A hot air balloon is rising vertically at a constant rate of 8 meters per second. An observer on the ground is 100 meters away from the point on the ground directly beneath the balloon. How fast is the angle of elevation changing when the balloon is 60 meters above the ground?
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Let the angle of elevation be θ and let the height of the balloon above the ground be h at any time. We are given that dh/dt = 8 m/s and we need to find dθ/dt when h = 60 m.
We can use the tangent function to relate θ and h:
tan(θ) = h/100
⟹ θ = arctan(h/100)
Taking the derivative of both sides with respect to time t:
dθ/dt = d(arctan(h/100))/dt
Applying the chain rule:
dθ/dt = (1/(1 + (h/100)^2)) * d(h/100)/dt
Given that dh/dt = 8 m/s:
dθ/dt = (1/(1 + (h/100)^2)) * (8/100)
When h = 60 m:
dθ/dt = (1/(1 + (60/100)^2)) * (8/100)
= (1/(1 + 0.36)) * (8/100)
= (1/1.36) * (0.08)
≈ 0.0588 rad/s
Therefore, the angle of elevation is changing at a rate of approximately 0.0588 rad/s when the balloon is 60 meters above the ground.
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