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A hot air balloon rising vertically problem

A hot air balloon rising vertically is tracked by an observer located 4 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is pi/3 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?

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Brenda L. | Calculus, Diffy Q, Pre-Calc, Algebra, PhysicsCalculus, Diffy Q, Pre-Calc, Algebra, Ph...
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The person and the hot air balloon form a right triangle, with the lift-off point as the vertex that form the right angle. If you were to try to solve for the height of the balloon, you would need to use tangent to do so:
tanθ = opp/adj = y/x
y = xtanθ
What this problem ultimately wants to know, is how the distance of the balloon is changing with respect to time: dy/dt. So we need to take the derivative of the above equation (you need to product rule to do so):
dy/dt = (dx/dt)tanθ + xsec2θ(dθ/dt)
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.
Just plug all the numbers in:
dy/dt = (0)tan(π/3) + (4)sec2(π/3)(0.1)
dy/dt = 0 + 1.6
dy/dt = 1.6 mi/min