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An airplane is flying at 150 ft/s at an altitude of 2000 ft in a direction that will take it directly over an observer at ground level. Find the rate of change of the angle of elevation between the observer and the plane when the plane is directly over a point on the ground that is 2000 ft from the observer.

Respuesta :

Answer:

Explanation:

Let x be the horizontal distance of airplane . angle of elevation of airplane from observer = θ , altitude of airplane = 2000 ft ,

from the construction  θ = 45 degree. , x = 2000 ft .

Tanθ = 2000 / x

sec²θ dθ / dt = (2000 / x²)  dx / dt

dθ / dt = 2000 /(sec²θ x²) x dx / dt

dx / dt  = 150 ft /s

dθ / dt = 2000x 150 /(sec²θ x²)

= 300000 / sec²45 x 2000²

= .15 degree/ s

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