Answer:
[tex]V = \sqrt[4]{K1*t+K2}[/tex]
Step-by-step explanation:
Since:
V: volume
t: time
K: proporcional constant
dV/dt = K*V^3 ---> (V^-3)dV = Kdt
Integrating by parts: [tex]\int\limits^{V}_{Vo} V^{-3} \, dV =K* \int\limits^{t}_{to} {t} \, dt[/tex]
Then:
[tex](-1/4)*(V^{-4} -Vo^{-4})= K*(t-to)[/tex]
[tex](-1/4)*V^{-4}= K*(t-to) + (-1/4)*Vo^{-4}[/tex]
[tex]V^{-4}= (-4)*K*(t-to) + (-4)*(-1/4)*Vo^{-4} = -4K*(t-to) + Vo^{-4}[/tex]
If we call: K1 = -4K ; K2 = Vo^-4, (both constants) then:
[tex]V = \sqrt[4]{K1*(t-to)+K2}[/tex]
If chose select the initial time into = 0, then:
[tex]V = \sqrt[4]{K1*t+K2}[/tex]
The differential equation predicting the rate of change of volume will be [tex]\dfrac{dV}{dt}=kV^3[/tex] where k is the constant.
Given information:
The rate of change of volume V with respect to time t of water leaking from a tank is proportional to the cube of the volume.
The volume is leaking from the tank. So, the rate can be written as negative.
According to the given condition, the rate of change of volume can be written as,
[tex]\dfrac{dV}{dt}\propto V^3\\\dfrac{dV}{dt}=kV^3[/tex]
Here, k is the proportionality constant. Its value can be found by using the boundary conditions.
Therefore, the differential equation predicting the rate of change of volume will be [tex]\dfrac{dV}{dt}=kV^3[/tex] where k is the constant.
For more details, refer to the link:
https://brainly.com/question/6612989
