Answer:
0.43 radians/second
Step-by-step explanation:
The relationship between the horizontal distance to the balloon (x), the vertical height of the balloon (y), and the angle (θ) is given by ...
[tex]\dfrac{y}{x}=\tan{\theta}[/tex]
Differentiating with respect to time gives ...
[tex]\dfrac{y'}{x}=\dfrac{\theta '}{\cos^2{\theta}}\\\\\theta '=\dfrac{y'\cos^2{\theta}}{x}=\dfrac{xy'}{x^2+y^2}\\\\\theta '=\dfrac{12\cdot 8}{12^2+9^2}=\dfrac{96}{225}=0.42\overline{6}\approx 0.43 \quad\text{rad/s}[/tex]
The rate of change of the angle of elevation is about 0.43 radians per second when the balloon is 9 feet high.
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You can obtain a good numerical approximation of this value by looking at the angles 0.05 seconds before and after the balloon is 9 ft high. Then the rate of change is approximately that difference divided by the time difference of 0.10 seconds.
[tex]\theta '\approx\dfrac{\arctan{\dfrac{9.4}{12}}-\arctan{\dfrac{8.6}{12}}}{0.1}\approx\dfrac{0.664495-0.621824}{0.1}\\\\\theta '\approx\dfrac{0.042671}{0.1}=0.42671\approx 0.43 \quad\text{rad/s}[/tex]
The key to a good approximation is to use ballon height values both below and above the nominal height. The smaller the time interval, the better the approximation. For this problem a time interval of 0.1 seconds is small enough to get sufficient accuracy for a 2 decimal place answer.
Some calculators will let you define a function that gives angle in terms of balloon height, then can calculate the numerical value of the derivative for you. (see attached)
