calculus parametric curves *need help asap please!

Question

 

Find  for the curve given by x = 2 cosΘ and y = sin θ. (4 points)

answer choices

	-2sec2θ

Answer 1

dy/dx = dy/dθ/(dx/dθ) = cosθ/(-2 sin θ) = (-x/2)/2y = -x/4y

Then, d²y/dx² = (-4y + 4xy')/16y² = (-4y + 4x(-x/4y))/16y² = (-4y - x^2/y)/16y² =

(-4y^2 - x^2)/y/(16y^2) = , substituting for x and y in the numerator,

(-4 sin^2θ - 4 cos^2θ)/(16y^3) = -4/(16y^3) = -1/(4y^3) =-1/(4 sin^3θ) = -1/4 csc³ θ