# Rate of change

1. May 9, 2004

Sorry to bore you all with this rate of change question, I really hate this area of maths and struggle to get my hea round it. Here it is, I'd be greatful for any help you have to offer:

Fluid flows out of a cylindrical tank with constant corss section. At the time t minutes (t is greater of equal to 0), the volume of fluid remaining in the tank is V m^3. The rate at which the fluid flows, in m^3min^-1, is proportional to the square root of V. Show that the depth h metre's of fluid in the tank satisfies the differential equation:

dh/dt = -kh^(1/2) where k is a +ve constant

Cheers

Pat

2. May 9, 2004

### arildno

dV/dt=-aV^(1/2), by assumption.

V(t)=pi*R^(2)*h(t)=b*h(t) (where R is the radius)
Rearranging a bit, and introducing k yields the desired result

3. May 9, 2004

thanks v much, great help

Pat

4. May 9, 2004

### HallsofIvy

In other words, the whole point of "rate of change problems" is to use a static equation (typically a geometry formula or something like that) that has no time variable, t, in it and then use the chain rule to differentiate both sides of the equation- giving a formula connecting the rates of change of the quantities.