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Question Number 49877 by ajfour last updated on 11/Dec/18

Commented by ajfour last updated on 11/Dec/18

$$Findmaximumareainblue.$$

Answered by mr W last updated on 12/Dec/18

$$\frac{\sin \alpha }{R}=\frac{\sin \theta }{l}$$$$\Rightarrow \sin \alpha =\frac{R}{l}\sin \theta =\frac{\sin \theta }{\lambda }$$$$\cos \alpha \frac{d\alpha }{d\theta }=\frac{1}{\lambda }\cos \theta$$$$A_{blue}=\frac{Rl}{2}\sin (\alpha +\theta )-\frac{R^2\theta }{2}$$$$A=\frac{R^2}{2}[\frac{l}{R}\sin (\alpha +\theta )-\theta ]$$$$A=\frac{R^2}{2}[\lambda \sin (\alpha +\theta )-\theta ]=\frac{R^2}{2}f\left(\theta \right)$$$$f\left(\theta \right)=\lambda \sin (\alpha +\theta )-\theta$$$$\frac{df\left(\theta \right)}{d\theta }=\lambda \cos (\alpha +\theta )(\frac{d\alpha }{d\theta }+1)-1=\lambda \cos (\alpha +\theta )(\frac{1}{\lambda \cos \alpha }\cos \theta +1)-1=0$$$$\Rightarrow \cos (\alpha +\theta )(\frac{\cos \theta }{\cos \alpha }+\lambda )-1=0$$$$\Rightarrow (\cos \alpha \cos \theta -\sin \alpha \sin \theta )(\frac{\cos \theta }{\cos \alpha }+\lambda )-1=0$$$$\Rightarrow (\cos \theta -\tan \alpha \sin \theta )(\cos \theta +\lambda \cos \alpha )-1=0$$$$\Rightarrow (\cos \theta -\frac{{\sin }^2\theta }{\sqrt{{\lambda }^2-{\sin }^2\theta }})(1+\frac{\sqrt{{\lambda }^2-{\sin }^2\theta }}{{\lambda }^2})-1=0$$$$\Rightarrow \theta =.....$$$$with\lambda =2/1=2,\Rightarrow \theta =47.06°\Rightarrow A=0.5199$$

Commented by ajfour last updated on 11/Dec/18

$$ThanksSir!$$

Answered by MJS last updated on 11/Dec/18

$$R=1$$$$A\left(p\right)=\mathrm{blue}\mathrm{area}=\mathrm{triangle}-\mathrm{part}\mathrm{of}\mathrm{circle}$$$$q=\sqrt{l^2-{\left(\sqrt{1-p^2}\right)}^2}=\sqrt{l^2+p^2-1}$$$$A\left(p\right)=\frac{1}{2}q\sqrt{1-p^2}-\underset{p}{\stackrel{1}{\int }}\sqrt{1-x^2}dx$$$$A\left(p\right)=\frac{1}{2}((p+\sqrt{l^2+p^2-1})\sqrt{1-p^2}-\arccos p)$$$$A^{\prime }\left(p\right)=-\frac{1}{2}\times \frac{2p^3+(l^2-2)p+2(p^2-1)\sqrt{l^2+p^2-1}}{\sqrt{1-p^2}\sqrt{l^2+p^2-1}}$$$$A^{\prime }\left(p\right)=0$$$$2p^3+(l^2-2)p+2(p^2-1)\sqrt{l^2+p^2-1}=0$$$$\Rightarrow p^4+\frac{l^4+4l^2-8}{4}p-l^2+1=0$$$$p=\sqrt{1-\frac{l^4}{8}-\frac{l^2}{2}+\frac{l^3\sqrt{l^2+8}}{8}}$$$$A\left(p\right)\mathrm{can}\mathrm{be}\mathrm{calculated}\mathrm{for}\mathrm{given}l$$

Commented by ajfour last updated on 11/Dec/18

$$ThankyouSir,quiteamazingway.$$$$whatifl=2,R=1?$$

Commented by MJS last updated on 11/Dec/18

$$l=2\Rightarrow p=\sqrt{2\sqrt{3}-3}\approx .681250$$$$A\left(p\right)\approx .519941$$$$\mathrm{which}\mathrm{values}\mathrm{do}\mathrm{you}\mathrm{get}?$$

Commented by mr W last updated on 12/Dec/18

$$IgotthesamevalueA\approx 0.5199.$$

Answered by ajfour last updated on 12/Dec/18

$$letR\sin \theta =l\sin \varphi =y$$$$\Rightarrow R\cos \theta =l\cos \varphi \frac{d\varphi }{d\theta }$$$$let\frac{l}{R}=\lambda \Rightarrow$$$$\lambda \frac{d\varphi }{d\theta }=\frac{\cos \theta }{\cos \varphi }\mathrm{\&}\sin \varphi =\frac{\sin \theta }{\lambda }...\left(i\right)$$$$A=\frac{R\sin \theta }{2}(R\cos \theta +l\cos \varphi )-\frac{R^2\theta }{2}$$$$2A=\sin \theta (\cos \theta +\lambda \cos \varphi )-\theta$$$$\frac{d\left(2A\right)}{d\theta }=\cos \theta (\cos \theta +\lambda \cos \varphi )$$$$-\sin \theta (\sin \theta +\lambda \sin \varphi \frac{d\varphi }{d\theta })-1$$$$\frac{dA}{d\theta }=0\Rightarrow$$$$\cos {}^2\theta +\lambda \cos \theta \cos \varphi =1+\sin {}^2\theta$$$$+\sin \theta \sin \varphi \frac{\cos \theta }{\cos \varphi }$$$$\Rightarrow 2\sin {}^2\theta =\cos \theta (\lambda \cos \varphi -\frac{\sin {}^2\theta }{\lambda \cos \varphi })$$$$\Rightarrow 4\sin {}^4\theta$$$$=\cos {}^2\theta ({\lambda }^2\cos {}^2\varphi +\frac{\sin {}^4\theta }{{\lambda }^2\cos {}^2\varphi }-2\sin {}^2\theta )$$$$let\sin {}^2\theta =t$$$$\Rightarrow 4t^2=(1-t)({\lambda }^2-t+\frac{t^2}{{\lambda }^2-t}-2t)$$$$\Rightarrow 4t^2({\lambda }^2-t)=(1-t)(4t^2-4{\lambda }^{2}t+{\lambda }^4)$$$$\Rightarrow 4t^2{\lambda }^2-4t^3=4t^2-4{\lambda }^{2}t+{\lambda }^4$$$$-4t^3+4{\lambda }^2{t}^2-{\lambda }^{4}t$$$$\Rightarrow 4t^2-{\lambda }^2(4+{\lambda }^2)t+{\lambda }^4=0$$$$t=\frac{{\lambda }^2(4+{\lambda }^2)\pm \sqrt{{\lambda }^4{(4+{\lambda }^2)}^2-16{\lambda }^4}}{8}$$$$for\lambda =2$$$$\sin \theta =\sqrt{4-2\sqrt{3}}$$$$$$

Commented by ajfour last updated on 12/Dec/18

$$ThanksforcheckingitSir!$$

Commented by mr W last updated on 12/Dec/18

$$\sin \theta =\sqrt{4-2\sqrt{3}}\Rightarrow \theta =47.0586°\Rightarrow A=0.5199$$$$thisiscorrectsir.$$

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