Differential equation tank problem

[mabr0406]

I need to find the volume of water in a conical tank as a function of time. I am given that the cone is standing on its point.
The radius at the top of the cone is 1m and the height of the cone is 4m.
The height of the water initially is h(0)=0 and h(2)=3.
I am also told that the top of the cone is open and we are to assume that the water evaporates according to the following model:dV/dt=kA where k is a constant of proportionality and A is the exposed surface area.

I think I am stuck because I am used to working with things likeV'=kV and in the above model we have V's and A's. Any help would be appreciated. Thanks Jacob

 

[neutrino]

Given a cone, can you find a relation between its volume and the area of its base?

 

[mabr0406]

I did that... V=1/48pi(h^3) and A=1/16pi(h^2). But I don't know how to put these in terms of t...

 

[neutrino]

I was looking for the more general relation, V = \frac{1}{3}\pi r^2h. Can you "find" the area in that?

The height of the water initially is h(0)=0 and h(2)=3.

I'm not sure I understand that. How can the height of the water increase if we are supposed to assume that water is evaporating??

 

[HallsofIvy]

Neutrino, the conical tank has radius 1/4 the height. The cone of water at each level will also have radius 1/4 the height. In that case, V= \frac{1}{3}\pi r^2h becomes V=\frac{1}{3}\frac{1}{16}\pi h^2 h= \frac{1}{48}\pi h^3.
Differentiating that with respect to t gives
\frac{dV}{dt}= \frac{1}{16}\pi h^2 \frac{dh}{dt}
Of course the base area is A= \pi r^2= \frac{1}{16}\pi h^2.

Your \frac{dV}{dt}= kA becomes
\frac{1}{16}\pi h^2 \frac{dh}{dt}= \pi r^2= k\frac{1}{16}\pi h^2
which reduce to just \frac{dh}{dt}= k!
Solve that for h as a function of t and plug into V= \frac{1}{48}\pi h^3 to get V as a function of t.

(Like neutrino, I have no idea what you mean by "h(0)= 0 and h(2)= 3".)

 

[mabr0406]

h(0)=4... sorry, those were the initial conditions for the height of the water in the tank