A television camera is positioned 4000f1 from the base of a rocket launching pad

Question

A television camera is positioned 4000f1 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. A so, 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 600f t/s when it has risen 3000ft. How fast the distance from they television camera to the rocket changing at that moment? If the television camera is always kept lamed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

Explanation / Answer

let camera at distance y=4000 ft from launch pad

let x be the rocket height =3000ft

distance between rocket and camera z=(x2+y2)

z=(30002+40002) =5000ft

dx/dt =600, dy/dt =0(as camera position is constant

z2=x2+y2 by pythogorus theorem

differentiate with respect to time t

2z dz/dt =2xdx/dt +2ydy/dt
z dz/dt =xdx/dt

dz/dt =(x/z)dx/dt

dz/dt =(3000/5000)600

dz/dt =360

distance between camera and rocket increasing at 360ft/sec

b)tan u =x/y

differentiate with respect to time t

sec2u du/dt =(dx/dt)/y

(5000/4000)2du/dt =(600)/4000

du/dt =(3/20)(16/25)

du/dt =0.096

angle changing at 0.096 rad/s

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