Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

The air in a small room 12 ft by 8 ft by 8 ft is 3% carbon monoxide. Starting at t = 0, fresh air containing no carbon monoxide is blown into the room at a rate of $100 f t^{3} / m i n$ . If air in the room flows out through a vent at the same rate, when will the air in the room be 0.01% carbon monoxide?

The amount of carbon monoxide in the air will reach 0.01% after 43.8 min.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Analyzing the given statement

It can view the room as a compartment containing air. If we let $x (t)$ denote the amount of carbon monoxide in the room at a time t, we can determine the concentration of carbon monoxide in the room by dividing $x (t)$ by the volume of the room at a time t. one uses the mathematical model described by the following equation to solve for x(t),

$\frac{d x}{d t} =$ Input rate – Output rate ......................(1)

Step 2: To determine the volume of air in the room

The dimensions of the room are 12 ft by 8 ft by 8 ft. So, the volume of the room is $(12 f t) \cdot (8 f t) \cdot (8 f t) = 768 f t^{3}$

Step 3: To determine the input rate of carbon monoxide in the room

We are given that at t = 0 fresh air containing no carbon monoxide is blown into the room at a rate of $100 f t^{3} / m i n$ . So, one concludes that the input rate of carbon monoxide in the room is,

$(100 f t^{3} / m i n) \cdot (0) = 0$

Step 4: To determine the output rate of carbon monoxide from the room

We are given that the air in the room flows out through a vent at the same rate as it is blown into the room i.e., $100 f t^{3} / m i n$ . So, one concludes that the output rate of carbon monoxide from the room is,

$(100) \cdot \frac{x (t)}{768} = \frac{25 \cdot x (t)}{192}$

Step 5: Determining the initial value problem

The air initially contained 3% carbon monoxide. So,

$\frac{x (0)}{768} = 3 % \frac{x (0)}{768} = \frac{3}{100} x (0) = \frac{(768) (3)}{100} x (0) = 23.04$

So, one set the initial value $x (0) = 23.04$ .

Substituting the input and output rates from step 3 and step 4 into the equation (1), one will get the following initial value problem as a mathematical model for the mixing problem,

i.e., $\frac{d x}{d t} = 0 - \frac{25 x (t)}{192}, x (0) = 23.04$

i.e; $\frac{d x}{d t} + \frac{25 x (t)}{192} = 0, x (0) = 23.04$

Step 6: To find the solution to the initial value problem obtained in step 4 to find the amount of carbon monoxide in the air at the time t min

The differential equation obtained in step 4 is

$\frac{d x}{d t} + \frac{25 x (t)}{192} = 0$ …… (2)

Integrating factor, I.F.= $e^{\int \frac{25}{192} d t} = e^{\frac{25}{192} t}$ .

Multiplying both sides of equation (2) by $e^{\frac{25}{192} t}$ ,

$e^{\frac{25}{192} t} \cdot \frac{d x}{d t} + e^{\frac{25}{192} t} \cdot \frac{25 \cdot x (t)}{192} = 0 \frac{d}{d t} (x e^{\frac{25}{192} t}) = 0$

Now, integrating both sides,

$x \cdot e^{\frac{25}{192} t} = C$

where C is an arbitrary constant.

$x = C e^{- \frac{25}{192} t}$ …… (3)

At t=0, x=23.04

Therefore, from equation (3),

$C = 23.04$

Substituting this value of C in equation (3),

$x = (23.04) e^{- \frac{25}{192} t}$

So, the amount of carbon monoxide in the room after the time, t min is $x (t) = (23.04) e^{- \frac{25}{192} t}$

Step 6: To determine the time after which the amount of carbon monoxide in the air will reach 0.01%

Firstly, we will find the amount of 0.01% of carbon monoxide by multiplying 0.01% by the volume of the room,

$(\frac{0.01}{100}) (768) = 0.0768$

Substituting this value of $x (t) = 0.0768$ in equation (4),

role="math" localid="1664260534765" $0.0768 = (23.04) e^{- \frac{25}{192} t} e^{\frac{25}{192} t} = \frac{23.04}{0.0768} e^{\frac{25}{192} t} = 300 \frac{25}{192} t = l n 300 t = \frac{(192) l n 300}{25} t = 43.8 m i n$

Hence, the amount of carbon monoxide in the air will reach 0.01% after 43.8 min.

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1: Analyzing the given statement

Step 2: To determine the volume of air in the room

Step 3: To determine the input rate of carbon monoxide in the room

Step 4: To determine the output rate of carbon monoxide from the room

Step 5: Determining the initial value problem

Step 6: To find the solution to the initial value problem obtained in step 4 to find the amount of carbon monoxide in the air at the time t min

Step 6: To determine the time after which the amount of carbon monoxide in the air will reach 0.01%

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

The air in a small room 12 ft by 8 ft by 8 ft is 3% carbon monoxide. Starting at t = 0, fresh air containing no carbon monoxide is blown into the room at a rate of 100ft3/min. If air in the room flows out through a vent at the same rate, when will the air in the room be 0.01% carbon monoxide?

Step by Step Solution

Step 1: Analyzing the given statement

Step 2: To determine the volume of air in the room

Step 3: To determine the input rate of carbon monoxide in the room

Step 4: To determine the output rate of carbon monoxide from the room

Step 5: Determining the initial value problem

Step 6: To find the solution to the initial value problem obtained in step 4 to find the amount of carbon monoxide in the air at the time t min

Step 6: To determine the time after which the amount of carbon monoxide in the air will reach 0.01%

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