1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Difficult integral - Falling mass

  1. Mar 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve the differential equation, dt/dv= 1/g (1/1-a^2*v^2) where a = (k/mg)^1/2 to yield v= 1/a(1-e^-2agt/1+e^-2agt).

    2. Relevant equations
    F=ma
    Newton's 2nd Law
    Integration Laws

    3. The attempt at a solution
    See image. I think I'm getting messed up on the integration laws.
     

    Attached Files:

  2. jcsd
  3. Mar 31, 2016 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Please post threads in the proper forum by subject.

    Solving differential equations and calculating integrals are by nature calculus problems, not pre-calculus.
     
  4. Mar 31, 2016 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your differential equation
    [tex] \frac{dt}{dv} = \frac{1}{g} \left( \frac{1}{1} - a^2 v^2 \right) = \frac{1}{g} ( 1 - a^2 v^2) [/tex]
    (which is what you wrote) does not have the solution
    [tex] v = \frac{1}{a} \left( 1 - e^{-2} \frac{agt}{1} + e^{-2} agt \right) [/tex]
    which is what you wrote. If these are not what you mean, you need to start using parentheses.

    I did not look at the attached image; I look only at typed work, which is the preferred format in this Forum.
     
  5. Mar 31, 2016 #4
    Sorry for the misconception. I meant to write:
    dt/dv = 1/g ((1)/(1-a^2v^2))
    V = 1/a ((1-e^(-2agt))/(1+e^(-2agt)))

    Hopefully this is clearer.
     
  6. Mar 31, 2016 #5

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Try to use LaTeX (https://www.physicsforums.com/help/latexhelp/ )
    Do you mean:
    ##\displaystyle \frac{dt}{dv}= \frac{1}{g}\frac{1}{1-a²v²}##
    ##\displaystyle v=\frac{1}{a}\frac{1-e^{-2agt}}{1+e^{-2agt}}##?
     
  7. Mar 31, 2016 #6
    [itex] \frac{dt} {dv} = \frac{1} {g} ( \frac{1} {1 - a^2 v^2} ) [/itex]

    [itex] v = \frac{1} {a} ( \frac{1 - e^(-2agt)} {1 + e^(-2agt)} ) [/itex]

    The (-2agt) is to the power of e
    e^(-2agt)
    Sorry for the confusion guys!! :(
     
  8. Mar 31, 2016 #7
    [itex] \frac{dt} {dv} = \frac{1} {g} ( \frac{1} {1 - a^2 v^2} ) [/itex]

    [itex] v = \frac{1} {a} ( \frac{1 - e^{-2agt}} {1 + e^{-2agt}} ) [/itex]

    This is what I mean to write.
    Sorry for the confusion guys!! :(
     
  9. Mar 31, 2016 #8

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Great!

    Now back to the exercise:

    integral.jpg

    Something went wrong in the part I circled in red.
    The derivative of ##\log (1-a²v²)## is ##\frac{-2av}{1-a²v²}##.

    EDIT: To fix the error above: the derivative of ##\log (1-a²v²)## is ##\frac{-2a²v}{1-a²v²}##.
     
    Last edited: Mar 31, 2016
  10. Mar 31, 2016 #9
    Yeh, I know there is probably an issue here in my work, but I wrote: ##\ln (1 - a^2 v^2)## when derived, this gives ##\frac{1}{1-a^2 v^2}## doesn't it?
     
  11. Mar 31, 2016 #10

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    No, it gives ##\frac{-2a²v}{1-a^2 v^2}## (I forgot the a² in my previous post).

    You have to apply the chain rule.
    If ##f(x)=\ln g(x)##, then ##f'(x)=\frac{g'(x)}{g(x)}##.
     
  12. Mar 31, 2016 #11
    Yeahh I'm really stuck on this one. I don't know how I can get my ##g'(x) = 1## whilst the bottom ##= 1 - a^2 v^2##??
     
  13. Mar 31, 2016 #12

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

  14. Mar 31, 2016 #13
  15. Mar 31, 2016 #14

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    So now you can solve that to get v in function of t.
    Remember that ##\ln \frac{b}{c}=\ln b -\ln c##.
     
  16. Apr 18, 2016 #15
    Hi Cincin, Nice to see you are working on your assignment.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Difficult integral - Falling mass
  1. Difficult Integral (Replies: 10)

  2. Difficult Integral (Replies: 31)

  3. Difficult integral (Replies: 7)

  4. Difficult integral. (Replies: 5)

  5. Difficult Integration (Replies: 22)

Loading...