Introduction to Parametric Equations
Welcome to the world of Parametric Equations! Up until now, you have likely been used to seeing graphs defined by a direct relationship between \(x\) and \(y\), like \(y = x^2\). This is called a Cartesian equation.
In this chapter, we introduce a "middleman" called a parameter (usually denoted as \(t\) or \(\theta\)). Instead of \(y\) depending directly on \(x\), both \(x\) and \(y\) will depend on this third variable. Don't worry if this seems tricky at first—think of it like a GPS tracking a car: the car's horizontal position (\(x\)) and vertical position (\(y\)) both change based on the time (\(t\)) that has passed. By the end of these notes, you'll be able to move between these two worlds with ease!
1. Understanding the Parameter
A parameter is an independent variable that links two other variables. In coordinate geometry, we use it to define the coordinates of a point on a curve separately.
The Definition:
If \(x = f(t)\) and \(y = g(t)\), these are called parametric equations. The variable \(t\) is the parameter. Each value of \(t\) gives you a specific point \((x, y)\) on the graph.
Example:
Imagine \(x = t + 1\) and \(y = 2t\).
If \(t = 0\), then \(x = 1\) and \(y = 0\). The point is \((1, 0)\).
If \(t = 1\), then \(x = 2\) and \(y = 2\). The point is \((2, 2)\).
Did you know?
Parametric equations are used extensively in video games! When a character jumps, the game calculates their horizontal position and vertical height separately based on the time elapsed since the jump began.
Key Takeaway:
A parameter is just a third variable that tells \(x\) and \(y\) where to be at any given moment.
2. Converting Between Forms
To understand the "shape" of a parametric curve, we often want to turn it back into a standard Cartesian equation (an equation with only \(x\) and \(y\)). This process is called eliminating the parameter.
Method A: Substitution
This is the most common method when the equations are algebraic (using \(t\)).
Step 1: Rearrange one equation (usually the simplest one) to make \(t\) the subject.
Step 2: Substitute this expression for \(t\) into the other equation.
Step 3: Simplify the resulting equation.
Example:
Given \(x = t - 3\) and \(y = t^2\).
1. From the first equation, \(t = x + 3\).
2. Substitute into the second: \(y = (x + 3)^2\).
3. This is a standard parabola shifted to the left!
Method B: Using Trigonometric Identities
When the parameter is an angle (\(\theta\)), we use trig identities to "get rid" of the trig functions.
Common Trick: Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\).
If you have \(x = a \cos \theta\) and \(y = a \sin \theta\), then \(\frac{x}{a} = \cos \theta\) and \(\frac{y}{a} = \sin \theta\).
Squaring and adding them gives: \((\frac{x}{a})^2 + (\frac{y}{a})^2 = 1\), which simplifies to \(x^2 + y^2 = a^2\).
Quick Review - Common Mistakes:
Watch out! When converting, always check the domain. If the parameter \(t\) is restricted (e.g., \(t > 0\)), then your resulting Cartesian graph might only be a part of the full curve. For MEI, you must be careful about these restricted domains.
3. The Parametric Equation of a Circle
According to the syllabus (Ref g14), you need to recognize the parametric form of a circle. This is one of the most elegant parts of coordinate geometry!
The Formula:
For a circle with center \((0, 0)\) and radius \(r\):
\(x = r \cos t\)
\(y = r \sin t\)
(where \(0 \leq t < 2\pi\))
Memory Aid:
Think of the unit circle from your trig lessons. \(x\) is always "across" (cosine) and \(y\) is always "up" (sine). The radius \(r\) just scales the circle up or down.
4. Finding the Gradient (Differentiation)
Sometimes we need the gradient (\(\frac{dy}{dx}\)) of a parametric curve without converting it to Cartesian form first. We use a version of the Chain Rule (Ref g15).
The Rule:
\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)
Step-by-Step Process:
1. Differentiate \(y\) with respect to \(t\) to find \(\frac{dy}{dt}\).
2. Differentiate \(x\) with respect to \(t\) to find \(\frac{dx}{dt}\).
3. Divide the first result by the second.
4. If you need the gradient at a specific point, substitute the value of \(t\) for that point into your final \(\frac{dy}{dx}\) expression.
Example:
Find the gradient of \(x = t^2\) and \(y = t^3\).
\(\frac{dx}{dt} = 2t\)
\(\frac{dy}{dt} = 3t^2\)
\(\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t\)
Summary Table for Gradient:
If you want... -> Do this...
\(\frac{dy}{dt}\) -> Differentiate the \(y\) equation.
\(\frac{dx}{dt}\) -> Differentiate the \(x\) equation.
\(\frac{dy}{dx}\) -> Divide them!
5. Modelling with Parametric Equations
In the real world, things move! Kinematics and Projectiles are the perfect use cases for parametric equations (Ref g16). In these models, the parameter is almost always time (\(t\)).
Projectiles Example:
A ball is kicked. Its horizontal distance might be \(x = 20t\) and its vertical height might be \(y = 15t - 4.9t^2\).
By eliminating \(t\), you can find the trajectory (the path) of the ball, which is usually a quadratic curve (parabola).
Important Note: In modelling, the domain of \(t\) is restricted by the context. For example, \(t \geq 0\) (because time can't be negative) and \(y \geq 0\) (because the ball stops when it hits the ground).
Key Takeaway:
In modelling, parametric equations allow us to treat horizontal and vertical motion independently, which makes complex physics much easier to solve!
Final Chapter Summary
1. Parameters: A third variable \(t\) that defines \(x\) and \(y\) separately.
2. Conversion: Use substitution for algebra, or trig identities (like \(\sin^2 + \cos^2 = 1\)) for angles.
3. Circles: Remember \(x = r \cos t\) and \(y = r \sin t\).
4. Differentiation: Find the gradient using \(\frac{dy/dt}{dx/dt}\).
5. Modelling: Use \(t\) as time to track objects in 2D space.
Keep practicing! Parametric equations are just a different way of looking at the same curves you already know. You've got this!