6. A rocket is rising vertically. An observer on the ground is standing a certain distance from the rocket's launch point. As the observer watches the rocket.the angle of elevation is increasing. Relate the rates of change of the angle of elevation with the speed of the rocket.

To relate the rate of change of the angle of elevation (\( \frac{d\theta}{dt} \)) with the speed of the rocket (\( v \)), we can use trigonometry and related rates.<br /><br />Let's denote:<br />- \( \theta \) as the angle of elevation.<br />- \( x \) as the horizontal distance between the observer and the rocket's launch point.<br />- \( y \) as the vertical distance of the rocket from the ground.<br /><br />Given that the rocket is rising vertically, the relationship between \( x \) and \( y \) can be expressed using the tangent function:<br />\[ \tan(\theta) = \frac{y}{x} \]<br /><br />To find the rate of change of the angle of elevation, we differentiate both sides of the equation with respect to time \( t \):<br />\[ \frac{d}{dt} [\tan(\theta)] = \frac{d}{dt} \left( \frac{y}{x} \right) \]<br /><br />Using the chain rule on the left side:<br />\[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{x} \cdot \frac{dy}{dt} \]<br /><br />We know that \( \frac{dy}{dt} = v \), the speed of the rocket. Rearranging the equation to solve for \( \frac{d\theta}{dt} \):<br />\[ \frac{d\theta}{dt} = \frac{v}{x \cdot \sec^2(\theta)} \]<br /><br />Since \( \sec^2(\theta) = \frac{1}{\cos^2(\theta)} \), we can rewrite the equation as:<br />\[ \frac{d\theta}{dt} = \frac{v}{x \cdot \frac{1}{\cos^2(\theta)}} \]<br />\[ \frac{d\theta}{dt} = \frac{v \cdot \cos^2(\theta)}{x} \]<br /><br />Therefore, the rate of change of the angle of elevation (\( \frac{d\theta}{dt} \)) is directly proportional to the speed of the rocket (\( v \)) and inversely proportional to the horizontal distance (\( x \)) between the observer and the rocket's launch point. The relationship is given by:<br />\[ \frac{d\theta}{dt} = \frac{v \cdot \cos^2(\theta)}{x} \]