Derivative word problem.

**Hibijibi** · March 2nd 2008, 08:28 PM

No idea what to do here, and all help is appreciated.

A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate of 10 cm^3/min. When the depth of the water in the cone is 8 cm, the depth is decreasing at 2 cm/min. What is the ratio of the height of the cone to its radius?

Help.

**mr fantastic** · March 2nd 2008, 09:03 PM

Originally Posted by Hibijibi

No idea what to do here, and all help is appreciated.

A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate of 10 cm^3/min. When the depth of the water in the cone is 8 cm, the depth is decreasing at 2 cm/min. What is the ratio of the height of the cone to its radius?

Help.

A very similar question was asked here. Read over the reply and then try your question again.

**Hibijibi** · March 2nd 2008, 09:10 PM

That problem kind of confused me. :/

**mr fantastic** · March 2nd 2008, 09:13 PM

Originally Posted by Hibijibi

That problem kind of confused me. :/

Then I'm not sure how a solution to your specific problem will be any less confusing for you .....?

**Hibijibi** · March 2nd 2008, 09:15 PM

Well, I understood what was happening, but I don't know how to take what was applied to that problem and use it to find a radius, which was given in that one. Please also bear in mind I am no where near your level of math education, and my perception into all of this is much more limited. I apologize for such things.

**mr fantastic** · March 3rd 2008, 12:20 AM

Originally Posted by Hibijibi

Well, I understood what was happening, but I don't know how to take what was applied to that problem and use it to find a radius, which was given in that one. Please also bear in mind I am no where near your level of math education, and my perception into all of this is much more limited. I apologize for such things.

Since you're dealing with a related rates problem, the chain rule is the key:

$\frac{dV}{dt} = \frac{dV}{dh} \, \frac{dh}{dt}$ .

When h = 8, $\frac{dV}{dt} = -10 \,$ cubic cm per minute and $\frac{dh}{dt} = -2 \,$ cm per minute.

Therefore when h = 8: $10 = \frac{dV}{dh} \, 2 \Rightarrow \frac{dV}{dh} = 5$ .

Let the radius of the conical volume be x when h = 8. Draw a side view of the filter and use similar triangles:

$\frac{x}{8} = \frac{r}{h} \Rightarrow r = \frac{hx}{8}$ .

Therefore:

$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left( \frac{hx}{8} \right)^2 h = \pi \, x^2 \, \frac{h^3}{192} \Rightarrow \frac{dV}{dh} = \pi x^2 \frac{h^2}{64}$ .

Therefore, when h = 8, $\frac{dV}{dh} = 5 \Rightarrow \pi x^2 \frac{8^2}{64} = 5 \Rightarrow \pi x^2 = 5 \Rightarrow x^2 = \frac{5}{\pi}$ .

Substitute $x^2 = \frac{5}{\pi}$ into $\frac{x}{8} = \frac{r}{h} \Rightarrow \frac{x^2}{64} = \left( \frac{r}{h} \right)^2$ and solve for the required ratio.

There are probably careless mistakes in the above - check for them. The method is what matters.

**Hibijibi** · March 3rd 2008, 06:46 AM

After looking at it, it seems so easy. Much appreciate the help! I finally get it, sort of!

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