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The second derivative question

  1. Jan 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Given (d^2)x/ (dt^2) = 2 for all t>= 0 and dx/dt= -5 and x= 4 when t= 0 find t when

    a) dx/ dy= 0
    b) x= 0

    2. The attempt at a solution

    a) Given (d^2)x/ (dt^2) = 2, then
    dx/dt= 2x + c = -5

    given dx/ dt= 0 then 2x + c= 0

    c= -2.5 using 2x + c= 0 and 2x + c = -5

    using dx/dt= 2x + c, t= x^2 + cx + d
    where 0= 16 - 2.5 x 4 + d (using x= 4 when t= 0 )
    where d= -6

    Thus,

    t= x^2 - 2.5x - 6

    when dx/dt= 0 then x= 1.25 in 2x + c= 0 (c= -2.5)

    subsituting x= 1.25 into x^2 - 2.5x - 6 gives t= -121/16,

    however the answer seems to differ giving t= 2.5 when dx/dt= 0
    How is this so?

    b) not to sure as confused about a).
     
  2. jcsd
  3. Jan 11, 2008 #2
    Perhaps you meant dx/dt=0?

    x and its derivatives are a function of t here, so from the second-order equation, you'll end-up at
    [tex]\frac{dx}{dt} = 2t+C[/tex]



    Was it? The question tells you what [itex]\frac{dx}{dt}[/itex] is at t=0. Use this condition to find the constant of integration. With one more integration, you can find x as a function of t. The rest is just algebra.


    Part b should become clear once you rework part a.
     
  4. Jan 11, 2008 #3
    I think what you mean is:

    Given [tex] \frac{d^2 x}{dt^2} = 2 , \quad \forall t \geq 0[/tex] and [tex] \frac{dx}{dt}(0) = -5 [/tex] and [tex] x(0) = 4 [/tex]. You will then find that c is not -2.5.

    And I now see that neutrino already told you that...
     
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