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)?

Answer 1

always draw a diagram. It should be clear here that
θ = A-B where
A is the angle to the top of the screen
B is the angle to the bottom of the screen
when you are x meters away,
tanA = 40/x
tanB = 12/x
tanθ = tan(A-B) = (40/x - 12/x)/(1 + 40/x * 12/x) = 28x/(x^2+480)
sec^2θ dθ/dt = 28(480-x^2)/(480+x^2)^2 dx/dt
so now just find sec^2θ when x=3, and plug in your numbers.

Answer 2

i got 0.055 for the final answer but its wrong
what did i do wrong??

Answer 3

hard to say, since you don't show what you did ...
However,
It looks like I made the mistake. I said x=3, but in fact dx/dt = 3
You should have noted that x=30
Make the fix, and if it still turns out wrong, come back with what you worked out.

Answer 4

i got -0.014 in the end