A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is \( \frac{\pi}{3} \), and it is changing at a rate of \( 0.1 \mathrm{rad} / \mathrm{min} \). How fast is the balloon rising at this moment? miles \( / \mathrm{min} \)

To find how fast the balloon is rising, we can use related rates in calculus. Let \( h \) be the height of the balloon above the ground, and \( \theta \) be the angle between the observer's line of sight and the horizontal. From the observer's perspective, using basic trigonometry in a right triangle, we have: \[ \tan(\theta) = \frac{h}{2} \] Differentiating both sides with respect to time \( t \) gives us: \[ \sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{2} \frac{dh}{dt} \] We know: - \(\theta = \frac{\pi}{3}\) - \(\frac{d\theta}{dt} = 0.1 \, \text{rad/min}\) First, let's calculate \(\sec^2(\theta)\): \[ \sec\left(\frac{\pi}{3}\right) = \frac{1}{\cos\left(\frac{\pi}{3}\right)} = \frac{1}{\frac{1}{2}} = 2 \implies \sec^2\left(\frac{\pi}{3}\right) = 2^2 = 4 \] Now substitute the values into the differentiated equation: \[ 4 \cdot 0.1 = \frac{1}{2} \frac{dh}{dt} \] \[ 0.4 = \frac{1}{2} \frac{dh}{dt} \] To find \(\frac{dh}{dt}\): \[ \frac{dh}{dt} = 0.4 \cdot 2 = 0.8 \, \text{miles/min} \] So, the balloon is rising at a rate of \(\boxed{0.8}\) miles per minute.