Rate of Change of the Angle of Elevation

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SUMMARY

The discussion focuses on calculating the rate of change of the angle of elevation of a balloon rising at 8 feet per second from a point 60 feet away from an observer. The angle of elevation when the balloon reaches 25 feet is approximately 22.6 degrees. Participants clarify that the goal is to determine how many degrees the angle changes per second, utilizing the formula V*t/D = tan(θ) and the derivative of θ with respect to time. The correct approach involves evaluating the derivative at the time when the balloon is at 25 feet, specifically at t = 25/8 seconds.

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  • Understanding of trigonometric functions, specifically tangent.
  • Knowledge of derivatives and their application in physics.
  • Familiarity with rates of change in calculus.
  • Basic understanding of kinematics, particularly vertical motion.
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A balloon rises at the fate of 8 feet per second from a point on the ground 60 ft. from an observer. Find the rate of change of the angle of elevation when the balloon is 25 feet above the ground

I converted 8 ft/s to 2.44 m/s2 to make it easier. I also figured the angle of elevation when the balloon was at 25 feet is around 22.6 degrees. I'm just not really sure what they're looking for. Do they want to know how many degrees the angle changes per second? If so, how do I find it? If not, what are they looking for and how do I find that?
 
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Yes what they want is the rate of change of the angle.

Your formula for elevation angle is

V*t/D = tanθ

But what is the rate of change of θ ? Isn't it the derivative of θ with respect to time?

So Tanθ*dθ/dt = ... ?

To find the magnitude of the change then evaluate the result at t = 25/8 seconds right?
 

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