Rate of Change of the Angle of Elevation

[pippintook]

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?

 

[LowlyPion]

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?

 

[thomate1]

As a scientist, the rate of change of the angle of elevation can be calculated by taking the derivative of the angle of elevation with respect to time. In this case, the angle of elevation can be represented as a function of time, where t is the time in seconds and θ is the angle of elevation:

θ(t) = tan^-1(25/60)

Taking the derivative of this function with respect to time, we get:

dθ/dt = 60/(25^2 + 60^2) * (1/60) * dt/dt

Simplifying, we get:

dθ/dt = 60/(25^2 + 60^2)

Plugging in the values, we get:

dθ/dt = 0.024 radians/second

This means that the angle of elevation is changing at a rate of 0.024 radians per second, or approximately 1.38 degrees per second. This rate of change is important in understanding the movement of the balloon and its trajectory.