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?
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
To find the rate of change of the viewing angle θ, we need to use trigonometry and related rates.
Let's denote the distance from the wall to your current position as x, and the height of your eyes from the ground as h. We know that the bottom of the screen is 12 ft above your eye level, so we have:
h = 12 ft
At any given position, we can create a right triangle connecting the top of the screen, your eyes, and your current position. The height of the screen is given as 40 ft above your eye level, so the height of the triangle is:
H = h + 40 ft
Now, let's consider the tangent of the viewing angle θ. The tangent is defined as the opposite side divided by the adjacent side in a right triangle. In this case, the height of the triangle serves as the opposite side, and the distance from your current position to the wall serves as the adjacent side. So we have:
tan(θ) = H / x
To find the rate of change of θ, we can take the derivative of this equation with respect to time t:
d(tan(θ)) / dt = d(H / x) / dt
Now, let's differentiate the right-hand side using the quotient rule. The derivative of H with respect to t is 0 since it is a constant. The derivative of x with respect to t is the given rate at which you are walking, 3 ft/s:
d(tan(θ)) / dt = (0 * x - H * 1) / x^2
Simplifying further, we get:
d(tan(θ)) / dt = -H / x^2
Now, we can substitute the known values:
H = h + 40 ft = 12 ft + 40 ft = 52 ft
x = 30 ft
Plugging these values into the equation, we get:
d(tan(θ)) / dt = -52 ft / (30 ft)^2
Simplifying, we find:
d(tan(θ)) / dt = -52 / 900 ft^(-1)
Therefore, the rate of change of the viewing angle θ when you are 30 ft from the wall is approximately -0.0578 ft^(-1) or -0.0578 radians per foot. Note that the negative sign indicates that the viewing angle is decreasing as you move away from the screen.
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)?