Drill. Find the integral using the given substitution. Lesson 6.5: Logistic Growth. Day #1: P. 369: 117 (odd) Day #2: P. 369/70: 1929; odd, 37. Partial Fraction Decomposition with Distinct Linear Denominators.
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Lesson 6.5: Logistic Growth
Day #1: P. 369: 117 (odd)
Day #2: P. 369/70: 1929; odd, 37
Because the numerator’s degree is larger than the denominator’s degree, we will need to use polynomial division.
3x2
+3

3x4 3x2

3x2 +1
3x2 3
4
When x = 1
A(0) + B(2) = 4; B =  2
When x = 1
A(2) + B(0) = 4; A= 2
When x = 2
B(4)(3)=36
12B = 36
B = 3
When x = 1
C(1)(3)=6
3C = 6
C = 2
When x = 2
A(4)(1)=4
4A=4
A=1
When x = 7, B(7) = 21, B = 3
When x = 0, 7A = 14, A = 2
Remember that exponential
growth can be modeled by
dP/dt = kP, for some k >0
If we want the growth rate
to approach 0 as P approaches
a maximal carrying capacity M,
we can introduce a limiting
factor of M – P.
Logistic Differential Equation
dP/dt = kP(MP)
is
Where A is a constant determined by an appropriate initial condition. The carrying capacity M and the growth constant k are positive constants.
When will the population first exceed 300,000?
Write a logistic equation in the form of dP/dt = kP(MP) that models the data.
We know that M = 316440.7 and Mk =.1026
Therefore 316440.7k=.1026, making k = 3.24 X 107
dP/dt= 3.24 X 107 P(316440.7P)
Remember that M is the carrying capacity, and according to the equation, M = 100 bears
The growth rate is always
maximized when the
population reaches half
the carrying capacity.
In this case, it is 50 bears.
dP/dt = .008P(100P)
dP/dt = .008(50)(10050)
dP/dt =20 bears per year
1000 moose
Generate a slope field. The window should be x: [0, 25] and y: [0, 1000]
We first need to separate the variables.
Then we will integrate one side by using partial fractions.
Multiply both sides by 1000 to clear decimals.
Multiply both sides by 1 to put 1000P on the top
Remember that ln(x) =y
Can be rewritten as ey = x
Solve for P