1. **Problem statement:** A hot air balloon is rising vertically from a point on a level field. A range finder is located 500 ft from the liftoff point. At the moment when the elevation angle $\theta$ of the range finder is $\frac{\pi}{4}$, the angle is increasing at a rate of $\frac{d\theta}{dt} = 0.14$ rad/min. We need to find how fast the balloon is rising, i.e., the rate of change of the balloon's height $\frac{dh}{dt}$ at that moment. 2. **Set up variables and formula:** Let $h$ be the height of the balloon at time $t$, and $\theta$ be the elevation angle from the range finder to the balloon. The horizontal distance from the range finder to the liftoff point is constant at 500 ft. From the right triangle formed, we have: $$\tan(\theta) = \frac{h}{500}$$ 3. **Differentiate with respect to time $t$:** Using implicit differentiation, $$\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{500} \cdot \frac{dh}{dt}$$ 4. **Solve for $\frac{dh}{dt}$:** $$\frac{dh}{dt} = 500 \cdot \sec^2(\theta) \cdot \frac{d\theta}{dt}$$ 5. **Evaluate at $\theta = \frac{\pi}{4}$ and $\frac{d\theta}{dt} = 0.14$ rad/min:** Since $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, $$\sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right)} = \sqrt{2}$$ Therefore, $$\sec^2\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2$$ Substitute values: $$\frac{dh}{dt} = 500 \times 2 \times 0.14 = 140$$ 6. **Interpretation:** The balloon is rising at 140 ft/min at the moment when the elevation angle is $\frac{\pi}{4}$ and increasing at 0.14 rad/min. **Final answer:** $$\boxed{140 \text{ ft/min}}$$