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Differentiate that equation for v(t)

  1. Nov 8, 2004 #1

    The question asks to differentiate that equation for v(t) and thus show that the acceleration is porportional to the speed at any time.

    I have no clue how to differentiate. Could someone start me off or give some clues? I'll find th rest in my text book. Thanks
  2. jcsd
  3. Nov 8, 2004 #2
  4. Nov 8, 2004 #3
    differentiate = take the derivative?

    wow...I must not be paying attention in math class
  5. Nov 8, 2004 #4
    I dont understand... the derivative of e^x = e^x?

    then the derivative of v=vi*e^(-ct) is:

    a=c*e^(-ct)+e^(-ct) * Vi?
    Last edited: Nov 8, 2004
  6. Nov 8, 2004 #5
    More like

    a = -c*vi*e^(-ct)

    Don't forget the vi.

    Now, anything in that expression look familiar? Could you fit v(t) in there somewhere?

  7. Nov 8, 2004 #6
    Where did the -c come from? Doesn't the derivative of Vi become c?
  8. Nov 8, 2004 #7
    [tex]\frac{d}{dt}v(t) = \frac{d}{dt} \left( v_i e^{-ct} \right)[/tex]

    [itex]v_i[/itex] is a constant, so

    [tex]\frac{d}{dt} v_i e^{-ct} = v_i \frac{d}{dt}e^{-ct}[/tex]

    Make a substitution u = -ct and apply the chain rule

    [tex]v_i \frac{d}{dt}e^{-ct} = v_i \left( \frac{d}{du} e^u \right) \left( \frac{du}{dt} \right) = v_i e^u \frac{d}{dt} \left( -ct \right) = v_i e^u \cdot (-c) = -c v_i e^{-ct}[/tex]

    Last edited: Nov 8, 2004
  9. Nov 8, 2004 #8
    thank you, that helps a lot. I was just wondering, my book gives an answer of a=-cv, where does the e go?
  10. Nov 8, 2004 #9
    [tex]v = v_i e^{-ct}[/tex]

    Plug v into the book's expression for a and see if it matches up with what you got. Find the e?

  11. Nov 8, 2004 #10
    [tex]a = -c v_i[/tex]

    [tex]v_i e^{-ct}= -c v_i[/tex]
    [tex]e^{-ct}= -c[/tex]
  12. Nov 8, 2004 #11
    You're confusing [itex]v_i[/itex] and v(t).

    [itex]v_i[/itex] is the initial velocity, i.e. the velocity at time = 0, or v(0).

    v is the velocity as a function of time, i.e. [itex]v_i e^{-ct}[/itex], or v(t).

    Remember that [itex]v_i[/itex] is a constant.

  13. Nov 8, 2004 #12
    okay, I understand, thanks for all your help.
    Last edited: Nov 8, 2004

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