Use app×
JEE MAIN 2024
JEE MAIN 2025 Foundation Course
NEET 2024 Crash Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
357 views
in Derivatives by (25.7k points)
closed by

If x = a (1 – cos θ), y =a (θ + sin θ), prove that \(\frac{d^2y}{dx^2}=-\frac{1}{a}\) at θ\(=\frac{\pi}{2}\)

1 Answer

+1 vote
by (27.7k points)
selected by
 
Best answer

 Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :\(\frac{dy}{dx}\)\(=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

Given,

y = a (θ + sin θ) ……equation 1

x = a (1– cos θ) ……equation 2

to prove :  \(\frac{d^2y}{dx^2}=-\frac{1}{a}\) at θ\(=\frac{\pi}{2}\) 

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

 Let’s find \(\frac{d^2y}{dx^2}\)

As, \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\)

So, lets first find dy/dx using parametric form and differentiate it again.

Similarly,

Differentiating again w.r.t x :

Using product rule and chain rule of differentiation together:

As we have to find\(\frac{d^2y}{dx^2}=-\frac{1}{a}\) at \(\theta=\frac{\pi}{2}\)

∴ put θ = π/2 in above equation:

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...