October 25th, 2014, 06:29 PM  #1 
Newbie Joined: Oct 2014 From: California Posts: 5 Thanks: 0  Related Rates Help
Can someone please help? I'm a Calculus Ab student in highschool and I'm having trouble figuring this problem out. Thanks in advance. A solution is passing through a conical filter 24 inches deep and 16 inches across the top into a cylindrical vessel of diameter 12 inches at a rate of 8 pi in^3/sec. a.) At what rate is the level of the solution in the filter changing when its depth is 12 inches? b.) At what rate is the level of the solution in the cylinder changing? 
October 25th, 2014, 07:13 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,678 Thanks: 1338  Quote:
$\displaystyle \frac{r}{h}=\frac{1}{3}$ $\displaystyle \frac{dV}{dt}=8 \, ft^3/min$ $\displaystyle V=\frac{\pi}{3} r^2h$ Solve for r in terms of h, sub into the cone volume formula, then take the derivative w/r to time. Sub in your known values & solve for dh/dt. Cylinder part should be easy ... Remember dV/dt will be + because it is filling up. Let's see what you can do.  
October 25th, 2014, 07:31 PM  #3 
Newbie Joined: Oct 2014 From: California Posts: 5 Thanks: 0 
Where did the 1/3 come from?

October 25th, 2014, 09:03 PM  #4 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 607 Thanks: 83 Math Focus: Electrical Engineering Applications 
The conical filter has a height of 24 inches and a radius of 8 inches so $\large \frac{r}{h}=\frac{1}{3}$.

October 25th, 2014, 09:04 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,678 Thanks: 1338  

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