Question

If \(x\) and \(y\) are connected parametrically by the equation,without eliminating the parameter,find \(\frac{dy}{dx}\).
\(x=a(cosθ+θsinθ),y=a(sinθ-θcosθ)\)

Answer 1

The correct answer is \(tan\,\theta\)
The given equations are \(x=a(cosθ+θsinθ),y=a(sinθ-θcosθ)\)
Then,\(\frac{dx}{dθ}=a[\frac{d}{dθ}(cosθ)+\frac{d}{dθ}(θsinθ)=a[-sinθ+θ\frac{d}{dθ}(sinθ)+sinθ\frac{d}{dθ}(θ)\)
\(=a[-sinθ+θcosθ+sinθ]=aθcosθ\)
\(\frac{dy}{dθ}=a[\frac{d}{dθ}(sinθ)-\frac{d}{dθ}(θcosθ)=a[cosθ-{θ\frac{d}{dθ}(cosθ)+cosθ.\frac{d}{dθ}(θ)}]\)
\(=a[cosθ+θsinθ-cosθ]=aθsinθ\)
\(∴\frac{dy}{dx}=\frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})}=\frac{aθsinθ}{aθcosθ}=tanθ\)