Asked by Mary

A hot air balloon, 50 feet from an observer, is rising at 20 ft/sec. At 5 seconds after lift off
1. How fast is the distance between the observer and the balloon changing?

2. How fast is the angle of elevation changing?

Answers

Answered by Steve
the distance z at height x is

z^2 = 50^2 + x^2
so,
2z dz/dt = 2x dx/dt

find z when x = 5*20=100, and plug and chug.

tanθ = x/50
sec^2 θ dθ/dt = 1/50 dx/dt
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