6. A plane is flying horizontally at a constant height of 4000 ft above a fixed observation point. At a certain instant the angle of elevation is \( \pi / 6 \) and decreasing, and the speed of the plane is 300 mph . How fast is the angle decreasing?

To find how fast the angle of elevation is decreasing, we can use related rates. Let \( h = 4000 \) ft be the height of the plane, \( x \) be the horizontal distance from the observation point to the plane, and \( \theta \) be the angle of elevation. Given that the angle of elevation is \( \frac{\pi}{6} \) radians when the plane's speed is 300 mph (which we'll convert to feet per second). First, we find the relation between \( h \), \( x \), and \( \theta \): \[ \tan(\theta) = \frac{h}{x} \] Differentiating both sides with respect to time \( t \), we have: \[ \sec^2(\theta) \frac{d\theta}{dt} = -\frac{h}{x^2} \frac{dx}{dt} \] At \( \theta = \frac{\pi}{6} \), \( \sec^2\left(\frac{\pi}{6}\right) = \frac{4}{3} \). The speed of the plane is \( \frac{dx}{dt} = 300 \text{ mph} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} = 440 \text{ ft/s} \). Next, we can find \( x \) when \( \theta = \frac{\pi}{6} \): \[ \tan\left(\frac{\pi}{6}\right) = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{4000}{x} \Rightarrow x = 4000\sqrt{3} \] Now we plug in everything we've found: \[ \frac{4}{3} \frac{d\theta}{dt} = -\frac{4000}{(4000\sqrt{3})^2} (440) \] \[ \frac{d\theta}{dt} = -\frac{440 \cdot 4000}{(4000\sqrt{3})^2} \cdot \frac{3}{4} \] Calculating this gives us: \[ \frac{d\theta}{dt} = -\frac{440 \cdot 3}{4 \cdot 4000 \cdot 3} = -\frac{440}{16000} = -\frac{11}{400} \text{ radians per second} \] Thus, the angle is decreasing at a rate of \( \frac{11}{400} \text{ radians per second} \).