Question

The bottom of a large theater screen is 3ft above your eye level and the top of the acreen is 10ft above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of 3ft/s while looking at the screen. What is the change of the viewing angle when you are 30ft from the wall on which the screen hangs assuming the floor is horizontal?

Answer 1

As usual, draw a diagram. You can easily see that if you are x away from the wall, 

the angle of elevation of the bottom of the screen (A) is 

cotA = x/3 
A = arccot(x/3) 

angle B to the top is 

cotB = x/10 
B = arccot(x/10) 

So, since θ = B-A 
dθ/dt = dB/dt - dA/dt 
= -3/(x^2+9) + 10/(x^2+100) 
= 7(x^2-30)/((x^2+9)(x^2+100)) 

so, at x=30 
dθ/dt = 203/30300