Re: Rate of change problem

You should know that the volume of a cone is $\displaystyle V_{cone} = \dfrac{1}{3} \pi r^3h$ and you are given $\displaystyle \dfrac{dr}{dt} = 3 \text{ and } h = 3r$

The question is asking you to find $\displaystyle \dfrac{dV}{dt}$.

From the chain rule we know that $\displaystyle \dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

Do you know how to proceed from there?

Re: Rate of change problem

I was stuck on H= Pi r^3. Basically, i am stuck at where you stopped. Pi R^4 works also.

Re: Rate of change problem

If $\displaystyle V = \pi r^4$ then what is $\displaystyle \dfrac{dV}{dr}$?

Re: Rate of change problem

Re: Rate of change problem

That is the answer to the final question: $\displaystyle \dfrac{dV}{dt} = 12\pi r^3 $.

To find the rate of change at 7 inch radius sub in $\displaystyle r=7$ into the above equation

Re: Rate of change problem

I did that before but it is different from the answer i have. The answer for the question is 441pi = 1385.4 from the final review sheet. I tried every way possible that I know of but couldnt get 441pi = 1385.4.

edit - what you did seem to be the correct way but the outcome is different from the answer on my final review sheet. I was doing your way before i came here, but my answer never match the solution on the review sheet.(Crying)

Re: Rate of change problem

the correct solution is $\displaystyle \frac{dV}{dt} = 4116\pi \, in^3/sec$

... looks like a digit was left off the end of the given "answer".

Re: Rate of change problem

4116 was what i got numerous time. I guess the Math department made a mistake. Thank you for all of your help I really appreciate it.