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Question Number 45187 by MrW3 last updated on 10/Oct/18

Commented by MrW3 last updated on 12/Oct/18

$$SolutiontoQ45122.$$$$FindthelengthofpathfromAtoB$$$$whichhasaconstantslope.$$$$$$$$ApointPonthepathcanbedescribed$$$$byparameters(\varphi ,\phi )with$$$$0⩽\varphi ⩽2\pi ,0⩽\phi ⩽\theta .$$$$$$$$z=R\sin \phi$$$$r=R\cos \phi$$$$$$$$theslopekofthepathisdefinedas$$$$ratiofromchangeinheighttochange$$$$inhorizontaldistancebetweentwo$$$$points:$$$$k=\tan \alpha =\frac{\mathrm{\Delta }h}{\mathrm{\Delta }l}$$$$let\lambda =\frac{1}{k}$$$$$$$$\mathrm{\Delta }h=dz=R\cos \phi d\phi$$$$\mathrm{\Delta }l=\sqrt{{\left(\mathrm{\Delta }{l}_{\varphi }\right)}^2+{\left(\mathrm{\Delta }{l}_r\right)}^2}$$$$\mathrm{\Delta }{l}_{\varphi }=rd\varphi =R\cos \phi d\varphi =R\cos \phi \frac{d\varphi }{d\phi }d\phi$$$$\mathrm{\Delta }{l}_r=dr=-R\sin \phi d\phi$$$$\Rightarrow k=\frac{R\cos \phi d\phi }{R\sqrt{{\cos }^2\phi {\left(\frac{d\varphi }{d\phi }\right)}^2{\left(d\phi \right)}^2+{\sin }^2\phi {\left(d\phi \right)}^2}}$$$$\Rightarrow k=\frac{\cos \phi }{\sqrt{{\cos }^2\phi {\left(\frac{d\varphi }{d\phi }\right)}^2+{\sin }^2\phi }}=\frac{1}{\sqrt{{\left(\frac{d\varphi }{d\phi }\right)}^2+{\tan }^2\phi }}$$$$\Rightarrow {\left(\frac{d\varphi }{d\phi }\right)}^2+{\tan }^2\phi =\frac{1}{k^2}={\lambda }^2$$$$\Rightarrow \frac{d\varphi }{d\phi }=\sqrt{{\lambda }^2-{\tan }^2\phi }$$$$\Rightarrow \varphi =\sqrt{1+{\lambda }^2}{\tan }^{-1}\left(\frac{\sqrt{1+{\lambda }^2}\tan \phi }{\sqrt{{\lambda }^2-{\tan }^2\phi }}\right)-{\sin }^{-1}\left(\frac{\tan \phi }{\lambda }\right)+C$$$$at\varphi =0:\phi =0\Rightarrow C=0$$$$\Rightarrow \varphi =\sqrt{1+{\lambda }^2}{\tan }^{-1}\left(\frac{\sqrt{1+{\lambda }^2}\tan \phi }{\sqrt{{\lambda }^2-{\tan }^2\phi }}\right)-{\sin }^{-1}\left(\frac{\tan \phi }{\lambda }\right)$$$$$$$$at\phi =\theta :\varphi =2\pi$$$$\Rightarrow 2\pi =\sqrt{1+{\lambda }^2}{\tan }^{-1}\left(\frac{\sqrt{1+{\lambda }^2}\tan \theta }{\sqrt{{\lambda }^2-{\tan }^2\theta }}\right)-{\sin }^{-1}\left(\frac{\tan \theta }{\lambda }\right)$$$$\Rightarrow weget\lambda =...intermsof\theta$$$$\left(thisisonlynumericallypossible\right)$$$$$$$$thelengthofpathfromAtoB:$$$$s=\frac{\sqrt{1+k^2}}{k}h=\sqrt{1+{\lambda }^2}h$$$$h=R\sin \theta$$$$\Rightarrow s=R\sqrt{1+{\lambda }^2}\sin \theta$$$$$$$$example:$$$$\theta =60°$$$$\Rightarrow \lambda =6.0546\Rightarrow k=0.1652\Rightarrow \alpha =9.38°$$$$\Rightarrow s=5.3145R$$$$$$$$limitcase:$$$$from$$$$2\pi =\sqrt{1+{\lambda }^2}{\tan }^{-1}\left(\frac{\sqrt{1+{\lambda }^2}\tan \phi }{\sqrt{{\lambda }^2-{\tan }^2\phi }}\right)-{\sin }^{-1}\left(\frac{\tan \phi }{\lambda }\right)$$$$weseethelimitcaseiswhen\lambda \to \tan \theta ,$$$$thenwehave$$$$2\pi =\sqrt{1+{\lambda }^2}\frac{\pi }{2}-\frac{\pi }{2}$$$$\sqrt{1+{\lambda }^2}=5$$$$\Rightarrow {\lambda }_{max}=\tan {\theta }_{max}=\sqrt{24}$$$$\Rightarrow {\theta }_{max}={\tan }^{-1}\sqrt{24}=78.46°$$$$\Rightarrow k_{max}=\frac{1}{\sqrt{24}}=0.204$$$$\Rightarrow {\alpha }_{max}={\tan }^{-1}\left(\frac{1}{\sqrt{24}}\right)=11.54°$$

Commented by MJS last updated on 10/Oct/18

$$\mathrm{great}!$$$$\mathrm{another}\mathrm{aspect}(\mathrm{sorry}\mathrm{if}\mathrm{it}\mathrm{seems}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{always}$$$$\mathrm{looking}\mathrm{for}\mathrm{problems},\mathrm{borders}\mathrm{or}\mathrm{mistakes},$$$$\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{really}\mathrm{interested})$$$$\mathrm{you}\mathrm{cannot}\mathrm{reach}\mathrm{some}\mathrm{regions}\mathrm{above}\mathrm{a}\mathrm{certain}$$$$\theta \mathrm{within}\mathrm{a}\mathrm{turn}\mathrm{of}2\pi \mathrm{because}\mathrm{the}\mathrm{slope}\mathrm{is}$$$$\mathrm{limited}\mathrm{by}\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{sphere}.$$$$\mathrm{Can}\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{maximum}\theta \mathrm{for}\mathrm{this}\mathrm{case}?$$

Commented by ajfour last updated on 10/Oct/18

$$WonderfulanalysisSir,$$$$buthowwastheintegrationdone$$$$Sir?$$$${\int }_0^{\theta }\sqrt{{\lambda }^2-\tan {}^2\phi }d\phi$$

Commented by MJS last updated on 10/Oct/18

$$...\mathrm{you}\mathrm{cannot}\mathrm{reach}\mathrm{the}\mathrm{peak}\mathrm{with}\mathrm{a}\mathrm{constant}$$$$\mathrm{slope}...$$

Commented by MrW3 last updated on 10/Oct/18

$$MJSsir,youareabsolutelyright.$$$$withaconstantslopeonecanreach$$$$apointatmax.\theta \approx 78.5°$$$$Ihaveaddedthisintomyworking$$$$above.pleasereview,thankyousir.$$

Commented by MrW3 last updated on 10/Oct/18

$$Toajfoursir:$$$$forintegralpleaseseeQ45213.$$

Answered by ajitbh last updated on 10/Oct/18

$$letmetake\theta inquestionas\mathit{\alpha }.$$$$\frac{dz}{rd\varphi }=\frac{R\cos \theta d\theta }{R\cos \theta d\varphi }=\mathit{k}$$$$\Rightarrow {\int }_0^{\alpha }d\theta ={\int }_0^{2\pi }\mathit{k}d\varphi$$$$\Rightarrow \mathit{k}=\mathit{\alpha }/2\mathit{\pi }$$$$\int \mathit{d}\mathit{s}=\int \sqrt{{\left(R\cos \theta d\varphi \right)}^2+{\left(R\cos \theta d\theta \right)}^2}$$$$={\int }_0^{\alpha }\sqrt{1+\frac{1}{k^2}}\left(R\cos \theta \right)d\theta$$$$\mathit{s}=\mathit{R}\sqrt{1+\frac{4{\mathit{\pi }}^2}{{\mathit{\alpha }}^2}}\sin \mathit{\alpha }.$$

Commented by ajitbh last updated on 10/Oct/18

$$Sir,mytinkutaraappiscrashing$$$$ichangedid,yettheproblem$$$$persists..$$

Commented by MrW3 last updated on 10/Oct/18

$$sir,Isee,hopetheappproblemwillbe$$$$solvedsoon.$$

Commented by MrW3 last updated on 10/Oct/18

$$onequestiontoyoursolution:$$$$in\frac{dz}{rd\varphi }=\frac{R\cos \theta d\theta }{R\cos \theta d\varphi }=\mathit{k}$$$$youassumedthatrisconstant.but$$$$ifrisnotconstant,asinthiscase,$$$$dl\ne rd\varphi ,butdl=\sqrt{{\left(rd\varphi \right)}^2+{\left(dr\right)}^2},as$$$$weknowfrompolarcoordinatesystem$$$$whencalculatinglengthofcurve.$$$$canyoupleasechecksir.thankyou.$$

Commented by ajfour last updated on 10/Oct/18

$$checkedsir,butthesolutioncant$$$$bedeletedverysoon..!$$

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