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damnson2000
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Guys I have the following homework problem to solve:
There are 2 given points in a plane. If we take a point-like object with mass m and take it to the "higher" point what path should it go on to reach the other point in the shortest possible time. Only gravitational force affects our point-like object.
My attempt to solution:
Let point A has (x1, f(x1)) and point B (x2, f(x2)) coordinates.
As we all know average velocity = whole distance/whole time, therefore whole time = whole distance/average velocity.
So we just have to calculate the components of the equation than plug in.
Average velocity:
First we need velocity in the function of position which can be calculated from potential energy difference.
ΔEpotential=ΔEkinetic
m*g*(f(x1)-f(x))=1/2*m*v2
so v2=2*g*(f(x1)-f(x))
therefore v or more likely v(x) =√(2*g*(f(x1)-f(x))) .
Now we have the velocity in the function of position, all we have to do is to calculate the average value of this function, which is
vaverage=∫x1x2√(2*g*(f(x1)-f(x))) dx / (x2 - x1)
Now we need the arc length of our f(x) curve in the intervall of x1 and x2.
so
swhole= ∫x1x2√(1+f2(x)) dx
Now divide the whole distance with average velocity:
twhole = swhole/vaverage
now we would have to take the quotient of these two expressions and solve its derivate for zero.
There are 2 given points in a plane. If we take a point-like object with mass m and take it to the "higher" point what path should it go on to reach the other point in the shortest possible time. Only gravitational force affects our point-like object.
My attempt to solution:
Let point A has (x1, f(x1)) and point B (x2, f(x2)) coordinates.
As we all know average velocity = whole distance/whole time, therefore whole time = whole distance/average velocity.
So we just have to calculate the components of the equation than plug in.
Average velocity:
First we need velocity in the function of position which can be calculated from potential energy difference.
ΔEpotential=ΔEkinetic
m*g*(f(x1)-f(x))=1/2*m*v2
so v2=2*g*(f(x1)-f(x))
therefore v or more likely v(x) =√(2*g*(f(x1)-f(x))) .
Now we have the velocity in the function of position, all we have to do is to calculate the average value of this function, which is
vaverage=∫x1x2√(2*g*(f(x1)-f(x))) dx / (x2 - x1)
Now we need the arc length of our f(x) curve in the intervall of x1 and x2.
so
swhole= ∫x1x2√(1+f2(x)) dx
Now divide the whole distance with average velocity:
twhole = swhole/vaverage
now we would have to take the quotient of these two expressions and solve its derivate for zero.
∫x1x2√(1+f2(x)) dx * (x2 - x1)
d __________________________________ = 0∫x1x2√(2*g*(f(x1)-f(x))) dx
_______
dx
But u haven't managed to find the solution of this equation and I am not even sure if my solution is wether correct or not.
If anyone has an idea how to solve this equation or have found a problem in my solution I am listening with open ears.
dx
But u haven't managed to find the solution of this equation and I am not even sure if my solution is wether correct or not.
If anyone has an idea how to solve this equation or have found a problem in my solution I am listening with open ears.
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