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.

Question

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 1000 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?

(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?

Answer 1

(a) D = √(4000² + H²); dD/dt = H(dH/dt)/√(4000² + H²) = 3000ft·1000ft/s/50000ft = 600 ft/s

(b) tanθ = H/4000; sec²θdθ/dt = (dH/dt)/4000; dθ/dt = (dH/dt)cos²θ/4000 = 1000ft/s(4/5)²/4000ft = 0.16 rad/s

Answer 2

Let x = distance (in ft) travelled by the rocket t seconds after launch

y = distance between the camera and the rocket at time t

Draw a right triangle with horizontal side 4000 ft, vertical side x and hypotenuse y.

Let θ be the angle formed by the hypotenuse and the horizontal side

a. Find dy/dt when x = 3000.

Given dx/dt = 1000

When x = 3000, y² = 3000² + 4000². So, y = 5000.

x² + 4000² = y²

2x(dx/dt) = 2y(dy/dt) So, dy/dt = (x/y)(dx/dt) = (3000/5000)(1000) = 600.000 ft/sec

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b. Find dθ/dt when x = 3000

tanθ = x / 4000

sec²θ(dθ/dt) = (1/4000)(dx/dt)

dθ/dt = (1/4000)(cos²θ)(dx/dt) = (1/4000ft)(4/5)²(1000ft/sec) = 0.160 radians / sec