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Question Number 132809 by mr W last updated on 17/Feb/21

Commented by mr W last updated on 18/Feb/21

$$amountainhastheshapeofaright$$$$circularconeasshown(h>\sqrt{3}r).$$$$frompointAtopointBtwo$$$$sightseeingroadsonetimearound$$$$themountainshouldbebuilt:$$$$roadIwiththeshortestlength.$$$$roadIIwithaconstantslope.$$$$1.findthelengthesofbothroads.$$$$2.dothebothroadscrossinthemiddle?$$$$ifyes,where?$$

Commented by bramlexs22 last updated on 17/Feb/21

$$\mathrm{s}\theta =2\pi \mathrm{r}\Rightarrow \theta =\frac{2\pi \mathrm{r}}{\mathrm{s}};\mathrm{s}=\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2}$$$$\theta =\frac{2\pi \mathrm{r}}{\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2}}.\mathrm{let}\mathrm{AB}=\mathrm{x}$$$${\mathrm{AB}}^2={(\mathrm{s}-\mathrm{x})}^2+{\mathrm{s}}^2-2\mathrm{s}.(\mathrm{s}-\mathrm{x})\cos \left(\frac{2\pi \mathrm{r}}{\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2}}\right)$$$$$$

Commented by mr W last updated on 17/Feb/21

$$thisisforroadI.thanks!$$$$givealsoatryforroadII?$$

Commented by ajfour last updated on 17/Feb/21

$$needtoknow-howthe{}^{\prime }{slope}^{\prime }in$$$${}^{\prime }constant{slope}^{\prime }isdefined;$$$$twooptions,ithink:$$$$\frac{dz}{rd\theta }(dz\ne dl);r=\sqrt{x^2+y^2}OR$$$$\frac{dz}{\sqrt{{\left(rd\theta \right)}^2+{\left(dl\right)}^2}};l=\sqrt{x^2+y^2+{(h-z)}^2}$$$$$$

Commented by mr W last updated on 17/Feb/21

$$definitionofslope:$$$$from(x,y,z)to(x+\mathrm{\Delta }x,y+\mathrm{\Delta }h,z+\mathrm{\Delta }z)$$$$slope=\frac{\mathrm{\Delta }z}{\sqrt{{\left(\mathrm{\Delta }x\right)}^2+{\left(\mathrm{\Delta }y\right)}^2}}=\frac{\mathrm{\Delta }z}{\sqrt{{\left(\mathrm{\Delta }l\right)}^2-{\left(\mathrm{\Delta }z\right)}^2}}$$$$with\mathrm{\Delta }l=\sqrt{{\left(\mathrm{\Delta }x\right)}^2+{\left(\mathrm{\Delta }y\right)}^2+{\left(\mathrm{\Delta }z\right)}^2}$$

Commented by mr W last updated on 18/Feb/21

$$couldyougowiththisdefinitionsir?$$

Answered by mr W last updated on 18/Feb/21

Commented by mr W last updated on 17/Feb/21

$$P=peakofmountain$$$$let\lambda =\frac{r}{h}$$$$\gamma ={\tan }^{-1}\frac{r}{h}={\tan }^{-1}\lambda$$$$s_A=PA$$$$s_B=PB$$$$s_A=\sqrt{h^2+r^2}=h\sqrt{1+{\lambda }^2}$$$$\frac{s_B}{\sin (\frac{\pi }{2}-\theta )}=\frac{h}{\sin (\frac{\pi }{2}-\theta +\gamma )}$$$$s_B=\frac{\cos \theta h}{\cos (\theta -\gamma )}=\frac{h}{\cos \gamma +\sin \gamma \tan \theta }$$$$s_B=\frac{h\sqrt{1+{\lambda }^2}}{1+\lambda \tan \theta }$$$$$$$$whenthesurfaceoftheconeis$$$$unfolded,theshortestroadfrom$$$$AtoBisthestraightlineAB.$$$${\stackrel{⌢}{AA}}^{\prime }=2\pi r$$$$\phi =\frac{2\pi r}{s_A}=\frac{2\pi r}{\sqrt{h^2+r^2}}=\frac{2\pi \lambda }{\sqrt{1+{\lambda }^2}}$$$$(assume\phi <\pi ,i.e.\lambda <\frac{1}{\sqrt{3}}or\gamma <30°)$$$$$$$$lengthofroad=L_I$$$$L_I=AB=\sqrt{s_A^2+s_B^2-2s_A{s}_B\cos \phi }$$$$L_I=\sqrt{h^2+r^2+\frac{h^2(1+{\lambda }^2)}{{(1+\lambda \tan \theta )}^2}-\frac{2h\sqrt{(1+{\lambda }^2)(h^2+r^2)}}{1+\lambda \tan \theta }\cos \frac{2\pi \lambda }{\sqrt{1+{\lambda }^2}}}$$$$\frac{L_I}{h}=\sqrt{(1+{\lambda }^2)[1+\frac{1}{{(1+\lambda \tan \theta )}^2}-\frac{2}{1+\lambda \tan \theta }\cos \frac{2\pi \lambda }{\sqrt{1+{\lambda }^2}}]}$$$$$$$$example:$$$$\lambda =\frac{1}{3},\theta =45°$$$$\frac{L_I}{h}=\frac{5}{12}\sqrt{10(1-\frac{24}{25}\cos \frac{\sqrt{10}\pi }{5})}\approx 1.552354$$$$$$$$weseebyroadI,thepointBisnot$$$$alwaysthehighestpointontheroad.$$$$thehighestpointontheroadhasthe$$$$closestdistancetothepeakP,this$$$$isthepointM.$$$$\frac{\sin (\phi +\alpha )}{\sin \alpha }=\frac{s_A}{s_B}$$$$\frac{\sin \phi }{\tan \alpha }+\cos \phi =1+\lambda \tan \theta$$$$\tan \alpha =\frac{\sin \phi }{1+\lambda \tan \theta -\cos \phi }$$$$s_M=PM=s_A\sin \alpha$$$$\Rightarrow s_M=\frac{h\sqrt{1+{\lambda }^2}\sin \phi }{\sqrt{1+(1+\lambda \tan \theta )(1+\lambda \tan \theta -2\cos \phi )}}$$

Commented by MJS_new last updated on 17/Feb/21

$$\mathrm{just}\mathrm{saying},\mathrm{the}\mathrm{shortest}\mathrm{distance}\mathrm{is}\mathrm{straight}$$$$\mathrm{up}\mathrm{the}\mathrm{mountain}\mathrm{and}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{also}\mathrm{a}\mathrm{route}\mathrm{with}$$$$\mathrm{constant}\mathrm{slope};-)$$

Commented by mr W last updated on 17/Feb/21

$$theroadsshouldbearoundthe$$$$mountain,thereforethedirectway$$$$rightupfromAtoBisnotvalid:($$

Commented by mr W last updated on 17/Feb/21

Commented by mr W last updated on 19/Feb/21

$$\lambda =\frac{r}{h}$$$$s_A=\sqrt{h^2+r^2}=h\sqrt{1+{\lambda }^2}$$$$s_B=\frac{h\sqrt{1+{\lambda }^2}}{1+\lambda \tan \theta }$$$$\phi =\frac{2\pi \lambda }{\sqrt{1+{\lambda }^2}}$$$$\alpha ={\tan }^{-1}\frac{\sin \phi }{1+\lambda \tan \theta -\cos \phi }$$$$s=PC$$$$\frac{s}{\sin \alpha }=\frac{s_A}{\sin (\alpha +\delta )}$$$$\Rightarrow s=\frac{h\sqrt{1+{\lambda }^2}}{\cos \delta +\frac{\sin \delta }{\tan \alpha }}$$$$\frac{ds}{d\delta }=\frac{s(\sin \delta -\frac{\cos \delta }{\tan \alpha })}{\cos \delta +\frac{\sin \delta }{\tan \alpha }}=-\frac{s}{\tan (\alpha +\delta )}$$$$\varphi r=\delta s_A$$$$\Rightarrow \varphi =\frac{\delta \sqrt{1+{\lambda }^2}}{\lambda }$$$$z=h-s\cos \gamma =h-\frac{s}{\sqrt{1+{\lambda }^2}}$$$$\rho =s\sin \gamma =\frac{\lambda s}{\sqrt{1+{\lambda }^2}}$$$$slopek=\frac{dz}{\sqrt{{\left(d\rho \right)}^2+{\left(\rho d\varphi \right)}^2}}$$$$k=\frac{-\frac{dz}{ds}}{\sqrt{{\left(\frac{d\rho }{ds}\right)}^2+{\left(\frac{\rho \times \frac{d\varphi }{d\delta }}{\frac{ds}{d\delta }}\right)}^2}}$$$$k=\frac{\frac{1}{\sqrt{1+{\lambda }^2}}}{\sqrt{{\left(\frac{\lambda }{\sqrt{1+{\lambda }^2}}\right)}^2+{\left(\frac{\frac{\lambda s}{\sqrt{1+{\lambda }^2}}\times \frac{\sqrt{1+{\lambda }^2}}{\lambda }}{-\frac{s}{\tan (\alpha +\delta )}}\right)}^2}}$$$$\Rightarrow k=\frac{1}{\sqrt{{\lambda }^2+(1+{\lambda }^2){\tan }^2(\alpha +\delta )}}$$$$$$$$weseethemaximumslopeisat$$$$\delta =0,i.e.atpointA,orat\delta =\phi ,i.e.$$$$atpointB.$$$$k_A=\frac{1}{\sqrt{{\lambda }^2+(1+{\lambda }^2){\tan }^2\left(\alpha \right)}}\left(uphill\right)$$$$k_B=\frac{1}{\sqrt{{\lambda }^2+(1+{\lambda }^2){\tan }^2(\alpha +\phi )}}\left(downhill\right)$$$$$$$$thesmallestslopeiszerowhen$$$$\delta +\alpha =\frac{\pi }{2}$$$$thisisatthepointM,thehighest$$$$piointontheroad.$$$$$$$$example:$$$$\lambda =\frac{1}{3},\theta =45°$$$$\phi =\frac{2\pi \lambda }{\sqrt{1+{\lambda }^2}}=\frac{2\pi }{\sqrt{10}}$$$$\tan \alpha =\frac{\sin \phi }{1+\lambda \tan \theta -\cos \phi }=\frac{\sin \frac{2\pi }{\sqrt{10}}}{\frac{4}{3}-\cos \frac{2\pi }{\sqrt{10}}}$$$$k_{max}=\frac{3}{\sqrt{1+10{\left(\frac{\sin \frac{2\pi }{\sqrt{10}}}{\frac{4}{3}-\cos \frac{2\pi }{\sqrt{10}}}\right)}^2}}=1.544858$$

Commented by MJS_new last updated on 17/Feb/21

$$\mathrm{the}\mathrm{concept}\mathrm{of}\mathrm{the}\mathrm{shortest}\mathrm{distance}\mathrm{around}$$$$\mathrm{the}\mathrm{cone}\mathrm{is}\mathrm{fuzzy}...$$$$\left(1\right)\mathrm{if}B=P{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{more}‘‘\mathrm{around}\mathrm{the}{\mathrm{cone}}^{″}$$$$\left(2\right)\mathrm{if}\phi =180°{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{straight}\mathrm{up}\mathrm{over}\mathrm{the}\mathrm{peak}$$$$\left(3\right)\mathrm{if}\phi >180°\mathrm{where}\mathrm{is}\mathrm{it}?$$$$\left(4\right)\mathrm{let}B=A\mathrm{in}\mathrm{cases}\left(2\right)\mathrm{and}\left(3\right)$$$$\mathrm{this}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{criticism},{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{just}\mathrm{playing}\mathrm{around}$$$$\mathrm{with}\mathrm{it}$$$$\mathrm{if}B\mathrm{is}\mathrm{not}360°\mathrm{away}\mathrm{from}A\mathrm{it}\mathrm{works}\mathrm{better}$$$$\mathrm{because}\mathrm{we}\mathrm{could}\mathrm{go}\mathrm{the}\mathrm{other}\mathrm{direction}...$$

Commented by mr W last updated on 18/Feb/21

$$ihaveaddedarestriction:h>\sqrt{3}r.$$$$actually{i}^{\prime }malsoplayingwithit,so$$$$allareallowedtoplayintheirways,$$$$andyouinyourway.$$

Answered by mr W last updated on 18/Feb/21

Commented by mr W last updated on 18/Feb/21

Commented by mr W last updated on 19/Feb/21

$$wehavegotfromroadI:$$$$\tan \gamma =\frac{r}{h}=\lambda$$$$s_A=h\sqrt{1+{\lambda }^2}$$$$s_B=\frac{h\sqrt{1+{\lambda }^2}}{1+\lambda \tan \theta }$$$$\phi =\frac{2\pi \lambda }{\sqrt{1+{\lambda }^2}}$$$$s=PC$$$$\varphi r=\delta s_A$$$$\varphi =\frac{\delta \sqrt{1+{\lambda }^2}}{\lambda }$$$$z=C^{\prime }C=h-s\cos \gamma =h-\frac{s}{\sqrt{1+{\lambda }^2}}$$$$\rho ={OC}^{\prime }=s\sin \gamma =\frac{\lambda s}{\sqrt{1+{\lambda }^2}}$$$$slopek=\frac{dz}{\sqrt{{\left(d\rho \right)}^2+{\left(\rho d\varphi \right)}^2}}$$$$k=\frac{-\frac{dz}{ds}}{\sqrt{{\left(\frac{d\rho }{ds}\right)}^2+{\left(\frac{\rho \times \frac{d\varphi }{d\delta }}{\frac{ds}{d\delta }}\right)}^2}}$$$$k=\frac{\frac{1}{\sqrt{1+{\lambda }^2}}}{\sqrt{{\left(\frac{\lambda }{\sqrt{1+{\lambda }^2}}\right)}^2+{\left(\frac{\frac{\lambda s}{\sqrt{1+{\lambda }^2}}\times \frac{\sqrt{1+{\lambda }^2}}{\lambda }}{\frac{ds}{d\delta }}\right)}^2}}$$$$k=\frac{1}{\sqrt{{\lambda }^2+(1+{\lambda }^2){\left(\frac{s}{\frac{ds}{d\delta }}\right)}^2}}$$$$\frac{ds}{d\delta }=-\frac{s}{\sqrt{\frac{1}{1+{\lambda }^2}(\frac{1}{k^2}-{\lambda }^2)}}$$$$let\xi =\frac{1}{\sqrt{\frac{1}{1+{\lambda }^2}(\frac{1}{k^2}-{\lambda }^2)}}$$$${\int }_{s_A}^s\frac{ds}{s}=-\xi {\int }_0^{\delta }d\delta$$$$\ln s-\ln s_A=-\xi \delta$$$$s=s_A{e}^{-\xi \delta }$$$$\Rightarrow s=h\sqrt{1+{\lambda }^2}{e}^{-\xi \delta }$$$$s_B=h\sqrt{1+{\lambda }^2}{e}^{-\xi \phi }=\frac{h\sqrt{1+{\lambda }^2}}{1+\lambda \tan \theta }$$$$e^{-\xi \phi }=\frac{1}{1+\lambda \tan \theta }$$$$\xi =\frac{\ln (1+\lambda \tan \theta )}{\phi }$$$$\frac{1}{\sqrt{\frac{1}{1+{\lambda }^2}(\frac{1}{k^2}-{\lambda }^2)}}=\frac{\ln (1+\lambda \tan \theta )}{\phi }$$$$\frac{1}{\frac{1}{k^2}-{\lambda }^2}=\frac{{\ln }^2(1+\lambda \tan \theta )}{(1+{\lambda }^2){\phi }^2}$$$$\frac{1}{k^2}={\lambda }^2+\frac{(1+{\lambda }^2){\phi }^2}{{\ln }^2(1+\lambda \tan \theta )}={\lambda }^2+\frac{4{\pi }^2{\lambda }^2}{{\ln }^2(1+\lambda \tan \theta )}$$$$\Rightarrow k=\frac{1}{\lambda \sqrt{1+{\left[\frac{2\pi }{\ln (1+\lambda \tan \theta )}\right]}^2}}$$$$$$$$lengthofroad{L}_{II}$$$$L_{II}={\int }_0^{\phi }\sqrt{s^2+{\left(\frac{ds}{d\delta }\right)}^2}d\delta$$$$L_{II}=h\sqrt{1+{\lambda }^2}{\int }_0^{\phi }{e}^{-\xi \delta }\sqrt{1+{\xi }^2}d\delta$$$$\frac{L_{II}}{h}=\frac{(1-e^{-\xi \phi })\sqrt{(1+{\lambda }^2)(1+{\xi }^2)}}{\xi }$$$$\Rightarrow \frac{L_{II}}{h}=(1-\frac{1}{1+\lambda \tan \theta })\sqrt{1+{\lambda }^2+{\left[\frac{2\pi \lambda }{\ln (1+\lambda \tan \theta )}\right]}^2}$$$$or$$$$L_{II}=\frac{\sqrt{1+k^2}}{k}(s_A-s_B)\cos \gamma$$$$L_{II}=h\sqrt{1+\frac{1}{k^2}}(1-\frac{1}{1+\lambda \tan \theta })$$$$\Rightarrow \frac{L_{II}}{h}=(1-\frac{1}{1+\lambda \tan \theta })\sqrt{1+{\lambda }^2+{\left[\frac{2\pi \lambda }{\ln (1+\lambda \tan \theta )}\right]}^2}$$$$$$$$example:$$$$\lambda =\frac{1}{3},\theta =45°$$$$\frac{L_{II}}{h}=\frac{1}{12}\sqrt{10+{\left(\frac{2\pi }{\ln \frac{4}{3}}\right)}^2}\approx 1.839039058$$$$k=\frac{3}{\sqrt{1+{\left(\frac{2\pi }{\ln \frac{4}{3}}\right)}^2}}=0.137214$$

Commented by mr W last updated on 18/Feb/21

Commented by mr W last updated on 18/Feb/21

Commented by mr W last updated on 18/Feb/21

$$weseethatbothroutes{don}^{\prime }tcrossin$$$$themiddle.$$

Commented by otchereabdullai@gmail.com last updated on 23/Feb/21

$$\mathrm{the}\mathrm{supper}\mathrm{prof}!$$

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