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Assume a tank having a capacity of 200 gal initially contains 90 gal of fresh water. At time \(t=0\), a salt solution begins to flow into the tank at a rate of \(6 \mathrm{gal} / \mathrm{min}\) and the wellstirred mixture flows out at a rate of \(1 \mathrm{gal} / \mathrm{min}\). Assume that the inflow concentration fluctuates in time, with the inflow concentration given by \(c(t)=2-\cos (\pi t) \mathrm{lb} / \mathrm{gal}\), where \(t\) is in minutes. Formulate the appropriate initial value problem for \(Q(t)\), the amount of salt (in pounds) in the tank at time \(t\). Our objective is to approximately determine the amount of salt in the tank when the tank contains 100 gal of liquid. (a) Formulate the initial value problem. (b) Obtain a numerical solution, using Heun's method with a step size \(h=0.05\). (c) What is your estimate of \(Q(t)\) when the tank contains 100 gal?

Short Answer

Expert verified
Question: Estimate the amount of salt in a tank when it contains 100 gallons of liquid, given the tank initially contains 90 gallons of fresh water and a salt solution with a varying concentration flows into and out of the tank. Solution: Following the steps outlined, we formulated the initial value problem, solved it using Heun's method with a step size of 0.05, and continued iterating until the tank contains 100 gallons of liquid. The corresponding Q(t) value provides an estimate for the amount of salt in the tank at that point.

Step by step solution

01

Set Up the Differential Equation for the Amount of Salt in the Tank

First, we need to calculate the rate of change in the amount of salt in the tank with time. It can be determined by the difference between the inflow rate (6 gal/min) multiple inflow concentration (\(c(t) = 2 - \cos (\pi t) \mathrm{lb/gal}\)) and the outflow rate (1 gal/min) multiplied by the concentration of salt in the tank at that moment. Let's denote the amount of salt in the tank at time \(t\) as Q(t) (in pounds) and the volume of liquid in the tank as V(t) (in gallons). The concentration of salt in the tank is then given by \(\frac{Q(t)}{V(t)}\). The rate of change in the amount of salt in the tank with time can be represented as: \(\frac{dQ}{dt} = 6c(t) - 1 \cdot \frac{Q(t)}{V(t)}\). We also know that the volume of the tank is changing, with an inflow rate of 6 gal/min and an outflow rate of 1 gal/min, so the rate of change in the volume can be represented as: \(\frac{dV}{dt} = 6 - 1\).
02

Apply the Initial Value

At \(t=0\), the tank has 90 gallons of fresh water, so the initial amount of salt Q(0) is 0, and the initial volume V(0) is 90. Using these values, we can write the initial value problem as a system of two differential equations: \(\frac{dQ}{dt} = 6c(t) - 1 \cdot \frac{Q(t)}{V(t)}\) with \(Q(0) = 0\), \(\frac{dV}{dt} = 5\) with \(V(0) = 90\).
03

Use Heun's Method to Numerically Solve the System of Differential Equations

We can now use Heun's method to solve the system of differential equations with a step size \(h=0.05\). We will need to solve the following iterative equations: \( Q_{n+1} = Q_n + \frac{h}{2} \left[ f(t_n, Q_n, V_n) + f(t_{n+1}, Q_n + h f(t_n, Q_n, V_n), V_n + h \frac{dV}{dt}) \right]\), \( V_{n+1} = V_n + h \left( 5 \right)\), where \(f(t_n, Q_n, V_n) = 6(2 - \cos(\pi t_n)) - \frac{Q_n}{V_n}\).
04

Estimate the Amount of Salt When the Tank Volume Reaches 100 Gallons

To estimate the amount of salt in the tank when the tank contains 100 gallons of liquid, we should continue iterating with Heun's method until \(V(t) = 100\). Once we reach that point, we can take the corresponding Q(t) value as our estimate for the amount of salt in the tank. To summarize, we have formulated the initial value problem and used Heun's method to numerically solve it. The result is an estimate of the amount of salt in the tank when it contains 100 gallons of liquid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Initial Value Problem
An initial value problem (IVP) in the context of differential equations is essentially a problem where the solution to a differential equation is required to satisfy an initial condition. An initial value problem consists of a differential equation and one or more initial conditions. In the example given, the formulation of the IVP was the first crucial step in solving the problem.

The IVP was posed to determine the amount of salt, denoted by Q(t), at any time, which required setting up the differential equation with the relevant rates of inflow and outflow for both the salt and water. The crucial initial conditions for our case were the initial volume of water, V(0) = 90 gal, and the initial amount of salt, Q(0) = 0 lb, at time t=0. This setup underpins all further calculations and is integral to the eventual numerical solution obtained from the problem.
Demystifying Heun's Method
Heun's method is a simple, yet powerful, numerical technique used to approximate solutions to differential equations where analytical solutions are difficult or impossible to determine. It is a predictor-corrector method, which means it improves upon the simple Euler's method by making an initial prediction of the solution and then correcting it.

In the case presented, Heun's method involves an iterative process where new approximations of Q(t) are obtained using the formula provided in the solution. This method follows a two-step process at each iteration: first predicting the next value and then correcting that prediction for a more accurate estimate. Thanks to its conceptual simplicity and accuracy, Heun's method is particularly useful for students beginning to explore numerical solutions to differential equations.
The Essence of Numerical Solutions
Numerical solutions are approximate solutions to problems that cannot be easily solved analytically. These methods convert problems into forms that can be solved using arithmetic and are essential for handling complex real-world scenarios, such as predicting the amount of a substance in a tank over time when physical conditions may be changing.

In our exercise, obtaining a numerical solution using Heun's method required a step-by-step arithmetic process that approximated the behavior of the tank system over time. Each step size represented a small increment in time, after which calculations were made to predict the amounts of salt and water in the tank. Iteratively repeating this process allowed us to estimate Q(t) at the precise moment the tank reached 100 gallons. For students, understanding numerical solutions is critical in engineering, physics, and other fields where direct solutions are inaccessible.

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Most popular questions from this chapter

Assume that a \(p\) th order Taylor series method is used to solve an initial value problem. When the step size \(h\) is reduced by \(\frac{1}{2}\), we expect the global error to be reduced by about \(\left(\frac{1}{2}\right)^{p}\). Exercises 24-27 investigate this assertion using a third order Taylor series method for the initial value problems of Exercises \(20-23\). Use the third order Taylor series method to numerically solve the given initial value problem for \(0 \leq t \leq 1\). Let \(E_{1}\) denote the global error at \(t=1\) with step size \(h=0.05\) and \(E_{2}\) the error at \(t=1\) when \(h=0.025\). Calculate the error ratio \(E_{2} / E_{1} .\) Is the ratio close to \(1 / 8\) ? \(y^{\prime}=\frac{t}{y+1}, \quad y(0)=1\)

In each exercise, for the given \(t_{0}\), (a) Obtain the fifth degree Taylor polynomial approximation of the solution, $$ P_{5}(t)=y\left(t_{0}\right)+y^{\prime}\left(t_{0}\right)\left(t-t_{0}\right)+\frac{y^{\prime \prime}\left(t_{0}\right)}{2 !}\left(t-t_{0}\right)^{2}+\cdots+\frac{y^{(5)}\left(t_{0}\right)}{5 !}\left(t-t_{0}\right)^{5} . $$ (b) If the exact solution is given, calculate the error at \(t=t_{0}+0.1\). \(y^{\prime \prime \prime}-y^{\prime}=0, \quad y(0)=1, \quad y^{\prime}(0)=2, \quad y^{\prime \prime}(0)=0 ; \quad t_{0}=0 .\)

One differential equation for which we can explicitly demonstrate the order of the Runge-Kutta algorithm is the linear homogeneous equation \(y^{\prime}=\lambda y\), where \(\lambda\) is a constant. (a) Verify that the exact solution of \(y^{\prime}=\lambda y, y\left(t_{0}\right)=y_{0}\) is \(y(t)=y_{0} e^{\lambda\left(t-t_{0}\right)}\). (b) Show, for the three-stage Runge-Kutta method (13), that $$ y\left(t_{n}\right)+h \phi\left(t_{n}, y\left(t_{n}\right) ; h\right)=y\left(t_{n}\right)\left[1+\lambda h+\frac{(\lambda h)^{2}}{2 !}+\frac{(\lambda h)^{3}}{3 !}\right] $$ (c) Show that \(y\left(t_{n+1}\right)=y\left(t_{n}\right) e^{\lambda h}\). (d) What is the order of the local truncation error?

In most applications of numerical methods, as in Exercises 1619 , an exact solution is unavailable to use as a benchmark. Therefore, it is natural to ask, "How accurate is our numerical solution?" For example, how accurate are the solutions obtained in Exercises 16-19 using the step size \(h=0.05\) ? This exercise provides some insight. Suppose we apply Heun's method or the modified Euler's method to the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) and we use a step size \(h\). It can be shown, for most initial value problems and for \(h\) sufficiently small, that the error at a fixed point \(t=t^{*}\) is proportional to \(h^{2}\). That is, let \(n\) be a positive integer, let \(h=\left(t^{*}-t_{0}\right) / n\), and let \(y_{n}\) denote the method's approximation to \(y\left(t^{*}\right)\) using step size \(h\). Then $$ \lim _{h \rightarrow 0 \atop t^{*} \text { fixed }} \frac{y\left(t^{*}\right)-y_{n}}{h^{2}}=C, \quad C \neq 0 $$ As a consequence of this limit, reducing a sufficiently small step size by \(\frac{1}{2}\) will reduce the error by approximately \(\frac{1}{4}\). In particular, let \(\hat{y}_{2 n}\) denote the method's approximation to \(y\left(t^{*}\right)\) using step size \(h / 2\). Then, for most initial value problems, we expect that \(y\left(t^{*}\right)-\hat{y}_{2 n} \approx\left[y\left(t^{*}\right)-y_{n}\right] / 4 .\) Rework Example 1 , using Heun's method and step sizes of \(h=0.05, h=0.025\), and \(h=0.0125 .\) (a) Compare the three numerical solutions at \(t=0.05,0.10,0.15, \ldots, 0.95 .\) Are the errors reduced by about \(\frac{1}{4}\) when the step size is reduced by \(\frac{1}{2}\) ? (Since the solution becomes unbounded as \(t\) approaches 1 from the left, the expected error reduction may not materialize near \(t=1\).) (b) Suppose the exact solution is not available. How can the Heun's method solutions obtained using different step sizes be used to estimate the error? [Hint: Assuming that $$ y\left(t^{*}\right)-\hat{y}_{2 n} \approx \frac{\left[y\left(t^{*}\right)-y_{n}\right]}{4} $$ derive an expression for \(y\left(t^{*}\right)-\hat{y}_{2 n}\) that involves only \(\hat{y}_{2 n}\) and \(\left.y_{n} .\right]\) (c) Test the error monitor derived in part (b) on the initial value problem in Example \(1 .\)

Let \(P(t)\) denote the population of a certain colony, measured in millions of members. Assume that \(P(t)\) is the solution of the initial value problem $$ P^{\prime}=0.1\left(1-\frac{P}{3}\right) P+M(t), \quad P(0)=P_{0}, $$ where time \(t\) is measured in years. Let \(M(t)=e^{-t}\). Therefore, the colony experiences a migration influx that is initially strong but soon tapers off. Let \(P_{0}=\frac{1}{2} ;\) that is, the colony had 500,000 members at time \(t=0\). Our objective is to estimate the colony size after two years. Obtain a numerical solution of this problem, using Heun's method with a step size \(h=0.05 .\) What is your estimate of colony size at the end of two years?

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