1. Rate of change problem

the radius of a cone is increasing at a rate of 3 in/sec, and the height of the cone is 3 times the radius. Find the rate of change for the volume of that cone when the radius is 7 inches.

The textbook is no help and the prof doesnt teach.

any help will be appreciated

2. Re: Rate of change problem

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

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

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

Do you know how to proceed from there?

Spoiler:
Substitute 3r for h in the equation for the volume of a cone - this will now become $V = \dfrac{1}{3}\pi r^3 \cdot 3r = \pi r^4$

3. 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.

4. Re: Rate of change problem

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

Pi 4R^3 x 3.

6. Re: Rate of change problem

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

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

7. 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.

8. Re: Rate of change problem

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

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

9. 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.