# Math Help - Differential Equation Help

1. ## Differential Equation Help

Hey everyone! I'm stuck on the following:

A person is trying to fill a bucket with water. Water is flowing into the bucket from the tap at a constant rate of 0.12 litres/sec. However there is a hole in the bottom of the bucket and water is flowing out of the bucket at a rate proportional to the square of the volume of water present in the bucket. If V(t) is the volume of water (in litres) present in the bucket at time t (in secs) and the bucket initially contains 2 litres of water, write down, but do not solve, the differential equation for V(t) along with its initial condition.

Okay here is what I have:

Rate leaving is proportional to (sorry don't know how to write the symbol here) $V^2$, therefore:

$Rate = kV^2$ or should it be $ktV^2$

So with that I get a function:

$V(t) = 2 + 0.12t - kV^2$

Okay my question is am I right in putting the rate it's leaving straight into that equation?

Can I just then sub in $dV/dt$ for V?

$dV/dt = 2 + 0.12t - k(dV/dt)^2$

Any help would be great!

2. Originally Posted by falconed
Hey everyone! I'm stuck on the following:

A person is trying to fill a bucket with water. Water is flowing into the bucket from the tap at a constant rate of 0.12 litres/sec. However there is a hole in the bottom of the bucket and water is flowing out of the bucket at a rate proportional to the square of the volume of water present in the bucket. If V(t) is the volume of water (in litres) present in the bucket at time t (in secs) and the bucket initially contains 2 litres of water, write down, but do not solve, the differential equation for V(t) along with its initial condition.

Okay here is what I have:

Rate leaving is proportional to (sorry don't know how to write the symbol here) $V^2$, therefore:

$Rate = kV^2$ or should it be $ktV^2$

So with that I get a function:

$V(t) = 2 + 0.12t - kV^2$

Okay my question is am I right in putting the rate it's leaving straight into that equation?

Can I just then sub in $dV/dt$ for V?

$dV/dt = 2 + 0.12t - k(dV/dt)^2$

Any help would be great!
dV/dt = rate in - rate out = 0.12 - kV^2 subject to the boundary condition V(0) = 2.

3. Originally Posted by mr fantastic
dV/dt = rate in - rate out = 0.12 - kV^2 subject to the boundary condition V(0) = 2.
So that is all I needed? I guess I was thinking it would be a bit more complicated.

4. lol the griffith assignment 3 haha its due thursday

5. for the rate flowing out why is it kV^2

6. Originally Posted by mark18
for the rate flowing out why is it kV^2