Question Detail
Question
Given that
y = (3sinθ)/(2sinθ+2cosθ) and dy/dθ = K / (1+sin(2θ)), what is the value of K?
Correct Answer
Option A is the correct answer.
Detailed Explanation
Given
y = (3sinθ)/(2sinθ+2cosθ), we first find dy/dθ using the quotient rule (u/v)' = (u'v - uv')/v².Let u = 3sinθ and v = 2sinθ+2cosθ.- Find the derivatives of
uandvwith respect toθ:u' = d/dθ(3sinθ) = 3cosθv' = d/dθ(2sinθ+2cosθ) = 2cosθ - 2sinθ - Apply the quotient rule:
dy/dθ = [ (3cosθ)(2sinθ+2cosθ) - (3sinθ)(2cosθ-2sinθ) ] / (2sinθ+2cosθ)²dy/dθ = [ (6sinθcosθ + 6cos²θ) - (6sinθcosθ - 6sin²θ) ] / (2(sinθ+cosθ))²dy/dθ = [ 6sinθcosθ + 6cos²θ - 6sinθcosθ + 6sin²θ ] / (4(sinθ+cosθ)²)dy/dθ = [ 6cos²θ + 6sin²θ ] / (4(sinθ+cosθ)²)dy/dθ = 6(cos²θ + sin²θ) / (4(sinθ+cosθ)²) - Use the identity
cos²θ + sin²θ = 1and(sinθ+cosθ)² = sin²θ + cos²θ + 2sinθcosθ = 1 + sin(2θ):dy/dθ = 6(1) / (4(1 + sin(2θ)))dy/dθ = 6 / (4(1 + sin(2θ)))dy/dθ = 3 / (2(1 + sin(2θ))) - Compare this with the given form
dy/dθ = K / (1+sin(2θ)):K / (1+sin(2θ)) = 3 / (2(1 + sin(2θ)))Therefore,K = 3/2 = 1.5.
Hint
Use the quotient rule to find
dy/dθ. Simplify the resulting expression using trigonometric identities like sin²θ + cos²θ = 1 and the expansion of (sinθ + cosθ)² to relate it to sin(2θ).