A residential home for elderly people has been investigating the performance of two different makes of Cleaner to use daily on the shower bases.
of the harmful
''Aquatec''was found to kill 98 percent of the harmful Bacteria present but another 2450 new Bacteria appeared daily.
''Magiclean'' was found to kill 96 present but another 2400 new Bacteria appeared daily.
Compare the effectiveness, in the long-term, of these two products as far as the eradication of harmful bacteria is concerned.
0
Help with recurrence relations question
Started by gary, Oct 21 2004 05:17 PM
6 replies to this topic
#2
-
- Members
-
- 131 posts
Site Swot
Posted 21 October 2004 - 05:49 PM
Let a = aquatec and m = magiclean.
an+1 = 0.02an + 2450
Since -1 < 0.02 < 1 there is a limit.
Let an+1 = mn = L
L = 0.02L + 2450
L - 0.02L = 2450
0.98L = 2450
L = 2450/0.98
L = ...
mn+1 = 0.04mn + 2400
Since -1 < 0.04 < 1 there is a limit.
Let mn+1 = mn = L
L = 0.04L + 2400
L - 0.04L = 2400
0.96L = 2400
L = 2400/0.96
L = ...
Since [insert name of better cleaner] converges at a lower value (and thus keeps the amount of bacteria at a lower amount), this is a more effective cleaner than [insert name of poorer cleaner]
an+1 = 0.02an + 2450
Since -1 < 0.02 < 1 there is a limit.
Let an+1 = mn = L
L = 0.02L + 2450
L - 0.02L = 2450
0.98L = 2450
L = 2450/0.98
L = ...
mn+1 = 0.04mn + 2400
Since -1 < 0.04 < 1 there is a limit.
Let mn+1 = mn = L
L = 0.04L + 2400
L - 0.04L = 2400
0.96L = 2400
L = 2400/0.96
L = ...
Since [insert name of better cleaner] converges at a lower value (and thus keeps the amount of bacteria at a lower amount), this is a more effective cleaner than [insert name of poorer cleaner]
#4
-
- Members
-
- 541 posts
Child Prodigy
- Location:Glasgow
- Interests:Anime, gaming, dvds, computers.
- Gender:Male
Posted 21 October 2004 - 09:37 PM
Ah, recurrence relations. I liked those questions. They couldn't really ever change it so you knew what the question would be.
#5
-
- Members
-
- 131 posts
Site Swot
Posted 21 October 2004 - 11:31 PM
Yeah they could hardly really change it. Linked recurrence relations takes a bit of time to get the hang of though. I didn't really want one of them coming up.
Come to think of it...there wasn't a recurrence relations question in the 2004 exam.
Come to think of it...there wasn't a recurrence relations question in the 2004 exam.
#6
-
- Members
-
- 8 posts
Newbie
- Location:Falkirk
- Interests:Titanic, Rangers, piano, mates and politics
Posted 21 November 2004 - 05:48 PM
good frickin god you lot are way clever!!!!!! hate recurrence relations questions. tend just to make it up!!!!!!
#7
-
- Members
-
- 131 posts
Site Swot
Posted 22 November 2004 - 10:50 PM
Look at the working for that carefully and take every line in. Don't understand it, ask.
Basically what you're doing it finding a formula for the number of bacteria.
With that there will be a limit if the value you multiply the original by is between 1 and -1. This is because eventually there will be a value by which the number you multiply by added to whatever random number afterwards will be the same as the previous and won't go any higher.
With this you find the limit. Since at the limit, the initial u value will equal that of the next, we can let both u's be the same, ie. call it L. Then it's just a case of basic algebra to get the limit L.
For the most effective cleaner here, it'll be the one that allows for less bacteria to live.
Basically what you're doing it finding a formula for the number of bacteria.
With that there will be a limit if the value you multiply the original by is between 1 and -1. This is because eventually there will be a value by which the number you multiply by added to whatever random number afterwards will be the same as the previous and won't go any higher.
With this you find the limit. Since at the limit, the initial u value will equal that of the next, we can let both u's be the same, ie. call it L. Then it's just a case of basic algebra to get the limit L.
For the most effective cleaner here, it'll be the one that allows for less bacteria to live.
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
