1. ## Conical tank...related rates

Hi,

I've explained everything here and have written my thoughts for the problem as well:

http://i44.tinypic.com/241v2m8.jpg

2. can anyone help wth even parts of my question if it's not all of it?

-thanks!!

3. Hi

Note that $\frac{h}{r} = \frac{H}{R}$

At instant t, the height is h
At instant (t+dt), the height is (h+dh)
The corresponding volume between t and (t+dt) is

$dV = V(h+dh) - V(h) = \frac{\pi}{3}\:\left(\frac{R^2}{H^2}\:\left(h+dh\r ight)^3 - \frac{R^2}{H^2}\:h^3\right) = \frac{\pi\:R^2}{3\:H^2}\:3\:h^2\:dh = \frac{\pi\:R^2\:h^2}{H^2}\:dh$

The flowrate is

$Q = \frac{dV}{dt} = \frac{\pi\:R^2\:h^2}{H^2}\:\frac{dh}{dt}$

Therefore $\frac{dh}{dt} = \frac{Q\:H^2}{\pi\:R^2\:h^2} = \frac{289}{18\pi}$

4. Hello, janedoe!

A conical tank is 12 ft across the top and 17 ft deep.
If water is flowing into the tank at a rate of 8 ft³/min,
find the rate of change of the depth when the water is 2 ft deep.
Code:
              :   6   :
- *-------+-------*
:  \      |      /
:   \     |  r  /
:    *----+----*
17     \   |   /
:      \ h|  /
:       \ | /
:        \|/
-         *

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

From the similar right triangles: . $\frac{r}{h} \:=\:\frac{6}{17} \quad\Rightarrow\quad r \:=\:\frac{6}{17}\,h$

Substitute into [1]: . $V \;=\;\frac{\pi}{3}\left(\frac{6}{17}\,h\right)\!\! ^2h \quad\Rightarrow\quad V \:=\:\frac{12\pi}{289}\,h^3$

Differentiate with respect to time: . $\frac{dV}{dt} \:=\:\frac{36\pi}{289}\,h^2\,\frac{dh}{dt}$

Since $\frac{dV}{dt} = 8,\;h = 2$. we have: . $8 \:=\:\frac{36\pi}{289}(2^2)\frac{dh}{dt}$

. . Therefore: . $\frac{dh}{dt} \:=\:\frac{289}{18\pi}$ ft/min.

And thank you, thank you for sharing your work and your thoughts!

You put in a lot of work before asking for help.
It is an extremely rare occurence . . .We truly appreciate it!

5. It's the least I could do!

Actually I did have that proportion r/h=6/17, but I just plugged in 2 for h... I see that I can't b/c that's a variable right?

I think I'm getting this now...I'm teaching myself calc so let's hope I do! haha
Thanks for the help!!!