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Suppose that a room containing 1800 cubic feet of air is originally free of carbon monoxide (CO).

0

$begingroup$

Suppose that a room containing 1800 cubic feet of air is originally free of carbon monoxide (CO). Beginning at time $t=0$, cigarette smoke containing 5% CO is introduced into the room at a rate of $0.7$ feet$^3$/minute. The well-circulated smoke and air mixture is allowed to leave the room at the same rate.

Let $A(t)$ represent the amount of CO in the room (in feet$^3$) after $t$ minutes.

(A) Write the DE model for the time rate of change of CO in the room. Also state the initial condition.

$fracdAdt$=?
$A(0)=$?

(B) Solve the IVP to find the amount of CO in the room at any time $t>0$.

$A(t)=$?

(C) Extended exposure to a CO concentration as low as 0.00012 is harmful to the human body. Find the time at which this concentration is reached.

$t=$?

ordinary-differential-equations

share|cite|improve this question

edited Oct 12 '15 at 7:35

Epsilon

516314

asked Oct 12 '15 at 6:52

PetePete

165

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey Pete. Welcome at SE. If you ask a question here, it is wanted that you show your own ideas for solving it first. You should not just state an exercise you are supposed to answer. So what have you tried to solve the exercise? Where exactly are your problems?
  $endgroup$
  – Marcel
  Oct 12 '15 at 7:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i honestly dont know where to start as far as finding dA/dt. It is obvious what A(0) is but beyond that I am not sure where to go. It is confusing to me that the smoke and air mixture can leave the room at the same rate it goes in. If that is the case would it not always be the same amount of CO in the room???
  $endgroup$
  – Pete
  Oct 12 '15 at 20:42
  

  
  
  
  

add a comment | 

0

$begingroup$

Suppose that a room containing 1800 cubic feet of air is originally free of carbon monoxide (CO). Beginning at time $t=0$, cigarette smoke containing 5% CO is introduced into the room at a rate of $0.7$ feet$^3$/minute. The well-circulated smoke and air mixture is allowed to leave the room at the same rate.

Let $A(t)$ represent the amount of CO in the room (in feet$^3$) after $t$ minutes.

(A) Write the DE model for the time rate of change of CO in the room. Also state the initial condition.

$fracdAdt$=?
$A(0)=$?

(B) Solve the IVP to find the amount of CO in the room at any time $t>0$.

$A(t)=$?

(C) Extended exposure to a CO concentration as low as 0.00012 is harmful to the human body. Find the time at which this concentration is reached.

$t=$?

ordinary-differential-equations

share|cite|improve this question

edited Oct 12 '15 at 7:35

Epsilon

516314

asked Oct 12 '15 at 6:52

PetePete

165

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey Pete. Welcome at SE. If you ask a question here, it is wanted that you show your own ideas for solving it first. You should not just state an exercise you are supposed to answer. So what have you tried to solve the exercise? Where exactly are your problems?
  $endgroup$
  – Marcel
  Oct 12 '15 at 7:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i honestly dont know where to start as far as finding dA/dt. It is obvious what A(0) is but beyond that I am not sure where to go. It is confusing to me that the smoke and air mixture can leave the room at the same rate it goes in. If that is the case would it not always be the same amount of CO in the room???
  $endgroup$
  – Pete
  Oct 12 '15 at 20:42
  

  
  
  
  

add a comment | 

$begingroup$

Suppose that a room containing 1800 cubic feet of air is originally free of carbon monoxide (CO). Beginning at time $t=0$, cigarette smoke containing 5% CO is introduced into the room at a rate of $0.7$ feet$^3$/minute. The well-circulated smoke and air mixture is allowed to leave the room at the same rate.

Let $A(t)$ represent the amount of CO in the room (in feet$^3$) after $t$ minutes.

(A) Write the DE model for the time rate of change of CO in the room. Also state the initial condition.

$fracdAdt$=?
$A(0)=$?

(B) Solve the IVP to find the amount of CO in the room at any time $t>0$.

$A(t)=$?

(C) Extended exposure to a CO concentration as low as 0.00012 is harmful to the human body. Find the time at which this concentration is reached.

$t=$?

ordinary-differential-equations

share|cite|improve this question

edited Oct 12 '15 at 7:35

Epsilon

516314

asked Oct 12 '15 at 6:52

PetePete

165

$endgroup$

share|cite|improve this question

edited Oct 12 '15 at 7:35

Epsilon

516314

asked Oct 12 '15 at 6:52

PetePete

165

share|cite|improve this question

edited Oct 12 '15 at 7:35

Epsilon

516314

asked Oct 12 '15 at 6:52

PetePete

165

edited Oct 12 '15 at 7:35

Epsilon

516314

edited Oct 12 '15 at 7:35

Epsilon

516314

asked Oct 12 '15 at 6:52

PetePete

165

asked Oct 12 '15 at 6:52

PetePete

165

1 Answer
1

active

oldest

votes

0

$begingroup$

$A$ represents the amount of CO in the room. For writing down a model for the time rate of change we have to ask how A changes over time?

Every minute, $0.7*0.05 text feet^3$ CO is coming into the room and $0.7*A text feet^3$ CO is leaving the room. Therefore, we get $fracdAdt = 0.7*0.05 - 0.7*A$. This is the differential equation for $A$.

Now, just solve this DE and calculate the time $t$ with $A(t)=0.00012$.

share|cite|improve this answer

answered Oct 13 '15 at 9:16

MarcelMarcel

573422

$endgroup$

add a comment | 

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0

$begingroup$

$A$ represents the amount of CO in the room. For writing down a model for the time rate of change we have to ask how A changes over time?

Every minute, $0.7*0.05 text feet^3$ CO is coming into the room and $0.7*A text feet^3$ CO is leaving the room. Therefore, we get $fracdAdt = 0.7*0.05 - 0.7*A$. This is the differential equation for $A$.

Now, just solve this DE and calculate the time $t$ with $A(t)=0.00012$.

share|cite|improve this answer

answered Oct 13 '15 at 9:16

MarcelMarcel

573422

$endgroup$

add a comment | 

0

$begingroup$

$A$ represents the amount of CO in the room. For writing down a model for the time rate of change we have to ask how A changes over time?

Every minute, $0.7*0.05 text feet^3$ CO is coming into the room and $0.7*A text feet^3$ CO is leaving the room. Therefore, we get $fracdAdt = 0.7*0.05 - 0.7*A$. This is the differential equation for $A$.

Now, just solve this DE and calculate the time $t$ with $A(t)=0.00012$.

share|cite|improve this answer

answered Oct 13 '15 at 9:16

MarcelMarcel

573422

$endgroup$

add a comment | 

$begingroup$

$A$ represents the amount of CO in the room. For writing down a model for the time rate of change we have to ask how A changes over time?

Every minute, $0.7*0.05 text feet^3$ CO is coming into the room and $0.7*A text feet^3$ CO is leaving the room. Therefore, we get $fracdAdt = 0.7*0.05 - 0.7*A$. This is the differential equation for $A$.

Now, just solve this DE and calculate the time $t$ with $A(t)=0.00012$.

share|cite|improve this answer

answered Oct 13 '15 at 9:16

MarcelMarcel

573422

$endgroup$

share|cite|improve this answer

answered Oct 13 '15 at 9:16

MarcelMarcel

573422

answered Oct 13 '15 at 9:16

MarcelMarcel

573422

answered Oct 13 '15 at 9:16

MarcelMarcel

573422