Problem 26 - Advanced Engineering Mathematics

Question

(a) Initially the tank contains 300 gallons of solution. since brine is pumped in at a rate of 3 gal/min and the mixture is pumped out at a rate of 2 gal/min, the net change is an increase of 1 gal/min. Thus, in 100 minutes the tank will contain its capacity of 400 gallons. (b) The differential equation describing the amount of salt in the tank is \(A^{\prime}(t)=6-2 A /(300+t)\) with solution $$A(t)=600+2 t-\left(4.95 \times 10^{7}\right)(300+t)^{-2}, \quad 0 \leq t \leq 100$$ as noted in the discussion following Example 5 in the text. Thus, the amount of salt in the tank when it overflows is $$A(100)=800-\left(4.95 \times 10^{7}\right)(400)^{-2}=490.625 \mathrm{lbs}$$. (c) When the tank is overflowing the amount of salt in the tank is governed by the differential equation $$\begin{aligned} \frac{d A}{d t} &=(3 \mathrm{gal} / \mathrm{min})(2 \mathrm{lb} / \mathrm{gal})-\left(\frac{A}{400} \mathrm{lb} / \mathrm{gal}\right)(3 \mathrm{gal} / \mathrm{min}) \\ &=6-\frac{3 A}{400}, \quad A(100)=490.625 \end{aligned}$$ Solving the equation, we obtain \(A(t)=800+c e^{-3 t / 400} .\) The initial condition yields \(c=-654.947,\) so that $$A(t)=800-654.947 e^{-3 t / 400}$$. When \(t=150, A(150)=587.37\) lbs. (d) As \(t \rightarrow \infty,\) the amount of salt is 800 lbs, which is to be expected since \((400 \text { gal })(2 \mathrm{lb} / \mathrm{gal})=800 \mathrm{lbs}\) (e)

Step-by-Step Solution

Answer

The tank holds 490.625 lbs of salt at 100 minutes and approaches 800 lbs as time goes to infinity.

Step 1: Understand the Problem

We are given a scenario where a tank initially containing 300 gallons of solution has brine pumping in and out at different rates, leading to changes both in volume and salt content. We need to understand the change in volume and the solution to the differential equations provided.

Step 2: Analyze Part (a)

The problem states that the tank's volume changes due to a net increase of 1 gallon per minute, so in 100 minutes the tank goes from 300 gallons to 400 gallons. This part helps us understand the dynamics of volume change in the tank.

Step 3: Solve Differential Equation for Salt - Part (b)

The differential equation given is \(A'(t)=6-\frac{2A}{300+t}\) with the solution provided as \(A(t)=600+2t-(4.95 \times 10^{7})(300+t)^{-2}\). We use this solution to find the amount of salt at \(t=100\). Substitute \(t=100\) into the equation to get \(A(100)=490.625 \text{ lbs}\).

Step 4: Analyze Tank Overflow - Part (c)

Another differential equation is given for when the tank is full: \(\frac{dA}{dt}=6-\frac{3A}{400}\), with \(A(100)=490.625\). Solve this to find \(A(t)=800 - 654.947e^{-3t/400}\). At \(t=150\), \(A(150)=587.37\) lbs, indicating the amount of salt when the tank overflow continues for 50 minutes more.

Step 5: Examine Limit as Time Approaches Infinity - Part (d)

As \(t \to \infty\), the exponential term tends to zero, making the salt content stabilize at 800 lbs, which matches the theoretical expectation due to the brine concentration.

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation accompanied by conditions that specify the solution's starting point. In our exercise, we're dealing with a tank and salt dynamics, where the initial condition is that the tank starts with 300 gallons of brine. Understanding initial value problems requires connecting

• the differential equation that governs changes,
• the initial state or condition described in a problem.

These are important because they set the foundation for solving the equation and finding out subsequent values, such as how the salt concentration changes over time.
In the given situation, the initial value of salt concentration is influenced by the rate of brine coming in and out, and this determined how the solution develops over time.

Rate of Change

The rate of change in the context of differential equations refers to how a certain quantity, like salt in a tank, changes over time. For example, in our exercise,

• the tank is initially filled with 300 gallons, and this volume increases by 1 gallon per minute due to differing rates of brine induction and extraction.
• This results in a net influx, impacting the concentration of salt.

The governing differential equations, like the one given as \( \frac{dA}{dt} = 6-\frac{3A}{400} \), are mathematical expressions that describe these changes by relating the rate of salt increase or decrease in the tank to time. By solving these equations, you can predict and calculate future concentrations, as done here at specific moments like \( t=100 \) or \( t=150 \) minutes.

Exponential Decay

Exponential decay is a process where a quantity reduces at a consistent percentage rate over time. In the context of salt concentration dynamics, it manifests as the rate at which the concentration approaches a stable value. When solving the overflow scenario, the differential equation for salt concentration involves solving for a function containing an exponential term.

• The solution \( A(t) = 800 - 654.947 e^{-3t/400} \) exemplifies this concept.
• Here, the term \( e^{-3t/400} \) causes the salt concentration to taper off towards a steady state over time.

This behavior is characteristic of exponential decay, where the concentration approaches a limit as time tends toward infinity, eventually resting at a point predicted by the balance between inflow and outflow concentrations.

Salt Concentration Dynamics

Salt concentration dynamics refer to how the concentration of salt in a solution changes due to interactions and transformations within a tank. Consider the conditions of our tank:

• Initially, it has 300 gallons that increase to 400 gallons over time, affecting salt concentration.
• Differential equations are employed to model how these volumes and concentrations interact.

These equations allow us to compute precise concentrations, like why, at \(t=100\), the concentration is 490.625 lbs of salt. Moreover, as time progresses, these dynamics dictate that salt concentration aligns with long-term equilibrium values that rely on the total system's balance, such as the predicted 800 lbs as \( t \rightarrow \infty \). Understanding these dynamics is crucial in determining both immediate and future concentrations in varying conditions.