QuestionA balloon rises vertically from the ground 200 m away from an observer. It rises with a position function $h=50 t^{2}$ where $h$ is the height of the balloon in metres and $t$ is in seconds. How fast is the angle of elevation changing 2 seconds after the balloon leaves the ground?

Studdy Solution
Evaluate at $t = 2$ seconds: First, find $h(2) = 50(2)^2 = 200$ meters.
Calculate $\tan(\theta) = \frac{h(2)}{200} = \frac{200}{200} = 1$.
Thus, $\theta = \tan^{-1}(1) = \frac{\pi}{4}$.
Since $\sec^2(\theta) = 1 + \tan^2(\theta) = 2$, substitute into the derivative: $$\frac{d\theta}{dt} = \frac{2}{2 \times 2} = \frac{1}{2}$$
The rate of change of the angle of elevation at $t = 2$ seconds is:
$$\boxed{\frac{1}{2} \text{ radians per second}}$$