Transcript Ch 6.5 Logistic Growth

Ch 6.5 Logistic Growth
Calculus Graphical, Numerical, Algebraic by
Finney, Demana, Waits, Kennedy
Example of Partial Fraction
Decomposition
2x  16
x 8
dx
=
2
 x2  x  6
 x 2  x  6 dx
x 8
=2 
dx
 x  3 x  2 
=2
A
B
+
 x  3 x  2 dx
A(x-2) + B(x+3) = x + 8
at x = 2, B = 2 and at x = -3, A = -1
2
 1

= 2 
+
dx 
x2
 x 3

= 2   ln x  3 + 2ln x  2  + C
= 2 ln
 x  2
x 3
2
+C
Integrating with Partial Fractions
4
3x + 1
Find  2
dx
x -1
Integrating with Partial Fractions
3x 4 + 1
Find  2
dx
x -1
3x 2 + 3
x 2 - 1 3x 4
+1
3x 4 - 3x 2
3x 2 + 1
3x 2 - 3
4
3x 4 + 1
4 
 2
dx
=
3x
+
3
+
 dx
2
 x2 - 1
 
x - 1
B 
 A
= x 3 + 3x +  
+
 dx
 x-1 x+1 
where A(x+1) + B(x-1) = 4. Use x = 1, and x = -1 to get
-2 
 2
= x 3 + 3x + 
+
 dx
 x-1 x+1 
= x 3 + 3x + 2 ln | x-1| - 2 ln | x+1| + C
dP
= kP (M - P)
dt
dP
 P(M  P) =  k dt
A
B
+
 P M  P dP = kt + C
A  M  P  + BP = 1 at P= M, B =
1
1
, and at P = 0, A =
M
M
1 1
1
+
dP = kt + C
M  P MP
1
1
+
 P M  P dP = Mkt + C
ln P - ln M  P = Mkt + C \
ln M  P - ln P = -Mkt + C
MP
= -Mkt + C
P
MP
= e-Mkt + C
P
M
- 1 = C e-Mkt
P
M
= C e-Mkt + 1
P
P
1
=
M 1  C e-Mkt
M
P=
General Logistic Formula
1  C e-Mkt
ln
Gorilla Population
A certain wild animal preserve can support no more than
250 lowland gorillas. Twenty-eight gorillas were known
to be in the preserve in 1970. Assume that the rate of
growth of the population is
dP
= 0.0004 P (250 - P)
dt
Where time t is in years.
a) Find a formula for the gorilla population in terms of t.
b) How long will it take the gorilla population to reach the
carrying capacity of the preserve?
Gorilla Population
dP
= 0.0004 P (250 - P)
dt
dP
= .0004 dt
P(250  P)
A
B
 P + 250  P dp =  .0004 dt
A  250  P  + B  P  = 1
at P = 0, A = .004; at P = 250, B = .004
.004
.004
+
dp =  .0004 dt
P
250  P
.004  ln P - ln 250  P  = .0004t +C

ln P - ln 250  P = .1t + C
ln
250  P
= -.1t - C
P
250
- 1 = C e .1t
P
250
= 1 + C e .1t at (0,28), C = 7.92857
P
P
1
=
250
1 + 7.92857 e .1t
250
P=
1 + 7.92857 e .1t
M
Note: 7.92857 =
-1
P0
and
k = .1 = .0004 * M
Gorilla Population
A certain wild animal preserve can support no more than
250 lowland gorillas. Twenty-eight gorillas were known
to be in the preserve in 1970. Assume that the rate of
growth of the population is
250
dP
= 0.0004 P (250 - P) P(t) 
dt
1 + 7.9286e- 0.1 t
Where time t is in years.
a) Find a formula for the gorilla population in terms of t.
b) How long will it take the gorilla population to reach the
carrying capacity of the preserve?
83 years to reach 249.5  250