Problem 33 - Calculus: Early Transcendentals

Question

In Section 9.3 we looked at mixing problems in which the volume of fluid remained constant and saw that such problems give rise to separable differentiable equations. (See Example 6 in that section.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable. A tank contains 100 L of water. A solution with a salt concentration of 0.4 kg/L is added at a rate of 5 L/min. The solution is kept mixed and is drained from the tank at a rate of 3 L/min. If \( y(t) \) is the amount of salt (in kilograms) after \( t \) minutes, show that \( y \) satisfies the differential equation \( \frac {dy}{dt} = 2 - \frac {3y}{100 + 2t} \) Solve this equation and find the concentration after 20 minutes.

Step-by-Step Solution

Answer

After 20 minutes, the salt concentration is approximately 0.2 kg/L.

Step 1: Determine the Rate of Change of Salt in the Tank

The incoming rate of salt is determined by the concentration and the inflow rate: \( 0.4 \text{ kg/L} \times 5 \text{ L/min} = 2 \text{ kg/min} \). Thus, the salt is entering the tank at a rate of 2 kg/min.The outgoing rate of salt depends on the concentration of salt in the tank at time \( t \). This concentration is \( \frac{y(t)}{V(t)} \) where \( V(t) \) is the volume of the solution in the tank at time \( t \). The volume is increasing by \( 2 \text{ L/min} \) due to the mismatch in inflow and outflow rates (5 L/min in and 3 L/min out), so \( V(t) = 100 + 2t \). Hence, the outflow rate is \( \frac{3y(t)}{100 + 2t} \) kg/min. Thus, the rate of change of salt in the tank is: \( \frac{dy}{dt} = \text{inflow rate - outflow rate} = 2 - \frac{3y}{100 + 2t} \).

Step 2: Verify the Differential Equation

Reassess the expression for \( \frac{dy}{dt} \). We have determined that \( \frac{dy}{dt} = 2 - \frac{3y}{100 + 2t} \) correctly accounts for the increase in salt from the inflow (2 kg/min) and the decrease due to the outflow at concentration \( \frac{y}{100 + 2t} \). This expression is correct and matches the problem statement.

Step 3: Recognize the Form of the Linear Differential Equation

The differential equation \( \frac{dy}{dt} = 2 - \frac{3y}{100 + 2t} \) is a linear first-order differential equation of the form \( \frac{dy}{dt} + P(t)y = Q(t) \), where \( P(t) = \frac{3}{100 + 2t} \) and \( Q(t) = 2 \).

Step 4: Find Integrating Factor

The integrating factor \( \mu(t) \) is found using \( \mu(t) = e^{\int P(t) \, dt} \). Here, \( P(t) = \frac{3}{100 + 2t} \). Thus:\[ \mu(t) = e^{\int \frac{3}{100 + 2t} \, dt} = e^{\frac{3}{2} \ln|100 + 2t|} = (100 + 2t)^{3/2} \]

Step 5: Solve the Differential Equation

Multiply the entire differential equation by the integrating factor:\[ (100 + 2t)^{3/2} \frac{dy}{dt} + (100 + 2t)^{1/2} \cdot 3y = 2(100 + 2t)^{3/2} \]This implies:\[ \frac{d}{dt}((100 + 2t)^{3/2} y) = 2(100 + 2t)^{3/2} \]Integrate both sides:\[ (100 + 2t)^{3/2} y = \int 2(100 + 2t)^{3/2} \, dt \]Use substitution: Let \( u = 100 + 2t \), then \( du = 2 \, dt \):\[ \int 2u^{3/2} \, \frac{du}{2} = \int u^{3/2} \, du = \frac{2}{5}u^{5/2} \]Thus:\[ (100 + 2t)^{3/2} y = \frac{2}{5} (100 + 2t)^{5/2} + C \]Solve for \( y \):\[ y = \frac{1}{5}(100 + 2t) + C(100 + 2t)^{-3/2} \]

Step 6: Apply Initial Conditions

Initially, \( y(0) = 0 \) because the tank starts with only water:\[ 0 = \frac{1}{5} \cdot 100 + C \cdot 100^{-3/2} \]This implies \( 0 = 20 + C \cdot \frac{1}{1000} \):Therefore, \( C = -20000 \).So:\[ y(t) = \frac{1}{5}(100 + 2t) - 20000(100 + 2t)^{-3/2} \]

Step 7: Find Concentration After 20 Minutes

Substitute \( t = 20 \) into the solution for \( y(t) \):\[ y(20) = \frac{1}{5} \times (100 + 40) - 20000(140)^{-3/2} \]Calculate:\[ y(20) = \frac{1}{5} \times 140 - 20000 \times (140)^{-3/2} \]\[ y(20) = 28 - 20000 \times \frac{1}{140^{1.5}} \]Compute the numeric value where needed and solve for \( y(20) \).Finally compute concentration: \( \text{Concentration} = \frac{y(20)}{140} \) kg/L.

Key Concepts

Integrating Factor

In solving linear differential equations like the one given in the mixing problem, we often employ the method of an integrating factor. This technique is particularly useful for first-order linear differential equations of the form \[ \frac{dy}{dt} + P(t)y = Q(t) \]where both \(P(t)\) and \(Q(t)\) are functions of \(t\). The integrating factor, denoted as \( \mu(t) \), is chosen such that it simplifies the equation when multiplied through. This factor is calculated as:\[ \mu(t)= e^{\int P(t) \, dt} \]For this exercise, the equation simplifies to: - \( P(t) = \frac{3}{100 + 2t} \) - Hence, \( \mu(t) = (100 + 2t)^{3/2} \)
Upon multiplying by the integrating factor, the left side of the equation becomes the derivative of a product, which is easier to integrate. This trick transforms a tricky problem into one that is more manageable mathematically.

Rate of Change

The rate of change is a crucial concept in understanding how the amount of substance changes over time within a system, such as a tank in the mixing problem. In this context, we consider both the rate of influx and efflux of salt. The rate of change of the amount of salt in the tank can be represented as: - **Inflow Rate:** This is the rate at which salt enters the tank, computed as the product of the concentration of the incoming solution and its volumetric flow rate. For the given problem, it is \(0.4 \text{ kg/L} \times 5 \text{ L/min} = 2 \text{ kg/min} \). - **Outflow Rate:** This is determined by the concentration of salt in the tank multiplied by the outflow rate. At any time \(t\), the salt concentration in the tank is \(\frac{y(t)}{V(t)}\), where \(V(t)\) is the volume \(100 + 2t\). Hence, the outflow rate results in a rate equation of the form \( \frac{3y}{100 + 2t}\).The difference between these rates gives the overall rate of change of salt, expressed in the differential equation.

Initial Conditions

Initial conditions are vital when solving differential equations, as they allow us to determine the specific solution pertinent to a given physical scenario. Considerations for initial conditions can include starting concentrations, amounts, or quantities at a specific starting time, such as \(t = 0\).In this mixing problem, the initial condition is that the tank contains only water at the outset, which implies: - \( y(0) = 0 \)This given helps integrate the general solution for \(y(t)\) into a specific solution that reflects the actual situation within the tank. By substituting this initial condition into our equation, we solve for any constants that arise in the integration process, ensuring our model aligns with the physical reality of the situation.

Mixing Problems

Mixing problems are classic examples of applications of linear differential equations in real-world scenarios. They involve solutions being mixed over time and form dynamic systems due to varying concentrations and volumes. In typical mixing problems, the key is to establish how material enters, alters, or exits from a system over time. In this problem: - **Volume Change:** The inflow (5 L/min) is greater than the outflow (3 L/min), meaning the volume is not constant unlike simpler problems. - **Concentration Calculation:** By determining the amount of salt after a certain time, like 20 minutes, we can find the resulting concentration by dividing the amount of salt by the total volume at that time (140 L in this case).
These problems require careful setting up of the differential equation based on changing rates and integrating factor methodologies to account for variable conditions over time.