Asked by Ryan
A plane flies at a speed 500 km/hr at a constant height of 12 km. How rapidly is the angle of elevation to the plane changing when the plane is directly above a point 110 km away from the observer?
The angle of elevation is changing at
radians/hr (enter a positive value). Round your answer to 3 decimal places.
The angle of elevation is changing at
radians/hr (enter a positive value). Round your answer to 3 decimal places.
Answers
Answered by
oobleck
tanθ = 12/x
sec^2 θ dθ/dt = -12/x^2 dx/dt
so plug in your numbers, and note that sec^2 θ = 1 + tan^2 θ = 1 + 144/x^2
sec^2 θ dθ/dt = -12/x^2 dx/dt
so plug in your numbers, and note that sec^2 θ = 1 + tan^2 θ = 1 + 144/x^2
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