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CHAPTER 7 FirstOrder Differential Equations 7 7.1 MODELING WITH DIFFERENTIAL EQUATIONS 7.2 SEPARABLE DIFFERENTIAL EQUATIONS 7.3 DIRECTION FIELDS AND EULER’S METHOD 74 SYSTEMS OF FIRST ORDER DIFFERENTIAL 7.4 SYSTEMS OF FIRSTORDER DIFFERENTIAL EQUATIONS Slide 1 © The McGrawHill Companies, Inc. Permission required for reproduction or display.

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Microsoft PowerPoint - SmithMinton_Calculus_ET_4e_C07_S01.pptx7FirstOrder Differential Equations 7q
7.1 MODELING WITH DIFFERENTIAL EQUATIONS 7.2 SEPARABLE DIFFERENTIAL EQUATIONS 7.3 DIRECTION FIELDS AND EULER’S METHOD 7 4 SYSTEMS OF FIRST ORDER DIFFERENTIAL7.4 SYSTEMS OF FIRSTORDER DIFFERENTIAL
EQUATIONS
Slide 1© The McGrawHill Companies, Inc. Permission required for reproduction or display.
7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Exponential Growth The table indicates the  number of E. coli bacteria (in  millions of bacteria per ml) in  a laboratory culture measured  at halfhour intervals during  the course of an experiment.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Exponential Growth The plot appears to indicate that  the bacterial culture is growing  exponentially. Careful analysis of  experimental data has shown  that the rate at which the  bacterial culture grows is directly  proportional to the current  population.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Exponential Growth Let y(t) represent the number of bacteria in a culture at  time t, then the rate of change of the population with respect to time is y’(t). Thus, since y’(t) is proportional to  y(t), we have the following differential equation
where k is the growth constant. We wish to solve this  equation for y(t). We begin by rewriting and integrating.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Exponential Growth We may combine c1 and c2 as a single constant we lablel c.
Since y(t) represent a population, y(t) > 0.
Solving for y(t) and defining A = ec we obtain the solution.
We call this the general solution of the differentialWe call this the general solution of the differential  equation. For k > 0, it is called an exponential growth law  and for k < 0 it is an exponential decay law
Slide 5© The McGrawHill Companies, Inc. Permission required for reproduction or display.
and for k < 0, it is an exponential decay law.
7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
EQUATIONS 1.1 Exponential Growth in a Bacterial Colony
A freshly inoculated bacterial culture of Streptococcus A contains 100 cells. When the culture is checked 60  minutes later, it is determined that there are 450 cells  present.
Assuming exponential growth, determine the number of  cells present at any time t (measured in minutes) and find  the doubling time.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
EQUATIONS 1.1 Exponential Growth in a Bacterial Colony
Solution Exponential growth means                                                         where A and k are constants to be determined.
We were given the initial condition .  b i i hi i h l l i iSubstituting this into the general solution gives us
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
Solution We can use the second observation to find k.
We now have a formula representing the number of cells  present at any time t:present at any time t:
Slide 8© The McGrawHill Companies, Inc. Permission required for reproduction or display.
7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
EQUATIONS 1.1 Exponential Growth in a Bacterial Colony
Solution To find the doubling time, we wish to find t such that
We substitute this into y(t) and solve for t.
We see that the population  doubles about every 28
Slide 9© The McGrawHill Companies, Inc. Permission required for reproduction or display.
minutes.
7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Exponential Decay Experiments have shown that the rate at which a  radioactive element decays is directly proportional to the  amount present. Let y(t) represent the amount (mass) of a  radioactive element present at time t, then the rate of  decay satisfies
We know the general solution to this differential equation.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Exponential Decay It is common to discuss the decay rate of a radioactive  element in terms of its halflife, the time required for half  of the initial quantity to decay into other elements.
For instance, scientists have calculated that the halflife of  carbon14 (14C) is approximately 5730 years. That is, if you
14have 2 grams of 14C today and you come back in 5730  years, you will have approximately 1 gram of 14C remaining.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
If you have 50 grams of 14C today, how much will be left  in 100 years?
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
Solution Let y(t) be the mass (in grams) of 14C present at time t.  Then, we have The initial condition y(0) = 50 gives
Using the halflife, we can find the decay constant k.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
Solution The graph shows the incredibly  slow decay of 14C.
If we start with 50 grams, then  the amount left after 100 years is
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Newton’s Law of Cooling The rate at which the object cools (or warms) is not  proportional to its temperature, but rather, to the  difference in temperature between the object and its surroundings.
Symbolically, if we let y(t) be the  temperature of the object  at time t and let Ta be the temperature of the surroundings  (the ambient temperature, which we assume to be  constant), we have the differential equation
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Newton’s Law of Cooling W i h t l thi ti f (t) W b i bWe wish to solve this equation for y(t). We begin by  rewriting and integrating.
To evaluate the integral, we will use the substitution                u = y(t) − T and we assume y(t) − T > 0u = y(t)   Ta and we assume y(t)   Ta > 0.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Newton’s Law of Cooling W h hi h bWe now have                                                  which can be  written as                                       .
Solving for y(t) we obtain the general solution
where we define A = ecwhere we define A = e .
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
EQUATIONS 1.3 Newton’s Law of Cooling for a Cup of
C ffCoffee A cup of fastfood coffee is 180oF when freshly poured.  f 2 i i 0 h ff h l dAfter 2 minutes in a room at 70oF, the coffee has cooled
to 165oF. Find the temperature at any time t and find the  i hi h h ff h l d 120 Ftime at which the coffee has cooled to 120oF.
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
EQUATIONS 1.3 Newton’s Law of Cooling for a Cup of
C ff
Solution
Coffee
Let y(t) be the temperature of the coffee at time t, we  have
Using the initial condition y(0) = 180, gives us
Using the second measured temperature, we have
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONS
EXAMPLE
EQUATIONS 1.3 Newton’s Law of Cooling for a Cup of
C ff
We now solve this equation for k.
U i thi l f k lUsing this value for k, we can now solve
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7.1 MODELING WITH DIFFERENTIAL  EQUATIONSEQUATIONS
Compound Interest If a bank agrees to pay you 8% (annual) interest on an  investment of $10,000, then at the end of a year, you will have If a bank agrees to pay you interest twice a year at 8% annually then you will receive (8/2)% interest twice perannually, then you will receive (8/2)% interest twice per year, and at the end of a year you will have Slide 21© The McGrawHill Companies, Inc. Permission required for reproduction or display. 7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS Compound Interest Compounding interest monthly would pay (8/12)% each month resulting in a balance of Compounding interest daily would result in a balance of Slide 22© The McGrawHill Companies, Inc. Permission required for reproduction or display. 7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS Compound Interest Is there is a limit to how much interest can accrue on a given investment at a given interest rate? If n is the number of times per year that interest is compounded, we wish to calculate the annual percentage yield (APY) under continuous compounding, Slide 23© The McGrawHill Companies, Inc. Permission required for reproduction or display. 7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS Compound Interest Recall that If we make the change of variables n = 0.08m, we have Continuous compounding would earn approximately 8.3%. Slide 24© The McGrawHill Companies, Inc. Permission required for reproduction or display. Continuous compounding would earn approximately 8.3%. 7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS Compound Interest$Suppose that you invest $P at an annual interest rate r, compounded n times per year. Then the value of your investment after t years is Under continuous compounding (i.e., limit as n→ ∞) this bbecomes Slide 25© The McGrawHill Companies, Inc. Permission required for reproduction or display. 7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS Compound Interest If y(t) is the value of your investment after t years, with continuous compounding, the rate of change of y(t) is proportional to y(t). F i i i l i f$P hFor an initial investment of $P, we have Slide 26© The McGrawHill Companies, Inc. Permission required for reproduction or display. 7.1 MODELING WITH DIFFERENTIAL EQUATIONS EXAMPLE EQUATIONS 1.5 Depreciation of Assets (a) Suppose that the value of a$10,000 asset decreases  i l f 2 % i dcontinuously at a constant rate of 24% per year. Find
its worth after 10 years; after 20 years.
(b) Compare these values to a $10,000 asset that is d i d l i 20 i lidepreciated to no value in 20 years using linear depreciation. Slide 27© The McGrawHill Companies, Inc. Permission required for reproduction or display. 7.1 MODELING WITH DIFFERENTIAL EQUATIONS EXAMPLE EQUATIONS 1.5 Depreciation of Assets Solution The value v(t) satisfies v’ = rv, where r = −0.24. Thus, we havehave Using the initial value, we haveg t =10: t =20: Slide 28© The McGrawHill Companies, Inc. Permission required for reproduction or display. t =20: EXAMPLE Solution For linear depreciation we use the following information. Using this information we find t =10: v(10) =$5000 t 20 (20) $0 Slide 29© The McGrawHill Companies, Inc. Permission required for reproduction or display. t =20: v(20) =$0 ##### 7.1. Introduction: Markov Property 7.2. Examplessuper7/19011-20001/19541.pdf · 7. Markov Chains (Discrete-Time Markov Chains) 7.1. Introduction: Markov Property 7.2. Examples - Two
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