Question

Suppose that a large mixing tank initially holds 400 gallons of water in which 70 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 4 gal/min, and when the solution is well stirred, it is then pumped out at the same rate. Determine a differential equation for the amount of salt A(t) in the tank at time t > 0. What is A(0)

Answer 1

Answer:

$\dfrac{dA}{dt}+ \dfrac{A}{100}=0$

A(0) = 70  (i.e the pounds of salt dissolved into the tank)

Step-by-step explanation:

Given that:

a large mixing tank initially holds 400 gallons of water  in which 70 pounds of salt have been dissolved.

The rate at which pure water is pumped into the tank is 4 gal/min

After stirring; the pure water is then pumped out at the same rate.

The objective is to determine the differential equation  for the amount of salt A(t) in the tank at time t > 0. What is A(0)

Taking the differential of:

$\dfrac{dA}{dt}=R_{in}-R_{out}$     ---- (1)

where ;

$R_{in$ = 0

$R_{out} =\dfrac{A(t)}{400}*4$

$R_{out} =\dfrac{A}{100}$

replacing them into (1) ; we have:

$\dfrac{dA}{dt}=0 - \dfrac{A}{100}$

$\dfrac{dA}{dt}=- \dfrac{A}{100}$

$\dfrac{dA}{dt}+ \dfrac{A}{100}=0$

A(0) = 70 (i.e the pounds of salt dissolved into the tank)

Answer 2

The differential equation for the amount of salt A(t) in the tank at time t > 0 is $\frac{dA}{dt} + \frac{A}{100} = 0$

A (0)= 70

Given that,

• Suppose that a large mixing tank initially holds 400 gallons of water in which 70 pounds of salt have been dissolved.
• Pure water is pumped into the tank at a rate of 4 gal/min

The Calculation is as follows:

$dA\div dt=R_{in}-R_{out}$

Rin=0

$R_{out}=A(t)\div 400 \times (4)\\\\ =A\div 100$

so,

$dA\div dt=R_{in}-R_{out}\\\\dA\div dt=0- A\div 100$

So,

$dA\div dt +A\div 100=0$

A(0)=70

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