Ch 6.5 Logistic Growth
18
Ch 6.5 Logistic Growth Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy
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Ch 6.5 Logistic Growth. Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy. Example of Partial Fraction Decomposition. Integrating with Partial Fractions. Integrating with Partial Fractions. Gorilla Population. - PowerPoint PPT Presentation
Transcript of Ch 6.5 Logistic Growth
Ch 6.5 Logistic GrowthCalculus Graphical, Numerical, Algebraic byFinney, Demana, Waits, Kennedy
Example of Partial Fraction Decomposition
2 2
2x 16 x 8
x x 6 x x 6x 8
x 3 x 2
A B
x 3 x 2
dx = 2 dx
= 2 dx
= 2 + dx A(x-2) + B(x+3) = x + 8
at x =
2
2
1 2
x 3 x 2
ln x 3 x 2
x 2
x 3
2, B = and at x = -3, A = -1
= 2 + dx
= 2 + 2ln + C
= 2 ln + C
Integrating with Partial Fractions
4
2
3x + 1Find dx
x - 1
Integrating with Partial Fractions
4
2
3x + 1Find dx
x - 12
2 4
4 2
2
2
3x + 3
x - 1 3x + 1
3x - 3x
3x + 1
3x - 3
4
4 2
2 2
3
3
3x + 1 4 dx = 3x + 3 + dx
x - 1 x - 1
A B = x + 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 + 3x + +
x-1 x+
3
dx1
= x + 3x + 2 ln | x-1| - 2 ln | x+1| + C
dP
dtdP
P(M P)
A B
P M P1 1
M P ,M M
1 1 1
M P M P1 1
P M Pln P M P \
ln M P P
= kP (M - P)
= k dt
+ dP = kt + C
A + BP = 1 at P= M, B = and at P = 0, A =
+ dP = kt + C
+ dP = Mkt + C
- ln = Mkt + C
- ln =
M P
P
M P
PM
PM
PP 1
MP
M 1
1
-Mkt + C
-Mkt
-M
-Mkt
kt
-Mkt
-Mkt + C
ln = -Mkt + C
= e
- 1 = C e
= C e + 1
= General Logistic Formula C e
= C e
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
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?
dP = 0.0004 P (250 - P)
dt
Gorilla Population
dP
dtdP
P(250 P)
A B.0004 250 P P
P 250 P
.004 .004
P
= 0.0004 P (250 - P)
= .0004 dt
+ dp = dt A + B = 1
at P = 0, A = .004; at P = 250, B = .004
+
.1t
.1t
.1t
2
.0004250 P
.004 ln P 250 P
ln P 250 P
250 P
P
250
P250
PP 1
2550
P1
0
dp = dt
- ln = .0004t +C
- ln = .1t + C
ln = -.1t - C
- 1 = C e
= 1 + C e at (0,28), C = 7.92857
= 1 + 7.92857
= + 7.
e
92857 e .1t
0
M7.92857
PNote:
a
= - 1
k = .1 = .0004nd * M
Gorilla Population
dP = 0.0004 P (250 - P)
dt - 0.1 t
250P(t)
1 + 7.9286e
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
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
