# calculus

An observer is 20 m above the ground floor of a large hotel atrium looking at a glass-enclosed
elevator shaft that is 20 m horizontally from the observer (see figure). The angle of elevation of the elevator is the angle
that the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator
rises at a rate of 5 mês, what is the rate of change of the angle of elevation when the elevator is 10 m above the ground?
When the elevator is 40 m above the ground?

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1. when the elevator is x meters up,

tanθ = (x-20)/20

So,

sec^2θ dθ/dt = 1/10 dx/dt

when x=10, tanθ = -1/2, so sec^2θ = 5/4

5/4 dθ/dt = 1/10 (5)
dθ/dt = 2/5 m/s

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2. Steve is wrong the correct answer is

10m = 1/5 rad/s and 40m = 1/8 rad/s

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