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Simple rate of change problem?

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data
    The Function S(t) = 6t(t+1) describes the distance (in kilometers) that a car has traveled after a time t ( in hours), t E [0,6]

    Find the average velocity of the car from t= 2 to t= 2 + ∆t

    2. Relevant equations

    y'= f(x+h) - f(x)
    ------------
    h


    3. The attempt at a solution

    I know from the power rule that the derivative of this function is 12t + 6
    however when I try to use the rate of change formula I get 6t+ 20

    Here is my work

    s(t)' = s(2 + ∆t) - s(2)
    ----------------
    ∆t
    =6(2+∆t)(∆t+2+1) - 36
    ---------------------
    ∆t
    = (12+ 6∆t)(∆t+3) -36
    --------------------
    ∆t
    = 12∆t + 36 + 6∆t^2 + 18t - 36
    ------------------------------
    ∆t
    =6∆t^2 + 30t
    ------------
    ∆t

    = 6t + 30

    So where did I go wrong?
     
  2. jcsd
  3. Feb 16, 2013 #2

    Dick

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    You didn't do anything wrong except for some typos and you changed Δt into t. You should get 6Δt+30. To compare that with the derivative you need to let Δt->0. So the derivative S'(2) should come out to 30. If you put t=2 into the derivative you computed, 12t+6, you'll get the same number.
     
  4. Feb 16, 2013 #3

    Ray Vickson

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    S(t) = 6*t^2 + 6*t, so the instantaneous velocity is v(t) = dS(t)/dt = 12*t + 6, and for small Δt > 0 that will also be the average velocity over [t,t+Δt]. For t = 2 this gives average_velocity = 24 + 6 = 30 (km/hr). Your error was in writing 12Δt + 18Δt as 12Δt + 18t.
     
  5. Feb 16, 2013 #4
    oh okay I see, for some reason I thought using the first principle and finding Y' would give me the derivative, however, it just gives me the rate of change. Thanks for helping!
     
  6. Feb 16, 2013 #5

    Dick

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    Right. You can get the derivative from what you did by replacing 2 with t, working through the algebra again and then taking the limit as Δt->0.
     
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