An observer stands 100 meters from the launch pad of a hotÂair balloon. The balloon rises vertically at a constant speed of 3 m/s. How fast is the distance between the observer and the balloon changing 25 seconds after the launch?

Let D = the distance between the observer and the balloon

Let x = the distance between the observer and the launch pad

Let y = the distance the balloon rises into the air

Then,

D^2 = x^2 + y^2

Now take the derivative of both sides in terms of t which is the time

2D*dD/dt = 2x*dx/dt + 2y*dy/dt

at 25 seconds: D = 125 m, y = 75 m

therefore,

2(125)*dD/dt = 2(100)*(0) + 2(75)(3)

250*dD/dt = 0 + 450

dD/dt = 450/250 = 1.8 m/s

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