Answer:
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!!!
