All You Need To Know About Parametric Equations And How To Differentiate Them

Basic Concept of Parametric Equations

In traditional equations, \( y \) might be written as a function of \( x \), such as \( y = f(x) \). However, in parametric equations, both \( x \) and \( y \) are described by separate equations as functions of \( t \). For example:

\[
x = f(t)
\]
\[
y = g(t)
\]

Where \( t \) is a parameter that can vary over a range, and as it does, it traces out a curve in the coordinate plane.

Example

Consider the parametric equations:
\[
x(t) = 2t
\]
\[
y(t) = t^2
\]

Here, \( x \) and \( y \) are both expressed in terms of \( t \). As \( t \) changes, the values of \( x \) and \( y \) change, and this describes a curve in the plane. To eliminate the parameter \( t \) and obtain a more familiar Cartesian equation, you could solve for \( t \) in terms of \( x \) or \( y \).

In this case, since \( x = 2t \), we can solve for \( t \) as \( t = \frac{x}{2} \). Substituting this into the equation for \( y \), we get:
\[
y = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}
\]
So, the parametric equations describe the parabola \(y = \frac{x^2}{4} \).

Why Use Parametric Equations?

1. Curves and Motion: Parametric equations are particularly useful for describing curves that may not easily be expressed as functions of \( x \). For example, a circle cannot be written as \( y = f(x) \), but it can be easily parameterized using trigonometric functions:
   \[
   x(t) = r \cos(t)
   \]
   \[
   y(t) = r \sin(t)
   \]
   where \( r \) is the radius of the circle, and \( t \) is the parameter (angle).
2. Modeling Motion: Parametric equations are often used in physics and engineering to describe the motion of objects. For example, if an object is moving in a plane, its position at time \( t \) might be described by parametric equations for \( x(t) \) and \( y(t) \), where \( t \) represents time.
3. Complex Curves: Some curves are more naturally described using parametric equations. For example, a Lissajous curve, which is a complex pattern, is easily defined by parametric equations.

Example 2: A Circle

To describe a circle with radius \( r \) centered at the origin, the parametric equations are:
\[
x(t) = r \cos(t)
\]
\[
y(t) = r \sin(t)
\]
As \( t \) varies from \( 0 \) to \( 2\pi \), these parametric equations trace out a circle.

Parameter Range

The parameter \( t \) is often allowed to vary over a specific interval. For instance, for the circle example, \( t \) typically ranges from \( 0 \) to \( 2\pi \) to complete one full circle.

Conclusion

Parametric equations are powerful tools that allow you to describe a wide variety of curves and motions. They offer a flexible way to represent curves in the plane, especially when those curves cannot be easily described using traditional Cartesian equations.

Differentiating Parametric Equations

If \( x = f(t) \) and \( y = g(t) \), then the derivative of \( y \) with respect to \( x \) is found using the chain rule. Here’s the process:

1. Differentiate \( x \) and \( y \) with respect to \( t \):

• \[ \frac{dx}{dt} = f'(t) \]
• \[ \frac{dy}{dt} = g'(t) \]

2. Find \( \frac{dy}{dx} \) using the chain rule:

• Since \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \), the derivative of \( y \) with respect to \( x \) is:
• \[ \frac{dy}{dx} = \frac{g'(t)}{f'(t)} \]

This is the derivative of the parametric curve at a specific point determined by the parameter \( t \).

Example:

Suppose the parametric equations are:

• \( x(t) = t^2 + 1 \)
• \( y(t) = 2t + 3 \)

Step 1: Differentiate both equations with respect to \( t \):

• \[ \frac{dx}{dt} = 2t \]
• \[ \frac{dy}{dt} = 2 \]

Step 2: Apply the formula for \( \frac{dy}{dx} \):

• \[ \frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t} \]

Thus, the slope of the curve at any point \( t \) is \( \frac{1}{t} \).

Second Derivative

To find the second derivative \( \frac{d^2y}{dx^2} \), you can apply the following process:

1. Differentiate \( \frac{dy}{dx} \) with respect to \( t \):
   \[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) \]
2. Then divide by \( \frac{dx}{dt} \) to get \( \frac{d^2y}{dx^2} \):
   \[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \]