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 3 ​ft/s while looking at the screen. What is the rate of change of the viewing angle θ when you are 30 ft from the wall on which the screen​ hangs, assuming the floor is flat​?

3 answers

Sorry wrong 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 3 ​ft/s while looking at the screen. What is the rate of change of the viewing angle θ when you are 30 ft from the wall on which the screen​ hangs, assuming the floor is flat​ (see figure)?
The bottom of a large theater screen is 9 ft above your eye level and the top of the screen is 30 ft above your eye level. Assume you walk away from the screen​ (perpendicular to the​ screen) at a rate of 7 ​ft/s while looking at the screen. What is the rate of change of the viewing angle θ when you are 70 ft from the wall on which the screen​ hangs, assuming the floor is flat​ (see figure)?
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 = 30/x
tanB = 9/x
tanθ = tan(A-B) = (30/x - 9/x)/(1 + 30/x * 9/x) = 21/x(x^2+270)
so, now you know that
sec^2θ dθ/dt = 21(270-x^2)/(270+x^2)^2 dx/dt
Now plug in x=70 and dx/dt = 7