Question
A hot air ballon lifts off the earth 300m away from an observer and rises straight up at a rate of 80m/min. At what rate is the angle of inclination of the observers line of sight increasing at the time the balloon is 300, above the ground?
Answers
when the balloon is at height h,
tanθ = h/300
so,
sec^2θ dθ/dt = 1/300 dh/dt
you know that when h=300, θ=π/4.
You know dh/dt = 80, so just solve for dθ/dt
tanθ = h/300
so,
sec^2θ dθ/dt = 1/300 dh/dt
you know that when h=300, θ=π/4.
You know dh/dt = 80, so just solve for dθ/dt
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