question archive A television camera is positioned 4000 feet from the base of a rocket launching pad
A television camera is positioned 4000 feet from the base of a rocket launching pad
Subject:MathPrice: Bought3
A television camera is positioned 4000 feet from the base of a rocket
launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. In addition, the auto-focus of the camera has to take into account the increasing distance between the camera and the rocket. We assume that the rocket rises vertically.
(a) Draw a figure that summarizes the given situation. What parts of the picture are changing? What parts are constant? Introduce the appropriate variables to represent the quantities that are changing.
(b) Find an equation that relates the camera's angle of elevation to the height of the rocket, and then find an equation that relates the instantaneous rate of change of the camera's elevation angle to the instantaneous rate of change of the rocket's height (all rates of change are with respect to time).
(c) Find an equation that relates the distance from the camera to the rocket and the rocket's height, as well as an equation that relates the instantaneous rate of change of distance from the camera to the rocket and the instantaneous rate of change of the rocket's height
(d) Suppose that the rocket's speed is 600 ft/sec at the instant it has risen 3000 ft. How fast is the distance from the television camera to the rocket changing at that moment? If the camera is following the rocket, how fast is the camera's angle of elevation changing at that same moment?
(e) If from an elevation of 3000 ft onward, the rocket continues to rise at 600 ft/sec, will the rate of change of distance with respect to time be greater when the elevation is 4000 feet than it was at 3000 feet or less? Explain how you know.
