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Math Help - Speed

  1. #1
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    Speed

    A 20 foot ladder slides down a wall at 5ft/sec. At what speed is the bottom sliding out when the top is 10 feet from the floor (in ft/sec).
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  2. #2
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    Quote Originally Posted by DINOCALC09 View Post
    A 20 foot ladder slides down a wall at 5ft/sec. At what speed is the bottom sliding out when the top is 10 feet from the floor (in ft/sec).
    Quote Originally Posted by DINOCALC09 View Post
    A 20 foot ladder slides down a wall at 5ft/sec. At what speed is the bottom sliding out when the top is 10 feet from the floor (in ft/sec).
    let y be the height of the ladder on the wall
    using the pythagorean theorem we get

    x^2+y^2=400 note: we also get when y=10 that x=3\sqrt{10}

    taking the derivative with respect to time we get

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

    plugging in all the info gives

    2(3\sqrt{10})\frac{dx}{dt}+2(10)(5)=0 \iff \frac{dx}{dt}=\frac{50}{3\sqrt{10}}=\frac{5\sqrt{1  0}}{3}
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  3. #3
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    i got 5/sqrt(3)

    Taking the derivatives of the pythagorean theorem,
    $x\left(\frac {dx}{dt}\right) + y\left(\frac {dy}{dt}\right) = z\left(\frac {dz}{dt}\right)$
    But $\left(\frac {dz}{dt}\right) = 0$ so
    $x\left(\frac {dx}{dt}\right) = - y\left(\frac {dy}{dt}\right)$
    Then from pythagoras, $x^{2} + y^{2} = z^{2}$. We are looking for $x$ at $y = 10$ so $x = \sqrt {400 - 100} = \sqrt {300}$.
    Then we have
    $\sqrt {300}\left(\frac {dx}{dt}\right) = - 10(5)\Rightarrow\left(\frac {dx}{dt}\right) = \frac {50}{\sqrt {300}} = \frac {5}{\sqrt {3}}$
    The negative sign just gives direction so it isn't really important.
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