Exponential growth and modelling.

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abigail2834

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A certain type of bacteria replicates at a rate of 160% every four hours, but a hole in the dish allows 50,000 bacteria to escape every hour. How long does it take for the amount of bacteria to reach 10^12 if the initial amount of bacteria is 1 million? Using a spreadsheet, develop a model for the growth of this particular strain of bacteria.

I'd just like to know to do this, like how to get the amount of bacteria at time t=4 or t=8. I just can't figure this out!
 
A certain type of bacteria replicates at a rate of 160% every four hours, but a hole in the dish allows 50,000 bacteria to escape every hour. How long does it take for the amount of bacteria to reach 10^12 if the initial amount of bacteria is 1 million? Using a spreadsheet, develop a model for the growth of this particular strain of bacteria.
Click to expand...

Are you studying recursive formulas by any chance?

You initial value is

U(sub0) = 10^6

Subsequent terms can be found from

U(sub n)] = [U(sub n-1)](1.6) – 50,000(4)

Each term will describe population at 4 hour intervals.

That’s why a spreadsheet was recommended; they are good for repetitive calculations.

Hope that helps.

Edited to correct factor of 4 on the 50,000.
 
Easiest way is treat this as a $1,000,000 loan at rate of 160% per period,
with payments of $200,000 per period; period = 4 hours.
(the $200,000 representing 50,000*4 escaping every 4 hours).
The loan balance will increase, of course, since payment is less than interest.

1000000(2.6^n) - 200000(2.6^n - 1) / 1.6 = 10^12
Solve for n ; n = ~14.5985

14.5985 * 4 = 58.394 hours

Say you wanted to know the accumulation after 12 hours (use above formula) :
1000000(2.6^3) - 200000(2.6^3 - 1) / 1.6 = ?
3 is used (not 12) since 12 hours is 3 periods.

OK?
 
The answer in the back of the book is 121 hours. Could this be continuous instead of recursive?
 
abigail2834 said:
The answer in the back of the book is 121 hours. Could this be continuous instead of recursive?
Click to expand...

In continuous case, it will take less time - not more. Following Mark's recursive relation - I do get ~ 120 hours
 
But when I find the values for each of the intervals, make a graph in excel, and attach a trendline, the equation that shows up is y = 952742(e^(0.3999x)). When I set this equal to 10^12, I get 34.668 intervals, or about 138 hours. This is with 5 data points. But when I use 15 data points, I get 124.340 hours. When I use 20 data points, I get 123.25 hours. How many data points did you use or am I doing something wrong?

Interval: ---- Bacteria:
0 ------------- 1000000
1 ------------- 1400000
2 ------------- 2040000
3 ------------- 3064000
4 ------------- 4702400
....and so on

Thanks for all the help.
 
You guys have thoroughly confused me.

wjm11 says:
U(sub0) = 10^6
Subsequent terms can be found from
U(sub n)] = [U(sub n-1)](1.6) – 50,000(4)
Each term will describe population at 4 hour intervals.

Agree.
This means:
0: 1,000,000
1: 2,400,000
2: 6,040,000
.....
14: 564,462,403,653
15: 1,467,602,049,500 : > 10^12

So 10^12 is reached in a bit less than 15*4 = 60 hours
About what I got in my previous post: 14.5985 * 4 = 58.394 hours

Then Subhotosh says:
In continuous case, it will take less time - not more.
Following Mark's recursive relation - I do get ~ 120 hours

Agree with 1st statement, of course.
BUT HOW do you get ~120 hours?
By the way, you mean wjm11, not Mark, right?

By the way, if startoff number is 125,000.5, then I get close to 120 hours.
125000 * 1.6 = 200000; so startoff must > 125000.
YES: 125,000.5 ; not a typo!

With 1 million being the startoff, I get ~120 months at 12^18 instead of 10^12.
 
I guess there is difference in interpretation.

In Bill's analysis (not Mark's, sorry Bill) - the equivalent "loan-rate" or interest rate is 60%. So Denis - if you use 1.6 - instead of 2.6 - all the answeres will converge.

Code: 
Months	Bugs
0	1000000
4	1400000
8	2040000
12	3064000
16	4702400
20	7323840
24	11518144
28	18229030.4
32	28966448.64
36	46146317.82
40	73634108.52
44	117614573.6
48	187983317.8
52	300573308.5
56	480717293.6
60	768947669.7
64	1230116272
68	1967986035
72	3148577655
76	5037524248
80	8059838797
84	12895542076
88	20632667321
92	33012067714
96	52819108343
100	84510373349
104	1.35216E+11
108	2.16346E+11
112	3.46153E+11
116	5.53845E+11
120	8.86152E+11
124	1.41784E+12
 
"replicates at rate of 160%" means a 60% rate? :shock:

OK then; agree with your ending balance after 124 hours.
10^12 is actually reached at 30.25716 "4hour periods", so 121.0286 hours,
which is Abigail's answer from "the back of the book".
So all it is is a simple "loan amortization" schedule.
For which the formula is:
P(1+i)^n - D[(1+i)^n - 1]/i = F
1000000(1.6^n) - 200000(1.6^n - 1) / .6 = 10^12

Puzzling is WHY mention 50,000 per hour instead of simply 200,000 per 4 hours?
I feel that whoever set up that problem for Abigail needs to re-visit exponentiation.
 
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