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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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 = 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
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.
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
What is the bear population when the population is growing the fastest?
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
Solve the differential equation with the initial condition P(0) = 61
Remember that ln(x) =y
Can be rewritten as ey = x