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)?
i got 0.055 for the final answer but its wrong
what did i do wrong??
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.
i got -0.014 in the end
To find the rate of change of the viewing angle θ, we need to determine how the angle θ changes with respect to time.
Let's break down the problem and analyze the situation step by step:
1. Establish a right triangle: Draw a line from your eye, perpendicular to the floor, to the bottom of the theater screen. This line represents the hypotenuse of the right triangle, with the length of 30 ft (distance from the screen).
2. Determine the height of the theater screen: The bottom of the screen is 12 ft above your eye level, and the top of the screen is 40 ft above your eye level. Thus, the height of the theater screen is 40 - 12 = 28 ft.
3. Calculate the adjacent side of the triangle: The adjacent side is the distance between your eye level and the bottom of the theater screen. Since the screen is 12 ft above your eye level, the adjacent side is 12 ft.
4. Find the opposite side of the triangle: The opposite side is the height of the theater screen (28 ft).
5. Use the tangent function: We can use the tangent function to relate the opposite side, the adjacent side, and the angle θ:
tan(θ) = opposite/adjacent
tan(θ) = 28/12
6. Solve for θ: Taking the inverse tangent of both sides, we find:
θ = tan^(-1)(28/12)
7. Differentiate with respect to time: We are given that you are walking away from the screen at a rate of 3 ft/s. To find the rate of change of the viewing angle θ with respect to time, we need to differentiate the equation θ(t) = tan^(-1)(28/12) with respect to t:
dθ/dt = d/dt[tan^(-1)(28/12)]
8. Calculate the rate of change of the viewing angle: Evaluate the derivative of the inverse tangent function:
dθ/dt = (1/[(28/12)^2 + 1]) * [d/dt(28/12)]
Simplifying further:
dθ/dt = (1/[(28/12)^2 + 1]) * (0)
dθ/dt = 0
Therefore, the rate of change of the viewing angle θ when you are 30 ft from the wall is 0. The viewing angle remains constant as you walk away from the screen.
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.