What is the value of 4a cos3θ(d2y/dx2) if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)? 3/2 b) 3/2 c) 1/2 d) 1/2

What is the value of 4a cos3θ(d2y/dx2) if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)? 3/2 b) 3/2 c) 1/2 d) 1/2 Correct Answer 3/2

Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ) => x = 2a cos2θ*sin2θ and y = 2a sin2θ*cos2θ Now differentiating x and y with respect to θ we get, dx/dθ = 2a = 4a cosθ (cosθ cos2θ – sinθ sin2θ) = 4a cosθ cos(θ + 2θ) = 4a cosθ cos3θ dy/dθ = 2a = 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, (d2y/dx2) = d/dx(tanθ) = sec2θ*dθ/dx = sec2θ*(1/(dx/dθ)) = sec2θ*1/(4a cosθ cos3θ) Or, 4a cos3θ (d2y/dx2) = sec3θ = (sec2θ)3/2 = (1 + tan2θ)3/2 As, dy/dx = tanθ So, 4a cos3θ(d2y/dx2) = 3/2