A television camera is positioned 4000 ft from the base of a rocket launching pa

Question

A television camera is positioned 4000 ft 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. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 500 ft/s when it has risen 3000 ft. (Round your answers to three decimal places.) (a) How fast is the distance from the television camera to the rocket changing at that moment? ? = ft/s (b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment? ? = rad/s

Explanation / Answer

At any time during the rocket's ascent, the distance between the rocket and the TV camera is D^2 = 10000^2 + H^2 where D = distance of the rocket from the TV camera at any time H = height at which the rocket has risen Differentiating the above, 2D(dD/dt) = 0 + 2H(dH/dt) Simplifying, D(dD/dt) = H(dH/dt) Solving for dD/dt, dD/dt = (H/D)(dH/dt) --- call this Equation A At this point, D = sqrt(10000^2 + 24000^2) D = 26000 Substituting values in Equation A, dD/dt = (24000/26000)(500) dD/dt = 461.54 ft/sec.

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