# Differential Equation Help

• October 12th 2008, 11:23 PM
falconed
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!
• October 13th 2008, 02:31 AM
mr fantastic
Quote:

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.
• October 13th 2008, 07:39 PM
falconed
Quote:

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.
• October 19th 2008, 11:24 PM
mark18
lol the griffith assignment 3 haha its due thursday
• October 20th 2008, 03:06 AM
mark18
for the rate flowing out why is it kV^2
• October 20th 2008, 03:10 AM
mr fantastic
Quote:

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

Read the question more carefully:

Quote:

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) [snip]

• October 20th 2008, 03:16 AM
mark18
is it possible to find what k is or is more information required
• October 20th 2008, 03:19 AM
mr fantastic
Quote:

Originally Posted by mark18
is it possible to find what k is or is more information required