/*! 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 26 When people smoke, carbon monoxi... [FREE SOLUTION] | 91Ó°ÊÓ

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When people smoke, carbon monoxide is released into the air. In a room of volume \(60 \mathrm{m}^{3},\) air containing \(5 \%\) carbon monoxide is introduced at a rate of \(0.002 \mathrm{m}^{3} / \mathrm{min}\) (This means that \(5 \%\) of the volume of the incoming air is carbon monoxide.) The carbon monoxide mixes immediately with the rest of the air, and the mixture leaves the room at the same rate as it enters. (a) Write a differential equation for \(c(t),\) the concentration of carbon monoxide at time \(t,\) in minutes. (b) Solve the differential equation, assuming there is no carbon monoxide in the room initially. (c) What happens to the value of \(c(t)\) in the long run?

Short Answer

Expert verified
The concentration approaches 6000 ppm as time approaches infinity.

Step by step solution

01

Understand the Problem

We need to model the concentration of carbon monoxide in a room over time. The problem involves a room with a volume of \(60 \mathrm{m}^3\), where air containing \(5\%\) carbon monoxide enters at a rate of \(0.002 \mathrm{m}^3/\mathrm{min}\). Our task is to write and solve a differential equation describing how the concentration changes over time \(t\).
02

Formulate the Differential Equation

Since \(5\%\) of the incoming air is carbon monoxide, the rate of carbon monoxide entering the room is \(0.002 \times 0.05 = 0.0001 \mathrm{m}^3/\mathrm{min}\). Let \(c(t)\) represent the concentration of carbon monoxide in the room at time \(t\). The room mixes air so that the concentration changes both with incoming air and outgoing flow. The change in concentration is described by: \(\frac{dc}{dt} = \text{rate in} - \text{rate out}\). The rate out is given by the current concentration times the air outflow, which is the same as the inflow rate, so: \(\frac{dc}{dt} = 0.0001 - \frac{c(t)}{60}\times 0.002\). Simplifying: \(\frac{dc}{dt} = 0.0001 - \frac{0.002}{60}c(t)\).
03

Solve the Differential Equation

This is a first-order linear differential equation. We rewrite it: \(\frac{dc}{dt} + \frac{0.002}{60}c(t) = 0.0001\). The standard solution involves an integrating factor, here \(\mu(t) = e^{\frac{0.002}{60}t}\). Multiply through by this factor: \(e^{\frac{0.002}{60}t} \frac{dc}{dt} + \frac{0.002}{60}e^{\frac{0.002}{60}t}c(t) = 0.0001e^{\frac{0.002}{60}t}\). Recognize the left side as the derivative of \(c(t) e^{\frac{0.002}{60}t}\), so integrate both sides with respect to \(t\). Find:\[c(t) e^{\frac{0.002}{60}t} = \int 0.0001e^{\frac{0.002}{60}t} dt\]. Evaluate the integral and solve for the initial condition \(c(0) = 0\). Thus, \(c(t) = Ce^{-\frac{0.002}{60}t} + \frac{6000}{60} (1 - e^{-\frac{0.002}{60}t})\), find \(C = 0\), thus \(c(t) = 6000 (1 - e^{-\frac{0.002}{60}t})\).
04

Analyze Long-term Behavior

As \(t\) approaches infinity, the exponential term approaches zero, so \(c(t)\) approaches \(6000\) ppm. This steady state is consistent with the input concentration rate stabilized over time.

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Key Concepts

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

Carbon Monoxide Concentration
Carbon monoxide (CO) is a colorless, odorless toxic gas produced by burning carbon-containing materials. In scenarios like smoking indoors, CO levels can increase rapidly, posing health risks. In the given problem, we focus on the concentration of CO in a room filled with air over time.
The concentration of carbon monoxide describes how much CO is present in a given volume of air. To consider this mathematically, we use the variable \(c(t)\), which represents CO concentration at time \(t\). Here, changes in concentration are modeled by differential equations, which capture how different factors influence CO levels.
Since air with a certain percentage of CO enters and exits the room, this keeps affecting the CO concentration. By setting up a proper differential equation, one can understand how quickly CO accumulates and predict the long-term behavior in the room. Effective analysis helps ensure CO does not reach harmful levels, especially important in enclosed spaces like rooms without sufficient ventilation.
First-Order Linear Differential Equation
A first-order linear differential equation involves the first derivative of a function and is linear in nature. The general form can typically be written as: \(\frac{dy}{dt} + p(t)y = g(t)\). This kind of equation is widely used to solve problems where a rate of change is affected linearly.
In the exercise, we modeled the scenario by deriving a first-order linear differential equation to track CO concentration changes over time. The equation derived was \(\frac{dc}{dt} + \frac{0.002}{60}c(t) = 0.0001\).
This represents a classic application where the rate of change of one quantity (CO concentration) depends on its current value and an external input (inflow of CO). By framing the problem with this equation, it becomes easier to find how \(c(t)\) evolves, ensuring predictions on CO levels remain accurate. Such equations are fundamental in many real-world applications, such as physics, engineering, and biology.
Integrating Factor Technique
The integrating factor technique is a method used to solve first-order linear differential equations. It involves multiplying the entire differential equation by an integrating factor to simplify the equation into a form that can be integrated easily.
For the given CO concentration equation \(\frac{dc}{dt} + \frac{0.002}{60}c(t) = 0.0001\), the integrating factor \(\mu(t)\) is calculated as \(e^{\int \frac{0.002}{60} dt} = e^{\frac{0.002}{60}t}\). Applying this factor transforms the equation, making its left side the derivative of \(c(t)\times e^{\frac{0.002}{60}t}\).
By recognizing this structure, we can integrate directly, leading to a solution for \(c(t)\). The result is \(c(t) = 6000 (1 - e^{-\frac{0.002}{60}t})\), providing insights on CO concentration behavior over time. This technique not only simplifies solving the equation but also gives clear functional forms, aiding in understanding system dynamics.

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Most popular questions from this chapter

Many organ pipes in old European churches are made of tin. In cold climates such pipes can be affected with tin pest, when the tin becomes brittle and crumbles into a gray powder. This transformation can appear to take place very suddenly because the presence of the gray powder encourages the reaction to proceed. The rate of the reaction is proportional to the product of the amount of tin left and the quantity of gray powder, \(p,\) present at time \(t .\) Assume that when metallic tin is converted to gray powder, its mass does not change. (a) Write a differential equation for \(p .\) Let the total quantity of metallic tin present originally be \(B\) (b) Sketch a graph of the solution \(p=f(t)\) if there is a small quantity of powder initially. How much metallic tin has crumbled when it is crumbling fastest? (c) Suppose there is no gray powder initially. (For example, suppose the tin is completely new.) What does this model predict will happen? How do you reconcile this with the fact that many organ pipes do get tin pest?

Explain what is wrong with the statement. The line \(y=2\) is an equilibrium solution to the differential equation \(d y / d x=y^{3}-4 x y\).

(a) Find all equilibrium solutions for the differential equation $$\frac{d y}{d x}=0.5 y(y-4)(2+y)$$ (b) Draw a slope field and use it to determine whether each equilibrium solution is stable or unstable.

An object of mass \(m\) is thrown vertically upward from the surface of the earth with initial velocity \(v_{0} .\) We will calculate the value of \(v_{0},\) called the escape velocity, with which the object can escape the pull of the gravity and never return to earth. since the object is moving far from the surface of the earth, we must take into account the variation of gravity with altitude. If the acceleration due to gravity at sea level is \(g,\) and \(R\) is the radius of the earth, the gravitational force, \(F\), on the object of mass \(m\) at an altitude \(h\) above the surface of the earth is given by $$F=\frac{m g R^{2}}{(R+h)^{2}}$$. (a) The velocity of the object (measured upward) is \(v\) at time \(t .\) Use Newton's Second Law of Motion to show that $$\frac{d v}{d t}=-\frac{g R^{2}}{(R+h)^{2}}$$. (b) Rewrite this equation with \(h\) instead of \(t\) as the independent variable using the chain rule \(\frac{d v}{d t}=\frac{d v}{d h} \cdot \frac{d h}{d t}\) Hence, show that $$v \frac{d v}{d h}=-\frac{g R^{2}}{(R+h)^{2}}$$. (c) Solve the differential equation in part (b). (d) Find the escape velocity, the smallest value of \(v_{0}\) such that \(v\) is never zero.

According to an article in The New York Times,' ' pigweed has acquired resistance to the weedkiller Roundup. Let \(N\) be the number of acres, in millions, where Roundup-resistant pigweed is found. Suppose the relative growth rate, \((1 / N) d N / d t,\) was \(15 \%\) when \(N=5\) and \(14.5 \%\) when \(N=10 .\) Assuming the relative growth rate is a linear function of \(N,\) write a differential equation to model \(N\) as a function of time, and predict how many acres will eventually be afflicted before the spread of Roundup-resistant pigweed halts.

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