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Applications of Function Models

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Applications of Function Models

Suppose a company has recorded the number of employees it has had at the beginning of each month that it has been in business. It wants to model this data with a function, but it is not sure which type of function to use. Could the company get a sense of what type of function to use by plotting the data points? If so, what would the data points have to look like for a linear model to be used? How about for an exponential model or a quadratic model? In this Concept, you'll learn to choose a function model for data sets like this one.

Guidance

As you learn more and more mathematical methods and skills, it is important to think about the purpose of mathematics and how it works as part of a bigger picture. Mathematics is used to solve problems that often arise from real-life situations. Mathematical modeling is a process by which we start with a real-life situation and arrive at a quantitative solution.

Modeling involves creating a set of mathematical equations that describes a situation, solving those equations, and using them to understand the real-life problem.

Often the model needs to be adjusted because it does not describe the situation as well as we wish.

A mathematical model can be used to gain understanding of a real-life situation by learning how the system works, which variables are important in the system, and how they are related to each other. Models can also be used to predict and forecast what a system will do in the future or for different values of a parameter. Lastly, a model can be used to estimate quantities that are difficult to evaluate exactly.

Mathematical models are like other types of models. The goal is not to produce an exact copy of the “real” object but rather to give a representation of some aspect of the real thing. The modeling process can be summarized as a flow chart:

Notice that the modeling process is very similar to the problem-solving format we have been using throughout this book. One of the most difficult parts of the modeling process is determining which function best describes a situation. We often find that the function we choose is not appropriate. Then we must choose a different one.

Consider an experiment regarding the elasticity of a spring.

Example A

A spring is stretched as you attach more weight at its bottom. The following table shows the length of the spring in inches for different weights in ounces.

& \text{Weight (oz)} && 0 && 2 && 4 && 6 && 8 && 10 && 12 && 14 && 16 && 18 && 20\\& \text{Length (in)} && 2 && 2.4 && 2.8 && 3.2 && 3.5 && 3.9 && 4.1 && 4.4 && 4.6 && 4.7 && 4.8

a) Find the length of the spring as a function of the weight attached to it.

b) Find the length of the spring when you attach 5 ounces.

c) Find the length of the spring when you attach 19 ounces.

Solution:

Begin by graphing the data to get a visual of what the model may look like.

  • This data clearly does not fit a linear equation. It has a distinct curve that will not be modeled well by a straight line.
  • Nor does this graph seem to fit a parabolic shape; thus, it is not modeled by a quadratic equation.
  • The curve does not fit the exponential curves studied in previous Concepts.
  • By taking the third set of differences, the value is approximately equal. Use the methods learned in the previous Concept to find a cubic regression equation. Check by graphing to see if this model is a good fit.

Example B

A golf ball is hit down a straight fairway. The following table shows the height of the ball with respect to time. The ball is hit at an angle of 70^\circ with the horizontal with a speed of 40 meters/sec .

& \text{Time (sec)} && 0 && 0.5 && 1.0 && 1.5 && 2.0 && 2.5 && 3.0 && 3.5 && 4.0 && 4.5 && 5.0 && 5.5 && 6.0 && 6.5 && 7.0\\& \text{Height (meters)} && 0 && 17.2 && 31.5 && 42.9 && 51.6 && 57.7 && 61.2 && 62.3 && 61.0 && 57.2 && 51.0 && 42.6 && 31.9 && 19.0 && 4.1

a) Find the height of the ball as a function of time.

b) Find the height of the ball when t=2.4 \ \text{seconds} .

c) Find the height of the ball when t=6.2 \ \text{seconds} .

Solution:

Begin by graphing the data to visualize the model.

This data fits a parabolic curve quite well. We can therefore conclude the best model for this data is a quadratic equation.

To solve part a), use a graphing calculator to determine the quadratic regression line.

y=-4.92 x^2+34.7 x+1.2

b) The height of the ball when t=2.4 \ seconds is:

y=-4.92(2.4)^2+34.7(2.4)+1.2=56.1 \ meters

c) The height of the ball when t=6.2 \ seconds is

y=-4.92(6.2)^2+34.7(6.2)+1.2=27.2 \ meters

Example C

Find a model to fit the following table of values:

& \text{Dependent Variable} && -2 && -1 && 0 && 1 && 2 && 3\\&\text{Independent Variable} && 3.31 && 3.64 && 4 && 4.4 && 4.84 && 5.324

Solution:

Start by graphing the points, to get a sense of the shape. This can help you rule out certain models, and lead you in the direction of which models might be a good fit.

The graph almost looks straight, but has a slight curve, so it can't be linear. It could be quadratic or cubic, but let's check if it's exponential:

& \text{Dependent Variable} && -2 && -1 && 0 && 1 && 2 && 3\\& \text{Independent Variable} && 3.31 && 3.64 && 4 && 4.4 && 4.84 && 5.324\\& \text{Ratio of Values} &&  && \frac{3.64}{3.31}=1.1 &&  \frac{4}{3.64}=1.1 &&  \frac{4.4}{4}=1.1 &&  \frac{4.84}{4.4}=1.1 && \frac{5.324}{4.84}=1.1

Since the ratios are all the same and equal to 1.1, it is an exponential function with a growth factor of 1.1. Given the point (0,4), the initial value is 4. The function is:

f(x)=4\cdot 1.1^x.

Video Review

Guided Practice

The profits in dollars of a company are given in the table below. Find the model that describes the relationship to profit as a function of time in years:

& \text{Time in Years} && 0 && 1 && 2 && 3 && 4 \\&\text{Profit in Dollars} && 0 && 5000 && 20000 && 45000 && 80000

Solution:

Start by graphing the points, to get a sense of the shape.

This curve looks like it could be quadratic or exponential. If you check the ratios, you would see they are not the same. Checking the differences of differences yields:

& \text{Time in Years} && 0 && 1 && 2 && 3 && 4 \\&\text{Profit in Dollars} && 0 && 5000 && 20000 && 45000 && 80000\\&\text{Differences} && && 5000-0=5000 && 20000-5000=15000 && 45000-20000=25000 && 80000-45000=35000\\&\text{Differences of Differences} && && 15000-5000=10000 && 25000-15000=10000 && 35000-25000=10000

Since the differences of differences are the same, this is a quadratic model.

Practice

Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.

CK-12 Basic Algebra: Identifying Quadratic Models (8:05)

CK-12 Basic Algebra: Identifying Exponential Models (4:00)

CK-12 Basic Algebra: Quadratic Regression (9:17)

  1. A thin cylinder is filled with water to a height of 50 centimeters. The cylinder has a hole at the bottom that is covered with a stopper. The stopper is released at time t=0 \ \text{seconds} and allowed to empty. The following data shows the height of the water in the cylinder at different times. & \text{Time (sec)} && 0 && 2 && 4 && 6 && 8 && 10 && 12 && 14 && 16 && 18 && 20 && 22 && 24\\& \text{Height (cm)} && 50 && 42.5 && 35.7 && 29.5 && 23.8 && 18.8 && 14.3 && 10.5 && 7.2 && 4.6 && 2.5 && 1.1 && 0.2
    1. What seems to be the best model for this situation?
    2. Find the linear regression line and determine the height of the water at 4.2 seconds.
    3. Find a quadratic equation and determine the height of the water at 4.2 seconds.
    4. Find a cubic regression line and determine the height of the water at 4.2 seconds.
    5. Which of these seems to be the best fit?
    6. Using the function of best fit, find the height of the water when t=5 \ \text{seconds} .
    7. Using the function of best fit, find the height of the water when t=13 \ \text{seconds} .
  1. A scientist counts 2,000 fish in a lake. The fish population increases at a rate of 1.5 fish per generation but the lake has space and food for only 2,000,000 fish. The following table gives the number of fish (in thousands) in each generation. & \text{Generation} && 0 && 4 && 8 && 12 && 16 && 20 && 24 && 28\\& \text{Number (thousands)} && 2 && 15 && 75 && 343 && 1139 && 1864 && 1990 && 1999
    1. Which function seems to best fit the data: linear, quadratic, or exponential?
    2. Find the model for the function of best fit.
    3. Find the number of fish as a function of generation.
    4. Find the number of fish in generation 10.
    5. Find the number of fish in generation 25.
  1. Using the golf ball example, find the maximum height the ball reaches.
  2. Using the golf ball example, evaluate the height of the ball at 5.2 seconds.

Mixed Review

  1. Evaluate 2 \div 6 \cdot 5+3^2-11 \cdot 9^{\frac{1}{2}} .
  2. 60 shirts cost $812.00 to screen print. 115 shirts cost $1,126.00 to screen print. Assuming the relationship between the number of shirts and the total cost is linear, write an equation in point-slope form.
    1. What is the start-up cost (the cost to set up the screen)?
    2. What is the slope? What does it represent?
  1. Solve by graphing: y=x^2+3x-1 .
  2. Simplify \frac{\frac{6}{7}}{\frac{1}{2}} .
  3. Newton’s Second Law states F=m \cdot a . Rewrite this equation to solve for m . Use it to determine the mass if the force is 300 Newtons and the acceleration is 70 m/sec.
  4. The area of a square game board is 256 square inches. What is the length of one side?
  5. Write as a percent: \frac{3}{1000} .

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