Questions: A rocket is launched so that it rises vertically. A camera is positioned 5000 ft from the launch pad. When the rocket is 1000 ft above the launch pad, its velocity is 520 ft / sec. Find the necessary rate of change of the camera's angle as a function of time so that it stays focused on the rocket.

Solution Steps

Let \( y(t) \) be the height of the rocket above the launch pad at time \( t \), and let \( \theta(t) \) be the angle of elevation of the camera. The camera is positioned 5000 ft from the launch pad horizontally.

We have the following right triangle relationship: \[ \tan(\theta) = \frac{y(t)}{5000} \]

To find the rate of change of the angle \( \theta \) with respect to time, we differentiate both sides of the equation with respect to \( t \): \[ \frac{d}{dt} \left( \tan(\theta) \right) = \frac{d}{dt} \left( \frac{y(t)}{5000} \right) \]

Using the chain rule on the left side: \[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{5000} \cdot \frac{dy}{dt} \]

Rearrange the equation to solve for \(\frac{d\theta}{dt}\): \[ \frac{d\theta}{dt} = \frac{1}{5000} \cdot \frac{dy}{dt} \cdot \cos^2(\theta) \]

Given that \( y = 1000 \) ft and \(\frac{dy}{dt} = 520 \) ft/sec, we need to find \(\cos(\theta)\) when \( y = 1000 \) ft. From the right triangle: \[ \cos(\theta) = \frac{5000}{\sqrt{5000^2 + 1000^2}} = \frac{5000}{\sqrt{25000000 + 1000000}} = \frac{5000}{\sqrt{26000000}} = \frac{5000}{\sqrt{26} \cdot 1000} = \frac{5}{\sqrt{26}} \]

Thus, \[ \cos^2(\theta) = \left( \frac{5}{\sqrt{26}} \right)^2 = \frac{25}{26} \]

Substitute \(\cos^2(\theta)\) and \(\frac{dy}{dt}\) into the equation: \[ \frac{d\theta}{dt} = \frac{1}{5000} \cdot 520 \cdot \frac{25}{26} = \frac{520 \cdot 25}{5000 \cdot 26} = \frac{13000}{130000} = \frac{1}{10} \text{ radians/sec} \]

Final Answer