# A spherical raindrops evaporates at rate proportional to surface area?

i want to find V(t)
At first i found this problem was very simple but when i try to write differential equations i ended up with these

V' = kA thats for sure

then i confined the problem only to spherical shape and no other shapes of raindrops involved

as i cant express A in term of V alone( surface area of sphere = 4∏r2, volume of sphere is 4/3∏r3 ) then i have to

use chain rule, dV/dt= dVdrdrdt substitute dV/dt from

V'=k4∏r2

i get

4∏r2r'= K4∏r2

r'=k

r = kt+c
r3 = (kt+c)3
4∏r3/3 = (kt+c)34∏/r=V(t)
im i correct? the answer to this problem is V'=kV2/3 im not sure how they transform Area to variable V alone

Last edited:

pasmith
Homework Helper
At first i found this problem was very simple but when i try to write differential equations i ended up with these

V' = kA thats for sure

then i confined the problem only to spherical shape and no other shapes of raindrops involved

as i cant express A in term of V alone( surface area of sphere = 4∏r2, volume of sphere is 4/3∏r3 ) then i have to

use chain rule, dV/dt= dVdrdrdt substitute dV/dt from

V'=k4∏r2

i get

4∏r2r'= K4∏r2

r'=k

im i correct? the answer to this problem is v'=kV2/3 im not sure how they transform Area to variable V alone
$$\left(\frac{3V}{4\pi}\right)^{1/3} = r.$$ Now substitute this into $A = 4\pi r^2$.

$$\left(\frac{3V}{4\pi}\right)^{1/3} = r.$$ Now substitute this into $A = 4\pi r^2$.
then how can u solve for V(t) ?

pasmith
Homework Helper
then how can u solve for V(t) ?
The ODE $$V' = CV^{2/3}$$ for constant $C$ is separable. Divide both sides by $V^{2/3}$ and then integrate with respect to $t$.

The ODE $$V' = CV^{2/3}$$ for constant $C$ is separable. Divide both sides by $V^{2/3}$ and then integrate with respect to $t$.
i mean how they derive for v'=cv^2/3