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 that its speed is 800 ft/s when it has risen 4000 ft.
How fast is the distance from the camera to the rocket changing in ft/s at that moment?
If the televison camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing in radians/s at that moment
It would help if we knew the distance of the camera to the launch pad, which would give us a constant.
The way you have it involves 3 variables:
the angle, the height, and the distance between rocket and camera.
Are you sure there wasn't more information?
Sorry, I meant a television camera is positioned 2500 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 that its speed is 800 ft/s when it has risen 4000 ft.
How fast is the distance from the camera to the rocket changing in ft/s at that moment?
If the televison camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing in radians/s at that moment?
ok, that's better
make a sketch of the situation, labeling the angle as Ø and the height as h, and the hypotenuse c.
preliminary:
when h = 4000
c^2 = 2500^2 + 4000^2
c = 4716.99
tanØ = 4000/2500 = 1.6
Ø = 57.995°
at that moment: we have a similar triangle with base 5, height 8 and hypotenuse √89
cosØ = 5/√89 ----> needed later
sinØ = 8/√89 ----> needed later
secØ = √89/5 ----> needed later
tanØ = h/2500
2500tanØ = h
2500 sec^2Ø dØ/dt = dh/dt
2500(89/25) dØ/dt = 800
dØ/dt = 25(800)/(2500(89)) = .08988764
in general:
sinØ = h/c
csinØ = h
c cosØ dØ/dt + sinØ dc/dt = dh/dt
for our given:
4716.999(5/√89) (.08988764) + (8/√89) dc/dt = 800
I will leave it up to you to do the "button-pushing" to find dc/dt
also, please check my arithmetic
I haven't checked my answer against Reiny's, but I also took a stab at it in
http://www.jiskha.com/display.cgi?id=1372720314
To find the rate at which the distance from the camera to the rocket is changing, we can use the concept of related rates.
Let's denote the distance from the camera to the rocket as "d" and the time as "t". We are given that the rocket's speed is 800 ft/s, which means the rate at which the rocket is changing its height with respect to time is dh/dt = 800 ft/s.
At any given moment, the camera is located at the point where the rocket is. Therefore, the distance from the camera to the rocket is equal to the rocket's height, which is "h". So, d = h.
We need to find the rate at which the distance from the camera to the rocket is changing, or dd/dt. To do that, we can differentiate both sides of the equation d = h with respect to time:
dd/dt = dh/dt.
Since we know that dh/dt = 800 ft/s, the rate of change of the distance from the camera to the rocket at that moment is also 800 ft/s.
Now, let's find the rate at which the camera's angle of elevation is changing in radians/s.
The angle of elevation can be defined as the angle between the line of sight from the camera to the rocket and the horizontal line. Let's denote this angle as "θ".
To find dθ/dt, we need to differentiate the equation with respect to time:
tan(θ) = h/d.
Differentiating both sides of the equation, we get:
sec^2(θ) * dθ/dt = (dh/dt * d - h * dd/dt) / d^2.
We know that d = h and dd/dt = 800 ft/s, so we can substitute these values into the equation:
sec^2(θ) * dθ/dt = (dh/dt * h - h * 800 ft/s) / h^2.
Since dh/dt = 800 ft/s and d = h, we simplify the equation further:
sec^2(θ) * dθ/dt = (800 ft/s * h - h * 800 ft/s) / h^2.
After simplification, we get:
sec^2(θ) * dθ/dt = 0.
To find dθ/dt, we divide both sides by sec^2(θ):
dθ/dt = 0.
Therefore, the camera's angle of elevation is not changing at that moment, which means dθ/dt = 0 radians/s.
In summary:
- The distance from the camera to the rocket is changing at a rate of 800 ft/s.
- The camera's angle of elevation is not changing at that moment, so the rate of change is 0 radians/s.