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Differentiation of Function in Parametric Form - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Differentiation of Function in Parametric Form is considered one of the most asked concept.
25 Questions around this concept.

Solve by difficulty

$\frac{d^{2}x}{dy^{2}}$ equals to

A

$\left ( \frac{d^{2}y}{dx^{2}} \right )\left ( \frac{dy}{dx} \right )^{-2}\;$

B

$\; \; -\left ( \frac{d^{2}y}{dx^{2}} \right )\left ( \frac{dy}{dx} \right )^{-3}\;$

C

$\; \; \left ( \frac{d^{2}y}{dx^{2}} \right )^{-1}\;$

D

$\; -\left ( \frac{d^{2}y}{dx^{2}} \right )^{-1}\left ( \frac{dy}{dx} \right )^{-3}$

Concepts Covered - 1

Differentiation of Function in Parametric Form

Differentiation of Function in Parametric Form

Sometimes, x and y are given as functions of a single variable, i.e., x = g(t) and y = f(t) are two functions and t is a variable. In such cases x and y are called parametric functions or parametric equations and t is called the parameter.

To find $\frac{dy}{dx}$ in such cases, first find the relationship between x and y by eliminating the parameter t and then differentiate with respect to t.

But sometimes it is not possible to eliminate t, then in that case use

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{f'(t)}{g'(t)}$

For example

If x = a(1 - cos Ө) and y = a(Ө - sin Ө), then dy/dx is

Solution.

$\\\mathrm{Given\;\;x=a(1-\cos\theta)\;\;and\;\;y=a(\theta+\sin\theta)}\\\Rightarrow \frac{dx}{d\theta}=a(\sin\theta)\;\;\;and\;\;\;\frac{dy}{d\theta}=a(1+\cos\theta)\\\\\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{1+\cos\theta}{\sin\theta}$

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Differentiation of Function in Parametric Form

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Differentiation of Function in Parametric Form

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 3.13

Line : 22

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