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Math Help - Application of differentiation

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    Application of differentiation

    A baseball diamond is a square with side 90 ft. A batter his the balls and runs toward first base with a speed of 24ft/s.
    a. At what rate is his distance from the second base decreasing when he is halfway to first base?
    b. At what rate is his distance from third base increasing at the same moment.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    A baseball diamond is a square with side 90 ft. A batter his the balls and runs toward first base with a speed of 24ft/s.
    a. At what rate is his distance from the second base decreasing when he is halfway to first base?
    b. At what rate is his distance from third base increasing at the same moment.
    a) I'm sure i've seen this problem before, but i can't find it now, so let's reinvent the wheel.

    As with all related rates problems, you want to begin by drawing a diagram to see what's going on (I see you have one already). I have attached one to show you how I am interpreting the problem. Then you want to write down what you know and don't know. I know the value for ds/dt but not dD/dt.

    Let the distance of the batter from 1st base as he runs be s. Let distance he is away from the second base be D. The rate at which the batter is running towards 1st, ds/dt, is equal to - 24 feet/sec (we make this negative becuase the distance between the runner and 1st is decreasing). Also notice that we have s and D changing, and not the distance between 1st and 2nd base, so we can keep the latter distance at a constant 90 feet.

    From the diagram, we realize that we have a right angled triangle connecting the runner, 2nd and 1st base. So our main formula will be Pythagoras' formula. Here goes.

    D^2 = 90^2 + s^2
    differentiate implicitly, we obtain:

    2D dD/dt = 2s ds/dt
    => dD/dt = (2s ds/dt)/2D
    The above is the formula we will be using, but we have to make sure that the only unknown in it is what we want to find, dD/dt.

    By pythagoras again, we realize, when the runner is halfway to 1st, he is 45 feet away from 1st. So we can find D, by D^2 = 45^2 + 90^2 => D = 100.62 feet.

    Now we know everything we need.
    Using s = 45, D = 100.62, ds/dt = - 24, we get:

    dD/dt = [2(45)(-24)]/[2(100.62)]
    = - 10.73 feet per sec


    b) Similar to how I constructed part a) we do part b).
    Skipping the details, we have

    here, x = 45 ft, E = 100.62, dx/dt = 24 ft/s ...this is positive now since the distance between the runner and home base is increasing

    so now
    E^2 = x^2 + 90^2
    => 2E dE/dt = 2x dx/dt
    => dE/dt = (2x dx/dt)/2E
    => dE/dt = [2(45)(24)]/[2(100.62)]
    = 10.73 feet per sec
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    sorry, here's the diagram i meant to post
    Attached Thumbnails Attached Thumbnails Application of differentiation-baseball2.gif  
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