Convert to Higher-Order Differential Equation
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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 -
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Exercise:
Convert the following systems of differential s o a single higher-order differential . abcliste abc dot x_ - x_+x_ dot x_ x_ - x_ abc dot x_ x_ + x_ dot x_ - x_ + x_ dot x_ x_ - x_ abcliste
Solution:
abcliste abc We can express the second derivative of x_ as follows using the differential for x_: ddot x_ -dot x_+dot x_ -dot x_+x_-x_ labelx The differential for x_ can be solved for x_: x_ dot x_+x_ Substituting x_ in refx leads to ddot x_ -dot x_+x_-dot x_+x_ -dot x_+x_-dot x_-x_ result-dot x_-x_ This is a second-order differential for x_. vspace.cm Alternatively we could also derive a second-order differential for x_: ddot x_ dot x_-dot x_ - x_+x_-dot x_ - x_+x_-dot x_ -dot x_+ x_+x_-dot x_ - dot x_-x_ This is the same differential as before. abc Deriving the differential for x_ and plugging in the expressions for dot x_ and dot x_ yields ddot x_ dot x_ + dot x_ x_ + x_ - x_ + x_ x_ + x_ Deriving again we find dddot x_ dot x_ + dot x_ x_ + x_ + x_ - x_ x_ + x_ - x_ We can express the term x_ - x_ in terms of dot x_ and ddot x_: dot x_-ddot x_ x_ + x_ - x_ - x_ x_ - x_ Upon substituting this expression in the for dddot x_ we find dddot x_ result-ddot x_ + dot x_ + x_ abcliste
Convert the following systems of differential s o a single higher-order differential . abcliste abc dot x_ - x_+x_ dot x_ x_ - x_ abc dot x_ x_ + x_ dot x_ - x_ + x_ dot x_ x_ - x_ abcliste
Solution:
abcliste abc We can express the second derivative of x_ as follows using the differential for x_: ddot x_ -dot x_+dot x_ -dot x_+x_-x_ labelx The differential for x_ can be solved for x_: x_ dot x_+x_ Substituting x_ in refx leads to ddot x_ -dot x_+x_-dot x_+x_ -dot x_+x_-dot x_-x_ result-dot x_-x_ This is a second-order differential for x_. vspace.cm Alternatively we could also derive a second-order differential for x_: ddot x_ dot x_-dot x_ - x_+x_-dot x_ - x_+x_-dot x_ -dot x_+ x_+x_-dot x_ - dot x_-x_ This is the same differential as before. abc Deriving the differential for x_ and plugging in the expressions for dot x_ and dot x_ yields ddot x_ dot x_ + dot x_ x_ + x_ - x_ + x_ x_ + x_ Deriving again we find dddot x_ dot x_ + dot x_ x_ + x_ + x_ - x_ x_ + x_ - x_ We can express the term x_ - x_ in terms of dot x_ and ddot x_: dot x_-ddot x_ x_ + x_ - x_ - x_ x_ - x_ Upon substituting this expression in the for dddot x_ we find dddot x_ result-ddot x_ + dot x_ + x_ abcliste
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Exercise:
Convert the following systems of differential s o a single higher-order differential . abcliste abc dot x_ - x_+x_ dot x_ x_ - x_ abc dot x_ x_ + x_ dot x_ - x_ + x_ dot x_ x_ - x_ abcliste
Solution:
abcliste abc We can express the second derivative of x_ as follows using the differential for x_: ddot x_ -dot x_+dot x_ -dot x_+x_-x_ labelx The differential for x_ can be solved for x_: x_ dot x_+x_ Substituting x_ in refx leads to ddot x_ -dot x_+x_-dot x_+x_ -dot x_+x_-dot x_-x_ result-dot x_-x_ This is a second-order differential for x_. vspace.cm Alternatively we could also derive a second-order differential for x_: ddot x_ dot x_-dot x_ - x_+x_-dot x_ - x_+x_-dot x_ -dot x_+ x_+x_-dot x_ - dot x_-x_ This is the same differential as before. abc Deriving the differential for x_ and plugging in the expressions for dot x_ and dot x_ yields ddot x_ dot x_ + dot x_ x_ + x_ - x_ + x_ x_ + x_ Deriving again we find dddot x_ dot x_ + dot x_ x_ + x_ + x_ - x_ x_ + x_ - x_ We can express the term x_ - x_ in terms of dot x_ and ddot x_: dot x_-ddot x_ x_ + x_ - x_ - x_ x_ - x_ Upon substituting this expression in the for dddot x_ we find dddot x_ result-ddot x_ + dot x_ + x_ abcliste
Convert the following systems of differential s o a single higher-order differential . abcliste abc dot x_ - x_+x_ dot x_ x_ - x_ abc dot x_ x_ + x_ dot x_ - x_ + x_ dot x_ x_ - x_ abcliste
Solution:
abcliste abc We can express the second derivative of x_ as follows using the differential for x_: ddot x_ -dot x_+dot x_ -dot x_+x_-x_ labelx The differential for x_ can be solved for x_: x_ dot x_+x_ Substituting x_ in refx leads to ddot x_ -dot x_+x_-dot x_+x_ -dot x_+x_-dot x_-x_ result-dot x_-x_ This is a second-order differential for x_. vspace.cm Alternatively we could also derive a second-order differential for x_: ddot x_ dot x_-dot x_ - x_+x_-dot x_ - x_+x_-dot x_ -dot x_+ x_+x_-dot x_ - dot x_-x_ This is the same differential as before. abc Deriving the differential for x_ and plugging in the expressions for dot x_ and dot x_ yields ddot x_ dot x_ + dot x_ x_ + x_ - x_ + x_ x_ + x_ Deriving again we find dddot x_ dot x_ + dot x_ x_ + x_ + x_ - x_ x_ + x_ - x_ We can express the term x_ - x_ in terms of dot x_ and ddot x_: dot x_-ddot x_ x_ + x_ - x_ - x_ x_ - x_ Upon substituting this expression in the for dddot x_ we find dddot x_ result-ddot x_ + dot x_ + x_ abcliste
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