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 Mathman97 October 25th, 2014 05:29 PM

Related Rates Help

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?

 skeeter October 25th, 2014 06:13 PM

Quote:
 Originally Posted by Mathman97 (Post 210495) 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?
For the cone ...

$\displaystyle \frac{r}{h}=\frac{1}{3}$

$\displaystyle \frac{dV}{dt}=-8 \, ft^3/min$

$\displaystyle V=\frac{\pi}{3} r^2h$

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.

 Mathman97 October 25th, 2014 06:31 PM

Where did the 1/3 come from?

 jks October 25th, 2014 08:03 PM

The conical filter has a height of 24 inches and a radius of 8 inches so $\large \frac{r}{h}=\frac{1}{3}$.

 skeeter October 25th, 2014 08:04 PM

Quote:
 Originally Posted by Mathman97 (Post 210497) Where did the 1/3 come from?
What is the cone's radius & height?

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