7) A TV camera is positioned level with and 4000 feet from a rocket launching pa

Question

7) A TV camera is positioned level with and 4000 feet from a rocket launching pad. Assume that the rocket travels straight upward. At the moment when the rocket is 3000 feet above the ground and traveling 600 ft/s, determine: a) Determine the rate of change of the distance between the camera and the rocket. b) Determine the rate of change of the angle of elevation of the camera. Basically, how fast does the camera angle have to change to keep the rocket in the picture? In problem 7, if the camera is on a cart that is moving away from the launching pad, how fast would the cart have to move in order to keep the angle fixed?

Explanation / Answer

a)Let the height of the rocket at any time be y

Distance between camera and rocket d=sqrt(4000^2+y^2)

dd/dt=y(dy/dt)/sqrt(4000^2+y^2)

At y=3000,

dd/dt=3000*600/sqrt(4000^2+3000^2)=360ft/s

b)Theta=Tan^-1(y/4000)

d(theta)/dt=4000*(dy/dt)/(y^2+4000^2)=4000*600/(3000^2+4000^2)=0.096rad/s=5.50 deg/s

c)Theta=Tan^-1(y/(4000+x))

Since theta is constant, tan theta is also constant.

Tan(theta)=y/(4000+x)=c

y=c(4000+x)

Differentiating wrt t,

dy/dt=c*dx/dt

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