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Homework Help: Seperable Differentiation. Help, please. =\

  1. Apr 13, 2005 #1
    I blanked out. Any help is greatly appreciated.

    Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>0. After her parachute opens, her velocity satisfies the differential equation: dv/dt= -2v-32, with initial condition v(0)=-50.

    (a) Use separation of variables to find an expression for v in terms of t, where t is measured in seconds.

    I got up to this far:

    integral [dv/(-2v-32)]= integral (dt)
    u= -2v-32; du= -2dv.

    Help. :cry:
     
  2. jcsd
  3. Apr 13, 2005 #2

    dextercioby

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    Good.May i suggest integrating with corresponding limits...?

    [tex] \int_{-50}^{v} \frac{dv}{-2v-32} = \int_{0}^{t} \ dt [/tex]

    The LHS integral is doable.With a substitution that u already did...So carry on.Pay attention to the way the limits are transformed.


    Daniel.
     
  4. Apr 13, 2005 #3
    Using the substitution that I did, I obtained:

    (-1/2) integral(1/u)du = integral (dt)

    = (-1/2) ln|u| + C = t+ C

    I'm not sure on what to do next. =\
     
  5. Apr 13, 2005 #4
    substitue the original variables back for the u substitutions. then use the initial conditon to find out what C is.
     
  6. Apr 13, 2005 #5
    I know you plug in 0 for t, but what'll happen to the -50?
    And do I have to exponentiate both sides?
     
  7. Apr 13, 2005 #6

    dextercioby

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    From the integration i get

    [tex] v(t)=-34e^{-2t}-16 [/tex]


    Daniel.
     
  8. Apr 13, 2005 #7
    That is the answer, but I don't know how to get it.
     
  9. Apr 13, 2005 #8

    dextercioby

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    The RHS is 't',okay??

    Now,the LHS can be written

    [tex]-\frac{1}{2} \int _{-50}^{v} \frac{d(-v-16)}{-v-16} = -\frac{1}{2}\left[\ln (-v-16)\right]\left |_{-50}^{v}\right = -\frac{1}{2} \ln \left(\frac{-v-16}{34}\right) [/tex]

    Therefore

    [tex] -2t=\ln\left(\frac{-v-16}{34}\right)\Rightarrow v(t)=-34e^{-2t}-16 [/tex]

    Daniel.
     
    Last edited: Apr 13, 2005
  10. Apr 13, 2005 #9
    Wow, thanks.
     
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