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https://texercises.com/exercise/shortest-path-on-a-cone/
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Exercise:
Given two pos AB on a cone with apex angle fracpi we aim to find the shortest path connecting those two pos. abcliste abc Asing that the path is parametrized by zmapstop zcosphi zsinphi z show that its length is _z_^z_sqrt+z^theta'^ddz abc Find the Euler-Lagrange s for this functional. Calculate a general solution. abcliste

Solution:
abcliste abc The length of any path tmapsto gammat is _z_^z_ ||dotgamma||ddt. In our case this gives: fracdddz pmatrix zcosthetaz zsinthetaz z pmatrix pmatrix costhetaz-ztheta'zsinthetaz sinthetaz+ztheta'zcosthetaz pmatrix &Rightarrow sqrtcos^thetaz-theta'zzsinthetazcosthetaz+z^theta'z^sin^thetaz+theta'zzsinthetazcosthetaz+z^theta'z^cos^thetaz+ sqrt+z^theta'^ With this we find the length of _z_^z_sqrt+z^theta'^ddz abc Here the functional is _z_^z_sz thetaz theta'zddz with sz thetaz theta'z sqrt+z^theta'^. This leads to the Euler-Lagrange s: fracpartial spartial theta fractextdddzfracpartial spartial theta' Rightarrow fractextdddzleftfracz^theta'zsqrt+z^theta'^right We can new find fracz^theta'zsqrt+z^theta'^ C which leads to z^theta'z^ C^+z^theta'^ theta'z^z^-C^z^ C^ theta'z fracsqrtCsqrtz^-C^z^. Integrate and substitute zCu to find thetaz fracsqrtC fracddzsqrtu^-u^+D fracsqrtCarctan leftsqrtfracz^C^-+Dright In principle we could now use thetaz_theta_ and thetaz_theta_ to determine the two constants CD. abcliste
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Exercise:
Given two pos AB on a cone with apex angle fracpi we aim to find the shortest path connecting those two pos. abcliste abc Asing that the path is parametrized by zmapstop zcosphi zsinphi z show that its length is _z_^z_sqrt+z^theta'^ddz abc Find the Euler-Lagrange s for this functional. Calculate a general solution. abcliste

Solution:
abcliste abc The length of any path tmapsto gammat is _z_^z_ ||dotgamma||ddt. In our case this gives: fracdddz pmatrix zcosthetaz zsinthetaz z pmatrix pmatrix costhetaz-ztheta'zsinthetaz sinthetaz+ztheta'zcosthetaz pmatrix &Rightarrow sqrtcos^thetaz-theta'zzsinthetazcosthetaz+z^theta'z^sin^thetaz+theta'zzsinthetazcosthetaz+z^theta'z^cos^thetaz+ sqrt+z^theta'^ With this we find the length of _z_^z_sqrt+z^theta'^ddz abc Here the functional is _z_^z_sz thetaz theta'zddz with sz thetaz theta'z sqrt+z^theta'^. This leads to the Euler-Lagrange s: fracpartial spartial theta fractextdddzfracpartial spartial theta' Rightarrow fractextdddzleftfracz^theta'zsqrt+z^theta'^right We can new find fracz^theta'zsqrt+z^theta'^ C which leads to z^theta'z^ C^+z^theta'^ theta'z^z^-C^z^ C^ theta'z fracsqrtCsqrtz^-C^z^. Integrate and substitute zCu to find thetaz fracsqrtC fracddzsqrtu^-u^+D fracsqrtCarctan leftsqrtfracz^C^-+Dright In principle we could now use thetaz_theta_ and thetaz_theta_ to determine the two constants CD. abcliste
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am23, euler-lagrange equations, hs23, lagrangian
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