Answer:
dy/dt = (0)tan(π/3) + (4)sec2(π/3)(0.1)
dy/dt = 0 + 1.6
dy/dt = 1.6 mi/min
Step-by-step explanation:
We need to find out values for all of the variables:
The observer is not moving, therefore dx/dt = 0.
The observer is standing 4 miles away from the lift-off point, so x = 4.
The angle between the horizontal and the observer's line of sight is π/3, therefore θ = π/3.
The rate of change for the angle, dθ/dt, is 0.1.
Let θ be the angle of elevation of the line of sight from the horizontal. Since the hot air balloon is rising at a speed of 100 m/min, its vertical distance in time, t is d = vt = 100t.
Since the point of the observer is 200 m away( the horizontal distance), then tanθ = vertical rise/horizontal distance = 100t/200 = t/2
tanθ = t/2
To find the rate of change of the angle of elevation, we differentiate with respect to t.
So, dtanθ/dt = d(t/2)/dt
sec²θdθ/dt = 1/2
dθ/dt = 1/(2sec²θ)
At 600 m high, tanθ = vertical rise/horizontal distance = 600 m/200 m = 3
Given that tan²θ + 1 = sec²θ
dθ/dt = 1/(2sec²θ)
dθ/dt = 1/(2[tan²θ + 1])
Since tanθ = 3
dθ/dt = 1/(2[tan²θ + 1])
dθ/dt = 1/(2[3² + 1])
dθ/dt = 1/(2[9 + 1])
dθ/dt = 1/(2[10])
dθ/dt = 1/20
dθ/dt = 0.05 rad/min
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