College Algebra

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Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Solve applied problems involving exponential growth and decay.
- Solve applied problems involving compound interest.


Population Growth

The function `P(t) = P_0e^(k*t)`, where `P_0` is the population at time zero, P is the population at time t, and k is the exponential growth rate, is an effective model for population growth. From the graph, you can see that the function P(t) increases exponentially as t approaches infinity, and the y-intercept is at the point `(0, P_0)`.



New York - Newark - Jersey City metropolitan area is the first largest metropolitan area in the United States. In 2010, the population of this area was 19.6 million, and the exponential growth rate was 3.14% per year.

The exponential growth function is `P(t) = P_0e^(kt)`, where `k > 0`. We're given that `P_0 = 19.6" million"`. We're also given the growth rate, which is 3.14%. We can convert this to decimal form, it'll be `0.0314`. `t` represents the number of years after 2010, so the exponential growth function is `P(t) = 19.6e^((0.0314)t)`.

Now in order to estimate the population of this area in 2020, we have to plug `t = 2020-2010`, which is equal to 10 years. So we have to do `P(10) = 19.6e^((0.0314)(10))`. And this is equal to 26.8 million.

Now for the doubling time, called big `T`. This means that `P(t)` is going to equal `2*P_0`, or `2*19.6`. So if we plug all of this into our function, we have `2*19.6 = 19.6e^((0.0314)T)`. So if we divide both sides by 19.6, that cancels, and then we're left with `2 = e^((0.0314)T)`, the doubling time. Now, let's take the natural log of both sides. `ln2 = ln(e^((0.0314)T))`. Remember that ln of e to a power is just equal to the power. So we have `ln2 = 0.0314*T`. And then we have `T = (ln2)/(0.0314)`. This is equal to about 22 years. So the doubling time is 22 years.



Interest Compounded Continuously.

When interest is paid on interest, it's called compound interest. The formula is `P(t) = P_0e^(k*t)`. It's the same formula as population growth. `P_0` is the amount that's invested, `P(t)` is the amount in the account after t years, and k is the interest rate.



Suppose that $5,000 is invested at an interest rate of 4.5% per year, compounded continuously.

So first, let's find the exponential function that describes the amount in the account after time, t, in years. We have `P(t) = P_0e^(kt)`. Since it says that $5,000 is invested, this is the amount in the account at time `t = 0`. So this is our `P_0`. And then they give us an interest rate of 4.5%. So k is going to be `.045`. So we have `P(t) = 5000e^((0.045)t)`.

Now let's find what the balance is after 10 years. So that means that `t=10`. `P(10) = 5000e^((0.045)(10))`. This is equal to 7841.56. So the balance after 10 years is $7,841.56.

Now let's find doubling time, big `T`. That means `P(t) = 2*P_0`, or `2*5000`. So if we plug this into our function, we have `2*5000 = 5000e^((0.045)T)`. We can divide both sides by 5000, and then take ln of both sides. So we get `ln2 = lne^((0.045)T)`. Now remember ln of e to an exponent is just equal to the exponent, so we have `ln2 = (0.045)T`, or `T = (ln2)/(0.045)`. This is equal to about 15 years.



Growth rate and doubling time.

The growth rate, k, and doubling time, big `T`, are related by `kT = ln(2)`. Notice that the relationship between the growth rate and the doubling time does not depend on the initial population.



Models of Limited Growth.

Some populations cannot experience unlimited growth, because of factors such as limited food, living space, or other natural resources.

One model of such growth is called the Logistic Function. `P(t) = a/(1+b*e^(-k*t))`. This function increases toward a limiting value as t increases toward infinity.

`y = a` is the horizontal asymptote of our Logistic Function.



Widespread disease overtakes a town which has a population of 5300. The number of people, big `N`, infected t days after the disease has begun is given by the function, `N(t) = 5300/(1+19.9e^(-0.6t))`. How many are infected after 5 days; how many after 20 days?

So in order to figure out how many are infected after 5 days, we need to find what N(5) is. It's equal to `5300/(1+19.9e^(-0.6*5)`. It's equal to 2,662.3. So about 2,662 people are infected after 5 days.

Now let's see how many after 20 days. `N(20) = 5300/(1+19.9e^(-0.6*20))`. This is equal to 5,299.35. So about 5,299 people are infected after 20 days.



Exponential Decay.

The function `P(t) = P_0e^(-kt)`, is an effective model for the decline of a population, such as the decay of a radioactive substance. `P_0` is the population at time `t = 0`. P(t) is equal to the amount of substance left after time, t, and k is the decay rate.

Notice that the graph of the function for exponential decay approaches zero as time approaches infinity, and the y-intercept occurs at the point `(0, P_0)`.



The half-life of a radioactive substance is the time it takes for half of the amount of substance to decay.

The decay rate, k, and half-life, `T`, are related by `k*T = ln(2)`. Notice that this is the same relationship between decay rate and half-life, as that between growth rate and doubling time.



In 2015, Ezeiza, Argentina, a farmer discovered a fossilized Glyptodon. Scientists determined that the fossil lost 69.8% of its Carbon-14. How old was the fossilized Glyptodon at the time it was discovered?

We're given the function for exponential decay is `P(t) = P_0e^(-0.00012t)`. Since the fossil lost 69.8% of its Carbon-14, that means that the amount present after time t, is equal to `30.2%*P_0`. So using this, we can solve for t. We have `0.302P_0 = P_0e^((-0.00012)(t))`. So if we divide both sides by `P_0`, those cancel, and then if we take the natural log of both sides, we have `ln(0.302) = ln(e^(-0.00012*t))`. So we're left with `ln(0.302) = -0.00012*t`, or `t = (ln(0.302))/(-0.00012)`, which is about 9,977.74. So the fossil was about 9,978 years old at the time of discovery.