Rate of evaporation from a dish fo water

[azzarule]

Homework Statement

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.

Homework Equations

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 can't get started on forming the equation from this. Maybe volume is needed?

 

[tiny-tim]

hi azzarule!

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

initially I have:
expressed x as a function of h
with x=(h^3/2-y^2)^1/2

forget x

use r ​

 

[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

r_t² = dh/dt * 1/pi ??

 

[tiny-tim]

wouldnt the height of the water be equivalent to the y value? and r would be equivalent to x?

no, the height is h (= z)

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

 

[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

 

[tiny-tim]

h(t) = ∫∏r_(t)² dt

= ∏r_(t)²(∫dt)

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

write h as a function of r

 

[azzarule]

dh = ∏r²(t) dt

h = ∏r²(t²/2)

 

[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 can't solve for either of these because I don't know r??

 

[azzarule]

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