Aspotlight 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.)

Draw a right angled triangle ABC, where <B = 90 degrees 

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

Draw a perpendicular DE from a point D on hypotenuse, AC to a point E on BC 

DE represents walking man and DE = 2 m 

Now AB represents height of shadow, let AB = y 

let EC = x 

from similar triangles, ABC and DEC 

AB/DE = BC/EC 

y/2 = 12/x 

y = 24/x 

differentiating with respect to t 

dy/dt = - (24/x^2) dx/dt 

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

dy/dt = - (24 / 64)(1.9) 

= -0.7125 

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

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 

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