Is there a reason

  • Thread starter LENIN
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  • #1
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Main Question or Discussion Point

I found this exercise in an old physics schoolbook. I maeged to solve it but I don't really understand why I have to solve it in this excact order. Before I start I would just like to add that I have very little experiance with differential equations (they are not in our highschool year plan), and therefor any explanations will be welcomed.

So here it goes.

t=?
s=10
v=2s+1
v=ds/dt
dt=(1/v)ds
t=Iteg[1/v]ds
t=1/2Ln(2s+1)
This is the way it was solved in the book (I might have memorized the integration of v wrong).

But I tried to solve it like this
t=?
s=10
v=2s+1
v=ds/dt
ds=v*dt
s=(2s+1)t
s/(2s+1)=t

The solutions are clearly different. I don't understend why. The only idea I got was that I cant integrate v by dt becouse t is not a veriable in v and v is therfore threted as a constent what changes the whole thing (is this true). But if there are any beter explanations I would be glad to hear them. Thanks.
 

Answers and Replies

  • #2
HallsofIvy
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Neither v nor t is a constant here nor can be treated as one. When you write
"ds= v dt so s= (2s+1)t" you are treating v as a constant- and it is not. It depends upon t. Since you don't yet know v or s as a function of t, the only way to integrate
ds= (2s+1)dt is as ds/(2s+1)= dt which is exactly the book's solution.
 
  • #3
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So in general I should always integrate in such a way that the expresion is not threated as a constant, if posibel of course. Right.
 
  • #4
HallsofIvy
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Not just "if possible". You should not assume a variable is a constant when it is not!

What you are doing there is separating the variables- so that you have only s on one side of the equation and t on the other. If that is NOT possible then it is not a "separable" equation and you will have to use some other method. If x is a function of t then you cannot integrate [tex]\int f(x,t)dt[/tex] without knowing exact what function it is!
 
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  • #5
100
1
Thank you! Your explanation is realy very helpful. :smile:
 

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