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 = 9.16Ā°/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 Īø
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.
dĀ² = hĀ² + 4000Ā²
Take the time derivative of both sides
(d/dt) (dĀ²) = (d/dt) [hĀ² + 4000Ā²]
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
dĀ² = hĀ² + 4000Ā² (but h = 3000 ft)
dĀ² = 3000Ā² + 4000Ā²
d = 5000 ft
2d (dd/dt) = 2h (dh/dt) + 0
(dd/dt) = (h/d) Ć (dh/dt)
(dd/dt) = (3000/5000) Ć (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, Īø, h and the initial distance of the camera from the base of the rocket are related through the trigonometric relation
Tan Īø = (h/4000) = 0.00025h
Taking the time derivative of both sides
(d/dt) (Tan Īø) = (d/dt) (0.00025h)
(SecĀ² Īø) (dĪø/dt) = 0.00025 (dh/dt)
At the point where h = 3000 ft, we can calculate the corresponding Īø at that point
Tan Īø = (3000/4000)
Īø = tanā»Ā¹ (0.75) = 0.6435 rad
(SecĀ² Īø) (dĪø/dt) = 0.00025 (dh/dt)
(SecĀ² 0.6435) (dĪø/dt) = 0.00025 (1000)
1.5625 (dĪø/dt) = 0.25
(dĪø/dt) = (0.25/1.5625) = 0.16 rad/s
Hope this Helps!!!
