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Optimum values and w/ chain rule

  1. Jan 6, 2007 #1
    A boat leaves dock at 2:00 PM, heading west at 15 km/h. Another boat heads south at 12 km/h and reaches the same dock at 3:00 PM. When were the boats closest to each other?

    I've solved other problems similar to this one, but I can't seem to figure this one out. I'm not sure what I have to do with the times that are given... I know you guys like to see work done to show that I've tried it, but all I have is a crappy diagram...

    thanks.
     
  2. jcsd
  3. Jan 6, 2007 #2
    On the assumption that this is parametric, that would make boat A:
    x(t) = -15t
    y(t) = 0

    And Boat B:
    x(t) = 0
    y(t) = -12t+12

    therefore,
    d = ((x_a(t)-x_b(t))^2+(x_a(t)-x_b(t))^2)^0.5

    Plug in the equations and simplify this(I have pity on you if you don't have a graphing calculator for this) and find the derivative of the distance with respect to time, which you can graph as you would any other derivative function. At that point you should know how to use the first derivative test for local extrema.

    I got 21 minutes, 36 seconds. I'd definitely doublecheck, though, since I'm still in AP Cal AB and could very well be pulling a great deal of this out of my ass.
     
    Last edited: Jan 7, 2007
  4. Jan 7, 2007 #3

    HallsofIvy

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    Since the derivative of x2 with respect to t is 2x(dx/dt), minimizing the distance is the same as minimizing the square of the distance. You don't need the 1/2 power.
     
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