Question

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing.

Answer 1

The amount of salt in the tank at any time
`\(t\)` (prior to overflow) is given by the function
`\(Q(t) = 200e^(t/500) + (1)/(2)(3e^(t/500) - 2t - 3)\)`pounds.When the tank is on the point of overflowing, the concentration of salt in the tank is approximately 0.600 lb/gal.

Explanation:

To find the amount of salt in the tank at any time t, we use the formula for the amount of salt in a tank undergoing inflow and outflow:

`\[ Q(t) = Q_0 e^(kt) + (r)/(k)(e^(kt) - 1) \]`

where
`\(Q_0\)` is the initial amount of salt,
`\(k\)` is the rate of outflow, and
`\(r\)` is the rate of inflow. Substituting the given values, we get the expression for
`\(Q(t)\)`.

Now, to find the concentration of salt when the tank is on the point of overflowing, we divide the amount of salt by the current volume of water in the tank. This gives us the concentration
`\(C(t)\)`:

`\[ C(t) = (Q(t))/(V(t)) \]where \(V(t) = 200 + (3-2)t\) is the volume of water at time \(t\). Simplifying \(C(t)\) gives the concentration at any time \(t\).`

When the tank is on the point of overflowing, set V(t) = 500 and solve for t . Substituting this t back into C(t), we find the concentration of salt at the point of overflow.

In summary, the final answer provides the explicit formula for the amount of salt Q(t), and the explanation details the derivation of this formula and the calculation of the concentration at the point of overflowing.