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Math Help - Word Problem Calculus!

  1. #1
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    Word Problem Calculus!

    Help! Hey guys, please help me out in this word problem:
    Carlo is perched precariously on top of a 10-foot ladder leaning against the back wall of an apartment building (spying on an enemy of his) when it starts to slide down the wall at a rate of 4 ft per minute. Carlo's accomplice, Alex, is standing on the groud 6 ft. away from the wall, with the bottom of the ladder between the wall and Alex. How fast is the base of the ladder moving when it hits Alex?
    So far... this is by solution.

    dy/dt = 4
    x = 6

    and my equation is (but obviously wrong):
    x^2 + y^2 = 10^ / 100.
    6^2 + y^2 = 100
    y^2 = 100-36
    y = 8

    But i don't really know what to do afterwards... Please help me out!
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  2. #2
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    Quote Originally Posted by anfieldgirl View Post
    I posted this in the calculus forum. I'm so sorry for posting again, but i didn't see this forum until afterwards. I'm in urgent need of help. Please please help me out.

    Help! Hey guys, please help me out in this word problem:


    So far... this is by solution.

    dy/dt = 4
    x = 6

    and my equation is (but obviously wrong):
    x^2 + y^2 = 10^ / 100.
    6^2 + y^2 = 100
    y^2 = 100-36
    y = 8

    But i don't really know what to do afterwards... Please help me out!

    x^2 + Y^2 =100
    diff bth sides wrt. t
    we get
    2x dx/dt +2y dy/dt =0
    dy/dt=-4 (given)
    x=6 y=8
    find dx/dt.........
    Last edited by sumit2009; January 21st 2009 at 07:15 AM. Reason: type error
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  3. #3
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    Thank you so much! Really appreciate the help.
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  4. #4
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    Lexington, MA (USA)
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    Hello, anfieldgirl!

    Sounds like "Home Alone" . . .


    Carlo is perched precariously on top of a 10-foot ladder leaning
    against the back wall of an apartment building (spying on an enemy of his)
    when it starts to slide down the wall at a rate of 4 ft per minute.
    Carlo's accomplice, Alex, is standing on the ground 6 ft away from the wall.
    How fast is the base of the ladder moving when it hits Alex?
    Code:
          |
          *
          |\
          | \
          |  \
        y |   \ 10
          |    \
          |     \
          |      \
          * - - - * - - *
              x         A

    We are given: . \frac{dy}{dt} \:=\:\text{-}4 ft/min.

    We know that: . x^2 + y^2 \:=\:10^2

    Differentiate with respect to time: . 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \:=\:0 \quad\Rightarrow\quad \frac{dx}{dt} \:=\:-\frac{y}{x}\,\frac{dy}{dt}


    When x = 6,\:y = 8

    So we have: . \frac{dx}{dt} \;=\;-\frac{8}{6}(-4) \;=\;\frac{16}{3}


    The base of the ladder hits Alex at 5\tfrac{1}{3} ft/min.

    Last edited by mr fantastic; January 21st 2009 at 11:41 PM. Reason: No edit - just flagging this reply as having been moved from a duplicate post.
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