Question

Viewing angle, The bottom of a large theater screen is 3 ft above your eye level and the top of the screen is 10 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 0 when you are 30 ft from the wall on which the screen hangs, assuming the floor is flat

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

Answer 2

You would do most parts in Steve's answer. However, the question uses the time derivative (in relations to t), not the x derivative. He forgot to take into account that dx/dt = -3ft/sec. The given rate (dx/dt) would be negative in the question's context.

Since Steve typed most of the steps already, lets skip to θ = B - A
dθ/dt = (dB/dt - dA/dt) dx/dt

We'll solve for (dB/dt - dA/dt) first
= -3/(x^2+9) + 10/(x^2+100)
= 7(x^2-30)/((x^2+9)(x^2+100))

At x = 30
= 7(30^2-30)/((30^2+9)(30^2+100))
= 203/30300

Now you have
dθ/dt = (203/30300)dx/dt

Lets substitute dx/dt in the equation and solve for dθ/dt
dθ/dt = (203/30300)(-3)
= -203/10100 or -0.02009 . . .

Answer 3

How can dx/dt be negative when you are walking away from the screen and adding more distance between you and the screen?? The answer to the problem is correct as stated -0.02 radians per second. The negative sign comes from the differential d(alpha)/dt = -dx/dt*(tan(alpha))/(x *(sec(alpha)*sec(alpha)) the other equation is similar but with the angle Beta. Arriving at d(theta)/dt = d(Beta)/dt - d(alpha)/dt
d(theta)/dt = -0.03 - (-0.01) giving us d(theta)/dt = -0.02