Question

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 θ when you are 30 ft from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

Accepted Answer

First, understand the setup: You have a right triangle where the horizontal distance from the screen is the base, and the height of the screen forms the vertical side. The angle θ is the angle of elevation from your eye level to the top of the screen. Define the variables: Let x be the distance from the screen, which changes over time. The height of the screen above your eye level is constant at 10 ft - 3 ft = 7 ft. Use trigonometry to relate the angle θ to the distance x. The tangent of the angle θ is given by the ratio of the opposite side (height of the screen) to the adjacent side (distance from the screen): tan(θ) = 7/x. Differentiate the equation tan(θ) = 7/x with respect to time t to find the rate of change of θ. Use implicit differentiation: d/dt [tan(θ)] = d/dt [7/x]. Apply the chain rule: sec²(θ) * (dθ/dt) = -7/x² * (dx/dt). Substitute dx/dt = 3 ft/s and x = 30 ft into the equation to solve for dθ/dt, the rate of change of the viewing angle.

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Key Concepts

Related Rates

Trigonometric Functions

Implicit Differentiation