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' = -k*Area thats for sure

but i cant express A in terms of V in order to integrate and find V(t)

anyways , my book answers V'=-kV^(2/3) i dont how they transform A into V^(2/3)

please help me

 

[JohnZ]

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' = -k*Area thats for sure

but i cant express A in terms of V in order to integrate and find V(t)

anyways , my book answers V'=-kV^(2/3) i dont how they transform A into V^(2/3)

please help me

volume of sphere is V=4/3*pi*r^3 => r=(4/3*pi*V)^(1/3)=(4/3*pi)^(1/3)*V^(1/3)=k1*V^(1/3)
surface area of sphere is A=4*pi*r^2
A/V=3*r^(-1)
A=3*r^(-1)*V

V'=-k*A=-k*3*r^(-1)*V=-k*3*k1^(-1)*V^(2/3)=-K*V^(2/3)

 

[Deleted member 4993]

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' = -k*Area thats for sure

but i cant express A in terms of V in order to integrate and find V(t)

anyways , my book answers V'=-kV^(2/3) i dont how they transform A into V^(2/3)

please help me

It is very easy if you know the relationship between the volume, area and the radius of a sphere!

A = 4 * π * r² → r = √(A/4/π)

V = 4/3 * π * r³ → r = {3*V/(4π)}^1/3

A = 4 * π * r² → A = 4 * π * [{3*V/(4π)}^1/3]² = C * V^2/3

 

[HallsofIvy]

Be sure you understand that the constant, k, in "\(\displaystyle V'= kA\)" is NOT the same number as the k in "\(\displaystyle V'= kV^{2/3}\). As subhotosh Kahn said, \(\displaystyle V= \frac{4}{3}\pi r^3\) while \(\displaystyle A= 4\pi r^2\). So \(\displaystyle r= \left(\frac{A}{4\pi}\right)^{1/2}\). \(\displaystyle V=\frac{4}{3}\pi\left(\frac{A}{4\pi}\right)^{3/2}= \)\(\displaystyle \frac{4}{3}\frac{1}{4^{3}{2}\pi^{3/2}}A^{3/2}\)\(\displaystyle = \frac{1}{6\pi^{3/2}}A^{3/2}\).

So if \(\displaystyle V'= kA\) and \(\displaystyle V'= KV^{3/2}\) then \(\displaystyle K= \frac{1}{6\pi^{3/2}}k\).

 

[ahorn]

volume of sphere is V=4/3*pi*r^3 => r=(\(\displaystyle {\color{red}{\text{4/3*pi}}}\)*V)^(1/3)=(\(\displaystyle {\color{red}{\text{4/3*pi}}}\))^(1/3)*V^(1/3)=k1*V^(1/3)
surface area of sphere is A=4*pi*r^2
A/V=3*r^(-1)
A=3*r^(-1)*V

V'=-k*A=-k*3*r^(-1)*V=-k*3*k1^(-1)*V^(2/3)=-K*V^(2/3)

JohnZ, I believe that your calculation should read

r=(3/(4*pi)*V)^(1/3)

However, this only affects k1 and the final answer is still -K*V^(2/3)

So if \(\displaystyle V'= kA\) and \(\displaystyle V'= KV^{{\color{red}{3/2}}}\) then \(\displaystyle K={\color{red}{ \frac{1}{6\pi^{3/2}}}}k\).

HallsofIvy, note that \(\displaystyle K=k\left(6\pi^{3/2}\right)^{2/3}=k\pi 6^{2/3}\)