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Homework Help: Instantaneous velocity electron

  1. Sep 2, 2010 #1
    1. The problem statement, all variables and given/known data

    an electron moving along the x axis has a position given by x=19te^-4t m, where t is in seconds. how far is the electron from the origin when it momentarily stops.

    2. Relevant equations

    velocty=delta x / delta t

    3. The attempt at a solution

    i think its a instantaneous velocity question but im not sure. i think you have to take the derivative of the equation. If somebody could just head me in the right direction then i could attempt to solve it.
     
  2. jcsd
  3. Sep 2, 2010 #2

    Doc Al

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    Staff: Mentor

    So far, so good. Keep going.
     
  4. Sep 2, 2010 #3
    so would the derivative of by x=19te^-4t m be:

    -4(19te)^-3 which would be 76te^-3.

    so i think i know what im doing this far but im not sure what the question is asking when it says
    "how far is the electron from the origin when it momentarily stops."

    how do i know the time when it momentarily stops. i mean if i know the time i can just substitute the time in for t and solve the equation to get the answer.
     
  5. Sep 2, 2010 #4

    Doc Al

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    Staff: Mentor

    Not exactly. Hint: You'll need the product rule. (I assume the expression is 19t e-4t.)
    Hint: What's the speed when it 'momentarily stops'? Use that to solve for the time.
     
  6. Sep 2, 2010 #5
    Hint: What's the speed when it 'momentarily stops'? Use that to solve for the time.[/QUOTE]

    im guessing it would be zero.


    and there is no gap in between the 19t and the e^-4t. its just one whole expression.
     
  7. Sep 2, 2010 #6

    Doc Al

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    Staff: Mentor

    Exactly.
    Right. I wrote it with a gap for clarity, so the t and e wouldn't blur together. Redo your derivative.
     
  8. Sep 2, 2010 #7
    Doc Al included the gap to make it clear that you need to use the product rule
     
  9. Sep 2, 2010 #8
    using the product rule i got.

    f(x)= 19t
    g(x)= e^-4t

    so would it be (im not sure of the derivative of e^-4t)

    19e^-4t + 19te^-4t
     
  10. Sep 2, 2010 #9
    Have to use the chain rule with e^(-4t)

    Your answer is incorrect but close
     
  11. Sep 2, 2010 #10
    using the chain rule i got the derivative of e^-4t to be ln(t)t^4

    f(y)=e^4
    g(x)=-4ln(t)

    (ln(t))e^-4ln(t)
    (ln(t))t^4
     
  12. Sep 2, 2010 #11
    To use the chain rule you need to view the function, g(t) = e^-4t, in two parts like this:
    g(u) = e^u, with u(t)=-4t

    then using the chain rule to to get the derivative of g(t), dg/dt, means using this rule:
    dg/dt = dg/du * du/dt
     
  13. Sep 2, 2010 #12
    sorry but im not really great at calculus, since im biology major.

    don't really know what u mean or what to do to use the chain rule.
     
  14. Sep 2, 2010 #13
    well let's do it step by step, what is the derivative of:
    e^u
     
  15. Sep 2, 2010 #14
    not sure what u is but:

    ln u
     
  16. Sep 2, 2010 #15
    I just used 'u' here instead of the usual 'x'. It was meant to be the same question as "what is the derivative of e^x".

    The derivative of a function is: the change of the value of the function per change in its variable.

    So the "derivative of e^u", means the change of e^u per change in u.

    For simple functions like f(x)=3x, it is clear that the change of f(x) per change in x is 3. since whenever x change by 1, f(x) changes by 3.

    The notation I used above for the derivative of f(x) was df/dx, which you could read as "the change of f(x) per change in x", which is also often written as f'(x).

    Anyway, for some functions the derivative is not immediately clear and you may just need to know what it is. The function f(x) = e^x, is a very special function because its derivative, df/dx or f'(x) is simply equal to the function itself.
     
  17. Sep 2, 2010 #16
    so the derivative of g(u)=e^u is just e

    and the derivative of u(t)=-4t is -4
     
  18. Sep 2, 2010 #17
    To use the chain rule you need to view the function, g(t) = e^-4t, in two parts like this:
    g(u) = e^u, with u(t)=-4t

    so using the chainrule here you would get the derivative of g(t) as
    g'(u) * u'(t)

    I already told you that g'(u)=g(u) so that is just e^u=e^-4t.

    now you just need to multiply this by the derivative of u(t)=-4t
     
  19. Sep 2, 2010 #18
    the derivative of g(u)=e^u is e^u, the derivative of u(t) is correct
     
  20. Sep 2, 2010 #19
     
  21. Sep 2, 2010 #20
    The derivative of the position function is not correct, you were going in the right direction here in an earlier post:
    It is just the second term that is not correct here, it is not 19te^-4t, it should be the 19t times the derivative of e^-4t
     
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