RE: help guys!! (Math related)
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RE: help guys!! (Math related)

 
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RE: help guys!! (Math related) - 10/2/2008 9:00:34 PM   
P360onBugs


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Status: offline 		actually you guys are wrong. on number two

in 12 hours. it will be 1.17

in 24 it will not be  134percent

again you guys are thinking linear. not expontial growth.

i can  explain more. but its not 72 years. its less.

< Message edited by P360onBugs -- 10/2/2008 9:09:35 PM >

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Post #: 31
RE: help guys!! (Math related) - 10/2/2008 9:09:45 PM   
matprice


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Status: offline 		f(t)=(%g/t) where t = time and %g= percent growth we are given the population increase in size by 17% every 12 hours.... so keep trying dude and use your brain to try and figure this out on your own it helps you none for me to just give you answer if you have a specific answer i will help you with it... but once you get this concept you can do any of these type problems and i am assuming if your taking calc survey your a business major and your gonna need to understand this concept in the business world 

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Post #: 32
RE: help guys!! (Math related) - 10/2/2008 9:13:13 PM   
OUTLAWED450S


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quote:

ORIGINAL: P360onBugs

actually you guys are wrong. on number two

in 12 hours. it will be 1.17

in 24 it will not be  134percent

again you guys are thinking linear. not expontial growth.

i can  explain more. but its not 72 years. its less.


yeah your right. its just like compound intrest. i see what your saying now. that pisses me off, i thought i was looking smart.

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Post #: 33
RE: help guys!! (Math related) - 10/2/2008 9:13:25 PM   
HONDA 4x4


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Status: offline 		do you set it up as: 200=.17/t ??

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Post #: 34
RE: help guys!! (Math related) - 10/2/2008 9:15:30 PM   
HONDA 4x4


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Status: offline 		seriously i have tried. so if anybody has the right answer and are sure of it, let me have it. these 10 bonus points will seriously boost my grade.

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Post #: 35
RE: help guys!! (Math related) - 10/2/2008 9:16:59 PM   
P360onBugs


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Status: offline 		no its exponential growth

so again lets say you start with one gram and you want to get to 2 grams

then your formula will be

2=1(1.17) to the x      you are using 1.17 because to find the next year  you multiply the previous year by 1.17.  and rember x is the number of 12 hour cycles

use logs to solve it and tell me what you get.

but to explain why you use this formula look at it this way..

lets say you start with one gram.... after 12 hours it will be 1*1.17=1.17

after 24 hours it will be 1.17*1.17=1.389

after 36 hours it will be 1.389*1.17=1.601613

make  sense. you have growth on top of growth.



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Post #: 36
RE: help guys!! (Math related) - 10/2/2008 9:18:23 PM   
P360onBugs


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Status: offline 		im sure of mine  method. no guess. dont make me pull out my BS  in math. lol

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Post #: 37
RE: help guys!! (Math related) - 10/2/2008 9:22:34 PM   
matprice


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Status: offline 		listen to p360 he is right on the money...

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Post #: 38
RE: help guys!! (Math related) - 10/2/2008 9:30:35 PM   
CodyD

 

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quote:

ORIGINAL: HONDA 4x4

#1


A town's population is 2000 in 1990.  By 1995, the population had decayed exponentially to 1248 people.
a) Write a function that models the population of the town as a function of the number of years since 1990.
b) What was the population in 2000?
c) When did the population fall below 1000?

#2
A bacteria population grows by 17% every 12 hours.  How long will it take to double?




Call me crazy but doesn't exponential decay involve e. If you care for my advice I will try and help.

1a. Setup an equation that looks like this

1248=2000e^(-5x)     Solve for x.
This should present the equation...

P=2000e^(-0.094t) where P is population and t is time in years.

1b. 2000e^(-0.094 x 10years)= 781.256    *round to your liking.

1c. 1000 = 2000e^(-0.094t)    Solve for t.

t= 7.374
Add this value to the original year and you get 1997.


**Note e^(-0.094)= .91  using (.91)^t does work. But if you would like your professor to see how you got that well there you go.

2. There is a formula for doubling time.

T=log(2) / ( log( 1 + ( r / 100 ) ) )

Where r is a percentage value and T is a doubling time constant.

Plug in for r as 17 and you get T = 4.41484

Multiply 4.41484 x 12 and you get the doubling time to be 52.9777 or 53 hours.


From one college student to another wikipedia is your friend. I would have directed you straight to the wikipedia pages but if you haven't had differential equations some of the information could be rough.

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Post #: 39
RE: help guys!! (Math related) - 10/2/2008 9:32:51 PM   
HONDA 4x4


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Status: offline 		okay is it 53.112 hours?
i did the log of 2 and the log of 1.17^x and got .301=.068x
divide .301 by .068 =4.426
x12 =53.112

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Post #: 40
RE: help guys!! (Math related) - 10/2/2008 9:33:00 PM   
CodyD

 

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Status: offline 		I believe I just said the same thing as P360 it just took a while to type.

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Post #: 41
RE: help guys!! (Math related) - 10/2/2008 9:35:06 PM   
OUTLAWED450S


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quote:

ORIGINAL: CodyD

I believe I just said the same thing as P360 it just took a while to type.


i hate when that happens

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Post #: 42
RE: help guys!! (Math related) - 10/2/2008 9:35:12 PM   
P360onBugs


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Status: offline 		actually your right cody. true exponential decay and exponetial growth for that matter is continous. therfore it does  involve. e. thats why i said that earlier. but i think what they meant(might be wrong) is that it decayed eponentually every year.  however if you do continous decay. you MUST use the formula you used.

Chris

< Message edited by P360onBugs -- 10/2/2008 9:36:46 PM >

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Post #: 43
RE: help guys!! (Math related) - 10/2/2008 9:36:50 PM   
matprice


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quote:

ORIGINAL: CodyD

quote:

ORIGINAL: HONDA 4x4

#1


A town's population is 2000 in 1990.  By 1995, the population had decayed exponentially to 1248 people.
a) Write a function that models the population of the town as a function of the number of years since 1990.
b) What was the population in 2000?
c) When did the population fall below 1000?

#2
A bacteria population grows by 17% every 12 hours.  How long will it take to double?




Call me crazy but doesn't exponential decay involve e. If you care for my advice I will try and help.

1a. Setup an equation that looks like this

1248=2000e^(-5x)     Solve for x.
This should present the equation...

P=2000e^(-0.094t) where P is population and t is time in years.

1b. 2000e^(-0.094 x 10years)= 781.256    *round to your liking.

1c. 1000 = 2000e^(-0.094t)    Solve for t.

t= 7.374
Add this value to the original year and you get 1997.


**Note e^(-0.094)= .91  using (.91)^t does work. But if you would like your professor to see how you got that well there you go.

2. There is a formula for doubling time.

T=log(2) / ( log( 1 + ( r / 100 ) ) )

Where r is a percentage value and T is a doubling time constant.

Plug in for r as 17 and you get T = 4.41484

Multiply 4.41484 x 12 and you get the doubling time to be 52.9777 or 53 hours.


From one college student to another wikipedia is your friend. I would have directed you straight to the wikipedia pages but if you haven't had differential equations some of the information could be rough.

yep you did and that is the correct way to write a Function doubling time or growth you just have to plug the right numbers in the right spots.... its all in knowing what you have and figgerin out what you need you will use this in chem till your blue in the face

< Message edited by matprice -- 10/2/2008 9:38:40 PM >

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Post #: 44
RE: help guys!! (Math related) - 10/2/2008 9:37:54 PM   
CodyD

 

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Status: offline 		From reading the question I assumed continuous decay...he gave no information that indicated otherwise.

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Post #: 45
RE: help guys!! (Math related) - 10/2/2008 9:39:16 PM   
CodyD

 

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Post #: 46
RE: help guys!! (Math related) - 10/2/2008 9:40:11 PM   
matprice


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Status: offline 		now he doesnt ask this in #2 but can you right a function using that formula in f(x) form?

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Post #: 47
RE: help guys!! (Math related) - 10/2/2008 9:41:07 PM   
P360onBugs


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Status: offline 		and technically cody. the second problem should be continous growth. and if it is.....it must have e in there as well.

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Post #: 48
RE: help guys!! (Math related) - 10/2/2008 9:42:22 PM   
P360onBugs


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Status: offline 		you need to have carlifed if this is continous decay and growth. until then. we cant anser the questions....

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Post #: 49
RE: help guys!! (Math related) - 10/2/2008 9:43:25 PM   
matprice


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Status: offline 		tech guys it would look like this
G (generation time) = (time, in minutes or hours)/n(number of generations)
G = t/n
t = time interval in hours or minutes
B = number of bacteria at the beginning of a time interval
b = number of bacteria at the end of the time interval
n = number of generations (number of times the cell population doubles during the time interval)
b = B x 2n (This equation is an expression of growth by binary fission)
Solve for n:
logb = logB + nlog2
n = logb - logB
           log2
n = logb - logB
           .301
n = 3.3 logb/B
G = t/n
Solve for G
G =        t
       3.3 log b/B

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Post #: 50
RE: help guys!! (Math related) - 10/2/2008 9:45:35 PM   
P360onBugs


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Status: offline 		so after all that go around. im not sure of any of my answers  till it  gets clarified. lol

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Post #: 51
RE: help guys!! (Math related) - 10/2/2008 9:45:37 PM   
matprice


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Status: offline 		but i think the question is just academic in nature and since it is just math problem from a math class continues growth isnt implied... cause TECHNICALY then we have to start talking about saturation points and decay   

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Post #: 52
RE: help guys!! (Math related) - 10/2/2008 9:47:15 PM   
matprice


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Status: offline 		p360 you had the right idea wrong calculation cody has it using the formula doubling time... this is the answer the proff is looking for IMO 

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Post #: 53
RE: help guys!! (Math related) - 10/2/2008 9:47:42 PM   
CodyD

 

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quote:

ORIGINAL: P360onBugs

and technically cody. the second problem should be continous growth. and if it is.....it must have e in there as well.


True...I didn't catch that. Would substituting natural logs solve that?

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Post #: 54
RE: help guys!! (Math related) - 10/2/2008 9:50:31 PM   
P360onBugs


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Status: offline 		no i had the right idea and formula. go  back and solve my formula . you will get 53 hours. and that is right. assuming that you dont have continous growth.

just like the orinial poster. he is right in his numbers  and formula if you dont have continous decay. and if you have continous decay. you must use e

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Post #: 55
RE: help guys!! (Math related) - 10/2/2008 9:50:52 PM   
CodyD

 

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quote:

ORIGINAL: matprice

but i think the question is just academic in nature and since it is just math problem from a math class continues growth isnt implied... cause TECHNICALY then we have to start talking about saturation points and decay   


And eventually if the growth rate changes slightly you begin to see bifurcation points and the the math gets horrendous. For this class, whatever it is...I don't think these problems are meant to be that complex.

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Post #: 56
RE: help guys!! (Math related) - 10/2/2008 9:51:58 PM   
P360onBugs


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quote:

ORIGINAL: CodyD

quote:

ORIGINAL: P360onBugs

and technically cody. the second problem should be continous growth. and if it is.....it must have e in there as well.


True...I didn't catch that. Would substituting natural logs solve that?


no natual logs will get you the same numbers. you have to  use the continous expontenail growth formula

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Post #: 57
RE: help guys!! (Math related) - 10/2/2008 9:52:15 PM   
matprice


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Status: offline 		i am not trying to throw a wrench in yalls thought process but bacterial growth even in lab environments is not continues growth... in truth it will plateau the bacterial colony will kill a % of its self due dying in its own waste... heheh just being complicated :)   

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Post #: 58
RE: help guys!! (Math related) - 10/2/2008 9:54:56 PM   
P360onBugs


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Status: offline 		yeah i know mat.  thing is bateria growth is closer to conintious than it is to really segmented like the original problem suggests/

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Post #: 59
RE: help guys!! (Math related) - 10/2/2008 9:56:20 PM   
matprice


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quote:

ORIGINAL: CodyD

quote:

ORIGINAL: matprice

but i think the question is just academic in nature and since it is just math problem from a math class continues growth isnt implied... cause TECHNICALY then we have to start talking about saturation points and decay   


And eventually if the growth rate changes slightly you begin to see bifurcation points and the the math gets horrendous. For this class, whatever it is...I don't think these problems are meant to be that complex.


exactly cody... and i dont even care to try to expalin that... poor guy all he wanted was help with his homework hehehehehe

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