The bottom of a large theater screen is 12 ft above your eye level and the top of the screen is 40 ft above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of 2 ft/s while looking at the screen. What is the rate of change of the viewing angle θ when you are 20 ft from the wall on which the screen hangs, assuming the floor is flat

Question

The bottom of a large theater screen is 12 ft above your eye level and the top of the screen is 40 ft above your eye level. Assume you walk away from the screen​ (perpendicular to the​ screen) at a rate of 2 ​ft/s while looking at the screen. What is the rate of change of the viewing angle θ when you are 20 ft from the wall on which the screen​ hangs, assuming the floor is flat​​

oobleck · Answer

Draw the diagram. Let
A = angle to top of screen
B = angle to bottom of screen
Then the viewing angle θ = A-B
When you are x ft away from the screen,
tanA = 40/x
tanB = 12/x
tanθ = tan(A-B) = (40/x - 12/x)/(1 + 40/x * 12/x) = 28x/(x^2+480)
so, now you know that
sec^2θ dθ/dt = 28(480-x^2)/(480+x^2)^2 dx/dt
Now plug in x=20 and dx/dt = 2