🚀 Further Pure Mathematics 1 (9231) Study Notes: Polar Coordinates (Topic 1.5)

Welcome to the world of Polar Coordinates! If you’re used to navigating the Cartesian plane using \(x\) and \(y\) coordinates (like moving along a street grid), Polar Coordinates offer a different, often simpler, way to describe curves.

Think of it like a radar system: instead of saying "go 3 units right and 4 units up," you say "go 5 units out at a 53-degree angle." This system is powerful in advanced physics, engineering, and calculus, making this chapter a crucial step in your Further Mathematics journey. Don't worry if it looks tricky—we’ll break it down!

1. Understanding the Polar System: \((r, \theta)\)

In the Cartesian system, a point \(P\) is defined by \((x, y)\). In the Polar system, \(P\) is defined by \((r, \theta)\).

1.1 Key Definitions

  • Radial Distance (\(r\)): This is the straight-line distance from the origin to the point \(P\).
  • Angle (\(\theta\)): This is the angle measured counter-clockwise from the positive x-axis.
  • The Pole: The origin \((0, 0)\) in the Cartesian system is called the Pole in the Polar system.
  • The Initial Line: The positive x-axis is called the Initial Line.

1.2 The Crucial Convention (\(r \ge 0\))

The syllabus strictly follows the convention that the radial distance \(r\) must be non-negative (\(r \ge 0\)). This simplifies the geometry and avoids confusion with negative distances.

The angle \(\theta\) usually lies in one of two standard intervals (or subsets thereof):

\(0 \le \theta < 2\pi\) (A full rotation, often used for general sketching)

or

\(-\pi < \theta < \pi\) (Symmetrical interval around the initial line)

🔑 Quick Review: Polar vs. Cartesian

Cartesian: Location based on two perpendicular distances \((x, y)\).
Polar: Location based on distance and direction \((r, \theta)\).

2. Converting Between Coordinate Systems

The link between the two systems comes directly from trigonometry applied to a right-angled triangle formed by the pole, the point \(P\), and the projection of \(P\) onto the initial line.

2.1 Polar to Cartesian (Finding \(x\) and \(y\) from \(r\) and \(\theta\))

Use the definitions of sine and cosine:
Formulas:
\(x = r \cos \theta\)
\(y = r \sin \theta\)

Example: If a point is at \((r, \theta) = (4, \frac{\pi}{6})\), then:
\(x = 4 \cos(\frac{\pi}{6}) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}\)
\(y = 4 \sin(\frac{\pi}{6}) = 4 \times \frac{1}{2} = 2\)

2.2 Cartesian to Polar (Finding \(r\) and \(\theta\) from \(x\) and \(y\))

Use Pythagoras and the tangent ratio:
Formulas:
\(r^2 = x^2 + y^2\) (Since \(r \ge 0\), \(r = \sqrt{x^2 + y^2}\))
\(\tan \theta = \frac{y}{x}\) (Remember to use the quadrant of \((x, y)\) to find the correct \(\theta\))

2.3 Converting Equations

The main challenge is often substituting these basic relations into complex equations.

Case 1: Cartesian to Polar

Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into the Cartesian equation.

Example: Convert the circle \(x^2 + y^2 = 9\).

Since \(x^2 + y^2 = r^2\), the equation becomes:

\(r^2 = 9 \implies r = 3\) (A simple circle centered at the pole!)

Case 2: Polar to Cartesian

Substitute \(r = \sqrt{x^2 + y^2}\) and \(\cos \theta = \frac{x}{r}\) (and \(\sin \theta = \frac{y}{r}\)). Often, you need to multiply by \(r\) first to simplify.

Example: Convert the line \(r = 3 \sec \theta\).

Recall that \(\sec \theta = \frac{1}{\cos \theta}\).

\(r = \frac{3}{\cos \theta}\)

Multiply by \(\cos \theta\): \(r \cos \theta = 3\)

Substitute \(r \cos \theta = x\): \(x = 3\) (This is a vertical line.)

💡 Memory Aid: The "R" Trick

If you see a curve \(r = f(\theta)\) that seems impossible to convert directly, try multiplying by \(r\) first. It helps create the identifiable Cartesian terms \(r^2\), \(r \cos \theta\), and \(r \sin \theta\).
e.g., \(r = a \cos \theta\) becomes \(r^2 = ar \cos \theta\), which is \(x^2 + y^2 = ax\).

Key Takeaway: Conversions rely entirely on the four substitution relationships. Practice rearranging and simplifying equations until you isolate \(x\) and \(y\) terms, or \(r\) and \(\theta\) terms.

3. Sketching Simple Polar Curves

Detailed plotting isn't required, but you must show the significant features of the curve.

3.1 Analyzing Key Features Step-by-Step

Step 1: Check for Symmetry

Symmetry helps you draw half the curve and reflect it.

  • Symmetry about the Initial Line (\(\theta = 0\)): If replacing \(\theta\) with \(-\theta\) results in the same equation, the curve is symmetric about the initial line (x-axis).
    Example: If \(r = 2(1 + \cos \theta)\), since \(\cos(-\theta) = \cos \theta\), the equation remains the same. Symmetric!
  • Symmetry about the Line \(\theta = \frac{\pi}{2}\) (y-axis): If replacing \(\theta\) with \(\pi - \theta\) results in the same equation, it is symmetric about the y-axis.
    Example: If \(r = 2 \sin 2\theta\), since \(\sin(2(\pi - \theta)) = \sin(2\pi - 2\theta) = -\sin 2\theta\), this is NOT symmetric.
Step 2: Find Intersections with the Initial Line (\(\theta = 0\) and \(\theta = \pi\))

This tells you where the curve crosses the positive and negative x-axes.

  • Set \(\theta = 0\) to find the maximum positive \(r\).
  • Set \(\theta = \pi\) to find the intersection on the negative x-axis (or another key point).
Step 3: Find Intersections at the Pole (The Pole Test)

The curve passes through the Pole if \(r = 0\) for some angle \(\theta\).

If \(r = 0\) when \(\theta = \alpha\), then the curve approaches the pole tangentially along the line \(\theta = \alpha\).

Step 4: Find Least and Greatest Values of \(r\) (Maximum/Minimum Extent)

Since \(r\) is often a function of \(\sin \theta\) or \(\cos \theta\), you can find the maximum and minimum values of \(r\) by knowing that \(-1 \le \sin \theta \le 1\) and \(-1 \le \cos \theta \le 1\).

To be thorough, you can check \(\frac{dr}{d\theta} = 0\) to find turning points for \(r\), but usually, inspection of the trigonometric function is enough.

🛑 Common Mistake Alert!

Remember the convention: \(r \ge 0\). When sketching, if your equation gives a negative \(r\), those points are not part of the curve under the Cambridge syllabus convention! Ensure you restrict \(\theta\) such that \(r\) stays positive.

Key Takeaway: When sketching, identify symmetry first, then check the maximum distance (\(r_{max}\)) and the angles where the curve meets the pole (\(r=0\)).

4. Calculating the Area of a Sector

Just as we use integration in Cartesian coordinates to find the area under a curve, we use a specific formula in polar coordinates to find the area bounded by a polar curve and two radial lines.

Imagine slicing the area into tiny sectors, like cutting a pizza into infinitesimally thin slices.

4.1 The Area Formula

The area \(A\) of a sector bounded by the curve \(r = f(\theta)\) and the radial lines \(\theta = \alpha\) and \(\theta = \beta\) is given by:

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

4.2 Step-by-Step Calculation Process

This formula is required for use in simple cases, meaning the integration should be straightforward after substituting \(r^2\).

  1. Express \(r^2\) in terms of \(\theta\). If \(r = f(\theta)\), then \(r^2 = [f(\theta)]^2\). If \(r^2\) involves terms like \(\cos^2 \theta\) or \(\sin^2 \theta\), you will often need to use the double angle identities to simplify the integration:

    \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\)

    \(\sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta)\)

  2. Identify the Limits (\(\alpha\) and \(\beta\)). These are the angles that define the boundaries of the area you are calculating. If you are calculating the entire area enclosed by a closed loop curve, the limits will usually be the angles where \(r=0\).
  3. Integrate the expression. Integrate \(\frac{1}{2} r^2\) with respect to \(\theta\).
  4. Substitute the Limits. Calculate \([F(\beta)] - [F(\alpha)]\), where \(F(\theta)\) is the integrated expression.

Did you know? The coefficient \(\frac{1}{2}\) in the formula comes from the basic area formula for a circular sector: \(A = \frac{1}{2} r^2 \theta\). We are essentially summing up infinitely many tiny sectors \(dA = \frac{1}{2} r^2 d\theta\).

4.3 Example Context: Area of a Cardioid

Consider the curve \(r = a(1 + \cos \theta)\). This is a cardioid (heart shape).

To find the total area enclosed, you need to integrate over the range where the curve is traced once. Since \(\cos \theta\) covers all values from 1 to -1 and back to 1 over \(0\) to \(2\pi\), and \(r \ge 0\), the limits are \(\alpha = 0\) and \(\beta = 2\pi\).

We set up the integral:
$$A = \frac{1}{2} \int_{0}^{2\pi} [a(1 + \cos \theta)]^2 d\theta$$
$$A = \frac{a^2}{2} \int_{0}^{2\pi} (1 + 2 \cos \theta + \cos^2 \theta) d\theta$$
(You would then use the identity \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\) to solve this integral.)

Key Takeaway: The area formula \(A = \frac{1}{2} \int r^2 d\theta\) is non-negotiable. Ensure you correctly square \(r\) and use double angle formulas if necessary before integrating.

🎓 Final Quick Review and Checklist

H5 Topics Covered (Syllabus 1.5)

  • Conversions: Can you switch easily between \((r, \theta)\) and \((x, y)\) equations?
  • Convention: Are you strictly applying \(r \ge 0\)?
  • Sketching: Can you identify symmetry, intersections with the initial line, points at the pole (\(r=0\)), and the max/min values of \(r\)?
  • Area: Can you set up and calculate the definite integral \(A = \frac{1}{2} \int r^2 d\theta\) using correct limits?

Keep practicing those conversions and sketching simple standard curves (like circles, lines, or cardioids). You've got this!