Shortest path on a cone
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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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
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
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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