Differentiation of Parametric Functions

Differentiation of parametric functions involves finding the derivative of a function that is defined in terms of a parameter, rather than as an explicit function of (x). In parametric differentiation, the coordinates (x) and (y) are given as functions of a third variable, typically (t), called the parameter.

If (x) and (y) are given as: [ x = f(t) \quad \text{and} \quad y = g(t), ] where both (x) and (y) are expressed in terms of the parameter (t), then the derivative of (y) with respect to (x), denoted as (\frac{dy}{dx}), can be found using the following steps:

Formula for Parametric Differentiation

To differentiate parametric functions, use the formula: [ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, ] where:

• (\frac{dy}{dt}) is the derivative of (y) with respect to (t).
• (\frac{dx}{dt}) is the derivative of (x) with respect to (t).

Steps for Differentiating Parametric Functions

1. Differentiate (x) with respect to (t): Find (\frac{dx}{dt}).
2. Differentiate (y) with respect to (t): Find (\frac{dy}{dt}).
3. Divide (\frac{dy}{dt}) by (\frac{dx}{dt}): This gives (\frac{dy}{dx}), the derivative of (y) with respect to (x).

Example 1: Parametric Differentiation

Given the parametric equations: [ x = t^2 + 2t, \quad y = 3t - 4, ] find (\frac{dy}{dx}).

1. Differentiate (x) with respect to (t): [ \frac{dx}{dt} = 2t + 2. ]

2. Differentiate (y) with respect to (t): [ \frac{dy}{dt} = 3. ]

3. Divide (\frac{dy}{dt}) by (\frac{dx}{dt}): [ \frac{dy}{dx} = \frac{3}{2t + 2}. ]

Example 2: Finding the Second Derivative

To find the second derivative (\frac{d^2y}{dx^2}), first find the derivative of (\frac{dy}{dx}) with respect to (t) and then divide by (\frac{dx}{dt}): [ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}. ]

Using the previous example:

1. Differentiate (\frac{dy}{dx} = \frac{3}{2t + 2}) with respect to (t): [ \frac{d}{dt}\left(\frac{3}{2t + 2}\right) = \frac{-6}{(2t + 2)^2}. ]

2. Divide by (\frac{dx}{dt}): [ \frac{d^2y}{dx^2} = \frac{\frac{-6}{(2t + 2)^2}}{2t + 2} = \frac{-6}{(2t + 2)^3}. ]

Example 3: Parametric Equations Involving Trigonometric Functions

Given: [ x = \cos(t), \quad y = \sin(t), ] find (\frac{dy}{dx}).

1. Differentiate (x) with respect to (t): [ \frac{dx}{dt} = -\sin(t). ]

2. Differentiate (y) with respect to (t): [ \frac{dy}{dt} = \cos(t). ]

3. Divide (\frac{dy}{dt}) by (\frac{dx}{dt}): [ \frac{dy}{dx} = \frac{\cos(t)}{-\sin(t)} = -\cot(t). ]

Parametric differentiation is especially useful in physics and engineering, where variables are often described in terms of a parameter, such as time. This method allows for analyzing the rate of change and curvature of parametric curves.