# Help with application: 'Cylinder filling with water, rate of sqrt(x), derive formula'

• Oct 31st 2009, 01:38 AM
Phatmat
Help with application: 'Cylinder filling with water, rate of sqrt(x), derive formula'
Hey guys, I'm completley stuck on question 2)a)i).

Everything I do leads me in a circle, of either the x's cancelling, or me getting no where. Can anyone direct me the right way, how should I set up the problem? Thanks.

Here's the question, I need help on Q2)a)i):
http://img99.imageshack.us/img99/1986/helpm.png
• Oct 31st 2009, 07:33 AM
Hello Phatmat
Quote:

Originally Posted by Phatmat
Hey guys, I'm completley stuck on question 2)a)i).

Everything I do leads me in a circle, of either the x's cancelling, or me getting no where. Can anyone direct me the right way, how should I set up the problem? Thanks.

Here's the question, I need help on Q2)a)i):
http://img99.imageshack.us/img99/1986/helpm.png

When the depth of water is $x$ metres, suppose that the volume of water is $V \text{m}^3$. Then if the time $t$ is measured in minutes, the rate of increase of $V$ with respect to $t$ is $\frac{dV}{dt} \text{m}^3/\text{min}$. This is the rate at which water is filling the tank.

So $\frac{dV}{dt}=\sqrt x$

The volume of a cylinder $= \pi r^2 h$. So when $r = 2$ and $h = x, V = 4\pi x$

$\Rightarrow \frac{dV}{dx}=4\pi$

$\Rightarrow \frac{dV}{dt}=\frac{dV}{dx}\cdot\frac{dx}{dt}=4\pi \frac{dx}{dt}$

$\Rightarrow 4\pi\frac{dx}{dt}=\sqrt x$

Can you continue from here?