# Thread: CAN ANY1 HELP ME!!! Math C - Rate's of Change

1. ## CAN ANY1 HELP ME!!! Math C - Rate's of Change

I'm New at math C and this question is bugging me, can any1 solve this equation so i know how to do it?

The question says:

The rate of loss of water from a rainwater tank with a leak in the side depends on the pressure at the hole. The pressure depends on the height of water above the leak.
A cylindrical tank has a leak 20cm from the top, which has not been fixed because it is only a short distance form the top. Five hours after the tank is filled, the level has dropped by 15cm. Using a differential equation, develop a model to describe the height of water above the hole as a function of time. If another hole is made 20cm from the bottom of the tank, which is 2 metres high, how long will it take before half the water has leaked away?

2. Originally Posted by sterny0
I'm New at math C and this question is bugging me, can any1 solve this equation so i know how to do it?

The question says:

The rate of loss of water from a rainwater tank with a leak in the side depends on the pressure at the hole. The pressure depends on the height of water above the leak.
A cylindrical tank has a leak 20cm from the top, which has not been fixed because it is only a short distance form the top. Five hours after the tank is filled, the level has dropped by 15cm. Using a differential equation, develop a model to describe the height of water above the hole as a function of time. If another hole is made 20cm from the bottom of the tank, which is 2 metres high, how long will it take before half the water has leaked away?
The rate of loss of water is a rate of change of volume, and it is proportional to the head $\displaystyle h$ (the height of the surface above the hole).

So:

$\displaystyle \frac{dV}{dt}=-kh$

and $\displaystyle V=(H+h)A$, where $\displaystyle H$ is the height of the hole from the bottom of the tank and $\displaystyle A$ is the crossectional area of the tank, so:

$\displaystyle \frac{dV}{dt}=A\frac{dh}{dt}$

putting there together we have:

$\displaystyle \frac{dh}{dt}=-\frac{k}{A}h$

RonL