# 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 angle of elevation of the sun is Pi/6?

kjcdb8er | Teacher | (Level 1) Associate Educator

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To answer this question consider the right triangle with sides h, r and L, where the angle between L and r is A. h is the height of the building, 400 ft.

The relation between these values is :

L * tan(A)  = 400 ________(1)

Lets note now that when A = pi/6, L = 400*sqrt3 = 692 ft

We also know that dA/dt = .25 rad/h.

Take the derivative of (1) with respect to time (use chain rule) :

dL/dt * tan(A) + L *d/dt(tan A) = 0

dL/dt * tan(A) + L*sec(A)^2 * dA/dt = 0

dL/dt * (1/sqrt3) + 692 ft * (2/sqrt3)^2 * 0.25 rad/h = 0

dL/dt = 400 ft/hr

neela | High School Teacher | (Level 3) Valedictorian

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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.