A spherical raindrops evaporates at rate proportional to surface area?

[kochibacha]

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 that's for sure

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

as i can't 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= dV_drdr_dt 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'=kV^2/3 I am not sure how they transform Area to variable V alone

 

[pasmith]

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

V' = kA that's for sure

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

as i can't 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= dV_drdr_dt 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'=kV^2/3 I am not sure how they transform Area to variable V alone

$$\left(\frac{3V}{4\pi}\right)^{1/3} = r.$$ Now substitute this into $A = 4\pi r^2$.

 

[kochibacha]

$$\left(\frac{3V}{4\pi}\right)^{1/3} = r.$$ Now substitute this into $A = 4\pi r^2$.

then how can u solve for V(t) ?

 

[pasmith]

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 $t$.

 

[kochibacha]

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