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Question Number 17302 by ajfour last updated on 03/Jul/17

$$\mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}$$$$\rho =\mathrm{a}(1-\cos \theta ).$$$$\rho =\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2},\theta ={\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right).$$

Commented by ajfour last updated on 03/Jul/17

$$\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{cardioid},\mathrm{Sir}.$$$$\mathrm{easy}\mathrm{question}\mathrm{sir},\mathrm{not}\mathrm{perhaps}$$$$\mathrm{for}\mathrm{you}..\mathrm{My}\mathrm{Answer}:\mathrm{s}=8\mathrm{a}$$$$\mathrm{ds}=\sqrt{{\rho }^2+{\left(\frac{\mathrm{d}\rho }{\mathrm{d}\theta }\right)}^2}\mathrm{d}\theta .$$$${\mathrm{Book}}^{\prime }\mathrm{s}\mathrm{answer}:\mathrm{not}\mathrm{given}.$$

Commented by mrW1 last updated on 03/Jul/17

$$\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{sir}!$$

Answered by mrW1 last updated on 03/Jul/17

$$\mathrm{dL}=\sqrt{{\left(\mathrm{d}\rho \right)}^2+{\left(\rho \mathrm{d}\theta \right)}^2}=\sqrt{{\rho }^2+{\left(\frac{\mathrm{d}\rho }{\mathrm{d}\theta }\right)}^2}\mathrm{d}\theta$$$$\frac{\mathrm{d}\rho }{\mathrm{d}\theta }=\mathrm{asin}\theta$$$${\rho }^2+{\left(\frac{\mathrm{d}\rho }{\mathrm{d}\theta }\right)}^2={\mathrm{a}}^2[{(1-\cos \theta )}^2+{\sin }^2\theta ]$$$$={2\mathrm{a}}^2[1-\cos \theta ]={4\mathrm{a}}^2{\sin }^2\frac{\theta }{2}$$$$\mathrm{L}=2\times 2\mathrm{a}{\int }_0^{\pi }\sin \frac{\theta }{2}\mathrm{d}\theta$$$$\mathrm{L}=8\mathrm{a}{\int }_0^{\pi }\sin \frac{\theta }{2}\mathrm{d}\frac{\theta }{2}$$$$\mathrm{L}=8\mathrm{a}{[-\cos \frac{\theta }{2}]}_0^{\pi }=8\mathrm{a}$$

Commented by ajfour last updated on 03/Jul/17

$$\mathrm{yes}\mathrm{sir},\mathrm{thanks}\mathrm{for}\mathrm{confirming}.$$

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