A tank contains $150$ 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. Find the number $A(t)$ of grams of salt in the tank at time $t$.

Neetesh Kumar | November 07, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh3Sec01 (Homework) Question - 4
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Step-by-Step-Solution:

To find the amount of salt in the tank at time $t$, we can set up a differential equation based on the rate of salt entering and leaving the tank.

1. Initial Conditions:

   • The initial amount of salt is $A(0) = 20$ grams.
   • The volume of the tank is constant at $150$ liters.

2. Rate of Salt Removal: Since pure water is entering the tank, the concentration of salt in the tank is given by:

   $\text{Concentration of salt} = \frac{A(t)}{150} \text{ grams per liter}$

   The rate at which salt leaves the tank is the product of this concentration and the outflow rate:

   $\text{Rate of salt leaving} = \frac{A(t)}{150} \times 5$

3. Setting Up the Differential Equation: The rate of change of the amount of salt in the tank is given by the rate of salt coming in minus the rate of salt going out. Since no salt is entering the tank (pure water is being added), the equation is:

   $\frac{dA}{dt} = -\frac{A(t)}{150} \times 5$

   This simplifies to:

   $\frac{dA}{dt} = -\frac{1}{30} A(t)$

4. Solving the Differential Equation: This is a first-order linear differential equation. We can separate variables and integrate:

   $\frac{dA}{A} = -\frac{1}{30} dt$

   Integrating both sides:

   $\ln |A| = -\frac{t}{30} + C$

   Exponentiating gives:

   $A(t) = e^{-\frac{t}{30} + C} = e^{C} e^{-\frac{t}{30} }$

   Let $A_0 = e^{C}$, thus:

   $A(t) = A_0 e^{-\frac{t}{30}}$

5. Applying Initial Conditions: We know that $A(0) = 20$ grams, so:

   $20 = A_0 e^{0}$

   Thus, $A_0 = 20$. Therefore:

   $A(t) = 20 e^{-\frac{t}{30} }$

Final Answer

The number of grams of salt in the tank at time $t$ is:

$A(t) = \boxed{20 e^{-\frac{t}{30}}} \text{ g}$

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