} i.e. r sin θ - 2 r^{2} {cos}^{2} θ = 0$

Therefore $r (sin θ - 2 r {cos}^{2} θ) = 0$ . Either $r = 0$ (which is a single point, the origin, and is clearly not a parabola) or

$sin θ - 2 r {cos}^{2} θ = 0 giving, finally r = \frac{1}{2} tan θ sec θ .$

This is the polar equation of this particular parabola, $y = 2 x^{2}$ .