Question: An image with a horizontal line, a point labelled "Aircraft" above the line and another point lablled "Anti-aircraft gun" on the right side of the horizontal line. This makes a right triangle with the angle at the point "Anti-aircraft gun" labelled "theta" and the triangle side adjacent to the point labelled "x" An aircraft is flying at 400 ft/s with an elevation of 180 ft from the ground, on a straight-line path that will take it directly over an anti-aircraft gun. How fast, in radians per second, will the gun have to turn to accurately track the aircraft when the plane is: 1000 feet away? radian/s 100 feet away? radian/s

1. **State the problem:** An aircraft is flying horizontally at a speed of $400$ ft/s at an elevation of $180$ ft above the ground. The anti-aircraft gun is tracking the aircraft by rotating an angle $\theta$ from the horizontal. We want to find how fast the gun must turn, i.e., the angular velocity $\frac{d\theta}{dt}$, when the aircraft is $1000$ ft and $100$ ft away horizontally. 2. **Set up the variables and formula:** Let $x$ be the horizontal distance from the gun to the aircraft. The elevation is constant at $180$ ft. The angle $\theta$ satisfies: $$\tan(\theta) = \frac{180}{x}$$ 3. **Differentiate with respect to time $t$:** Using implicit differentiation: $$\sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{180}{x^2} \cdot \frac{dx}{dt}$$ Since $\sec^2(\theta) = 1 + \tan^2(\theta)$, and $\tan(\theta) = \frac{180}{x}$, we have: $$\sec^2(\theta) = 1 + \left(\frac{180}{x}\right)^2 = 1 + \frac{32400}{x^2} = \frac{x^2 + 32400}{x^2}$$ 4. **Solve for $\frac{d\theta}{dt}$:** $$\frac{d\theta}{dt} = -\frac{180}{x^2} \cdot \frac{dx}{dt} \cdot \frac{1}{\sec^2(\theta)} = -\frac{180}{x^2} \cdot \frac{dx}{dt} \cdot \frac{x^2}{x^2 + 32400} = -\frac{180 \cdot \frac{dx}{dt}}{x^2 + 32400}$$ 5. **Plug in the values:** The aircraft is moving towards the gun, so $\frac{dx}{dt} = -400$ ft/s (distance decreasing). - When $x = 1000$ ft: $$\frac{d\theta}{dt} = -\frac{180 \times (-400)}{1000^2 + 32400} = \frac{72000}{1000000 + 32400} = \frac{72000}{1032400} \approx 0.0697 \text{ radians/s}$$ - When $x = 100$ ft: $$\frac{d\theta}{dt} = -\frac{180 \times (-400)}{100^2 + 32400} = \frac{72000}{10000 + 32400} = \frac{72000}{42400} \approx 1.698 \text{ radians/s}$$ 6. **Interpretation:** The gun must turn faster as the aircraft gets closer. **Final answers:** - At $1000$ ft away, angular velocity $\approx 0.0697$ radians/s. - At $100$ ft away, angular velocity $\approx 1.698$ radians/s.