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Math Help - rate of change

  1. #1
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    rate of change

    hi all,

    How about do I go to answer the following rate of change question.

    A ladder of length 10metres is placed with one end of the horizontal ground and the other end against a vertical wall. Given the ladder makes an angle \theta with the horizontal then it slips vertically down the wall at rate of 2 metres/sec. Find the rate at which the angle \theta is decreasing at the distant when  \theta = \pi/4 radian.

    To help I have attached a picture.

    I know the rate has to be minus (-2m/sec) as it is decreasing.

    Can someone please explain the process step by step as I am totally lost

    Thanks
    Attached Thumbnails Attached Thumbnails rate of change-ladder.gif  
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dadon
    hi all,

    How about do I go to answer the following rate of change question.

    A ladder of length 10metres is placed with one end of the horizontal ground and the other end against a vertical wall. Given the ladder makes an angle \theta with the horizontal then it slips vertically down the wall at rate of 2 metres/sec. Find the rate at which the angle \theta is decreasing at the distant when  \theta = \pi/4 radian.

    To help I have attached a picture.

    I know the rate has to be minus (-2m/sec) as it is decreasing.

    Can someone please explain the process step by step as I am totally lost

    Thanks
    Let the top end of the ladder be x m above the ground, then:

    <br />
\cos(\theta)=x/10<br />
.

    So differentiating this wrt t gives:

    <br />
-\sin(\theta) \frac{d\theta}{dt}=\frac{1}{10}\ \frac{dx}{dt}<br />
.

    So now plug the values of \theta and \frac{dx}{dt}=-2 in and solve for \frac{d\theta}{dt}.

    RonL
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  3. #3
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    CaptainBlack you are so clear in your explanations it's a pleasure to read your answers! really :-)
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  4. #4
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    Thumbs up

    Thanks for that captain black!

    Kind regards,

    dadon
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