Question

A tank contains 300 liters of fluid in which 30 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 6 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t..

Find the rate at which the number of grams of salt in the tank is changing at time t.
Find the number A(t) of grams of salt in the tank at time t.
A(t) =

Answer 1

The number of grams of salt in the tank remains constant over time, and the rate of change is zero as water is pumped in and out at the same rate.

Step-by-step explanation:

To find the number of grams of salt in the tank at time t, we need to consider the rate at which salt is added and the rate at which it is being removed. The initial amount of salt is 30 grams, and water is being pumped into the tank at a rate of 6 liters per minute. This means that for every liter of water added, 5 grams of salt are being added (since 30 grams / 6 liters = 5 grams per liter).

Since the water is being pumped out at the same rate it is being pumped in, the amount of salt in the tank remains constant over time. Therefore, A(t) = 30 grams for all values of t.

The rate at which the number of grams of salt in the tank is changing at time t is zero, since the amount of salt in the tank is not changing over time.