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 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?

a) The distance from the television camera to the rocket is changing at that moment at a speed of

600 ft/s

b) the camera's angle of elevation is changing at that same moment at a rate of

0.16 rad/s = 9.16°/s

Step-by-step explanation:

This is a trigonometry relation type of problem.

An image of when the rocket is 3000 ft from the ground is presented in the attached image.

Let the angle of elevation be θ

The height of the rocket at any time = h

The distance from the camera to the rocket = d

a) At any time, d, h and the initial distance from the camera to the rocket can be related using the Pythagoras theorem.

d² = h² + 4000²

Take the time derivative of both sides

(d/dt) (d²) = (d/dt) [h² + 4000²]

2d (dd/dt) = 2h (dh/dt) + 0

At a particular instant,

h = 3000 ft,

(dh/dt) = 1000 ft/s

d can be obtained using the same Pythagoras theorem

d² = h² + 4000² (but h = 3000 ft)

d² = 3000² + 4000²

d = 5000 ft

2d (dd/dt) = 2h (dh/dt) + 0

(dd/dt) = (h/d) × (dh/dt)

(dd/dt) = (3000/5000) × (1000)

(dd/dt) = 600 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?

At any moment in time, θ, h and the initial distance of the camera from the base of the rocket are related through the trigonometric relation

Tan θ = (h/4000) = 0.00025h

Taking the time derivative of both sides

(d/dt) (Tan θ) = (d/dt) (0.00025h)

(Sec² θ) (dθ/dt) = 0.00025 (dh/dt)

At the point where h = 3000 ft, we can calculate the corresponding θ at that point

Tan θ = (3000/4000)

θ = tan⁻¹ (0.75) = 0.6435 rad

(Sec² θ) (dθ/dt) = 0.00025 (dh/dt)

(Sec² 0.6435) (dθ/dt) = 0.00025 (1000)

1.5625 (dθ/dt) = 0.25

(dθ/dt) = (0.25/1.5625) = 0.16 rad/s

Hope this Helps!!!

4th option is correct xy=-10

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