The angle of elevation of the sun is decreasing at a rate of .25 rad/h.How fast is the shadow cast by a 400-ft-tall building increasing when the...

Let x be the angle of
elevation.

Let the house be  represented by BC the vertical
line , and the shadow be represented by BA, the horizontal
line.

Now consider the the triangle ABC in the vertical
plane. x = angle BAC is the angle of elevation.

Therefore
tanx = BC/AC.

Therefore AC = BC/
tanx.

Let AC = f(h) = BC/tanx =
400/tanx.

Differentiating with respect h, we
get:

 f'(h) =  d/dh {400/tanx} = d/dx {400/tanx}
*dx/dh

f'(h) =  [(-1)400/(tanx)^2] (secx)^2 *
dx/dh

f'(x) = -400[1+(tanx)^2]/(tanx)^2}{
dx/dh}

Given  x= pi/6, then (tanx)^2 = 1/3. Also given 
dx/dh = -0.25.

Therefore f'(pi/6)) = -400{1+1/3}/(1/3)
}(-0.25).

f'(pi/6) = -400
+(4*)(-0.25)

f'(pi/6) =
400.

Therefore , when the angle of elevation is x= pi/6
radians, the shadow is increasing at the rate of 400 times with respect to the time in
hour.