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Antiderivatives with Slope Fields
Consider: or then: It doesn’t matter whether the constant was 3 or -5, since when we take the derivative the constant disappears. However, when we try to reverse the operation: We don’t know what the constant is, so we put “C” in the answer to remind us that there might have been a constant. Given: find
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6.1 Antiderivatives with Slope Fields
Given: and when , find the equation for . This is called an initial value problem. We need the initial values to find the constant. An equation containing a derivative is called a differential equation. It becomes an initial value problem when you are given the initial condition and asked to find the original equation.
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6.1 Antiderivatives with Slope Fields
Definition Slope Field or Directional Field A slope field or a directional field for the first order differentiable equation is a plot of sort line segments with slopes f(x,y) for a lattice of points (x,y) in the plane. Initial value problems and differential equations can be illustrated with a slope field.
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6.1 Antiderivatives with Slope Fields
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6.1 Antiderivatives with Slope Fields
If you know an initial condition, such as (1,-2), you can sketch the curve. By following the slope field, you get a rough picture of what the curve looks like. In this case, it is a parabola.
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6.1 Antiderivatives with Slope Fields
Integrals such as are called definite integrals because we can find a definite value for the answer. The constant always cancels when finding a definite integral, so we leave it out!
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6.1 Antiderivatives with Slope Fields
Integrals such as are called indefinite integrals because we can not find a definite value for the answer. When finding indefinite integrals, we always include the “plus C”.
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6.1 Antiderivatives with Slope Fields
Definition Indefinite Integral The set of all antiderivatives of a function f(x) is the indefinite integral of f with respect to x and is denoted by
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6.1 Antiderivatives with Slope Fields
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6.1 Antiderivatives with Slope Fields
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6.1 Antiderivatives with Slope Fields
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6.1 Antiderivatives with Slope Fields
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6.1 Antiderivatives with Slope Fields
Find the position function for v(t) = t3 - 2t2 + t s(0) = 1 C = 1
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6.1 Antiderivatives with Slope Fields
log cabin houseboat log cabin + C
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6.2 Integration by Substitution
The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.
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6.2 Integration by Substitution
The variable of integration must match the variable in the expression. Don’t forget to substitute the value for u back into the problem!
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6.2 Integration by Substitution
One of the clues that we look for is if we can find a function and its derivative in the integrand. The derivative of is Note that this only worked because of the 2x in the original. Many integrals can not be done by substitution.
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6.2 Integration by Substitution
Solve for dx.
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6.2 Integration by Substitution
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6.2 Integration by Substitution
We solve for because we can find it in the integrand.
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6.2 Integration by Substitution
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6.2 Integration by Substitution
The technique is a little different for definite integrals. new limit We can find new limits, and then we don’t have to substitute back. new limit We could have substituted back and used the original limits.
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6.2 Integration by Substitution
Using the original limits: Leave the limits out until you substitute back. Wrong! The limits don’t match! This is usually more work than finding new limits
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6.2 Integration by Substitution
Don’t forget to use the new limits.
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6.2 Integration by Substitution
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6.2 Integration by Substitution
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6.2 Integration by Substitution
In another generation or so, we might be able to use the calculator to find all integrals. Until then, remember that half the AP exam and half the nation’s college professors do not allow calculators. You must practice finding integrals by hand until you are good at it!
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6.3 Integrating by Parts Start with the product rule:
This is the Integration by Parts formula.
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6.3 Integrating by Parts u differentiates to zero (usually).
dv is easy to integrate. The Integration by Parts formula is a “product rule” for integration. Choose u in this order: LIPET Logs, Inverse trig, Polynomial, Exponential, Trig
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6.3 Integrating by Parts LIPET polynomial factor
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6.3 Integrating by Parts LIPET logarithmic factor
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6.3 Integrating by Parts LIPET
This is still a product, so we need to use integration by parts again.
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6.3 Integrating by Parts A Shortcut: Tabular Integration
Tabular integration works for integrals of the form: where: Differentiates to zero in several steps. Integrates repeatedly. Also called tic-tac-toe method
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6.3 Integrating by Parts Compare this with the same problem done the other way:
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6.3 Integrating by Parts
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6.3 Integrating by Parts LIPET This is the expression we started with!
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6.3 Integrating by Parts
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6.3 Integrating by Parts The goal of integrating by parts is to go from an integral that we don’t see how to integrate to an integral that we can evaluate.
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6.4 Exponential Growth and Decay
The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present (at least for awhile.) So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account. If the rate of change is proportional to the amount present, the change can be modeled by:
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6.4 Exponential Growth and Decay
Rate of change is proportional to the amount present. Divide both sides by y. Integrate both sides.
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6.4 Exponential Growth and Decay
Exponentiate both sides. When multiplying like bases, add exponents. So added exponents can be written as multiplication.
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6.4 Exponential Growth and Decay
Since is a constant, let At , This is the solution to our original initial value problem.
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6.4 Exponential Growth and Decay
Exponential Change: If the constant k is positive then the equation represents growth. If k is negative then the equation represents decay.
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6.4 Exponential Growth and Decay
Continuously Compounded Interest If money is invested in a fixed-interest account where the interest is added to the account k times per year, the amount present after t years is: If the money is added back more frequently, you will make a little more money. The best you can do is if the interest is added continuously.
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6.4 Exponential Growth and Decay
Of course, the bank does not employ some clerk to continuously calculate your interest with an adding machine. We could calculate: Since the interest is proportional to the amount present, the equation becomes: You may also use: which is the same thing. Continuously Compounded Interest:
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6.4 Exponential Growth and Decay
Radioactive Decay The equation for the amount of a radioactive element left after time t is: The half-life is the time required for half the material to decay.
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6.4 Exponential Growth and Decay
Radioactive Decay 60 mg of radon, half-life of 1690 years. How much is left after 100 years?
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6.4 Exponential Growth and Decay
100 bacteria are present initially and double every 12 minutes. How long before there are 1,000,000
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6.4 Exponential Growth and Decay
Half-life Half-life:
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6.4 Exponential Growth and Decay
Espresso left in a cup will cool to the temperature of the surrounding air. The rate of cooling is proportional to the difference in temperature between the liquid and the air. If we solve the differential equation: Newton’s Law of Cooling where is the temperature of the surrounding medium, which is a constant. we get:
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6.5 Population Growth Bears Years
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6.5 Population Growth We have used the exponential growth equation
to represent population growth. The exponential growth equation occurs when the rate of growth is proportional to the amount present. If we use P to represent the population, the differential equation becomes: The constant k is called the relative growth rate.
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6.5 Population Growth The population growth model becomes:
However, real-life populations do not increase forever. There is some limiting factor such as food, living space or waste disposal. There is a maximum population, or carrying capacity, M.
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6.5 Population Growth A more realistic model is the logistic growth model where growth rate is proportional to both the amount present (P) and the fraction of the carrying capacity that remains:
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6.5 Population Growth The equation then becomes:
Logistics Differential Equation Our book writes it this way: We can solve this differential equation to find the logistics growth model.
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6.5 Population Growth Logistics Differential Equation Partial
Fractions
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6.5 Population Growth Logistics Differential Equation
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6.5 Population Growth Logistics Differential Equation
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6.5 Population Growth Logistics Growth Model
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6.5 Population Growth Logistic Growth Model
Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100?
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6.5 Population Growth Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100?
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6.5 Population Growth At time zero, the population is 10.
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6.5 Population Growth After 10 years, the population is 23.
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6.5 Population Growth Years Bears We can graph this equation and use “trace” to find the solutions. y=50 at 22 years y=75 at 33 years y=100 at 75 years
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6.6 Euler’s Method Leonhard Euler made a huge number of contributions to mathematics, almost half after he was totally blind. (When this portrait was made he had already lost most of the sight in his right eye.) Leonhard Euler
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6.6 Euler’s Method It was Euler who originated the following notations: (function notation) (base of natural log) (pi) (summation) (finite change) Leonhard Euler
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6.6 Euler’s Method There are many differential equations that can not be solved. We can still find an approximate solution. We will practice with an easy one that can be solved. Initial value:
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6.6 Euler’s Method Recall the formula for local linearization where
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6.6 Euler’s Method (0,1) (.5,1) (1,1.5)
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6.6 Euler’s Method (1.5,2.5) (2,4)
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6.6 Euler’s Method Exact Solution:
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6.6 Euler’s Method This is called Euler’s Method.
It is more accurate if a smaller value is used for dx. It gets less accurate as you move away from the initial value.
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6.6 Euler’s Method The book refers to an “Improved Euler’s Method”. We will not be using it, and you do not need to know it. The calculator also contains a similar but more complicated (and more accurate) formula called the Runge-Kutta method. You don’t need to know anything about it other than the fact that it is used more often in real life. This is the RK solution method on your calculator.
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