Metrikproblem in euklidschen Koordinaten
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:
Berechne die Länge des roten Vektors in den drei verschiedenen angegebenen Basen. Benutze dabei den Satz von Pythagoras und nimm fälschlich auch beim dritten Beispiel an dass die beiden Basisvektoren orthogonal sind! center tikzpicturevect/.stylstealth' scope tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB e_x tkzLabelSegmentleft colorgreen!!blackAC e_y scope scopedashed xshiftcm tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB tilde e_x tkzLabelSegmentleft colorgreen!!blackAC tilde e_y scope scopedotted xshiftcm tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB hat e_x tkzLabelSegmentleft colorgreen!!blackAC hat e_y scope scopexshiftcm tkzInitxmin xmax ymin ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDrawSegmentsvect colorred ultra thickAB scope tikzpicture center
Solution:
Die Länge des Vektors in in den drei Fällen: |L| sqrt |tilde L| frac sqrt |hat L| sqrt Falls der Vektor ein reales physikalisches Objekt ist darf das nicht sein. Die Länge müsste invariant sein. Da kommt die Metrik ins Spiel.
Berechne die Länge des roten Vektors in den drei verschiedenen angegebenen Basen. Benutze dabei den Satz von Pythagoras und nimm fälschlich auch beim dritten Beispiel an dass die beiden Basisvektoren orthogonal sind! center tikzpicturevect/.stylstealth' scope tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB e_x tkzLabelSegmentleft colorgreen!!blackAC e_y scope scopedashed xshiftcm tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB tilde e_x tkzLabelSegmentleft colorgreen!!blackAC tilde e_y scope scopedotted xshiftcm tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB hat e_x tkzLabelSegmentleft colorgreen!!blackAC hat e_y scope scopexshiftcm tkzInitxmin xmax ymin ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDrawSegmentsvect colorred ultra thickAB scope tikzpicture center
Solution:
Die Länge des Vektors in in den drei Fällen: |L| sqrt |tilde L| frac sqrt |hat L| sqrt Falls der Vektor ein reales physikalisches Objekt ist darf das nicht sein. Die Länge müsste invariant sein. Da kommt die Metrik ins Spiel.
Meta Information
Exercise:
Berechne die Länge des roten Vektors in den drei verschiedenen angegebenen Basen. Benutze dabei den Satz von Pythagoras und nimm fälschlich auch beim dritten Beispiel an dass die beiden Basisvektoren orthogonal sind! center tikzpicturevect/.stylstealth' scope tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB e_x tkzLabelSegmentleft colorgreen!!blackAC e_y scope scopedashed xshiftcm tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB tilde e_x tkzLabelSegmentleft colorgreen!!blackAC tilde e_y scope scopedotted xshiftcm tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB hat e_x tkzLabelSegmentleft colorgreen!!blackAC hat e_y scope scopexshiftcm tkzInitxmin xmax ymin ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDrawSegmentsvect colorred ultra thickAB scope tikzpicture center
Solution:
Die Länge des Vektors in in den drei Fällen: |L| sqrt |tilde L| frac sqrt |hat L| sqrt Falls der Vektor ein reales physikalisches Objekt ist darf das nicht sein. Die Länge müsste invariant sein. Da kommt die Metrik ins Spiel.
Berechne die Länge des roten Vektors in den drei verschiedenen angegebenen Basen. Benutze dabei den Satz von Pythagoras und nimm fälschlich auch beim dritten Beispiel an dass die beiden Basisvektoren orthogonal sind! center tikzpicturevect/.stylstealth' scope tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB e_x tkzLabelSegmentleft colorgreen!!blackAC e_y scope scopedashed xshiftcm tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB tilde e_x tkzLabelSegmentleft colorgreen!!blackAC tilde e_y scope scopedotted xshiftcm tkzInitxmax ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDefPoC tkzDrawSegmentsvect colorgreen!!black thickAB tkzDrawSegmentsvect colorgreen!!black thickAC tkzLabelSegmentbelow colorgreen!!blackAB hat e_x tkzLabelSegmentleft colorgreen!!blackAC hat e_y scope scopexshiftcm tkzInitxmin xmax ymin ymax tkzGridsub subxstep subystep tkzDefPoA tkzDefPoB tkzDrawSegmentsvect colorred ultra thickAB scope tikzpicture center
Solution:
Die Länge des Vektors in in den drei Fällen: |L| sqrt |tilde L| frac sqrt |hat L| sqrt Falls der Vektor ein reales physikalisches Objekt ist darf das nicht sein. Die Länge müsste invariant sein. Da kommt die Metrik ins Spiel.
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