A tank contains 300 liters of fluid in which 20 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 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.

Step-by-step explanation:

Given that a tank contains 300 liters of fluid in which 20 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate

Water at any time t = 300 L +incoming rate - outgoing rate

= 300L (since both are the same)

Quantity of salt at any time t is A(t)

A(0) = 20 gm (initial salt level)

Rate of change of salt level = incoming rate - outgoing rate

Incoming rate =0

Outgoing rate is Q(t) *5/200 at time t (assuming mixed well)

So

We separate the variables and solve

Use the fact A(0) = 20

We get

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step-by-step explanation:

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