# DiffEQs modelling/solving Q

1. Feb 3, 2013

### GreatEscapist

1. The problem statement, all variables and given/known data

A circular cone (height H = 2m, radius of the top R = 1m) with a vertical axis
and the vertex pointing down is fi lled with water. At t = 0 a small circular hole
is opened at the vertex. By Torricelli's law, the rate of change of the volume
V = V (t) of water in the cone is

(1) dV/dt= -2√y

where y = y(t) is the depth of water.
a) Express the volume of water V (t) in terms of the depth of water y(t). (Draw
a picture!)
b) Use equation (1) to fi nd an equation for the rate of change of y(t) and solve
this equation.
c) How long does it take for all water to drain from the tank?
2. Relevant equations

N/A?

3. The attempt at a solution

First of all, I tried putting it in terms of y. The only thing that I can think of is that V = Ay, so therefore dv/dt = (dA/dt)(dy/dt)...right? But I feel like that transformation does not truly help me that much. I am confused how to incorporate the m's in this, besides just A (is ther a way I can make dA/dt in terms of m?)

Thanks for hints

2. Feb 3, 2013

### GreatEscapist

Technically I know that A= .5(m2)∏...but I can't write that like I can dA/dt

I can write V in terms of m -> V=∏m3,and get an equation for dV/dm...does this help?

Last edited: Feb 3, 2013
3. Feb 3, 2013

### JPaquim

You should write the expression of the volume of the cone as $\frac{1}{3}y\pi r^2(y)$, where $r(y)=\frac{y}{2}$, according to the data they give you.