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

1. ## 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):

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

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

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

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

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

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

Can you continue from here?

Grandad