A television camera is positioned 10000 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 800 ft/s when it has risen 24000 ft. (Round the answers to three decimal places.) (a) How fast is the distance from the television camera to the rocket changing at that moment?
its easy....first let the camera and rocket in line position..,,,when the rocket has risen find the distance between rocket and camera by using phytagoras theorema.. \[\sqrt{10000^2+ 24000^2}\] = 676000000 ft then v =x \t t can be found,,then the velocity of rocket from camera view can be found by the same equation..
dr/dt = dr/dy * dy/dt or implicitly derive the pythagorean theorum and fill in what you know :)
Ill implicit it... x^2 + y^2 = r^2 10000^2 = r^2 - y^2 dx/dt 2(10000) = dr/dt 2r - dy/dt 2y divide it all by "2" to simplify: (dx/dt) 10000 = (dr/dt)r - (dy/dt)y dx/dt = 0 ; dy/dt = 800 ; y=24000; and r= sqrt(10000^2 - 24000^2) 0 = (dr/dt)sqrt(10000^2 - 24000^2) - (800)(24000) solve for dr/dt...
dr/dt = 800(24000) - sqrt(10000^2 - 24000^2) happy number crunching :)
Thank you
ack!! sign error lol
sqrt(10000^2 + 24000) sneaky little bugger it is :)
your smart; youll be able to read thru the typos lol
Just for eruditional purposes..... dr/dt = 800(24000) - sqrt(10000^2 + 24000^2)
...... murmur...... still got errors in it..... if you can work the algebra, be by guest :) 0 = (dr/dt)sqrt(10000^2 + 24000^2) - (800)(24000) (800)(24000) = (dr/dt)sqrt(10000^2 + 24000^2) (800)(24000) ------------------------ = (dr/dt) sqrt(10000^2 + 24000^2) Maybe? with a little luck? better double check it :)
sure, thanks yahoo, thank you so much, answer is correct
LOL .... that makes us BOTH happy :)
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