1. **State the problem:** A hot air balloon is 200 feet horizontally from an observer and ascends at 100 ft/min. We want to find how fast the angle of elevation $\theta$ is changing when the balloon is at 2500 feet altitude. 2. **Set up variables:** Let $x=200$ ft (horizontal distance), $y(t)$ be the altitude at time $t$, and $\theta(t)$ be the angle of elevation. 3. **Relationship:** From the right triangle formed, $\tan(\theta) = \frac{y}{x} = \frac{y}{200}$. 4. **Differentiate with respect to time $t$:** $$\frac{d}{dt}[\tan(\theta)] = \frac{d}{dt}\left(\frac{y}{200}\right)$$ Using chain rule: $$\sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{200} \frac{dy}{dt}$$ 5. **Solve for $\frac{d\theta}{dt}$:** $$\frac{d\theta}{dt} = \frac{1}{200} \frac{dy}{dt} \cos^2(\theta)$$ 6. **Find $\cos(\theta)$ when $y=2500$:** From $\tan(\theta) = \frac{2500}{200} = 12.5$, so $$\theta = \arctan(12.5)$$ Using the identity: $$\cos^2(\theta) = \frac{1}{1+\tan^2(\theta)} = \frac{1}{1+12.5^2} = \frac{1}{1+156.25} = \frac{1}{157.25}$$ 7. **Plug in values:** $$\frac{dy}{dt} = 100 \text{ ft/min}$$ $$\frac{d\theta}{dt} = \frac{1}{200} \times 100 \times \frac{1}{157.25} = \frac{100}{200 \times 157.25} = \frac{1}{2 \times 157.25} = \frac{1}{314.5} \approx 0.00318 \text{ radians per minute}$$ **Final answer:** The angle of elevation is changing at approximately $0.00318$ radians per minute when the balloon is at 2500 feet altitude.