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