Related Rates Problem

**lysserloo** · November 10th 2009, 08:35 AM

Problem:
A coffee filter in the shape of a right circular cone with a base radius of 6cm and depth 10cm contains water, which drips out through a hole at the bottom at a constant rate of $2cm^3$ per second. How fast is the water level falling when the depth is 8cm.

What I've Done:

So I know I have $\frac{dv}{dt}$ which is $2cm^3$ . I also know that I need $\frac{dh}{dt}$ when h = 8cm. And I know that I have to use $V = \frac{1}{3} \pi r^2h$ to find an equation in terms of h (the depth).

I'm also pretty sure I need to find a way to relate r and h so I can plug that in for r, and get a volume equation entirely in terms of h. Is that true?

This is what I've tried to relate r and h:

$tan(theta) = \frac{10}{6}$
$theta = 68.2 degrees$

so $r = \frac{h}{tan(68.2)}$

Then I plugged that into my volume equation for r:

$V = \frac{1}{3}\pi(\frac{h}{tan(68.2)})^2 h$

Before I continue trying to solve it this way, I'd like to know if I'm doing this correctly or not.

If not, can anybody help me with the problem?

Thank you very much!!

**Soroban** · November 10th 2009, 10:31 AM

Hello, lysserloo!

We can solve this without using Trig.

A coffee filter in the shape of a right circular cone with a base radius of 6cm and depth 10cm
contains water, which drips out through a hole at the bottom at a constant rate of 2 cm³/sec.
How fast is the water level falling when the depth is 8 cm?

Code:

      : - 6 - : - 6 - :
    - *-------+-------*
    :  \      |      /
    :   \     |  r  /
    :    \- - + - -/
   10     \   |   /
    :      \ h|  /
    :       \ | /
    :        \|/
    -         *

The volume of the water is: . $V \:=\:\tfrac{\pi}{3}r^2h$ .[1]

From the similar right triangles: . $\frac{r}{h} \:=\:\frac{6}{10} \quad\Rightarrow\quad r \:=\:\tfrac{3}{5}h$

Substitute into [1]: . $V \:=\:\tfrac{\pi}{3}\left(\tfrac{3}{5}h\right)^2h \quad\Rightarrow\quad V \:=\:\frac{3\pi}{25}h^3$

Differentiate with respect to time: . $\frac{dV}{dt} \:=\:\frac{9\pi}{25}h^2\left(\frac{dh}{dt}\right)$ .[2]

We are given: . $h = 8,\;\;\frac{dV}{dt} = \text{-}2$

Substitute into [2]: . $-2 \;=\;\frac{9\pi}{25}(8^2)\cdot\frac{dh}{dt}$

Therefore: . $\frac{dh}{dt} \:=\:-\frac{25}{288\pi}$ cm/sec.

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