Question:

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

Updated On: Sep 12, 2023
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Solution and Explanation

The correct answer is tanθtan\,\theta
The given equations are x=a(cosθ+θsinθ),y=a(sinθθcosθ)x=a(cosθ+θsinθ),y=a(sinθ-θcosθ)
Then,dxdθ=a[ddθ(cosθ)+ddθ(θsinθ)=a[sinθ+θddθ(sinθ)+sinθddθ(θ)\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=a[-sinθ+θcosθ+sinθ]=aθcosθ
dydθ=a[ddθ(sinθ)ddθ(θcosθ)=a[cosθθddθ(cosθ)+cosθ.ddθ(θ)]\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+θsinθcosθ]=aθsinθ=a[cosθ+θsinθ-cosθ]=aθsinθ
dydx=(dydθ)(dxdθ)=aθsinθaθcosθ=tanθ∴\frac{dy}{dx}=\frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})}=\frac{aθsinθ}{aθcosθ}=tanθ
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