Parametric Differentiation

Normally, when we define functions or equations of curves, we relate \( x \) and \( y \) directly, either in explicit form, such as \( y = f(x) \), or in implicit form, where \( x \) and \( y \) satisfy an equation. However, there is another possibility: instead of relating \( x \) and \( y \) directly, we may define them in terms of a third variable, called a parameter.

For example, suppose an object moves in a two-dimensional space. Its position along the \( x \)-axis may be given as \( x = f(t) \), and its position along the \( y \)-axis as \( y = g(t) \). Here, \( x \) and \( y \) are related to each other not directly, but through the parameter \( t \). If necessary, we can eliminate \( t \) to express \( y \) explicitly in terms of \( x \), but this is often impractical or too complicated to be useful.

For instance, if

\[ x = 2t, \quad y = t^2, \]

then eliminating \( t \) by solving for \( t \) in terms of \( x \),

\[ t = \frac{x}{2}, \]

substituting in \( y \),

\[ y = \left(\frac{x}{2} \right)^2 = \frac{x^2}{4}. \]

This expresses \( y \) explicitly in terms of \( x \), but in more complex cases, eliminating \( t \) is either infeasible or unnecessary. Instead, we differentiate directly in parametric form by computing \( \frac{dy}{dx} \) without eliminating \( t \).

Since \( x = f(t) \) and \( y = g(t) \), we define their derivatives as

\[ \frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}, \quad \frac{dy}{dt} = \lim_{\Delta t \to 0} \frac{\Delta y}{\Delta t}. \]

We want to compute

\[ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}. \]

Since both \( x \) and \( y \) depend on \( t \), we use the chain rule,

\[ \frac{dy}{dx} = \lim_{\Delta t \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta t \to 0} \frac{\frac{\Delta y}{\Delta t}}{\frac{\Delta x}{\Delta t}}. \]

Taking limits on both the numerator and denominator separately,

\[ \frac{dy}{dx} = \frac{\lim_{\Delta t \to 0} \frac{\Delta y}{\Delta t}}{\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}}. \]

Since these limits define the derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), we obtain the fundamental formula for parametric differentiation:

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

This allows us to compute \( \frac{dy}{dx} \) without needing to eliminate \( t \), making differentiation of parametric equations systematic and efficient.

Example

Given the parametric equations

\[ x = \theta - \sin \theta, \quad y = 1 - \cos \theta, \]

find \( \frac{dy}{dx} \).

Solution:

Using the formula for parametric differentiation,

\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}. \]

First, differentiating \( x \) with respect to \( \theta \),

\[ \frac{dx}{d\theta} = \frac{d}{d\theta} (\theta - \sin \theta) = 1 - \cos \theta. \]

Next, differentiating \( y \) with respect to \( \theta \),

\[ \frac{dy}{d\theta} = \frac{d}{d\theta} (1 - \cos \theta) = \sin \theta. \]

Now, substituting these into the formula,

\[ \frac{dy}{dx} = \frac{\sin \theta}{1 - \cos \theta}. \]

Thus,

\[ \frac{dy}{dx} = \frac{\sin \theta}{1 - \cos \theta}. \]

Differentiating a Function with Respect to Another Function

When given two functions, \( z = f(x) \) and \( y = g(x) \), we want to find the derivative of \( z \) with respect to \( y \), denoted as

\[ \frac{dz}{dy}. \]

Since both \( z \) and \( y \) are functions of \( x \), we differentiate them with respect to \( x \) and use the chain rule.

Using the definition of a derivative, we write

\[ \frac{dz}{dx} = f'(x), \quad \frac{dy}{dx} = g'(x). \]

Applying the chain rule,

\[ \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}. \]

Rearranging,

\[ \frac{dz}{dy} = \frac{\frac{dz}{dx}}{\frac{dy}{dx}} = \frac{f'(x)}{g'(x)}. \]

Thus, differentiating a function \( z = f(x) \) with respect to another function \( y = g(x) \) is given by

\[ \frac{dz}{dy} = \frac{f'(x)}{g'(x)}. \]

This result is useful when we want to express the rate of change of one function in terms of another without explicitly eliminating \( x \).

Example

Differentiate \( \sin x \) with respect to \( \cos x \).

Solution:

Let \( z = \sin x \) and \( y = \cos x \). We want to compute

\[ \frac{dz}{dy} = \frac{d}{d (\cos x)} (\sin x). \]

Using the formula

\[ \frac{dz}{dy} = \frac{\frac{dz}{dx}}{\frac{dy}{dx}}, \]

we first compute:

\[ \frac{dz}{dx} = \frac{d}{dx} \sin x = \cos x, \]

\[ \frac{dy}{dx} = \frac{d}{dx} \cos x = -\sin x. \]

Now, substituting these,

\[ \frac{dz}{dy} = \frac{\cos x}{- \sin x} = -\cot x. \]

Thus, the derivative of \( \sin x \) with respect to \( \cos x \) is

\[ \frac{d}{d (\cos x)} (\sin x) = -\cot x. \]