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Math Help - Relate Rates Problem

  1. #1
    mjo
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    Relate Rates Problem

    Here is the question I am having trouble with:
    The sun is passing over a 100 m tall building. The angle θ made by the sun with the ground is increasing at a rate of pi/20 rads/min. At what rate is the length of the shadow of the building changing when the shadow is 60 m long? Give your answer in exact values.
    So far I have got:
    100cosfata=x

    dx/dt=-100(cosfata)(dfata/dx)
    Dont no where to get cosfata from.
    Last edited by skeeter; December 4th 2012 at 03:13 PM. Reason: insert rate of change value
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  2. #2
    Newbie SujiCorp12345's Avatar
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    Re: Relate Rates Problem

    "Increasing at a rate of rads/min?"

    How many radians per minute?
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  3. #3
    mjo
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    Re: Relate Rates Problem

    Oh my gosh I have read this problem so many times I can just picture the numbers when they are not even there. Ugh.
    It is pi/20 rads/min
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  4. #4
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    Re: Relate Rates Problem

    The sun is passing over a 100 m tall building. The angle θ made by the sun with the ground is increasing at a rate of pi/20 rads/min. At what rate is the length of the shadow of the building changing when the shadow is 60 m long?
    \cot{\theta} = \frac{x}{100}

    -\csc^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dx}{dt}
    Attached Thumbnails Attached Thumbnails Relate Rates Problem-trigprob2.png  
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  5. #5
    mjo
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    Re: Relate Rates Problem

    But how do I find a value for -csc^2fata
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    Re: Relate Rates Problem

    Quote Originally Posted by mjo View Post
    But how do I find a value for -csc^2fata
    review your basic right triangle trig ...

    \csc{\theta} = \frac{hypotenuse}{opposite} = \frac{\sqrt{60^2+100^2}}{100}

    square the result and change its sign to get the value of
    -\csc^2{\theta}



    btw ... \theta is prounounced "theta" , not "fata"
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  7. #7
    mjo
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    Re: Relate Rates Problem

    I found my original problem. Thanks guys
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