Question

The volume of water remaining in a hot tub when it is being drained satisfies the differential equation dV/dt = −3(V)^1/2 , where V is the number of cubic feet of water that remain t minutes after the drain is opened. Find V if the tub initially contained 225 cubic feet of water.

Answer 1

The given function is a variable separable differential equation. Combine like terms, integrate, apply the appropriate limits, and express V in terms of t. This is done as follows:

dV/dt = -3(V)^1/2
dV/-3V^1/2 = dt
$\int\limits^V_m { \frac{1}{03 \sqrt{V} } } \, dV = \int\limits^t_0 {} \, dt$

m here is the initial V which is 225. Then after integrating,

-2/3 (√V - √225) = t
-2/3 (√V - 15) = t

$V= \sqrt{ \frac{-3}{2}t+15 }$

That is the expression for V at time t. I hope I was able to help. Have a good day.