Question

If \(x\) and \(y\) are connected parametrically by the equation,without eliminating the parameter,find \(\frac{dy}{dx}\).
\(x=cos\,θ-cos2θ,y=sin\,θ-sin2θ\)

Answer 1

The correct answer is \(\frac{cos\,θ-2cos\,2θ}{2sin\,2θ-sin\,θ}\)
The given equations are \(x=cos\,θ-cos2θ,y=sin\,θ-sin2θ\)
Then,\(\frac{dx}{dθ}=\frac{d}{dθ}(cos\,θ-cos\,2θ)=\frac{d}{dθ}(cos\,θ)-\frac{d}{dθ}(cos\,2θ)\)
\(=-sinθ-(-2sin\,2θ)=2sin\,2θ-sin\,θ\)
\(\frac{dy}{dθ}=\frac{d}{dθ}(sinθ-sin\,2θ)=\frac{d}{dθ}(sin\,θ)-\frac{d}{dθ}(sin\,2θ)\)
\(=cos\,θ-2cos\,2θ\)
\(∴\frac{dy}{dx}=\frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})}=\frac{cos\,θ-2cos\,2θ}{2sin\,2θ-sin\,θ}\)