• Support PF! Buy your school textbooks, materials and every day products Here!

Calc 2 differentials

  • Thread starter tnutty
  • Start date
  • #1
327
1
The air in a room with volume 180 m^3 contains 0.20% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 3 m^3/min and the mixed air flows out at the same rate.

(a) Find the percentage of carbon dioxide in the room as a function of time.




attempt :

dy/dx = rate in - rate out

rate in = 0.05 * 3

rate out = 3 * y/180

from that my answer turned out to be :

y(t) = - (e^ (-t/60) * e ^(ln(9) + 1/300 ) - 9 )

any help?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,793
922
What help are you asking for? If you want to know if that is the correct answer, no, it is not. For one thing, when t= 0, y should be 0.2(180)= 36 but your formula gives y(0)= eln(9)+ 1/300- 9. If you want to know what you did wrong, we will have to see what you did first!
 

Related Threads for: Calc 2 differentials

Replies
4
Views
3K
Replies
1
Views
2K
Replies
4
Views
3K
Replies
5
Views
6K
  • Last Post
Replies
5
Views
922
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
398
  • Last Post
Replies
4
Views
2K
Top