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Question Number 44120 by peter frank last updated on 21/Sep/18

Answered by ajfour last updated on 21/Sep/18

$$\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{a\sin \theta }{a+a\cos \theta }=\frac{\sin \theta }{1+\cos \theta }$$$$\frac{d^{2}y}{{dx}^2}=\left[\frac{d}{d\theta }\left(\frac{dy}{dx}\right)\right]/\left(\frac{dx}{d\theta }\right)$$$$=\frac{\cos \theta (1+\cos \theta )-\sin \theta (-\sin \theta )}{{(1+\cos \theta )}^2\times a(1+\cos \theta )}$$$$=\frac{\cos \theta +1}{{(1+\cos \theta )}^2\times a(1+\cos \theta )}$$$$\Rightarrow \frac{d^{2}y}{{dx}^2}=\frac{1}{a{(1+\cos \theta )}^2}$$$$(\mathrm{\forall }\theta \ne 2n\pi +\pi ).$$

Commented by peter frank last updated on 21/Sep/18

$$thesecondlineamnotunderstoodwhyyoudivide\frac{dx}{d\theta }?$$$$$$

Commented by ajfour last updated on 21/Sep/18

$$\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)\times \frac{d\theta }{dx}$$$$=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)/\left(\frac{dx}{d\theta }\right).$$

Commented by peter frank last updated on 21/Sep/18

$$\mathrm{okay}\mathrm{sir}\mathrm{thank}\mathrm{you}$$

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