Question
If x = a cos³ θ, y = a sin³ θ, then dy/dx at any θ
is:Solution
We are given the parametric equations: · x = a·cos³θ · y = a·sin³θ We are to find dy/dx in terms of θ. First, find dx/dθ and dy/dθ : · dx/dθ = a · d/dθ (cos³θ) = a · 3cos²θ (–sinθ) = –3a cos²θ sinθ · dy/dθ = a · d/dθ (sin³θ) = a · 3sin²θ (cosθ) = 3a sin²θ cosθ Now: dy/dx = (dy/dθ) / (dx/dθ) = [3a sin²θ cosθ] / [–3a cos²θ sinθ] Cancel 3a: = (sin²θ cosθ) / (–cos²θ sinθ) = – (sinθ / cosθ) = –tan θ
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