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# A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle betwee

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A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is pi/6 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?

Apr 12, 2014

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We have tanθ = h/2(miles)

And so, the height of the balloon at any time is given by

h = (2miles)* tanθ

What we're looking for is dh/dt (the change in ht. per min)

So we have

dh/dt = dh/dθ * dθ/dt

Note that dh/dθ = (2miles)*sec2θ and dθ/dt = .1 rad /min

(Also, note that the 2 miles is a constant...i.e., the "x" part of the tanget isn't changing, only the "y" part  - the height - is.)

Then dh/dt = (2miles)*sec2θ * .1 rad /min

And when θ = π/6, we have

dh/dt = (2miles)*sec2(π/6) * .1 rad /min

dh/dt = (2 miles) * (2/√(3))2  *  .1 rad / min

dh / dt = (2 miles) * (4/3) * .1 rad / min

dh /dt = (8 / 3) * (1 /10) miles / min

dh /dt = (8/30) miles / min = (4/15) miles / min

(Since rads are a "dimensionless quantity" they "fall out" of the answer)

I believe that's it...............   Apr 13, 2014