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Math Help - Calculus Word Problem Ladder

  1. #1
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    Calculus Word Problem Ladder

    A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.5 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? Evaluate your answer numerically. (Round the answer to three decimal places.)
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  2. #2
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    Quote Originally Posted by kashifzaidi View Post
    A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.5 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? Evaluate your answer numerically. (Round the answer to three decimal places.)
    make a sketch of the problem.

    let x = horizontal distance ... \frac{dx}{dt} = 1.5 ft/sec

    let \theta = angle


    determine the trigonometric relationship equation using x , \theta, and the 10 ft ladder.

    take the time derivative, sub in your given/calculated values, then determine \frac{d\theta}{dt}
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  3. #3
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    You want to find: \frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt}

    Note that:

    x=10cos(\theta)

    \frac{dx}{d\theta}=-10sin(\theta)

    \frac{d\theta}{dx}=\frac{-csc(\theta)}{10}

    I think that's the trick you're looking for. Just plug this into the equation for \frac{d\theta}{dt}.

    To find the numerical answer, you can use the fact that you are given 6=10cos(\theta) at this point.
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