PAGE 1

Modeling w/ 1st-order eqs. (continued)

Problem Setup (Last Time)

last time: tank w/ initially \( 100\text{ gal} \) pure water
inflow: \( 2\text{ lb/gal} \) at \( 3\text{ gal/min} \)
outflow: mixed salt water in tank at \( 3\text{ gal/min} \)

Concept Breakdown: Rate of Change The differential equation is derived from the conservation principle: \[ \frac{dS}{dt} = \text{Rate In} - \text{Rate Out} \] where \( \text{Rate In} = (2\text{ lb/gal})(3\text{ gal/min}) = 6\text{ lb/min} \) and \( \text{Rate Out} = \left(\frac{S}{100}\text{ lb/gal}\right)(3\text{ gal/min}) = \frac{3}{100}S\text{ lb/min} \).

\[ \frac{dS}{dt} = 6 - \frac{3}{100}S, \quad S(0) = 0 \]

\[ S(t) = 200(1 - e^{-0.03t}) \]

Asymptotic Analysis

Coordinate graph of salt amount S(t) versus time t asymptotically approaching 200 — Figure: Coordinate graph of salt amount \( S(t) \) versus time \( t \) asymptotically approaching 200 Show Details

Coordinate plane with horizontal time axis \( t \) and vertical salt amount axis \( S \). The curve starts at the origin \( (0, 0) \) and increases concavely downward toward a dashed horizontal asymptote at \( S = 200 \).

we note \( \lim_{t \to \infty} S(t) = 200 \)

\[ \frac{200\text{ lb}}{100\text{ gal}} = 2\text{ lb/gal} \]

tank concentration = inflow concentration

Physical Interpretation As \( t \to \infty \), the exponential decay term \( e^{-0.03t} \to 0 \). Thus, the total salt \( S(t) \to 200\text{ lb} \). Since the tank volume remains constant at \( 100\text{ gal} \), the long-term internal concentration matches the incoming concentration of \( 2\text{ lb/gal} \).

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Mixing Problems: Varying Flow Rates

Initially gets saltier from inflow.
As tank gets saltier we start to lose more from outflow.
They balance each other out eventually the same way gravity and drag do in the free fall problem.

Physical Analogy: In free fall with air resistance, downward gravitational force is constant, while upward drag increases with velocity until reaching a dynamic equilibrium (terminal velocity). Similarly, with constant inflow concentration, the amount of salt leaving increases with salt concentration until the rates reach equilibrium.

Problem Setup: Unequal Rates

Now let's try one w/ different flow rates:

Same initial set up: \(100\text{ gal}\) pure water
\(2\text{ lb/gal}\) at \(3\text{ gal/min}\) into the tank
Flows out at \(5\text{ gal/min}\)

Important Observation: The tank is losing liquid overall because the outflow rate (\(5\text{ gal/min}\)) exceeds the inflow rate (\(3\text{ gal/min}\)). The total volume of brine in the tank at time \(t\) is \(V(t) = 100 + (3 - 5)t = 100 - 2t\text{ gallons}\).

Differential Equation Formulation

\(S(t)\): amount of salt in the tank

diff. eq. for the rate of change:

\[ \frac{dS}{dt} = (\text{salt in}) - (\text{salt out}) \]

Conservation of Mass: The net rate of change of salt mass is \(\frac{dS}{dt} = \text{Rate}_{\text{in}} - \text{Rate}_{\text{out}}\), where each rate is calculated as: \((\text{concentration}) \times (\text{flow rate})\).

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Non-Constant Volume Mixing Problem

\[ \frac{dS}{dt} = \underbrace{(2\text{ lb/gal})(3\text{ gal/min})}_{\text{in}} - \underbrace{\left(\frac{S}{V}\text{ lb/gal}\right)(5\text{ gal/min})}_{\text{out}} \]

Concept Explanation: The rate of change of salt \(\frac{dS}{dt}\) is the inflow rate minus the outflow rate. The inflow rate is the incoming concentration multiplied by the flow rate: \((2\text{ lb/gal}) \times (3\text{ gal/min}) = 6\text{ lb/min}\). The outflow rate is the concentration in the tank \(\frac{S(t)}{V(t)}\) multiplied by the exit flow rate of \(5\text{ gal/min}\).

\(V\) is the volume which is no longer constant

\(3\text{ in},\ 5\text{ out} \rightarrow \text{net loss of } 2\text{ gal/min}\)

\[ V(t) = \underbrace{100}_{\text{initial}} - \underbrace{2}_{\substack{\text{net loss/} \\ \text{min}}} \overbrace{t}^{\substack{\text{time in} \\ \text{minutes}}} \]

Volume Function: Fluid enters at \(3\text{ gal/min}\) and leaves at \(5\text{ gal/min}\), resulting in a net loss of \(2\text{ gal/min}\). With an initial volume of \(100\text{ gal}\), the volume at time \(t\) is \(V(t) = 100 - 2t\).

\[ \frac{dS}{dt} = 6 - \frac{5}{100 - 2t}S, \quad S(0) = 0 \]

initial value problem to solve

linear but not separable

solve as linear

Classification & Strategy: In standard linear form \(\frac{dS}{dt} + P(t)S = Q(t)\), this equation is \(\frac{dS}{dt} + \frac{5}{100 - 2t}S = 6\). Because the variables cannot be separated cleanly into functions of \(S\) and \(t\) alone, it must be solved using an integrating factor \(\mu(t) = e^{\int P(t)\,dt}\).

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Linear Differential Equation: Integrating Factor Solution

Continuation of the tank mixing problem derivation.

Standard Form and Integrating Factor

Writing the mixing rate balance in standard first-order linear form:

\[ \frac{dS}{dt} + \underbrace{\frac{5}{100-2t}}_{p(t)} S = \underbrace{6}_{g(t)} \]

Identifying Standard Form:

A first-order linear differential equation has the general form \( \frac{dS}{dt} + p(t)S = g(t) \). Here, \( p(t) = \frac{5}{100 - 2t} \) represents the outflow concentration coefficient, and \( g(t) = 6 \) represents the constant solute inflow rate.

Compute the integrating factor \( \mu(t) \):

\[ \mu(t) = e^{\int p(t)\,dt} = \cdots = (100 - 2t)^{-5/2} \]

Step-by-step Integrating Factor Derivation:

\[ \int p(t)\,dt = \int \frac{5}{100-2t}\,dt = -\frac{5}{2} \ln|100-2t| \]

Exponentiating yields:

\[ \mu(t) = e^{-\frac{5}{2}\ln(100-2t)} = e^{\ln\bigl((100-2t)^{-5/2}\bigr)} = (100-2t)^{-5/2} \]

General Solution via Product Rule

Multiply the equation through by \( \mu(t) \) and apply the reverse product rule:

\[ \frac{d}{dt} \bigl[ \mu(t) S(t) \bigr] = \mu(t) \cdot 6 \] \[ \vdots \]

Intermediate Integration Steps:

Integrating both sides with respect to \( t \):

\[ \mu(t)S(t) = \int 6(100-2t)^{-5/2}\,dt = 6 \cdot \frac{(100-2t)^{-3/2}}{\left(-\frac{3}{2}\right)(-2)} + C = 2(100-2t)^{-3/2} + C \]

Dividing by \( \mu(t) = (100-2t)^{-5/2} \):

\[ S(t) = 2(100-2t) + C(100-2t)^{5/2} = 200 - 4t + C(100-2t)^{5/2} \]

General Solution

\[ S(t) = 200 - 4t + C(100 - 2t)^{5/2} \]

Initial Condition

\( S(0) = 0 \) (pure water initially)

\[ \Downarrow \]

\[ C = -\frac{1}{500} \]

Evaluation of Constant \( C \):

Substituting \( t = 0 \) into \( S(t) \): \[ S(0) = 200 - 4(0) + C(100 - 0)^{5/2} = 200 + 100{,}000 C = 0 \implies C = -\frac{200}{100{,}000} = -\frac{1}{500} \]

Particular Solution

\[ S(t) = 200 - 4t - \frac{1}{500}(100 - 2t)^{5/2} \]

Valid up to the moment tank is dry (\( t = 50 \))

\[ 0 \le t < 50 \]

\( t \ne 50 \) because it makes the diff. eq. invalid.

Why \( t < 50 \)?

The liquid volume at time \( t \) is \( V(t) = 100 - 2t \). At \( t = 50 \), \( V(50) = 0 \), which makes the denominator in \( p(t) = \frac{5}{100-2t} \) zero. Hence, the differential equation has a singular point at \( t = 50 \) and is physically valid only on the half-open interval \( [0, 50) \).

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Amount of Salt \( S(t) \) over Time

Graph of the amount of salt S(t) versus time t , peaking around t \approx 23 and returning to zero at t = 50 . — Figure: Graph of the amount of salt \( S(t) \) versus time \( t \), peaking around \( t \approx 23 \) and returning to zero at \( t = 50 \). Show Details

Plot of the salt amount function \( S(t) = 200 - 4t - \frac{1}{500}(100 - 2t)^{5/2} \) on the domain \( [0, 50] \). The curve starts at the origin \( (0, 0) \), rises smoothly to a peak value of approximately \( 65 \) near \( t \approx 23 \), and decreases monotonically back to \( 0 \) at \( t = 50 \). A vertical red dashed line at \( t = 50 \) indicates the point where the tank is completely empty.

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Analysis of Maximum Salt Amount

Note from the graph we notice the salt amount reaches a maximum.

Why?

Initially, no salt in the tank, gain salt from inflow and lose no salt from the outflow, this continues even as the tank gets saltier and loses salt from the outflow. Concentration in tank increases much faster because of the decreasing volume, the loss will eventually match and overtake the gain.

Conceptual Breakdown: At early times, the tank contains very little salt, so the rate of salt leaving via outflow is near zero while inflow constantly brings salt in. As the fluid volume decreases, concentration \( \frac{S(t)}{V(t)} \) rises at an accelerating rate. Eventually, the rate of salt leaving equals and exceeds the rate of salt entering, causing the total salt to peak and then decline.

What's the max?

Setting the derivative equal to zero to find the critical point:

\[ \frac{dS}{dt} = 0 \implies S_{\text{max}} \approx 65.15, \quad t \approx 22.86 \]

Critical Point: The maximum salt level \( S_{\text{max}} \approx 65.15 \) occurs when the inflow rate balances the outflow rate, which evaluates to time \( t \approx 22.86 \).

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Integrating Factor & Concentration Behavior

The integrating factor is what makes the solutions so different

\[ \begin{aligned} \frac{d}{dt}(\mu S) &= \mu g \\[6pt] \mu S &= \int \mu g \, dt + C \\[6pt] S &= \frac{1}{\mu} \left( \int \mu g \, dt + C \right) \end{aligned} \]

Behavior of \( \frac{1}{\mu} \):

Equal flow rate case: \( \frac{1}{\mu} \) is exponential (solution is negative exponential).
Outflow > inflow case: \( \frac{1}{\mu} \) is fractional power algebraic (solution behaves like power function).

Concept Explanation: When inflow and outflow rates are equal, the volume \( V(t) \) is constant, yielding a constant coefficient linear ODE with an integrating factor of exponential form \( \mu(t) = e^{kt} \). When the outflow exceeds inflow, the changing volume \( V(t) = V_0 + \Delta r \cdot t \) produces a variable coefficient, resulting in a polynomial/power-law integrating factor \( \mu(t) = (V_0 + \Delta r \cdot t)^n \).

Let's look at the concentration: \[ C(t) = \frac{S(t)}{V(t)} \]

\[ C(t) = \frac{200 - 4t - \frac{1}{500}(100 - 2t)^{5/2}}{100 - 2t} = 2 - \frac{1}{500}(100 - 2t)^{3/2}, \quad 0 \le t < 50 \]

As \( t \to 50 \), \( C \to 2 \) (matches inflow concentration)

Physical Interpretation: At time \( t = 50 \), the tank becomes completely empty since \( V(50) = 100 - 2(50) = 0 \). As the residual original liquid is flushed out, the concentration of the mixture in the tank approaches \( 2 \text{ kg/L} \), which exactly equals the concentration of the incoming stream.

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Salt Concentration \( C(t) \) over Time

Graph showing salt concentration C(t) increasing over time t toward incoming concentration C_{\text{in}} = 2 until tank is empty at t = 50 — Figure: Graph showing salt concentration \( C(t) \) increasing over time \( t \) toward incoming concentration \( C_{\text{in}} = 2 \) until tank is empty at \( t = 50 \) Show Details

Plot of salt concentration \( C(t) = 2 - \frac{1}{500}(100 - 2t)^{3/2} \) as a function of time \( t \). The horizontal axis denotes time \( t \in [0, 50] \) and the vertical axis denotes concentration \( C(t) \) (mass / volume) ranging from \( 0.00 \) to \( 2.00 \). The solid green curve starts at \( (0, 0.00) \) and rises monotonically to \( (50, 2.00) \). The horizontal dashed red line shows the incoming concentration \( C_{\text{in}} = 2 \), and the vertical dotted line indicates the empty-tank condition at \( t = 50 \).

Model Governing Salt Concentration:

\[ C(t) = 2 - \frac{1}{500}(100 - 2t)^{3/2} \]

Incoming Concentration: \( C_{\text{in}} = 2 \)
Initial Concentration: \( C(0) = 2 - \frac{1}{500}(100)^{3/2} = 2 - 2 = 0 \)
Final Concentration at Tank Empty (\( t = 50 \)): \( C(50) = 2 - 0 = 2 \)

Physical Interpretation: Brine with an incoming concentration of \( C_{\text{in}} = 2 \) enters a tank initially containing pure water (\( C(0) = 0 \)). Because the outflow exceeds the inflow, the liquid volume decreases over time, completely draining when \( t = 50 \). Over this interval \( 0 \le t \le 50 \), the salt concentration monotonically increases, approaching the inflow concentration and reaching exactly \( C(50) = 2 \) at the precise moment the tank empties.

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Cascading Tanks

Two tanks

Initial parameters and flow rates for cascading tanks
Tank	Initial Contents	Inflow	Outflow
Tank 1	\( 50\text{ L},\; 10\text{ kg salt} \)	Pure water, \( 5\text{ L/min} \)	\( 5\text{ L/min} \)
Tank 2	\( 50\text{ L pure water} \)	Outflow of tank 1	\( 5\text{ L/min} \)

General Principle: The rate of change of solute mass \( S(t) \) is governed by \( \frac{dS}{dt} = \text{Rate}_{\text{in}} - \text{Rate}_{\text{out}} \), where each rate is given by \( (\text{concentration}) \times (\text{flow rate}) \). Because both tanks have equal inflow and outflow rates of \( 5\text{ L/min} \), liquid volumes remain constant at \( 50\text{ L} \).

Diagram of two cascading tanks, \text{Tank 1} and \text{Tank 2} , with flow rates of 5\text{ L/min} . — Figure: Diagram of two cascading tanks, \( \text{Tank 1} \) and \( \text{Tank 2} \), with flow rates of \( 5\text{ L/min} \). Show Details

Schematic of two cascading cylindrical tanks: \( \text{Tank 1} \) receives an inflow of pure water at \( 5\text{ L/min} \). The well-mixed solution drains into \( \text{Tank 2} \) at \( 5\text{ L/min} \). Liquid from \( \text{Tank 2} \) exits the system via an outflow of \( 5\text{ L/min} \).

Steps

write diff. eq. for each tank

Tank 1

\[ \frac{dS_1}{dt} = 0 - \frac{S_1}{50}(5) \]

Tank 2

\[ \frac{dS_2}{dt} = \left(\frac{S_1}{50}\right)(5) - \left(\frac{S_2}{50}\right)(5) \]

then solve for \( S_1 \) and

plug into \( \frac{dS_2}{dt} \) and solve for \( S_2 \)

Step-by-Step Solution Breakdown:

Solve the uncoupled differential equation for Tank 1: \( \frac{dS_1}{dt} = -\frac{1}{10}S_1 \) with initial value \( S_1(0) = 10 \), which gives \( S_1(t) = 10e^{-t/10} \).
Substitute \( S_1(t) \) into Tank 2's equation: \( \frac{dS_2}{dt} + \frac{1}{10}S_2 = \frac{1}{10}(10e^{-t/10}) = e^{-t/10} \).
Solve this linear first-order differential equation using the integrating factor \( I(t) = e^{t/10} \) with the initial condition \( S_2(0) = 0 \).