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Math Help - calculus word problems

  1. #1
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    calculus word problems

    I am having a tough time getting these two problems on tonights calc homeowork. Help would be much appreciated.

    1. A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 25 ft/s.

    (a) At what rate is his distance from second base decreasing when he is halfway to first base? m/s

    (b) At what rate is his distance from third base increasing at the same moment? m/s

    2.
    A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 6 m from the dock?


    Answer in m/s-[IMG]file:///C:/DOCUME%7E1/User/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]
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  2. #2
    MHF Contributor
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    (1.a) At what rate is his distance from second base decreasing when he is halfway to first base? m/s

    Draw the figure.
    Place the the batter somewhere between homeplate and first base .
    Let x be his distance from the homeplate.
    Let D be his distance from the 2nd base.

    So, by Pythagorean theorem,
    D = sqrt[90^2 +(90 -x)^2]
    D^2 = 90^2 +(90 -x)^2
    Differentiate both sides with respect to time t,
    2D dD/dt = 2(90 -x)(-1) dx/dt
    D dD/dt = (x -90)dx/dt -----------**

    When x = 90/2 = 45m, we know dx/dt = 25m/sec, but we don't know D. So we find D.

    D = sqrt[90^2 +(90 -45)^2] = 100.623 m

    Hence,
    (100.623)dD/dt = (45 -90)(25)
    dD/dt = (-45)(25)/100.623 = -11.18 m/sec

    Meaning, when the batter is halfway to the first base, his distance from the second base is decreasing at the rate of 11.18 m/sec. --------answer.

    --------------------------------------------
    (1.b) At what rate is his distance from third base increasing at the same moment? m/s

    Use same figure for the part (1.a) above.
    Let S = distance of batter from 3rd base.

    S^2 = 90^2 +x^2
    2S dS/dt = 2x dx/dt
    S dS/dt = x dx/dt

    When x = 90/2 = 45m,
    S = sqrt[90^2 +45^2] = 100.623 m again

    (100.623)dS/dt = 45(25)
    dS/dt = 45(25)/100.623 = 11.18 m/sec

    Therefore, the distance of the batter from the 3rd base is increasing at the rate of 11.18 m/sec when he is at halfway from the homeplate to the 1st base. --------answer.

    ================================================== ======
    Draw the figure.
    It is a right triangle with these:
    >>>vertical leg = 1 meter
    >>>horizontal leg = x meters .....the distance of the boat's bow from the dock.
    >>>hypotenuse = y meters .....the length of the rope from the pulley to the boat's bow.

    Given: dy/dt = 1 m/sec
    Find: dx/dt when x = 6m.

    y^2 = 1^2 +x^2
    2y dy/dt = 2x dx/dt
    y dy/dt = x dx/dt

    y = sqrt[1^2 +x^2] = sqrt[1^2 +6^2] = sqrt(37) m
    So,
    sqrt(37)*(1) = 6 dx/dt
    dx/dt = sqrt(37) /6 = 1.0138 m/sec

    Therefore, at the instant the boat's bow is 6 meters from the dock, the boat is approaching the dock at the rate of 1.0138 m/sec. ------answer.
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