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Math Help - word prob

  1. #1
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    word prob

    The illumination of an object (in lumens) created by a certain light source is given by I = 80/D^2 where D is the distance (in meters) of the object from the light source. Use differentials to approximated the decrease in illumination as the distance from the light source changes from 2 meters to 2.5 meters.

    How to do this?
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  2. #2
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    Quote Originally Posted by kwivo View Post
    The illumination of an object (in lumens) created by a certain light source is given by I = 80/D^2 where D is the distance (in meters) of the object from the light source. Use differentials to approximated the decrease in illumination as the distance from the light source changes from 2 meters to 2.5 meters.

    How to do this?
    Hello,

    you have: I(D)=\frac{80}{D^2}

    you get: \frac{d(I(D))}{dD}=-\frac{160}{D^3}

    To estimate the decrease in illumination it is sufficient to use the following values: D = 2, \ dD = \frac12

    d(I(D))=-\frac{160}{D^3} \cdot dD~\implies~ d(I(2))=-20 \cdot \frac12 = -10
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