Thread: Calculus Related Rates :S

If the rocket shown Below is rising vertically at 268 m/s when it is 1220m up, how fast is the camera-to-rocket distance changing at that instant?

Theres a picture of a right angle triangle is below the question and the horizontal or base length is 915m

howww do u dooo thissss!!:S:S:S

Use ol' Pythagoras,



Differentiate:



Find D by using Pythagoras. You are given dy/dt, x, y

dx/dt is 0 because x is a constant.

Plug them in and solve for dD/dt

A word of advice. I assume you are dealing with related rates in calc right

now. Whenever you see a right triangle, the chances are Pythagoras is going

to be involved somehow. Think about it that way. Label all the knowns you

are given and then try to find what they are asking for. In this case, they

wanted D, the hypoteneuse of the triangle. Since they asked for how fast it

is changing, we know we need dD/dt.

Sometimes they may ask for how fast the angle is changing. That is another matter using tan, but one can still incorporate Pythagoras.

Ahh ic, yea we're doing related rates and optimization right now. that is a very helpful hint thank you very much.

one question, if the x value is 0, would that mean the whole x dx/dt, cancels out leaving just y dy/dt?

In this case, x=915. But the rocket is not moving horizontally, just vertically.

Therefore, dx/dt=0

Thus, we have

Did you use Pythagoras to find D?. You have x=915 and y=1220

sorry i didnt reply, i just noticed your question. uhm yes i did and i got 1525 as D

Yep.Now all you do is plug in all your knowns(you have them all) and find dD/dt.

Alright kewl and i got the right answer thank you! if i have another question should i start a new thread or could i ask you through this thread because it is still related rates:S im new tho this forumn sorry-_-

Start a new thread. That is best with a 'whole nuther' problem.

If you have another, post what you attempted so far and then we can see where you may have went astray.