Question

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt per gallon of water is pumped into the tank at the rate of 5 liters per minute, and the well-stirred mixture is pumped out at the same rate. Let A (t) represent the amount of salt in the tank at time t. The amount of salt in the tank at time t is:

Answer 1

Answer:

$A(t)=600-300e^t^/^3^0^0$

Step-by-step explanation:

let $A(t)$ represent the amount of salt in the tank at time $t.$ Then$A(t)=300 \ at \ t=0$ and

$\frac{dA}{dt}=rate \ in- \ rate \ out\\\frac{dA}{dt}=2-\frac{A}{300}\\\\\frac{dA}{dt}+\frac{A}{300}=2\\\\e^t^/^3^0^0 \frac{dA}{dt}+\frac{e^t^/^3^0^0}{300}A=2e^t^/^3^0^0\\\\e^t^/^3^0^0A=\int2e^t^/^3^0^0dt=600e^t^/^3^0^0+C\\\\A(t)=600+Ce^t^/^3^0^0$#To find $C$ we use  $A=300, \ when \ t=0$

Hence:

$300=600+Ce^t^/^3^0^0\\C=-300$

#The amount of salt in tank at time time is expressed as:

$A(t)=600-300e^t^/^3^0^0$