Problem 21 - Calculus Early Transcendentals

Question

At time \(t=0,\) a tank contains 25 oz of salt dissolved in 50 gal of water. Then brine containing 4 oz of salt per gallon of brine is allowed to enter the tank at a rate of 2 gal \(/ \mathrm{min}\) and the mixed solution is drained from the tank at the same rate. (a) How much salt is in the tank at an arbitrary time \(t ?\) (b) How much salt is in the tank after \(25 \mathrm{min} ?\)

Step-by-Step Solution

Answer

(a) $S(t) = 200 - 175e^{-t/25}$; (b) $S(25) \approx 135.6$ oz.

Step 1: Understanding the Problem

We are given a tank with brine flowing in and out, initially containing 25 oz of salt in 50 gallons of water. The inflow has a salt concentration of 4 oz/gal and both in and out rates are 2 gal/min. We need to determine the amount of salt at time $t$ and after 25 min.

Step 2: Define the Variables

Let $S(t)$ be the amount of salt (in oz) in the tank at time $t$. We know $S(0) = 25$ oz, the tank volume stays constant at 50 gal since the inflow rate equals the outflow rate.

Step 3: Set Up the Differential Equation

The rate of change of salt in the tank is the rate of salt coming in minus the rate of salt going out. The inflow rate of salt is \(4 \text{ oz/gal} \times 2 \text{ gal/min} = 8\) oz/min. The outflow rate is \(\frac{S(t)}{50}\) oz/gal \times 2 gal/min, which simplifies to \(\frac{2S(t)}{50}\) or \(\frac{S(t)}{25}\) oz/min.

Step 4: Write the Differential Equation

The differential equation is \( \frac{dS}{dt} = 8 - \frac{S(t)}{25} \). This describes how the salt changes over time in the tank.

Step 5: Solve the Differential Equation

This is a first order linear differential equation. Rearrange it as \( \frac{dS}{dt} + \frac{S}{25} = 8 \). The integrating factor is \( e^{\int \frac{1}{25} dt} = e^{t/25} \). Multiply through by this factor and integrate both sides.

Step 6: Perform Integration

The left side becomes \( e^{t/25} S(t) \), and integrating the right side, we get \( \int 8 \, e^{t/25} dt = 200 \, e^{t/25} + C \). Therefore, \( e^{t/25} S(t) = 200 \, e^{t/25} + C \). Divide by \( e^{t/25} \) to solve for \( S(t) \): \( S(t) = 200 + Ce^{-t/25} \).

Step 7: Determine the Constant

Use the initial condition \( S(0) = 25 \) for the constant \( C \). Substituting, we have \( 25 = 200 + C \times 1 \), so \( C = 25 - 200 = -175 \).

Step 8: Final Expression for Salt Amount

Substitute \( C \) back into the equation: \( S(t) = 200 - 175e^{-t/25} \). This is the formula for the amount of salt at any time \( t \).

Step 9: Calculate Salt Amount at 25 min

Substitute \( t = 25 \) into \( S(t) = 200 - 175e^{-t/25} \). Simplify to find \( S(25) = 200 - 175e^{-1} \). Calculate \( e^{-1} \approx 0.368 \), so \( S(25) \approx 200 - 175 \times 0.368 = 200 - 64.4 = 135.6 \).

Key Concepts

Understanding brine concentration

Brine is a solution comprised of salt and water. In this scenario, brine enters the tank with a specific salt concentration. It is important to know how much salt per gallon is coming in to understand changes in salt concentration over time. Here, the brine entering the tank contains 4 oz of salt per gallon. This concentration is pivotal, as it influences the overall salt balance in the system. The constant concentration of brine affects the differential equation that models the system.
Understanding this concentration helps in setting up the equation accurately, and any miscalculation could lead to incorrect predictions about salt amounts at future times.

Salt concentration dynamics

Salt concentration in the tank changes over time due to the interaction of incoming and outgoing brine. Initially, the tank has 25 oz of salt dissolved in 50 gallons of water. This starting point is the base concentration. As brine inflows with salt concentration of 4 oz/gal, it affects the overall solution composition.
The key is that the tank's volume remains constant, due to equal inflow and outflow rates. This means any change in salt comes purely from the added brine, minus what's exiting. The focus is on the difference between these rates to form a differential equation for salt amount over time. This dynamic plays a crucial role in predicting salt levels accurately as time progresses.

Inflow and outflow rates

The brine flows into the tank at a rate of 2 gallons per minute, matching the rate at which it leaves. When inflow and outflow rates are equal, the system's volume remains stable, simplifying the analysis. This is crucial for creating precise mathematical models. The brine's inflow carries new salt into the tank, while the outflow carries salt out of it, leading to changes in salt concentration.
Calculating the rate of salt coming in and going out is essential for setting up the differential equation. With a 4 oz/gal brine concentration and 2 gal/min flow rate, the inflow introduces 8 oz of salt each minute. The outgoing salt quantity depends on the current salt concentration in 50 gallons of the tank at any given time.

Integrating factor in differential equations

The solution of the differential equation involves using the integrating factor. This is a technique to solve first-order linear differential equations. Our equation takes the form \( \frac{dS}{dt} + \frac{S}{25} = 8 \). To solve this, find the integrating factor, which is \( e^{\int \frac{1}{25} dt} = e^{t/25} \).
Multiply the entire equation by this integrating factor to simplify solving. Once multiplied, the left side becomes a derivative of \( e^{t/25} S(t) \), which can then be integrated easily. This integration process yields the general solution. By using initial conditions, solve for any constants to determine the specific solution. This method is fundamental in handling differential equations efficiently and accurately.