Question

If $x$ and $y$ are connected parametrically by the equation,without eliminating the parameter,find $\frac{dy}{dx}$.
$x=a(θ-sin\,θ),y=a(1+cos\,θ)$

Answer 1

The correct answer is $-cot\frac{θ}{2}$
The given equations are $x=a(θ-sin\,θ),y=a(1+cos\,θ)$
Then,$\frac{dx}{dθ}=a[\frac{d}{dθ}(θ)-\frac{d}{dθ}(sin\,θ)]=a(1-cos\,θ)$
$\frac{dy}{dθ}=a[\frac{d}{dθ}(1)+\frac{d}{dθ}(cos\,θ)]=a[0+(-sin\,θ)]=-asin\,θ$
∴\frac{dy}{dx}=\frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})}=\frac{-asin\,θ}{a(1-cos\,θ)}=\frac{-2sin\,\frac{θ}{2}cos\,\frac{θ}{2}}{2sin^2\frac{θ}{2}}$$=\frac{-cos\frac{θ}{2}}{sin\frac{θ}{2}}=-cot\frac{θ}{2}