Initial Value Problem
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
Short
Video
\(\LaTeX\)
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Exercise:
A thredimensional system of differential s is characterised by the eigenvalues lambda_ - lambda_ lambda_ - and the corresponding eigenvectors vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix Derive the solution for the initial conditions pmatrix x y z pmatrix pmatrix - pmatrix
Solution:
The general solution is given by the vector pmatrix xt yt zt pmatrix a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t For t this can be written as pmatrix x y z pmatrix pmatrix - pmatrix pmatrix a_ + a_ a_ a_+a_+a_ pmatrix From the second component it follows immediately that a_ The first component then yields a_ -a_ - - and the last component a_ - - a_ - a_ - + - - The solution can therfore be written as xt -e^-t+e^t yt e^t zt -e^-t+e^t-e^-t
A thredimensional system of differential s is characterised by the eigenvalues lambda_ - lambda_ lambda_ - and the corresponding eigenvectors vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix Derive the solution for the initial conditions pmatrix x y z pmatrix pmatrix - pmatrix
Solution:
The general solution is given by the vector pmatrix xt yt zt pmatrix a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t For t this can be written as pmatrix x y z pmatrix pmatrix - pmatrix pmatrix a_ + a_ a_ a_+a_+a_ pmatrix From the second component it follows immediately that a_ The first component then yields a_ -a_ - - and the last component a_ - - a_ - a_ - + - - The solution can therfore be written as xt -e^-t+e^t yt e^t zt -e^-t+e^t-e^-t
Meta Information
Exercise:
A thredimensional system of differential s is characterised by the eigenvalues lambda_ - lambda_ lambda_ - and the corresponding eigenvectors vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix Derive the solution for the initial conditions pmatrix x y z pmatrix pmatrix - pmatrix
Solution:
The general solution is given by the vector pmatrix xt yt zt pmatrix a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t For t this can be written as pmatrix x y z pmatrix pmatrix - pmatrix pmatrix a_ + a_ a_ a_+a_+a_ pmatrix From the second component it follows immediately that a_ The first component then yields a_ -a_ - - and the last component a_ - - a_ - a_ - + - - The solution can therfore be written as xt -e^-t+e^t yt e^t zt -e^-t+e^t-e^-t
A thredimensional system of differential s is characterised by the eigenvalues lambda_ - lambda_ lambda_ - and the corresponding eigenvectors vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix Derive the solution for the initial conditions pmatrix x y z pmatrix pmatrix - pmatrix
Solution:
The general solution is given by the vector pmatrix xt yt zt pmatrix a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t For t this can be written as pmatrix x y z pmatrix pmatrix - pmatrix pmatrix a_ + a_ a_ a_+a_+a_ pmatrix From the second component it follows immediately that a_ The first component then yields a_ -a_ - - and the last component a_ - - a_ - a_ - + - - The solution can therfore be written as xt -e^-t+e^t yt e^t zt -e^-t+e^t-e^-t
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