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Can I undo a derivative to solve for a variable?

  1. Dec 30, 2015 #1
    If I have an equation where there is a derivative surrounding the variable, how do I undo the derivative and solve for the variable?

    Example would be-
    A= dx/dy when x=m*v*λ-2 and y=y
    Solving for v.

    I am a beginner so please explain thoroughly.
     
  2. jcsd
  3. Dec 30, 2015 #2

    SteamKing

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    It's not clear what dx/dy means when there is no explicit relationship between x and y.

    If you want to solve for v in the equation above, that can be done using simple algebra.
     
  4. Dec 30, 2015 #3
    Since dx/dy = A, there is some function x whose derivative with respect to y is the constant A. So think backwards: what function of y has the constant A as its derivative with respect to y?

    Then since x will be equal to this this function and also x = m*v*λ2, set the function you found by working backwards from the derivative equal to m*v*λ2 and solve for v using basic algebra.
     
  5. Dec 30, 2015 #4
    I'm afraid I'm not following your train of thought. My question actually was more of a physics question; however the equation I was referring to had derivatives in it and I wanted to know how to undo them. What I typed above was an example; however I now see that that may have let to confusion of what I was asking. So the equation I want to solve is,-
    Force = the derivative of momentum with respect to time. So it would look like this- F=dp/dt. However p, or momentum can be expanded into, m*v*λ. I would like to solve for v, but I don't know how to get it out of this derivative expression. I have not yet taken calculus so that's why I am asking.
     
  6. Dec 30, 2015 #5

    Erland

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    Isn't it so that λ also contains v: that λ=1-v2/c2?
     
  7. Dec 31, 2015 #6

    Ssnow

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    Before solving something you must decide what are constants or not. For example in your example the velocity ##v## can be view as a function of the time, the same for ##\lambda## (if ## \lambda=1-\frac{v^2}{c^2}##), and assuming constant mass, you equation will be:

    ##F=m\frac{d}{dt}\left(\frac{v(t)}{\sqrt{1-\frac{v(t)^2}{c^2}}}\right).##

    If ## \lambda## is constant the equation will be much simple ## \frac{F}{m}\lambda^{\frac{1}{2}}=\frac{d}{dt}v(t) ##.
     
  8. Dec 31, 2015 #7
    So if i follow your work correctly I see that you put all of the constants out of the derivative expression and then just used algebra to get the constants to the other side. But doesn't the final answer still contain one unknown, the t, and there still is a v in the λ expression (I apologize for not clarifying, it's not a constant). Also, why did you put a t in parentheses by the v, is this some sort of notation?
     
  9. Dec 31, 2015 #8

    Ssnow

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    So you must to solve this differential equation:

    ## F=m\frac{d}{dt}\left(\frac{v(t)}{1-\frac{v^2(t)}{c^2}}\right)##

    that gives you ##\frac{v(t)}{1-\frac{v^2(t)}{c^2}}=\frac{F}{m}t+c## with ##c## a constant. From this you can find ##v(t)## with algebraic operations...
     
  10. Dec 31, 2015 #9
    Alright, can you show me how you got to that point?
     
  11. Jan 1, 2016 #10

    Ssnow

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    you have that ##\frac{F}{m}=\frac{d}{dt}\left(\frac{v(t)}{1-\frac{v^{2}(t)}{c^2}}\right)## so integrating the constant ##\int \frac{F}{m} d\,t=\frac{F}{m}t+c##.
     
  12. Jan 1, 2016 #11
    Are you familiar with anitderivatives? The derivative take you one way and the antiderivative take you the other way. This is what Ssnow is trying to show you.
     
    Last edited: Jan 2, 2016
  13. Jan 2, 2016 #12
    I understand the concept of an anti derivative, that it undoes a derivative and that it is an integral I believe. So the integral of 2x would be x2, yeah iI understand the "reverse operation"; however, I don't understand how you can take the integral, or the anti derivative of a a variable.
     
  14. Jan 2, 2016 #13
    Are you asking if you just have x, how do you get the antiderivative?
    For something like 2x you would break it apart to 2∫xdx, which would then become 2 ∫ (xn+1/n+1)

    so 2∫xdx would break down to 2 ∫ (x1+1/1+1)
    2 ∫ (x2/2)
    put it back together for 2x2/2
    You would then do your cancellations, equaling x2

    for just x focus on the xn+1/n+1,
    remember that it's always understood to be x1 for any variable that doesn't have a superscript written and simply plug in 1 for n
    So you would get x1+1/1+1, which would come out to x2/2 with nothing to cancel
     
  15. Jan 2, 2016 #14

    Erland

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    If we assume that F and m are constants and that v is a function of t, then the solution is as Ssnow wrote.
     
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