Question

The rate of change of the volume V with respect to time t of water leaking from a tank is proportional to the cube of the volume. Which of the following is a differential equation that could describe this relationship?

Answer 1

Answer:

$V = \sqrt[4]{K1*t+K2}$

Step-by-step explanation:

Since:

V: volume

t: time

K: proporcional constant

dV/dt = K*V^3 ---> (V^-3)dV = Kdt

Integrating by parts: $\int\limits^{V}_{Vo} V^{-3} \, dV =K* \int\limits^{t}_{to} {t} \, dt$

Then:

$(-1/4)*(V^{-4} -Vo^{-4})= K*(t-to)$

$(-1/4)*V^{-4}= K*(t-to) + (-1/4)*Vo^{-4}$

$V^{-4}= (-4)*K*(t-to) + (-4)*(-1/4)*Vo^{-4} = -4K*(t-to) + Vo^{-4}$

If we call: K1 = -4K ; K2 = Vo^-4, (both constants) then:

$V = \sqrt[4]{K1*(t-to)+K2}$

If chose select the initial time into = 0, then:

$V = \sqrt[4]{K1*t+K2}$