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A Diving problem not sure what to call it.

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A diver leaps from the edge of a diving platform into a pool. At time t seconds after she leaps, the horizontal distance from the front edge of the platform to the diver's shoulders is given by x(t), and the vertical distance from the water surface to her shoulders is given by y(t), where x(t) and y(t) are measured in meters. Suppose that the diver's shoulders are 11.4 meters above the water when she makes her leap that dx/dt=0.8 and dy/dt=3.6-9.8t, for 0<or= t <or= A, where A is the time that the diver's shoulders ender the water.

    A) Find the max vert. distance from the water surface to the diver's shoulders.
    B) Find A, the time that the diver's shoulders enter the water
    C) Find the total distance traveled by the diver's shoulders from the time she leaps from the platform until the time her shoulders enter the water.
    D) Find the angle θ, 0 < θ < л(Pi)/2, between the path of the diver and the water at the instant the diver's shoulders enter the water.


    2. Relevant equations
    D = Sqaure root of ((dx/dt)² + (dy/dt)²) dt


    3. The attempt at a solution
    A)
    3.6 - 9.8t = 0
    t= .3673
    B)
    3.6t - 4.9t² + 11.4 = 0
    t = 1.9362
    C)
    square root of ((.8)² + (3.6 - 9.8t)²) dt
    and i'm not sure how to finish this...
    D)
    I don't even know where to start to try and figure this question out.
     
    Last edited: May 17, 2009
  2. jcsd
  3. May 17, 2009 #2

    zcd

    User Avatar

    Lol this was on the AP exam for Calc BC

    C) is solved by finding arc length of the parametric equations (so integrate Sqrt((dx/dt)^2+(dy/dt)^2) from t=0 to t=1.9362) this is equivalent to finding the length of the diver's path above the water, which is a parabola

    D) when the diver's path above the water is in the shape of a parabola. when he hits the water, there's a tangent line of the trajectory. because his trajectory is found by eliminating the parameter t, the slope of his tangent line would be dy/dx. from there, you can draw a right triangle with dy representing the vertical y component and dx representing the horizontal x component of dy/dx, and angle theta. tan(theta) = dy/dx, so theta = arctan (dy/dx)
     
  4. May 17, 2009 #3
    yeah my teacher is makeing me do all the problems from the calc BC AP exam. :(
     
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