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Math Help - Rates of Change of a 15ft ladder

  1. #1
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    Rates of Change of a 15ft ladder

    A ladder 15 feet long is leaning against a building so that end X is on level ground and end Y is on the wall as shown in the figure. The bottom of the ladder (X) is moved away from the building at a constant rate of foot per second.
    a) Find the rate in feet per second at which the length OY is changing when X is 9 feet from the building.

    b) Find the rate of change in square feet per second of the area of triangle XOY when X is 9 feet from the building.

    So I know I have to use the pythagorean theorem so I statred out with √(225-x(t)^2). Then I have truble finding its deriv. Can someoe help me and with (b) too? Do i use the general area equation A(t)=[x(t)*y(t)]2 first with that one?
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  2. #2
    Senior Member Pinkk's Avatar
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    x^{2}+y^{2}=15^{2}
    Then differentiate with respect to t:

    2x\frac{dx}{dt}+ 2y\frac{dy}{dt} = 0

    So the rate of change we are interested in is when x=9, then y=12 (9,12,15 triangle). And \frac{dx}{dt}=\frac{1}{2}

    Plug in the values for x,y, \frac{dx}{dt} and solve for \frac{dy}{dt}

    For the second part, A=\frac{1}{2}xy so therefore \frac{dA}{dt}=\frac{1}{2}x\frac{dy}{dt}+\frac{1}{2  }y\frac{dx}{dt} Plug in all your values and you will get your \frac{dA}{dt} value.
    Last edited by Pinkk; March 21st 2009 at 05:01 PM.
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  3. #3
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    ok for a.) I got -3/8 and b.) 21/16

    is that correct?
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  4. #4
    Senior Member Pinkk's Avatar
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    Your answers are correct.

    Edit: Since your problem included units, I would be careful to include those units in your answers if you're ever taking a test. So \frac{dy}{dt}=\frac{-3}{8} ft/sec and \frac{dA}{dt}=\frac{21}{16} ft^{2}/sec
    Last edited by Pinkk; March 21st 2009 at 06:15 PM.
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