Differentiation of Parametric Functions

1. Statement of the Concept

If both [x] and [y] are expressed as functions of another variable (parameter) [t], i.e.,
[x = f(t)] and [y = g(t)],
then the derivative of [y] with respect to [x] is:

[\dfrac{d}{dx}(y)][ = \dfrac{dy/dt}{dx/dt}], provided [\dfrac{dx}{dt} \ne 0].

2. Explanation and Derivation

Since [y = g(t)], differentiating both sides with respect to [t]:

[\dfrac{dy}{dt} = g'(t)]

Similarly, differentiating [x = f(t)] with respect to [t]:

[\dfrac{dx}{dt} = f'(t)]

Using the Chain Rule:

[\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}]

Thus, parametric differentiation converts the problem into two simpler derivatives with respect to [t].

3. Key Features

1. Useful when relations between x and y are complicated.
2. Helps in differentiating curves defined by non-functional relations.
3. Makes geometry-based curves easy (circle, ellipse, cycloid, etc.).
4. Simplifies velocity/acceleration in motion problems.
5. Allows slope calculation even for multi-valued curves.

4. Important Formulas to Remember

5. Solved Examples (Expanded)

Example 1

Find the derivative of [y] w.r.t [x]
[x = t^2 + 3], [y = 2t + 1]

Step-by-Step Solution:

1. [\dfrac{dx}{dt} = 2t]
2. [\dfrac{dy}{dt} = 2]
3. [\dfrac{dy}{dx}][ = \dfrac{2}{2t}][ = \dfrac{1}{t}]

Conclusion: [\dfrac{d}{dx}(y) = \dfrac{1}{t}]

Example 2

Find the derivative of [y] w.r.t [x]
[x = \sin t], [y = \cos t]

Steps:

1. [\dfrac{dx}{dt} = \cos t]
2. [\dfrac{dy}{dt} = -\sin t]
3. [\dfrac{dy}{dx} = \dfrac{-\sin t}{\cos t}][ = -\tan t]

Example 3

Find the derivative of [y] w.r.t [x]
[x = e^t], [y = e^{2t}]

Steps:

1. [\dfrac{dx}{dt} = e^t]
2. [\dfrac{dy}{dt} = 2e^{2t}]
3. [\dfrac{dy}{dx}][ = \dfrac{2e^{2t}}{e^t}][ = 2e^t]

Example 4

Find the derivative of [y] w.r.t [x]
[x = t^3], [y = t^2 + 4]

Steps:

1. [\dfrac{dx}{dt} = 3t^2]
2. [\dfrac{dy}{dt} = 2t]
3. [\dfrac{dy}{dx} = \dfrac{2t}{3t^2} = \dfrac{2}{3t}]

Example 5

Find the derivative of [y] w.r.t [x]
[x = a\cos t], [y = a\sin t]

Steps:

1. [\dfrac{dx}{dt} = -a\sin t]
2. [\dfrac{dy}{dt} = a\cos t]
3. [\dfrac{dy}{dx}][ = \dfrac{a\cos t}{-a\sin t}][ = -\cot t]

Example 6

Find the second order derivative of [y] w.r.t [x]
[x = t^2], [y = t^3]

Steps:

1. [\dfrac{dx}{dt} = 2t]
2. [\dfrac{dy}{dt} = 3t^2]
3. [\dfrac{dy}{dx}][ = \dfrac{3t^2}{2t}][ = \dfrac{3t}{2}]
4. Differentiate again w.r.t. t:
   [\dfrac{d}{dt}(\dfrac{3t}{2})][ = \dfrac{3}{2}]
5. Apply formula:
   [\dfrac{d^2y}{dx^2} ][= \dfrac{\dfrac{3}{2}}{2t} ][= \dfrac{3}{4t}]

7. Conceptual Questions with Solutions

8. FAQ / Common Misconceptions