the logistics of the logistic curve
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The Logistics of the Logistic Curve. Dan Kennedy Baylor School Chattanooga, TN [email protected]. A real-world problem from my high school days: - PowerPoint PPT PresentationTRANSCRIPT
A real-world problem from my high school days:
Under favorable conditions, a single cell of the bacterium Escherichia coli divides into two about every 20 minutes. If the same rate of division is maintained for 10 hours, how many organisms will be produced from a single cell?
Solution:
10 hours = 30 20-minute periods
There will be 1 ∙ 2^30 = 1,073,741,824 bacteria after 10 hours.
A problem that seems just as reasonable:
Under favorable conditions, a single cell of the bacterium Escherichia coli divides into two about every 20 minutes. If the same rate of division is maintained for 10 days, how many organisms will be produced from a single cell?
Solution:
10 days = 720 20-minute periods
There will be 1 ∙ 2^720 ≈ 5.5 ∙ 10^216 bacteria after 10 days.
Makes sense…
…until you consider that there are probably fewer than 10^80 atoms in the entire universe.
Real world Bizarro world
Why didn’t they tell us the truth? Most of those classical “exponential growth” problems should have been “logistic growth” problems!
Exponential Logistic
1991 / BC-6 (The Rumor)
A certain rumor spreads through a community at the rate
2 (1 )dy
y ydt
, where y is the proportion of the community that has
heard the rumor at time t. (a) What proportion of the community has heard the rumor when it is spreading the fastest? (b) If at time t = 0 ten percent of the people have heard the rumor, find y as a function of t. (c) At what time t is the rumor spreading the fastest?
This was the first logistic differential equation to appear in an AP FRQ.
The first part did not involve solving the differential equation.
The answer was the value of y that maximized 2y (1 – y) … equivalent to finding the vertex of an upside-down parabola!
0 1 y
2 (1 )dy
y ydt
(1/2, 1/2)
The max occurs at y = 1/2.
The next part required finding an antiderivative using partial fractions.
Separate the variables!
2 (1 )
2 (1 )
2(1 )
dyy y
dtdy
dty y
dydt
y y
Oliver Heaviside
Partial fractions in the AP course are best done by the Heaviside Method:
1
(1 ) 1
A B
y y y y
(1 ) ( ) 1A y B y
(1 ) )1 ( 11A B 1B (1 ) )0 ( 10A B 1A
1 1 1
(1 ) 1y y y y
Oliver Heaviside Oliver, on the heavy side
2 2
(Note 0 1)
12
(1 )
1 12
1
ln ln(1 ) 2
ln 21
1t C t
y
dy dty y
dy t Cy y
y y t C
yt C
y
ye Ae
y
2
0 Since 10% when 0
1
0.1
1 0.11
9
t
y t
yAe
y
Ae
A
21
1 9ty
ey
2
2
2
2
2
2
2
1 9
9 (1 )
9
1
9 1 9
t
t
t
t
t
t
t
y e
e
y
y e y
e y y e
ye e
Answer: y = 1/2
2
2
2 2
1
9 2
9 2
t
t
t t
e
e
e e
2
1ln9 l
9
2 ln9
n32
t
t
e
t
Exponential growth occurs when the growth rate is proportional to the size of the population.
dPkP
dt
ktP Cet
P(t)
Logistic growth occurs when the growth rate slows as the population approaches a maximal sustainable population M.
( )dP
kP M Pdt
t
P(t)
M
Before solving the differential equation, it is useful to observe how much information can be gleaned from the differential equation itself.
( )dP
kP M Pdt
• The graph of kP(M – P) is an upside- down parabola with zeros at P = 0 and P = M. The maximum is always at M/2.
P(t)
t
M
steepest growth
M/2
( )dP
kP M Pdt
• When P < M, the growth rate is positive. When P > M, the
growth rate is negative. In either case, the population approaches M as t increases.
We can see this in a slope field for the differential equation.
M
0
Now let’s look at one of the most infamous logistic problems of all:
2004 / BC-5
Suffice it to say that the BC students in 2004 were not amused.
2004 / BC-5 (Logistic Growth)
A population is modeled by a function P that satisfies the logistic differential equation
15 12
dP P P
dt
.
(a) If P(0) = 3, what is lim ( )
tP t
?
If P(0) = 20, what is lim ( )tP t
?
(b) If P(0) = 3, for what value of P is the population growing the fastest? (c) A different population is modeled by a function Y that satisfies the separable differential equation
1 .5 12
dY Y t
dt
Find Y(t) if Y(0) = 3. (d) For the function Y found in part (c), what is lim ( )
tY t
?
What the students were supposed to do was to answer (a), (b), and (c) using their rich knowledge of logistic functions.
(a) 12 ; 12 (b) 12/2 = 6
Then they had to solve a separable differential equation in (c). The equation was not logistic. In fact, part (d) was designed to let them savor the difference.
Sadly, many of them never made it past (a) and (b).
The logistic equation was:
15 12
PdP P
dt
The equation in part (c) was:
15 12
tdy P
dt
Notice that this population would grow until t = 12 and then crash fast as t got bigger and bigger.
Here’s the slope field for 15 12
tdy P
dt
12
And here’s the solution to the initial value problem in (c):
2
15 12
1
5 60
ln5 120
dy y t
dt
dy tdt
y
t ty C
2
2
2
5 120
5 120
5 120
(0)
3
3 3t
t tC
t
t
t
y e
y Ae
y A
y e
Then, finally, the limit in (d):
2
5 120lim 3 0.t t
te
As with the rumor problem, solving a logistic differential equation requires the use of partial fractions.
In fact, it can be done in the general case:
( )
( )
1 Mk t
dPkP M P
dt
MP
Ae
2004 / BC-5 (Logistic Growth)
A population is modeled by a function P that satisfies the logistic differential equation
15 12
dP P P
dt
.
(a) If P(0) = 3, what is lim ( )
tP t
?
If P(0) = 20, what is lim ( )tP t
?
(b) If P(0) = 3, for what value of P is the population growing the fastest? (c) A different population is modeled by a function Y that satisfies the separable differential equation
1 .5 12
dY Y t
dt
Find Y(t) if Y(0) = 3. (d) For the function Y found in part (c), what is lim ( )
tY t
?
In the case of 2004 / BC-5:
(1/ 5)
0.2
15 12
112
6012
112
if (0) 3.1 3
t
t
dP P P
dt
dPP P
dt
PAe
P Pe
John Kenelly Clemson University
Two neat things I learned about logistic functions from John Kenelly:
1. The Wal-Mart Story
2. The Infection Game
Sam Walton apparently knew that hot-selling items did not sell like hotcakes.
They sold like logistic functions.
So he tracked sales from all his stores…daily.
Daily sales are the differences in your total sales from day to day. Positive daily sales mean your total sales are going up. A big sales day means your sales curve has a steep slope that day.
But Sam was not looking at slope.
He was looking at differences in daily sales.
While his competitors were reacting to slope, Sam was reacting to concavity.
He was looking for the point of inflection.
Sam finds the first negative difference in daily sales.
Total sales look great, but Sam knows what is coming.
While his competitors are stocking up, Sam starts unloading his inventory. When demand lets up, his shelves are free to stock the next hot item!
The Infection Game is a great way to show students that logistic growth is a real phenomenon.
Give everyone in the class a number. (For small classes, make it a hand infection and give each hand a number.)
Enter this command and “infect” a random victim. From now on, each new ENTER command will spread the infection.
Press ENTER again. You now have two infected victims at time t = 2.
Press ENTER two times (since each of the two infected victims can spread the infection.)
Now press ENTER 4 times…and so on.
The infection spreads exponentially at first, because you press ENTER more often as it spreads. But soon it slows down because there are fewer new victims to infect.
It actually gets exciting when there are two healthy students left and they wonder who will go first and how many ENTERs they can survive.
I gathered some data from an actual class…
TimeTime 11 22 33 44 55 66 77 88 99 1100
1111
InfecteInfectedd
11 22 44 88 1133
2222
2266
2288
2299
2299
3030
