A hot air balloon rising vertically is tracked by an observer located 8 km from the lift-off point. At a certain moment, the angle between the observer's line of sight and the horizontal is (pi )/(5) and it is changing at a rate of 0.2rad/min How fast is the balloon rising at this moment? Let y be the height of the balloon (in km), t be time (in minutes), and Theta the angle between the line-of-sight and the horizontal (in radians). (Use decimal notation.Give your answer to two decimal places.) (d^2y)/(dt)approx square kwint

To determine how fast the balloon is rising at the given moment, we need to relate the height of the balloon \( y \) to the angle \( \Theta \) and use the given rate of change of the angle.<br /><br />Given:<br />- The observer is 8 km from the lift-off point.<br />- The angle \( \Theta = \frac{\pi}{5} \) radians.<br />- The rate of change of the angle \( \frac{d\Theta}{dt} = 0.2 \) rad/min.<br /><br />We can use trigonometry to relate \( y \), \( \Theta \), and the distance from the observer to the lift-off point. Specifically, we use the tangent function:<br /><br />\[ \tan(\Theta) = \frac{y}{8} \]<br /><br />Differentiating both sides with respect to time \( t \):<br /><br />\[ \sec^2(\Theta) \cdot \frac{d\Theta}{dt} = \frac{1}{8} \cdot \frac{dy}{dt} \]<br /><br />We need to find \( \frac{dy}{dt} \). First, let's calculate \( \sec^2(\Theta) \):<br /><br />\[ \sec(\Theta) = \frac{1}{\cos(\Theta)} \]<br /><br />Since \( \tan(\Theta) = \frac{y}{8} \), we have:<br /><br />\[ y = 8 \tan\left(\frac{\pi}{5}\right) \]<br /><br />Now, calculate \( \tan\left(\frac{\pi}{5}\right) \):<br /><br />\[ \tan\left(\frac{\pi}{5}\right) \approx 0.7265 \]<br /><br />Thus,<br /><br />\[ y \approx 8 \times 0.7265 = 5.812 \text{ km} \]<br /><br />Next, we calculate \( \sec^2\left(\frac{\pi}{5}\right) \):<br /><br />\[ \cos\left(\frac{\pi}{5}\right) \approx 0.8090 \]<br />\[ \sec\left(\frac{\pi}{5}\right) = \frac{1}{0.8090} \approx 1.236 \]<br />\[ \sec^2\left(\frac{\pi}{5}\right) \approx (1.236)^2 \approx 1.528 \]<br /><br />Now substitute these values into the differentiated equation:<br /><br />\[ 1.528 \cdot 0.2 = \frac{1}{8} \cdot \frac{dy}{dt} \]<br /><br />Solving for \( \frac{dy}{dt} \):<br /><br />\[ 0.3056 = \frac{1}{8} \cdot \frac{dy}{dt} \]<br />\[ \frac{dy}{dt} = 0.3056 \times 8 \]<br />\[ \frac{dy}{dt} \approx 2.44 \text{ km/min} \]<br /><br />Therefore, the balloon is rising at a rate of approximately \( 2.44 \) km/min at that moment.