1. **State the problem:** A camera is mounted 3000 ft from the base of a rocket launching pad. The rocket rises vertically at 880 ft/s when it is 4000 ft above the pad. We want to find how fast the camera's elevation angle $\theta$ must change at that instant to keep the camera aimed at the rocket. 2. **Set up variables and formula:** Let $x = 3000$ ft be the horizontal distance from the camera to the pad (constant), and $y(t)$ be the rocket's height at time $t$. The angle $\theta$ is the elevation angle from the camera to the rocket. We have: $$\tan(\theta) = \frac{y}{x}$$ 3. **Differentiate with respect to time $t$:** Using implicit differentiation, $$\sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{x} \frac{dy}{dt}$$ 4. **Solve for $\frac{d\theta}{dt}$:** $$\frac{d\theta}{dt} = \frac{1}{x} \frac{dy}{dt} \cos^2(\theta)$$ 5. **Calculate $\theta$ at the instant:** $$\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{4000}{3000}\right) = \arctan\left(\frac{4}{3}\right)$$ 6. **Calculate $\cos^2(\theta)$:** Using the identity $\cos^2(\theta) = \frac{1}{1+\tan^2(\theta)}$, $$\cos^2(\theta) = \frac{1}{1 + \left(\frac{4}{3}\right)^2} = \frac{1}{1 + \frac{16}{9}} = \frac{1}{\frac{25}{9}} = \frac{9}{25}$$ 7. **Plug in values:** $$x = 3000, \quad \frac{dy}{dt} = 880, \quad \cos^2(\theta) = \frac{9}{25}$$ $$\frac{d\theta}{dt} = \frac{1}{3000} \times 880 \times \frac{9}{25} = \frac{880 \times 9}{3000 \times 25} = \frac{7920}{75000} = 0.1056 \text{ radians per second}$$ **Final answer:** The camera's elevation angle must change at approximately $0.106$ radians per second to keep aimed at the rocket.