(1 point) A spotlight on the ground is shining on a wall 20m20m away. If a woman 2m2m tall walks from the spotlight toward the building at a

Question

(1 point) A spotlight on the ground is shining on a wall 20m20m away. If a woman 2m2m tall walks from the spotlight toward the building at a speed of 0.8m/s,0.8m/s, how fast is the length of her shadow on the building decreasing when she is 8m8m from the building

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King 5 years 2021-09-02T22:38:13+00:00 1 Answers 21 views 0

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  1. Thunguyet
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    2021-09-02T22:40:06+00:00
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    Answer:

    The answer to the question is;

    When she is 8 m from the building fast the length of her shadow on the building is decreasing at \frac{2}{9} m/s or 0.22 m/s.

    Explanation:

    We have

    Distance of the spotlight from the building = 20 m

    Distance  of woman from the building when her speed is measured = 8 m

    Height of the woman = 2 m

    Actual speed of the woman = 0.8 m/s

    Comparing the distance of the woman from the spotlight and the wall from the spotlight, we have when the woman is 8 m from the building she is 12 m from the spotlight    

    Therefore we have

    \frac{12}{20} = \frac{2}{y} where y is the shadow cast by the woman on the building = 10/3

    When the woman is x distance from the building, she is 20 – x meters from the spotlight

    Therefore the above equation can be written  as

    \frac{20-x}{20} = \frac{2}{y}  which gives 1 - \frac{1}{20}*x = 2* \frac{1}{y} finding the derivative of both sides gives

    -\frac{1}{20}dx =-2*\frac{1}{y^2}dy hence we have by dividing by dt gives -\frac{1}{20}\frac{dx}{dt} =-2*\frac{1}{y^2}\frac{dy}{dt}

    However we know that \frac{dx}{dt} = 0.8 m/s

    Therefore -\frac{0.8}{20} = -0.18\frac{dy}{dt}

    The rate of decrease of her shadow \frac{dy}{dt} is given by

    \frac{dy}{dt} = \frac{0.8}{3.6} =\frac{2}{9} m/s or 0.222 m/s.

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