Polar Coordinates and Parametric Equations

Cartesian coordinates, with their familiar \( x \) and \( y \) axes, are not always the most natural way to describe a curve. A spiral, the petals of a flower or the orbit of a satellite are much easier to handle using a distance and an angle, while the path of a cricket ball is best described by saying where it is at each moment in time. Polar coordinates and parametric equations give us these two alternative viewpoints, and both are essential in FSc, A Level and university calculus.

Polar Coordinates

In polar coordinates, a point is described by

• \( r \): its distance from the origin (called the pole), and
• \( \theta \): the angle measured anticlockwise from the positive \( x \)-axis.

The connection with Cartesian coordinates comes straight from a right-angled triangle:

$$ x = r\cos\theta, \qquad y = r\sin\theta $$

$$ r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x} $$

One important difference from Cartesian coordinates: a point has many polar descriptions. The point \( (2, \tfrac{\pi}{6}) \) is the same as \( (2, \tfrac{13\pi}{6}) \), because adding a full turn changes nothing. Allowing negative \( r \) (meaning "go backwards") gives even more options.

Worked Example 1: Converting Coordinates

(a) Convert the Cartesian point \( (3, 4) \) to polar form.

$$ r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$

$$ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.927 \text{ rad} \approx 53.13^\circ $$

The point lies in the first quadrant, so no adjustment is needed. The polar form is \( (5, 0.927) \).

(b) Convert the polar point \( (4, \tfrac{2\pi}{3}) \) to Cartesian form.

$$ x = 4\cos\frac{2\pi}{3} = 4 \times \left(-\frac{1}{2}\right) = -2 $$

$$ y = 4\sin\frac{2\pi}{3} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \approx 3.464 $$

The point is \( (-2, 2\sqrt{3}) \), in the second quadrant, as expected for an angle of 120°.

A word of warning: a calculator gives \( \tan^{-1}(y/x) \) only between \( -90^\circ \) and \( 90^\circ \). For a point like \( (-3, -4) \), you must add 180° to get the correct third-quadrant angle.

Famous Polar Curves

Equations that look messy in Cartesian form are often beautifully simple in polar form:

• \( r = a \): a circle of radius \( a \) centred at the pole.
• \( r = a\theta \): the Archimedean spiral, where each turn is the same distance from the last, like a rolled carpet.
• \( r = a(1 + \cos\theta) \): the cardioid, a heart-shaped curve.
• \( r = a\cos(k\theta) \): a rose curve, with \( k \) petals if \( k \) is odd and \( 2k \) petals if \( k \) is even.

Area in Polar Coordinates

For a region swept out by a polar curve, we add up thin sectors rather than thin rectangles. A sector of radius \( r \) and small angle \( d\theta \) has area \( \tfrac{1}{2}r^2\,d\theta \), so

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

Worked Example 2: Area of a Cardioid

Find the area enclosed by \( r = 1 + \cos\theta \).

The curve is traced once as \( \theta \) runs from 0 to \( 2\pi \):

$$ A = \frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2 \, d\theta = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \cos^2\theta\right) d\theta $$

Use the identity \( \cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta) \):

$$ A = \frac{1}{2}\int_0^{2\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right) d\theta $$

Over a full period, the integrals of \( \cos\theta \) and \( \cos 2\theta \) are both zero, so

$$ A = \frac{1}{2} \times \frac{3}{2} \times 2\pi = \frac{3\pi}{2} \approx 4.712 $$

Sanity check: the cardioid stretches from \( x = 0 \) to \( x = 2 \), roughly like a circle of radius 1.2. Such a circle has area about 4.5, close to our answer.

Parametric Equations

In parametric form, both coordinates are written as functions of a third variable, the parameter, often \( t \) for time:

$$ x = f(t), \qquad y = g(t) $$

The unit circle, for instance, is \( x = \cos t \), \( y = \sin t \) for \( 0 \le t < 2\pi \). Parametric equations carry extra information that a Cartesian equation loses: not just where the curve is, but when and in which direction it is traced.

Eliminating the Parameter

To find the Cartesian equation, solve one equation for \( t \) and substitute. For \( x = 2t + 1 \), \( y = t^2 \):

$$ t = \frac{x - 1}{2} \quad \Rightarrow \quad y = \frac{(x - 1)^2}{4} $$

which is a parabola with vertex at \( (1, 0) \).

Slopes of Parametric Curves

By the chain rule,

$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt} \ne 0 $$

Worked Example 3: A Projectile

A ball is struck so that its position in metres after \( t \) seconds is

$$ x = 15t, \qquad y = 20t - 4.9t^2 $$

(a) Time of flight. The ball lands when \( y = 0 \):

$$ t(20 - 4.9t) = 0 \quad \Rightarrow \quad t = \frac{20}{4.9} \approx 4.08 \text{ s} $$

(b) Range.

$$ x = 15 \times 4.08 \approx 61.2 \text{ m} $$

(c) Maximum height. The ball is highest when \( dy/dt = 20 - 9.8t = 0 \), so \( t = 20/9.8 \approx 2.04 \) s, exactly half the flight time. Then

$$ y = 20(2.041) - 4.9(2.041)^2 \approx 40.82 - 20.41 = 20.41 \text{ m} $$

(d) Direction at launch. At \( t = 0 \):

$$ \frac{dy}{dx} = \frac{20 - 9.8(0)}{15} = \frac{4}{3} $$

so the launch angle is \( \tan^{-1}(4/3) \approx 53.1^\circ \), the same angle as in Example 1, since the initial velocity vector is \( (15, 20) \), a 3-4-5 triangle scaled by 5.

Choosing the Right Description

How do you decide which system to use? A practical rule of thumb:

• Use Cartesian coordinates for straight lines, rectangles and graphs of functions \( y = f(x) \).
• Use polar coordinates whenever there is a natural centre: circles about the origin, spirals, orbits, radar and anything with rotational symmetry. Integrals over discs, which are awkward in \( x \) and \( y \), often become simple in \( r \) and \( \theta \).
• Use parametric equations when motion or timing matters, or when a curve loops back on itself so it cannot be written as a single function \( y = f(x) \). Computer graphics, robot arm paths and animation software rely heavily on parametric curves, such as the Bézier curves used to draw fonts on your screen.

Many exam questions mix the systems, for example asking you to convert a polar curve to parametric form using \( x = r(\theta)\cos\theta \) and \( y = r(\theta)\sin\theta \), with \( \theta \) playing the role of the parameter.

Did You Know?

• Isaac Newton described polar-type coordinates in work from around 1671, and Jacob Bernoulli used them in print in 1691.
• Jacob Bernoulli loved the logarithmic spiral \( r = ae^{b\theta} \) so much that he asked for it to be carved on his tombstone; the stonemason carved an Archimedean spiral by mistake.
• Radar screens and sonar displays plot targets in polar form, giving distance and bearing directly.

Key Takeaways

• Polar coordinates use distance \( r \) and angle \( \theta \), linked by \( x = r\cos\theta \) and \( y = r\sin\theta \).
• Always check the quadrant when finding \( \theta \) from \( \tan^{-1}(y/x) \).
• Area inside a polar curve is \( \tfrac{1}{2}\int r^2 \, d\theta \).
• Parametric equations describe both position and timing; eliminate \( t \) for the Cartesian form.
• The slope of a parametric curve is \( (dy/dt)/(dx/dt) \).

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