A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by $\theta=\tan ^{-1}\left(\frac{x}{2000}\right)$ where $x$ is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 3940 feet apart. \[ \theta^{\prime}= \] radians per foot. Round to five decimal places.

Step 1 ：We are given that the television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically. The angle of elevation of the camera can be found by \(\theta=\tan ^{-1}\left(\frac{x}{2000}\right)\) where \(x\) is the height of the rocket.

Step 2 ：We are asked to find the rate of change of the angle of elevation after launch when the camera and the rocket are 3940 feet apart. This is asking for the derivative of the angle with respect to time.

Step 3 ：We can use the Pythagorean theorem to express the height of the rocket in terms of the distance between the camera and the rocket, and then take the derivative of the angle with respect to time using the chain rule.

Step 4 ：However, we realize that we forgot to consider the rate of change of the height of the rocket with respect to time. We need to correct this in our calculations.

Step 5 ：By considering the rate of change of the height of the rocket with respect to time, we find that \(\theta^\prime = \frac{0.257672189440594}{t}\).

Step 6 ：Final Answer: The rate of change of the angle of elevation after launch when the camera and the rocket are 3940 feet apart is approximately \(\boxed{0.25767}\) radians per foot.