Posted by Robin on Wednesday, July 3, 2013 at 10:38am.
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