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Varying Mass:Raindrops

  • #1
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Homework Statement


xg=x*(dv/dt)+v^2 is a differential equation
Which has a solution of the form v=at
Where a is a constant.Find a

Homework Equations



dv/dt=v*(dv/dx)
V=dx/dt
A=dv/dt=d^2x/dt^2


The Attempt at a Solution


I assumed 'a' as a function of g
That is a=kg for some constant k
And proceeded
I did get the correct answer a=g/3
My question is since both a and g are accelerations and constants
Is it correct to assume one as a function of the other?
For 2 arbitrary constants a and b, we have (a=kb) for another constant 'k' where a and b are never equal to zero at the same time
I hope my approach is correct
Any confirmations and insights are much appreciated!



UchihaClan13[/B]
 

Answers and Replies

  • #2
145
12
I can post the rest of the solution if required
All I need is some confirmation




UchihaClan13
 
  • #3
haruspex
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Since they are constants, you have not really assumed one is a fixed function of the other. Rather, you have assumed that the ratio between them is also a constant, which is trivially true.
 
  • #4
145
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Yeah sorry
That's what I did
But y=kx (say)
Doesn't this lead to y=f (x)
For different values of x, won't we have different values of y
The only difference with this case and my original one is that
In my original one, both were constants
While in this one, both are variables
X being the independent variable and y being the dependent one

Is my reasoning correct then??


UchihaClan13
 
  • #5
haruspex
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Yeah sorry
That's what I did
But y=kx (say)
Doesn't this lead to y=f (x)
For different values of x, won't we have different values of y
The only difference with this case and my original one is that
In my original one, both were constants
While in this one, both are variables
X being the independent variable and y being the dependent one

Is my reasoning correct then??


UchihaClan13
Say we consider g to be a variable; we might run the same experiment in a different gravitational field. You say you assumed a=kg for some constant k, but you never used the assumption that it was constant. So instead, you could say let a, as function of g, be g times some other function of g, namely k(g). That would be quite valid and general.
But to solve the problem, you must assume dg/dt=0. You are given da/dt=0, so dk/dt=0. This allows you to substitute k(g)g for a and obtain k=1/3, a constant. So it turns out that k is constant even if g varies.
 
  • #6
145
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I see what you mean
Since the first derivative of g is zero,it implies g is a constant
And we already have da/dt=0 from the problem statement
So this would force dk/dt=0 meaning k would be a constant
Thanks a lot for your help













UchihaClan13
 

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