# Homework Help: Rate of evaporation from a dish fo water

1. Nov 6, 2012

### azzarule

1. The problem statement, all variables and given/known data
A dish has a shape described by the equation:
h=(x^2+y^2)^3/2
At time = 0 it is filled to a height of 20cm with a fluid that evaporates when exposed to air. The evaporation rate is proportional to the exposed surface area (that is decreasing) at any time t.
if h(t) is the height of the fluid at time t then
dh/dt is proportional to pir(t)^2, r(t) is the radius at time t. After 20 minutes the height of the fluid was 19.7cm.
im trying to make a differential equation that governs the height h(t) during the evaporation.

2. Relevant equations

3. The attempt at a solution
initially I have:
expressed x as a function of h
with x=(h^3/2-y^2)^1/2
now I cant get started on forming the equation from this. Maybe volume is needed?

Last edited by a moderator: Nov 6, 2012
2. Nov 6, 2012

### tiny-tim

hi azzarule!

(try using the X2 button just above the Reply box )
forget x

use r

3. Nov 6, 2012

### azzarule

got mixed up,
wouldnt the height of the water be equivalent to the y value? and r would be equivalent to x?

r2 is then

rt2 = dh/dt * 1/pi ??

4. Nov 6, 2012

### tiny-tim

no, the height is h (= z)

r is the radius at height h … you needn't bother with x and y

5. Nov 6, 2012

### azzarule

ok so,

dh/dt = ∏r(t)2

h(t) = ∫∏r(t)2 dt

= ∏r(t)2(∫dt)

=∏r(t)2t

so now h(t) = ∏r(t)2t +C

6. Nov 6, 2012

### tiny-tim

no, you can't take r2(t) outside the integral !!

write h as a function of r

7. Nov 6, 2012

### azzarule

dh = ∏r2(t) dt

h = ∏r2(t2/2)

Last edited: Nov 6, 2012
8. Nov 6, 2012

### azzarule

I went like this and then got stuck again,

dh/dt = A**r2(t)

dh= A*∏*r2(t) dt

h = A*∏*r2(t)*t+C

then use t=20 and h=19.7

do I solve for A or c, also I cant solve for either of these because I dont know r??

Last edited: Nov 7, 2012
9. Nov 6, 2012

### azzarule

Or h=pi*r^3(t)/3 +c