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Question Number 65367 by ajfour last updated on 29/Jul/19

Commented by ajfour last updated on 29/Jul/19

$$\mathrm{For}\mathrm{maximum}\mathrm{arc}\mathrm{length}\mathrm{of}\mathrm{an}$$$$\mathrm{origin}\mathrm{centered}\mathrm{circle}\mathrm{within}$$$$\mathrm{the}\mathrm{shown}\mathrm{unit}\mathrm{circle},\mathrm{what}\mathrm{is}$$$$\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}.$$

Answered by ajfour last updated on 29/Jul/19

$$x=r\cos \theta ,y=r\sin \theta$$$${(r\cos \theta -1)}^2+{(r\sin \theta -1)}^2=1$$$$\Rightarrow r^2-2\sqrt{2}r\cos (\theta -\frac{\pi }{4})+1=0$$$$L=2r\mid \theta -\frac{\pi }{4}\mid =2r{\cos }^{-1}\left(\frac{r^2+1}{2\mathit{r}\sqrt{2}}\right)$$$$\frac{dL}{dr}=0\Rightarrow$$$${\cos }^{-1}\left(\frac{r^2+1}{2\mathit{r}\sqrt{2}}\right)=\frac{r^2-1}{\sqrt{8r^2-{(r^2+1)}^2}}$$$$\Rightarrow r_0\approx 1.67376$$

Answered by mr W last updated on 29/Jul/19

$$lowerintersectionpointP:$$$$x_P=r\cos \alpha$$$$y_P=r\sin \alpha$$$${(r\cos \alpha -1)}^2+{(r\sin \alpha -1)}^2=1$$$$r^2-2r(\cos \alpha +\sin \alpha )+1=0$$$$\Rightarrow \cos \alpha +\sin \alpha =\frac{r^2+1}{2r}$$$$\Rightarrow \sin 2\alpha ={\left(\frac{r^2+1}{2r}\right)}^2-1={\left(\frac{r^2-1}{2r}\right)}^2=\cos (\frac{\pi }{2}-2\alpha )$$$$\Rightarrow \frac{\pi }{2}-2\alpha ={\cos }^{-1}{\left(\frac{r^2-1}{2r}\right)}^2$$$$L=r(\frac{\pi }{2}-2\alpha )=r{\cos }^{-1}{\left(\frac{r^2-1}{2r}\right)}^2$$$$\frac{dL}{dr}={\cos }^{-1}{\left(\frac{r^2-1}{2r}\right)}^2-\frac{r^4-1}{2r^2\sqrt{1-{\left(\frac{r^2-1}{2r}\right)}^4}}=0$$$$\Rightarrow {\cos }^{-1}{\left(\frac{r^2-1}{2r}\right)}^2=\frac{r^4-1}{2r^2\sqrt{1-{\left(\frac{r^2-1}{2r}\right)}^4}}$$$$\Rightarrow r=1.6738$$$$\Rightarrow L_{max}=2.1374$$

Commented by ajfour last updated on 29/Jul/19

$$thankssir.$$$$pleaseattemptthenewerone..$$

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