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## Drill

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**Drill**• Find the integral using the given substitution**Lesson 6.5: Logistic Growth**Day #1: P. 369: 1-17 (odd) Day #2: P. 369/70: 19-29; odd, 37**Partial Fraction Decomposition with Distinct Linear**Denominators • If f(x) = P(x)/Q(x) where P and Q are polynomials with the degree of P less than the degree of Q, and if Q(x) can be written as a product of distinct linear factors, then f(x) can be written as a sum of rational functions with distinct linear denominators.**Example: Finding a Partial Fraction Decomposition**• Write the function f(x)= (x-13)/(2x2 -7x+3) as a sum of rational functions with linear denominators.**Write the function f(x)= (2x+16)/(x2 +x-6) as a sum of**rational functions with linear denominators.**Example: Antidifferentiating with Partial Fractions**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**Finding Three Partial Fractions**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**Drill: Evaluate the Integral**When x = -7, B(-7) = -21, B = 3 When x = 0, 7A = 14, A = 2**The Logistic Differential Equation**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(M-P)**General Logistic Formula**• The solution of the general logistic differential equation is Where A is a constant determined by an appropriate initial condition. The carrying capacity M and the growth constant k are positive constants.**Example: Using Logistic Regression**• The population of Aurora, CO for selected years between 1950 and 2003**Use logistic regression to find a model.**• In order to this with a table, enter the data into the calculator (L1 = year, L2 = population) • Stat, Calc, B: Logistic • Based on the regression equation, what will the Aurora population approach in the long run? • 316,441 people (note that the carrying capacity is the numerator of the equation!)**When will the population first exceed 300,000?**Write a logistic equation in the form of dP/dt = kP(M-P) that models the data. We know that M = 316440.7 and Mk =.1026 Therefore 316440.7k=.1026, making k = 3.24 X 10-7 dP/dt= 3.24 X 10-7 P(316440.7-P) • Use the calculator to intersect. • When x = 59.12 or in the 59th year: 1950 + 59 = 2009**6.5, day #2**• Review the following notes, ON YOUR OWN • You can get extra credit if you complete the homework for day 2. (Will replace a missing homework, OR if you do have no missing assignments, it will push your homework grade over 100%) • Correct answers + work will give you extra credit on Friday’s quiz. • No work = No extra credit**Example**• The growth rate of a population P of bears in a newly established wildlife preserve is modeled by the differential equation dP/dt = .008P(100-P), where t is measured in years. • What is the carrying capacity for bears in this wildlife preserve? 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. • What is the rate of change of population when it is growing the fastest? dP/dt = .008P(100-P) dP/dt = .008(50)(100-50) dP/dt =20 bears per year**Example**• In 1985 and 1987, the Michigan Department of Natural Resources airlifted 61 moose from Algonquin Park, Ontario to Marquette County in the Upper Peninsula. It was originally hoped that the population P would reach carrying capacity in about 25 years with a growth rate of dP/dt = .0003P(1000-P) • What is the carrying capacity? 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. • Solve the differential equation with the initial condition P(0) = 61**Multiply both sides by 1000 to clear decimals.**Multiply both sides by -1 to put 1000-P 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