hw: p. 369 #23 – 26 (all) #31, 38, 41, 42 pick up one
TRANSCRIPT
Warm Up 1989 – AB 6
Name: _________________________ Oil is being pumped continuously from a certain oil well at a rate proportional to the
amount of oil left in the well; that is, kydt
dy , where y is the amount of oil left in the
well at any time t. Initially there were 1,000,000 (106) gallons of oil in the well, and 6 years later there were 500,000 gallons remaining. It will no longer be profitable to pump oil when there are fewer than 50,000 gallons remaining.
a) Write an equation for y, the amount of oil remaining in the well at any time t. b) At what rate is the amount of oil in the well decreasing when there are 600,000
gallons of oil remaining? c) In order not to lose money, at what time t should oil no longer be pumped from
the well?
HW:p. 369 #23 – 26 (all)
#31, 38, 41, 42
Pick up one
Logistic growth is slowed by population-limiting factors
M = Carrying capacity is the maximum population size that an environment can support
Exponential growth is unlimited growth.
We have used the exponential growth equationto represent population growth.
0kty y e
The exponential growth equation occurs when the rate of growth is proportional to the amount present.
If we use P to represent the population, the differential equation becomes: dP
kPdt
The constant k is called the relative growth rate.
/dP dtk
P
The population growth model becomes: 0ktP P e
However, real-life populations do not increase forever. There is some limiting factor such as food, living space or waste disposal.
There is a maximum population, or carrying capacity, M.
A more realistic model is the logistic growth model where
growth rate is proportional to both the amount present (P)
and the fraction of the carrying capacity that remains: M P
M
M
P1
The equation then becomes:
dP M PkP
dt M
Some books writes it this way:
Logistics Differential Equation
dP kP M P
dt M
We can solve this differential equation to find the logistics growth model.
PartialFractions
Logistics Differential Equation
dP kP M P
dt M
1 k
dP dtP M P M
1 A B
P M P P M P
1 A M P BP
1 AM AP BP
1 AM
1A
M
0 AP BP AP BP
A B1
BM
1 1 1 kdP dt
M P M P M
ln lnP M P kt C
lnP
kt CM P
Logistics Differential Equation
kt CPe
M P
kt CM Pe
P
1 kt CMe
P
1 kt CMe
P
1 kt C
MP
e
1 C kt
MP
e e
CLet A e
1 kt
MP
Ae
Logistics Growth Model
1 kt
MP
Ae
PMkPdt
dP
Logistics Differential Equation
dP kP M P
dt M
Logistics Differential Equation Logistics Growth Model
tMkAe
MP
1
Logistic Growth Model
Years
Bears
Example:
Logistic Growth Model
Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears.
Assuming a logistic growth model, when will the bear population reach 50? 75? 100?
Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears.
Assuming a logistic growth model, when will the bear population reach 50? 75? 100? (round answers to nearest year)
1 kt
MP
Ae 100M 0 10P 10 23P
1 kt
MP
Ae 100M 0 10P 10 23P
0
10010
1 Ae
10010
1 A
10 10 100A
10 90A
9A
At time zero, the population is 10.
100
1 9 ktP
e
1 kt
MP
Ae 100M 0 10P 10 23P
After 10 years, the population is 23.
100
1 9 ktP
e
10
10023
1 9 ke
10 1001 9
23ke
10 779
23ke
10 0.371981ke
10 0.988913k
0.098891k
0.1
100
1 9 tP
e
0.1
100
1 9 tP
e
Years
BearsWe can graph this equation and use “intersect” to find the solutions.
y=50 at 22 years
y=75 at 33 years
y=100 at 75 years
0.1
100
1 9 tP
e
Years
BearsWhat is the equation for the RATE of this logistic growth?
When does the rate of population growth start to decrease?
PMPM
k
dt
dP
When the second derivative is equal to zero AND changes
from positive to negative
PPdt
dP 100
100
1.0
PPdt
dP 100
100
1.0
dt
dPP
dt
Pd002.1.
2
2
P002.1.0
50PWhen does the rate of population growth start to decrease?
0.1
100
1 9 tP
e
50
Years
Bears
y=50 at 22 years
Y1
0.1
100
1 9 tP
e
Y2
Y3 50
2001.1.0 PPdt
dP
2001.1.0 YY Slopefield program
Positive on (0, 100)