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Zanzabum
3 years ago
5

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed

of 1.9 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building? (Round your answer to one decimal place.)
Mathematics
2 answers:
Harman [31]3 years ago
6 0
12 m
 -4 m
____
8 m

the answer to the problem is the subtract by the 8 -4=4 isnt it the work to show 





Stells [14]3 years ago
4 0
Draw a right angled triangle ABC, where <B = 90 degrees 

<span>BC is the base, and BC = 12 m and C represents spot light on the ground and B is foot of the wall </span>

<span>Draw a perpendicular DE from a point D on hypotenuse, AC to a point E on BC </span>

<span>DE represents walking man and DE = 2 m </span>

<span>Now AB represents height of shadow, let AB = y </span>

<span>let EC = x </span>

<span>from similar triangles, ABC and DEC </span>

<span>AB/DE = BC/EC </span>

<span>y/2 = 12/x </span>

<span>y = 24/x </span>

<span>differentiating with respect to t </span>

<span>dy/dt = - (24/x^2) dx/dt </span>

<span>substitute x = (12 - 4) = 8 and dx/dt = 1.9 m/s </span>

<span>dy/dt = - (24 / 64)(1.9) </span>

<span>= -0.7125 </span>

<span>length of shadow is decreasing at the rate of 0.7125 m /s</span>
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