Help cylinder volume optimization problem

**wair** · November 25th 2010, 10:08 AM

For the hollow cylinder shown, assume that R and r are increasing at a rate of 2 m/s, and h is decreasing at a rate of 3 m/s. at what rate is the volume changing when R= 7m and r =4m and h = 5m ?

please help me on the path to solving this problem. I'm sorry the picture isn't perfect.

**wair** · November 25th 2010, 10:09 AM

hmm does anyone know how to get the picture to show up ? with out having to click on it ?

**skeeter** · November 26th 2010, 10:08 AM

$V = \pi(R^2 - r^2)h$

take the derivative of the above equation w/r to time, then sub in your given values to determine $\frac{dV}{dt}$

**wair** · November 26th 2010, 12:35 PM

Thanks again skeeter. I really appreciate your help.

**wair** · November 27th 2010, 12:32 PM

I still can't solve this problem.. When I take the derivative of the equation I get 2pieR - 2pier but my H's are gone ? how do I take the derivative with respect to time correctly ?

**skeeter** · November 27th 2010, 12:34 PM

Originally Posted by wair

I still can't solve this problem.. When I take the derivative of the equation I get 2pieR - 2pier but my H's are gone ? how do I take the derivative with respect to time correctly ?

product rule ... R, r, and h are all variables.

**wair** · November 27th 2010, 12:41 PM

So I can't multiply everything out then take the derivative ? Also could you tell me how you came up with the equation ? ohh and what about the rates at which r, R, and h change that are given in the problem, are those just extra information ?

**skeeter** · November 27th 2010, 12:49 PM

Originally Posted by wair

So I can't multiply everything out then take the derivative ? Also could you tell me how you came up with the equation ? ohh and what about the rates at which r, R, and h change that are given in the problem, are those just extra information ?

you can multiply it all out but you still will have to use the product rule.

this solid is a large cylinder with a smaller cylindrical "hole" in it ... $\pi R^2 h - \pi r^2 h = \pi(R^2 - r^2)h$

as I stated earlier, once you get the derivative w/r to time sub in your given values for R, r, h, and their respective rates of change to determine dV/dt.

**wair** · November 27th 2010, 12:57 PM

Ok so once I take the derivative I get, 2hRpi + piR^2 - pir^2 + 2hrpi. but now if i substitute I have two values of R,r and h. I have values of rates and I have values of size. kind of like position and velocity. so how do i relate the two together ?

**skeeter** · November 27th 2010, 01:05 PM

Originally Posted by wair

Ok so once I take the derivative I get, 2hRpi + piR^2 - pir^2 + 2hrpi. but now if i substitute I have two values of R,r and h. I have values of rates and I have values of size. kind of like position and velocity. so how do i relate the two together ?

your derivative is incorrect. are you familiar with the technique of implicit differentiation in solving related rates problems? It is important that you clearly understand that R, r, and h are all implicit functions of time.

note the following derivative for the first term on the right side of the equation $V = \pi R^2h - \pi r^2 h$

$\displaystyle \frac{d}{dt} (\pi R^2 h) = \pi R^2 \cdot \frac{dh}{dt} + 2\pi R h \cdot \frac{dR}{dt}$

**wair** · November 27th 2010, 01:14 PM

Ok is dh/dt equivilant to writing h prime ? ok I understand that now, so when I solve for any related rates problem I use implicit differentiation. ok so now the rate values would be substituted with the R prime h prime and r prime variables, or the dR/dt, dh/dt and dr/dt correct ?

**wair** · November 27th 2010, 02:26 PM

I get -122.52 m^3/s for this problem. Could you please confirm if this is correct ?

**skeeter** · November 27th 2010, 02:46 PM

Originally Posted by wair

I get -122.52 m^3/s for this problem. Could you please confirm if this is correct ?

correct.

$\displaystyle \frac{dV}{dt} = -39\pi \, \frac{m^3}{s}$

**wair** · November 27th 2010, 02:51 PM

Thanks alot for all your help.

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