/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A tank contains \(100 \mathrm{ga... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 18

A tank contains \(100 \mathrm{gal}\) of a brine solution in which \(20 \mathrm{lb}\) of salt is initially dissolved. (a) Water (containing no salt) is then allowed to flow into the tank at a rate of \(4 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out of the tank at an equal rate of \(4 \mathrm{gal} / \mathrm{min}\). Determine the amount of salt \(y(t)\) at any time \(t\). What is the eventual concentration of the brine solution in the tank? (b) If instead of water a brine solution with concentration \(2 \mathrm{lb} /\) gal flows into the tank at a rate of \(4 \mathrm{gal} / \mathrm{min}\), what is the eventual concentration of the brine solution in the tank?

Short Answer

Expert verified
For (a), the eventual concentration is 0 lb/gal. For (b), the eventual concentration is 2 lb/gal.

Step by step solution

01

- Define the Problem

We start with a tank that initially contains 100 gallons of brine solution with 20 pounds of salt. We need to determine the amount of salt, y(t), at any time t when fresh water flows in at 4 gal/min and the mixture flows out at the same rate.
02

- Set Up the Differential Equation for Part (a)

Let y(t) be the amount of salt (in pounds) in the tank at time t (in minutes). The rate of change of the amount of salt is given by the differential equation:\[ \frac{dy}{dt} = \text{(Rate of salt coming in)} - \text{(Rate of salt going out)} \]Since fresh water is entering the tank, the rate of salt coming in is 0. The rate of salt going out is given by:\[ 4 \text{ gal/min} \times \frac{y(t)}{100 \text{ gal}} \]Thus, the differential equation becomes:\[ \frac{dy}{dt} = - \frac{4y}{100} = -\frac{y}{25} \]
03

- Solve the Differential Equation

This is a separable differential equation. Rearrange and integrate:\[ \int \frac{1}{y} dy = -\frac{1}{25}t dt \]Integrating both sides, we get:\[ \text{ln}|y| = -\frac{t}{25} + C \]Exponentiating both sides, we get:\[ y = Ce^{-t/25} \]To find the constant C, use the initial condition: y(0) = 20.\[ 20 = Ce^0 \] thus C = 20
04

- Find y(t)

Substitute C back into the equation to get the amount of salt at any time t:\[ y(t) = 20e^{-t/25} \]
05

- Determine the Eventual Concentration for Part (a)

As t approaches infinity, \( e^{-t/25} \) approaches 0. Therefore, the eventual concentration of salt in the brine solution is 0 pounds per gallon.
06

- Set Up the Differential Equation for Part (b)

Now, instead of fresh water, a brine solution with a salt concentration of 2 lb/gal flows into the tank at 4 gal/min. The rate of salt coming in is:\[ 4 \text{ gal/min} \times 2 \text{ lb/gal} = 8 \text{ lb/min} \]The rate of salt going out is still:\[ 4 \text{ gal/min} \times \frac{y(t)}{100 \text{ gal}} = \frac{4y}{100} = \frac{y}{25} \]Thus, the differential equation becomes:\[ \frac{dy}{dt} = 8 - \frac{y}{25} \]
07

- Solve the Differential Equation for Part (b)

This is a first-order linear differential equation. Rearrange it as follows:\[ \frac{dy}{dt} + \frac{y}{25} = 8 \]Multiply by the integrating factor \( e^{t/25} \):\[ e^{t/25} \frac{dy}{dt} + \frac{y}{25} e^{t/25} = 8 e^{t/25} \]Integrate both sides to get:\[ y(t)e^{t/25} = 25 \times 8 e^{t/25} + C \]\[ y(t) = 200 + Ce^{-t/25} \]Using the initial condition y(0) = 20, solve for C:\[ 20 = 200 + C \] thus C = -180.
08

- Find y(t) for Part (b)

Substitute C back into the equation to get the amount of salt at any time t:\[ y(t) = 200 - 180e^{-t/25} \]
09

- Determine the Eventual Concentration for Part (b)

As t approaches infinity, \( e^{-t/25} \) approaches 0. Therefore, the eventual concentration of salt in the brine solution is 200 lb in 100 gal, which is 2 lb/gal.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

brine solution
A brine solution is simply a mixture of salt and water. In many real-world scenarios, brine solutions are used in various processes, such as in food preservation or in chemical reactions. In this particular exercise, we start with a tank containing a brine solution. Initially, there is 100 gallons of this solution, within which 20 pounds of salt are dissolved. This sets up the initial conditions for our problem. Understanding how the brine solution changes over time involves tracking how salt moves in and out of the tank as water or other brine enters and exits.
rate of change
The concept of the rate of change is crucial in understanding differential equations. Here we are interested in how the amount of salt in the tank changes over time. The rate of change of salt in the tank, \(\frac{dy}{dt}\), is determined by the difference between the rate at which salt enters the tank and the rate at which salt leaves the tank. For example, when fresh water enters the tank (which contains no salt), the rate of salt coming in is 0. On the other hand, if a brine solution with a certain salt concentration enters the tank, the rate of salt entering is given by the flow rate of that brine solution multiplied by its salt concentration.
separable differential equation
A separable differential equation is a type of differential equation that can be rewritten so that all terms involving the unknown function and its differential are on one side of the equation, and all terms involving the independent variable are on the other side. In our exercise, we dealt with the equation \[ \frac{dy}{dt} = - \frac{y}{25} \] which is easily separable. By rearranging terms and integrating both sides, we obtained a solution that describes the amount of salt in the tank as a function of time. The separable form made it straightforward to solve.
initial conditions
Initial conditions are crucial for solving differential equations, as they allow us to find specific solutions that fit the context of a problem. In this exercise, the initial condition is given by the amount of salt in the tank at time t=0. Initially, there are 20 pounds of salt dissolved in the 100 gallons of the brine solution. This information helps us determine the constant of integration when solving the differential equation. It ensures that our solution accurately reflects the real-world scenario described in the problem.
concentration
Concentration refers to the amount of a substance (in this case, salt) present in a certain volume of solution. In our exercise, the concentration of salt in the brine solution changes over time as water or another brine solution enters and leaves the tank. By finding the amount of salt at any given time, we can then determine the concentration by dividing the amount of salt by the volume of the solution (which remains constant). For instance, as time goes to infinity, we found that the concentration of salt in the tank approached 0 lbs/gallon when fresh water was added and 2 lbs/gallon when a 2 lbs/gallon brine solution was added.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equilibrium solution \(y=c\) is classified as semistable if solutions on one side of \(y=\) \(c\) approach \(y=c\) as \(t \rightarrow \infty\) but on the other side of \(y=c\), they move away from \(y=c\) as \(t \rightarrow \infty\). Use this definition to classify the equilibrium solutions (as asymptotically stable, semistable, or unstable) of the following differential equations. (a) \(y^{\prime}=y(y-2)^{2}\) (b) \(y^{\prime}=y^{2}(y-1)\) (c) \(y^{\prime}=y-\sqrt{y}\) (d) \(y^{\prime}=y^{2}\left(9-y^{2}\right)\) (e) \(y^{\prime}=1+y^{3}\) (f) \(y^{\prime}=y-y^{3}\) (g) \(y^{\prime}=y^{2}-y^{3}\)

(a) Find the solution to the IVP \(d y / d t=\) \(-r(1-y / A) y, y(0)=y_{0}\), where \(r\) and \(A\) are positive constants. (b) If \(y_{0}A\), show that \(\lim _{t \rightarrow \infty} y(t)=\infty\). (d) Show that if \(y_{0}>A\), then \(y(t)\) has a vertical asymptote at \(t=\) \(\frac{1}{r} \ln \left(\frac{y_{0}}{y_{0}-A}\right)\).

(See Exercise 21) Determine the minimum initial velocity needed to launch the lunar module (used on early space missions) from the surface of the Earth's moon given that the moon's radius is \(R \approx 1080\) miles and the acceleration of gravity of the moon is \(16.5 \%\) of that of the Earth.

Suppose that an object of mass \(1 \mathrm{~kg}\) is thrown with a downward initial velocity of \(5 \mathrm{~m} / \mathrm{s}\) and is subjected to an air resistance equivalent to the instantaneous velocity. (a) Find the velocity of the object and the distance fallen at time \(t\). (b) How far does the object drop after \(5 \mathrm{~s}\) ?

At the request of his children, a father makes homemade popsicles. At 2:00 p.m., one of the children asks if the popsicles are frozen \(\left(0^{\circ} \mathrm{C}\right)\), at which time the father tests the temperature of a popsicle and finds it to be \(5^{\circ} \mathrm{C}\). If the father placed the popsicles with a temperature of \(15^{\circ} \mathrm{C}\) in the freezer at 12:00 p.m. and the temperature of the freezer is \(-2^{\circ} \mathrm{C}\), when will the popsicles be frozen?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.