Suppose that a large mixing tank initially holds 500 gallons of water in which 20 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 5 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 $A(0)$ ? (Use $A(0)$ for $A(t)$ .)

Question :

Suppose that a large mixing tank initially holds 500 gallons of water in which 20 pounds of salt have been dissolved. pure water is pumped into the tank at a rate of 5 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 $a(0)$ ? (use $a(0)$ for $a(t)$ .)

Suppose that a large mixing tank initially holds 500 gallons of water in which 2 | Doubtlet.com

Solution:

Neetesh Kumar | November 08, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh1Sec03 (Homework) Question - 4
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Step-by-Step-Solution:

Let $A(t)$ be the amount of salt in pounds at time $t$ . Initially, the tank contains 20 pounds of salt, so:

$A(0) = 20$

Step 1: Determine the Rate of Change of Salt

The rate of salt entering the tank is $0$ pounds per minute because only pure water is added.

The rate at which salt leaves the tank depends on the concentration of salt in the tank. The concentration of salt at any time $t$ can be calculated as follows:

The volume of the tank remains constant at 500 gallons, since the inflow and outflow rates are equal.
The concentration of salt in the tank is:

$\text{Concentration} = \frac{A(t)}{500} \, \text{pounds/gallon}$

The rate of salt leaving the tank is given by:

$\text{Rate out} = \text{Concentration} \times \text{Outflow rate} = \frac{A(t)}{500} \cdot 5$

Therefore, the rate of salt leaving the tank is:

$\frac{5A(t)}{500} = \frac{A(t)}{100} \, \text{pounds/min}$

Step 2: Write the Differential Equation

The change in the amount of salt in the tank over time can be described by the following differential equation:

$\frac{dA}{dt} = \text{Rate in} - \text{Rate out}$

Since the rate in is $0$ :

$\frac{dA}{dt} = 0 - \frac{A(t)}{100}$

Thus, the differential equation is:

$\frac{dA}{dt} = -\frac{A(t)}{100}$

Conclusion

The differential equation governing the amount of salt $A(t)$ in the tank is:

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

And the initial condition is:

$A(0) = \boxed{20}$

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