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,$\frac{dx}{dθ}=a[\frac{d}{dθ}(cosθ)+\frac{d}{dθ}(θsinθ)=a[-sinθ+θ\frac{d}{dθ}(sinθ)+sinθ\frac{d}{dθ}(θ)$
$=a[-sinθ+θcosθ+sinθ]=aθcosθ$
$\frac{dy}{dθ}=a[\frac{d}{dθ}(sinθ)-\frac{d}{dθ}(θcosθ)=a[cosθ-{θ\frac{d}{dθ}(cosθ)+cosθ.\frac{d}{dθ}(θ)}]$
$=a[cosθ+θsinθ-cosθ]=aθsinθ$
$∴\frac{dy}{dx}=\frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})}=\frac{aθsinθ}{aθcosθ}=tanθ$