Derivative of Parametric Functions

Parametric functions are a way of defining a function using a set of parameters rather than an explicit formula for y in terms of x. This approach is particularly useful in cases where it's difficult or impossible to express y as a function of x, or when dealing with curves in the plane that fail the vertical line test.

Understanding Parametric Functions

A parametric function is given by two functions:

$$ x = f(t) $$ $$ y = g(t) $$

where ( t ) is the parameter, often representing time. The functions ( f ) and ( g ) define the x and y coordinates of points on the curve, respectively.

Derivative of a Parametric Function

To find the derivative of a parametric function, we use the chain rule of calculus. The derivative of y with respect to x, denoted as ( \frac{dy}{dx} ), is found by dividing the derivative of y with respect to t by the derivative of x with respect to t:

$$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$

provided that ( \frac{dx}{dt} \neq 0 ).

Higher Derivatives

For the second derivative, we differentiate ( \frac{dy}{dx} ) with respect to t and divide by ( \frac{dx}{dt} ):

$$ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} $$

Table of Differences and Important Points

Aspect	Explicit Functions	Parametric Functions
Definition	Defined by an equation y = f(x)	Defined by a pair of equations x = f(t), y = g(t)
Representation	Represents y directly as a function of x	Represents both x and y as functions of a third variable t
Derivative	( \frac{dy}{dx} ) is found directly	( \frac{dy}{dx} ) is found using the chain rule
Application	Used when a relationship can be expressed as y in terms of x	Used in more complex curves, motion problems, and where x cannot be expressed as a function of y

Examples

Example 1: Basic Parametric Derivative

Consider the parametric equations:

$$ x = t^2 $$ $$ y = t^3 $$

To find ( \frac{dy}{dx} ), we first find ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ):

$$ \frac{dx}{dt} = 2t $$ $$ \frac{dy}{dt} = 3t^2 $$

Now, we can find ( \frac{dy}{dx} ):

$$ \frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t $$

Example 2: Higher Derivative

Using the same parametric equations as in Example 1, let's find the second derivative ( \frac{d^2y}{dx^2} ):

First, we find ( \frac{dy}{dx} ) as before:

$$ \frac{dy}{dx} = \frac{3}{2}t $$

Now, we differentiate ( \frac{dy}{dx} ) with respect to t:

$$ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} $$

Finally, we divide by ( \frac{dx}{dt} ):

$$ \frac{d^2y}{dx^2} = \frac{\frac{3}{2}}{2t} = \frac{3}{4t} $$

Example 3: Trigonometric Parametric Functions

Consider the parametric equations for a circle:

$$ x = r \cos(t) $$ $$ y = r \sin(t) $$

To find ( \frac{dy}{dx} ), we calculate:

$$ \frac{dx}{dt} = -r \sin(t) $$ $$ \frac{dy}{dt} = r \cos(t) $$

Now, ( \frac{dy}{dx} ) is:

$$ \frac{dy}{dx} = \frac{r \cos(t)}{-r \sin(t)} = -\cot(t) $$

In this example, the derivative ( \frac{dy}{dx} ) gives us the slope of the tangent to the circle at any point defined by the parameter t.

Conclusion

The derivative of parametric functions is a powerful tool in calculus, allowing us to analyze and understand the behavior of complex curves that cannot be easily expressed as explicit functions. By using the chain rule, we can find the rate of change of y with respect to x, even when both x and y are defined in terms of a third variable, t. This concept is widely used in physics, engineering, and computer graphics to model motion and curves.