A weather balloon is rising vertically at a speed of 5 meters per second. An observer on the ground 40 meters away from the launch point. Let h be the height of the balloon. 1. How fast is the distance x between the observer and the balloon increasing when that distance is 50 meters (x = 50)? dx/dt = [ Select ] meters per second 2. How fast is the angle of elevation ? changing at the same instant? d?/dt = [ Select ] radians per second
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We need to find the rate of change of the distance x between the observer and the balloon when x = 50 meters. This can be done using the Pythagorean theorem: x^2 = h^2 + 40^2, where h is the height of the balloon. Differentiating both sides with respect to time Show more…
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