Question

What is the value of 4a cos3θ(d^2y/dx^2) if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)?

(a) [1 + (d^2y/dx^2)^2]^3/2

(b) [1 – (d^2y/dx^2)^2]^3/2

(c) [1 + (d^2y/dx^2)^2]^1/2

(d) [1 – (d^2y/dx^2)^2]^1/2

I got this question during an interview.

My doubt is from Second Order Derivative topic in portion Limits and Derivatives of Mathematics – Class 11

Answer 1

The correct option is (a) [1 + (d^2y/dx^2)^2]^3/2

For explanation I would say: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)

=> x = 2a cos^2θ*sin2θ and y = 2a sin^2θ*cos2θ

Now differentiating x and y with respect to θ we get,

dx/dθ = 2a[cos^2θ*2cos2θ + sin2θ*2cos2θ cos2θ]

= 4a cosθ (cosθ cos2θ – sinθ sin2θ)

= 4a cosθ cos(θ + 2θ)

= 4a cosθ cos3θ

dy/dθ = 2a[cos2θ*2cosθ sinθ + sin^2θ (-2sin2θ)]

= 4a sinθ (cosθ cos2θ – sinθ sin2θ )

= 4a sinθ cos(θ + 2θ)

= 4a sinθ cos3θ

Thus, dy/dx = (dy/dθ)/(dx/dθ)

= (= 4a cosθ cos3θ)/( 4a sinθ cos3θ)

= tanθ

So, (d^2y/dx^2) = d/dx(tanθ)

= sec^2θ*dθ/dx

= sec^2θ*(1/(dx/dθ))

= sec^2θ*1/(4a cosθ cos3θ)

Or, 4a cos3θ (d^2y/dx^2) = sec^3θ

= (sec^2θ)^3/2

= (1 + tan2θ)^3/2

As, dy/dx = tanθ

So, 4a cos3θ(d^2y/dx^2) = [1 + (d^2y/dx^2)^2]^3/2