Related Rates: Television Camera and Rocket

A television camera is positioned 4000 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 600 ft/s when it has risen 3000 ft.
(a) How fast is the distance from the television camera to the rocket changing at that moment?
(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?

[@coltonbradshaw979]

very helpful, great video!

[@RubiMercuri]

i dont understand you step here @4:00 why did you put 4000 in front of sec squared theta?

[@Turnupmath]

I multiplied both sides by 4000.

[@eatmunchies6617]

why couldnt you use cos, since you do have adj and hypotnuse values

[@Turnupmath]

The point is to get an equation in terms of h and cos doesn't include h.

[@danieltsimoshkin4952]

why tangent ?

[@CrafticJMineCraft]

Very sloppy work. Nicely done however.

[@Turnupmath]

Thanks for the comment. Email me any other related rates problems that you have at [e-mail]

[@taybrazy5942]

what is a 345 triangle? my problem is very similar to yours. my base and height are also 3000 and 4000 but i dont understand how the hypotenuse is 5000. i did the pythagoren theorem and got 7000

[@Turnupmath]

A 345 triangle is a Pythagorean Triple. You can multiply by 1000 to get 3000 4000 and 5000 which is your triangle. The other way is to take sqrt(3000^2+4000^2).