Welcome to Polar Coordinates!
Hello! You've successfully navigated the familiar world of Cartesian coordinates \((x, y)\) and are now ready to explore an exciting, alternative way to pinpoint locations: Polar Coordinates.
This chapter (FP2.3) moves away from the rigid grid system you're used to and teaches you to describe points using a distance and an angle. This method is incredibly powerful for describing curves that naturally spiral or rotate, like orbits or complicated loops.
Don't worry if this seems tricky at first—it's just a new perspective! By the end of these notes, you'll be converting coordinates, sketching unusual curves, and calculating the areas they enclose. Let's dive in!
1. Defining Polar Coordinates \((r, \theta)\)
1.1 The Concept: Distance and Direction
In the Cartesian system, a point \((x, y)\) is found by moving sideways \(x\) and then vertically \(y\).
In the Polar system, a point \((r, \theta)\) is found using two components:
- \(r\) (The Radius/Modulus): This is the straight-line distance from the origin (which we now call the Pole) to the point.
- \(\theta\) (The Angle/Argument): This is the angle, measured counter-clockwise, from the positive x-axis (which we now call the Initial Line) to the line segment connecting the Pole and the point.
Key Convention (Syllabus Requirement)
The OxfordAQA syllabus uses the convention that \(r \geq 0\). This means the distance from the Pole must be zero or positive. Angles \(\theta\) are typically given in radians.
Analogy: Think of a robot arm. Cartesian coordinates tell the arm to move "3 units right, 4 units up." Polar coordinates tell the arm to move "extend 5 units at an angle of 53 degrees." The Polar system is much more intuitive for anything rotational!
Quick Review: Polar Basics
A point P is defined by \((r, \theta)\):
Pole: The origin \((0, 0)\).
Initial Line: The positive x-axis.
2. Converting Between Coordinate Systems
The relationship between Cartesian \((x, y)\) and Polar \((r, \theta)\) coordinates is based entirely on basic trigonometry and Pythagoras' theorem, applied to a right-angled triangle where \(r\) is the hypotenuse.
2.1 Polar to Cartesian Conversion
If you are given \((r, \theta)\), you can find \((x, y)\) using:
$$x = r \cos \theta$$ $$y = r \sin \theta$$
Example: Convert \((4, \frac{\pi}{3})\) to Cartesian.
- \(x = 4 \cos(\frac{\pi}{3}) = 4 \times \frac{1}{2} = 2\)
- \(y = 4 \sin(\frac{\pi}{3}) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}\)
2.2 Cartesian to Polar Conversion
If you are given \((x, y)\), you can find \((r, \theta)\) using:
$$r^2 = x^2 + y^2 \quad \text{or} \quad r = \sqrt{x^2 + y^2}$$ $$\tan \theta = \frac{y}{x}$$
Step-by-Step Guide: Finding \(\theta\)
Finding \(r\) is easy (just use Pythagoras), but finding the correct angle \(\theta\) is where most students make mistakes! You must use the quadrant of \((x, y)\) to adjust the principal value of \(\arctan(\frac{y}{x})\).
- Find \(r\). \(r = \sqrt{x^2 + y^2}\). (Always positive or zero).
- Find the reference angle \(\alpha\). Calculate \(\alpha = \arctan \left| \frac{y}{x} \right|\).
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Determine \(\theta\) based on the Quadrant:
- Q1 (x>0, y>0): \(\theta = \alpha\)
- Q2 (x<0, y>0): \(\theta = \pi - \alpha\)
- Q3 (x<0, y<0): \(\theta = \pi + \alpha\) (or \(-\pi + \alpha\))
- Q4 (x>0, y<0): \(\theta = 2\pi - \alpha\) (or \(-\alpha\))
Common Mistake to Avoid:
Using \(\theta = \arctan(\frac{y}{x})\) directly from your calculator usually only gives answers in Quadrant 1 or 4. If your point is in Quadrant 2 or 3, you MUST manually adjust the angle by adding or subtracting \(\pi\).
Key Takeaway: Converting coordinates is essential for moving between geometric representations. Polar curves are often simplest when left in terms of \(r\) and \(\theta\).
3. Sketching Polar Curves \(r = f(\theta)\)
Sketching a curve defined by \(r = f(\theta)\) (where \(r\) depends on the angle) is a vital skill. These curves can look very different from the Cartesian graphs you are used to!
3.1 Strategy for Sketching
Since \(r\) is a function of \(\theta\), the method is straightforward:
- Determine the range of \(\theta\): Usually, the curve will be traced out fully for \(0 \leq \theta \leq 2\pi\), but check if the function repeats sooner (e.g., if \(r\) involves \(\cos(2\theta)\)).
- Create a table of values: Pick key, simple angles (like \(0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \dots\)) and calculate the corresponding value of \(r\).
- Plot key points: Remember you are plotting a distance \(r\) at an angle \(\theta\).
- Check if the curve passes through the Pole: This happens when \(r = 0\). Solve \(f(\theta) = 0\) to find the angles where the curve hits the origin.
Important Tip: Plotting polar points is like tracing a radar screen. Start at \(\theta=0\) (the positive x-axis) and spin counter-clockwise, marking the distance \(r\) for each angle.
3.2 Common Shapes and Examples
Example 1: The Circle
$$r = a$$
When \(r\) is a constant \(a\), the distance from the Pole is always \(a\), regardless of the angle \(\theta\). This is simply a circle, centre the Pole, radius \(a\).
$$r = a \cos \theta$$
This is also a circle, but it passes through the Pole and has its centre on the initial line (x-axis).
Example 2: The Cardioid (Heart Shape)
$$r = a(1 + \cos \theta)$$
This curve is called a cardioid (from the Greek word for heart).
- At \(\theta = 0\): \(r = a(1+1) = 2a\). (Farthest point)
- At \(\theta = \frac{\pi}{2}\) or \(\frac{3\pi}{2}\): \(r = a(1+0) = a\).
- At \(\theta = \pi\): \(r = a(1-1) = 0\). (Passes through the Pole)
Did you know? Cardioid shapes are important in engineering, particularly in microphone design, as their shape dictates their sound pick-up pattern!
Example 3: The Rose Curve (or petal curves)
$$r = a \cos(n\theta)$$
These curves create petal shapes. The number of petals depends on \(n\):
- If \(n\) is odd, there are \(n\) petals.
- If \(n\) is even, there are \(2n\) petals.
Key Takeaway: Sketching
To sketch \(r=f(\theta)\), plot points for simple angles, especially where \(\cos\theta\) or \(\sin\theta\) are 0, \(\pm 1\), or \(\pm \frac{1}{2}\). Check where \(r=0\) (the Pole) and where \(r\) is maximum.
4. Calculating Area in Polar Coordinates
Just as we use integration in Cartesian coordinates to find the area under a curve, we use integration in polar coordinates to find the area swept out by a rotating line segment (a sector).
4.1 The Area Formula
The area enclosed by the curve \(r = f(\theta)\) and the radial lines \(\theta = \alpha\) and \(\theta = \beta\) is given by the formula:
$$\text{Area} = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta$$
Why does this work? The formula comes from approximating the area using very thin sectors (like tiny slices of pizza). The area of a sector is \(\frac{1}{2} r^2 \delta\theta\). By integrating (summing) these infinitesimally thin sectors from angle \(\alpha\) to \(\beta\), we get the total area.
4.2 Applying the Area Formula Step-by-Step
Finding the area often requires careful attention to the limits of integration, \(\alpha\) and \(\beta\).
- Identify the Curve and \(r\): Make sure the equation is in the form \(r = f(\theta)\).
- Square \(r\): Calculate \(r^2 = [f(\theta)]^2\). (Do not forget to square \(r\)! This is a common error.)
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Determine the Limits (\(\alpha, \beta\)): These angles define the boundary of the region.
- If you are finding the area between two given lines, \(\alpha\) and \(\beta\) are those angles.
- If you are finding the total area enclosed by a closed curve that passes through the Pole (e.g., a cardioid or rose curve), \(\alpha\) and \(\beta\) are the angles where \(r=0\).
- Set up the Integral: Substitute \(r^2\) and the limits into \( \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta \).
- Solve the Integral: You will likely need trigonometric identities, especially double-angle formulas (like \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\)), to integrate terms like \(\cos^2 \theta\) or \(\sin^2 \theta\).
Example of Finding Limits
Suppose we need the area of the upper loop of the curve \(r = 2 \cos(2\theta)\).
We first find where the curve passes through the Pole (where \(r=0\)):
\(0 = 2 \cos(2\theta)\)
\(\cos(2\theta) = 0\)
The solutions for \(2\theta\) are \(\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\)
Therefore, \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \dots\)
The first loop is traced out between \(\theta = -\frac{\pi}{4}\) and \(\theta = \frac{\pi}{4}\). These become your limits \(\alpha\) and \(\beta\).
Tip for Struggling Students: Limits
If a curve is symmetrical about the initial line (x-axis), it is often easier to calculate the area for the upper half (e.g., from \(\theta=0\) to \(\theta=\pi\)) and then multiply the result by two, provided the entire area is covered in that range. Always visualize or sketch the region first!
Chapter Summary: Key Takeaways
We only need three major relationships for Polar Coordinates:
- Conversion: \(x = r \cos \theta\) and \(y = r \sin \theta\). Use \(r^2 = x^2 + y^2\) and \( \tan \theta = \frac{y}{x} \) (adjusting for quadrant).
- Sketching: Plot \(r\) vs \(\theta\). Pay special attention to points where \(r\) is max/min, and where \(r=0\).
- Area: \( \text{Area} = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta \). Remember the \(\frac{1}{2}\) and remember to square \(r\)!
You do not need to know the formula relating the tangent angle \(\phi\) to the derivatives (\( \tan \phi = r \frac{d\theta}{dr} \)). Focus on the conversion and area calculation. You've got this!