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# 10.8: Problem Solving Strategies: Choose a Function Model

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Read and understand given problem situations
• Develop and use the strategy: Choose a Function
• Develop and use the strategy: Make a Model
• Plan and compare alternative approaches to solving problems
• Solve real-world problems using selected strategies as part of a plan

## Introduction

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 which 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 follows.

Notice that the modeling process is very similar to the problem solving format we have been using throughout this book. In this section, we will focus mostly on the assumptions we make and the validity of the model. Functions are an integral part of the modeling process because they are used to describe the mathematical relationship in a system. 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 chose is not appropriate. Then, we must choose a different one, or we findthat a function model is good for one set of parameters but we need to use another function for a different set of parameters. Often, for certain parameters, more than one function describes the situation well and using the simplest function is most practical.

Here we present some mathematical models arising from real-world applications.

Example 1 Stretching springs beyond the “elastic limit”

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

Weight (oz)Length (in)0222.442.863.283.5103.9124.1144.4164.6184.7204.8\begin{align*}& \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\end{align*}

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

Step 1 Understand the problem

Define x =\begin{align*}x \ =\end{align*} weight in ounces on the spring

y =\begin{align*}y \ =\end{align*} length in inches of the spring

Step 2 Devise a plan

Springs usually have a linear relationship between the weight on the spring and the stretched length of the spring. If we make a scatter plot, we notice that for lighter weights the points do seem to fit on a straight line (see graph). Assume that the function relating the length of the spring to the weight is linear.

Step 3 Solve

Find the equation of the line using points describing lighter weights:

(0, 2) and (4, 2.8).

The slope is m=.84=0.2\begin{align*} m = \frac{.8} {4} = 0.2\end{align*}

Using y=mx+b\begin{align*}y = mx + b\end{align*}

a) We obtain the function y=.2x+2\begin{align*}y = .2x + 2\end{align*}.

b) To find the length of the spring when the weight is 5 ounces, we plug in x=5\begin{align*}x = 5\end{align*}.

y=.2(5)+2=3 inches\begin{align*}y = .2(5) + 2 = 3 \ inches\end{align*}

c) To find the length of the spring when the weight is 19 ounces, we plug in x=19\begin{align*}x = 19\end{align*}.

y=.2(19)+2=5.8 inches\begin{align*}y = .2(19) + 2 = 5.8 \ inches\end{align*}

Step 4 Check

To check the validity of the solutions lets plot the answers to b) and c) on the scatter plot. We see that the answer to b) is close to the rest of the data, but the answer to c) does not seem to follow the trend.

We can conclude that for small weights, the relationship between the length of the spring and the weight is a linear function.

For larger weights, the spring does not seem to stretch as much for each added ounces. We must change our assumption. There must be a non-linear relationship between the length and the weight.

Step 5 Solve with New Assumptions

Let’s find the equation of the function by cubic regression with a graphing calculator.

a) We obtain the function y=.000145x3.000221x2+.202x+2.002\begin{align*}y = -.000145x^3 - .000221x^2 +.202x + 2.002\end{align*}.

b) To find the length of the spring when the weight is 5 ounces, we plug in x=5\begin{align*}x = 5\end{align*}.

y=.000145(5)3.000221(5)2+.202(5)+2.002=3 inches\begin{align*}y = -.000145(5)^3 - .000221(5)^2 + .202(5) + 2.002 = 3 \ inches\end{align*}

c) To find the length of the spring when the weight is 19 ounces, we plug in x=19\begin{align*}x = 19\end{align*}.

y=.000145(19)3.000221(19)2+.202(19)+2.002=4.77 inches\begin{align*}y = -.000145(19)^3 - .000221(19)^2 +.202(19) + 2.002 = 4.77 \ inches\end{align*}

Step 6 Check

To check the validity of the solutions lets plot the answers to b) and c) on the scatter plot. We see that the answer to both b) and c) are close to the rest of the data.

We conclude that a cubic function represents the stretching of the spring more accurately than a linear function. However, for small weights the linear function is an equally good representation, and it is much easier to use in most cases. In fact, the linear approximation usually allows us to easily solve many problems that would be very difficult to solve by using the cubic function.

Example 2 Water flow

A thin cylinder is filled with water to a height of 50 centimeters. The cylinder has a hole at the bottom which is covered with a stopper. The stopper is released at time t=0\begin{align*}t = 0\end{align*} seconds and allowed to empty. The following data shows the height of the water in the cylinder at different times.

Time(sec)Height(cm)050242.5435.7629.5823.81018.81214.31410.5167.2184.6202.5221.1240.2\begin{align*}& \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\end{align*}

a) Find the height (in centimeters) of water in the cylinder as a function of time in seconds.

b) Find the height of the water when t=5\begin{align*}t = 5\end{align*} seconds.

c) Find the height of the water when t=13\begin{align*}t = 13\end{align*} seconds.

Solution:

Step 1 Understand the problem

Define x =\begin{align*}x \ =\end{align*} the time in seconds

y =\begin{align*}y \ =\end{align*} height of the water in centimeters

Step 2 Devise a plan

Let’s make a scatter plot of our data with the time on the horizontal axis and the height of water on the vertical axis.

Notice that most of the points seem to fit on a straight line when the water level is high. Assume that a function relating the height of the water to the time is linear.

Step 3 Solve

Find the equation of the line using points describing lighter weights:

(0, 50) and (4, 35.7).

The slope is m=14.34=3.58\begin{align*} m = \frac{-14.3} {4} = -3.58\end{align*}

Using y=mx+b\begin{align*}y = mx + b\end{align*}

a) We obtain the function: y=3.58x+50\begin{align*}y = -3.58x + 50\end{align*}

b) The height of the water when t=5\begin{align*}t = 5\end{align*} seconds is

y=3.58(5)+50=32.1 centimeters\begin{align*}y = -3.58 (5) + 50 = 32.1 \ centimeters\end{align*}

c) The height of the water when t=13\begin{align*}t = 13\end{align*} seconds is

y=3.58(13)+50=3.46 centimeters\begin{align*}y = -3.58(13) + 50 = 3.46 \ centimeters\end{align*}

Step 4 Check

To check the validity of the solutions, let's plot the answers to b) and c) on the scatter plot. We see that the answer to b) is close to the rest of the data, but the answer to c) does not seem to follow the trend.

We can conclude that when the water level is high, the relationship between the height of the water and the time is a linear function. When the water level is low, we must change our assumption. There must be a non-linear relationship between the height and the time.

Step 5 Solve with new assumptions

Let's assume the relationship is quadratic and let’s find the equation of the function by quadratic regression with a graphing calculator.

a) We obtain the function y=.075x23.87x+50\begin{align*}y = .075x^2 - 3.87x + 50\end{align*}

b) The height of the water when t=5\begin{align*}t = 5 \end{align*} seconds is

y=.075(5)23.87(5)+50=32.53 centimeters\begin{align*}y = .075(5)^2 - 3.87 (5) + 50 = 32.53 \ centimeters\end{align*}

c) The height of the water when t=13\begin{align*}t = 13\end{align*} seconds is

y=.075(13)23.87(13)+50=12.37 centimeters\begin{align*}y = .075(13)^2 - 3.87(13) + 50 = 12.37 \ centimeters\end{align*}

Step 6: Check

To check the validity of the solutions lets plot the answers to b) and c) on the scatterplot. We see that the answer to both b) and c) are close to the rest of the data.

We conclude that a quadratic function represents the situation more accurately than a linear function. However, for high water levels the linear function is an equally good representation.

Example 3 Projectile motion

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 degrees with the horizontal with a speed of 40 meters/sec.

Time (sec)Height (meters)000.517.21.031.51.542.92.051.62.557.73.061.23.562.34.061.04.557.25.051.05.542.66.031.96.519.07.04.1\begin{align*}& \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\end{align*}

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

b) Find the height of the ball when t=2.4\begin{align*}t = 2.4\end{align*} seconds.

c) Find the height of the ball when t=6.2\begin{align*}t = 6.2\end{align*} seconds.

Solution

Step 1 Understand the problem

Define x =\begin{align*}x \ =\end{align*} the time in seconds

y =\begin{align*}y \ =\end{align*} height of the ball in meters

Step 2 Devise a plan

Let’s make a scatter plot of our data with the time on the horizontal axis and the height of the ball on the vertical axis. We know that a projectile follows a parabolic path, so we assume that the function relating height to time is quadratic.

Step 3 Solve

Let’s find the equation of the function by quadratic regression with a graphing calculator.

a) We obtain the function y=4.92x2+34.7x+1.2\begin{align*}y = -4.92x^2 + 34.7x + 1.2\end{align*}

b) The height of the ball when t=2.4\begin{align*}t = 2.4 \end{align*} seconds is:

y=4.92(2.4)2+34.7(2.4)+1.2=56.1 meters\begin{align*}y = -4.92(2.4)^2 + 34.7(2.4) + 1.2 = 56.1 \ meters\end{align*}

c) The height of the ball when t=6.2\begin{align*}t = 6.2\end{align*} seconds is:

y=4.92(6.2)2+34.7(6.2)+1.2=27.2 meters\begin{align*}y = -4.92(6.2)^2 + 34.7(6.2) + 1.2 = 27.2 \ meters\end{align*}

Step 4 Check

To check the validity of the solutions lets plot the answers to b) and c) on the scatterplot. We see that the answer to both b) and c) follow the trend very closely. The quadratic function is a very good model for this problem

Example 4 Population growth

A scientist counts two thousand 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.

GenerationNumber (thousands)0241587512343161139201864241990281999\begin{align*}& \text{Generation} & & 0 & & 4 & & 8 & & 12 & & 16 & & 20 & & 24 & & 28\\ & \text{Number (thousands)} & & 2 & & 15 & & 75 & & 343 & & 1139 & & 1864 & & 1990 & & 1999\end{align*}

a) Find the number of fish as a function of generation.

b) Find the number of fish in generation 10.

c) Find the number of fish in generation 25.

Solution:

Step 1 Understand the problem

Define x =\begin{align*}x \ =\end{align*} the generation number y =\begin{align*}y \ =\end{align*} the number of fish in the lake

Step 2 Devise a plan

Let’s make a scatterplot of our data with the generation number on the horizontal axis and the number of fish in the lake on the vertical axis. We know that a population can increase exponentially. So, we assume that we can use an exponential function to describe the relationship between the generation number and the number of fish.

Step 3 Solve

a) Since the population increases at a rate of 1.5 per generation, assume the function y=2(1.5)x\begin{align*}y = 2(1.5)x\end{align*}

b) The number of fish in generation 10 is: y=2(1.5)10=115\begin{align*}y = 2(1.5)^{10} = 115\end{align*} thousand fish

c) The number of fish in generation 25 is: y=2(1.5)25=50502\begin{align*}y = 2(1.5)^{25} = 50502\end{align*} thousand fish

Step 4 Check

To check the validity of the solutions, let's plot the answers to b) and c) on the scatter plot. We see that the answer to b) fits the data well but the answer to c) does not seem to follow the trend very closely. The result is not even on our graph!

When the population of fish is high, the fish compete for space and resources so they do not increase as fast. We must change our assumptions.

Step 5 Solve with new assumptions

When we try different regressions with the graphing calculator, we find that logistic regression fits the data the best.

a) We obtain the function y2023.61+1706.3(2.71)484x\begin{align*} y - \frac{2023.6} {1 + 1706.3(2.71)^{-484x}}\end{align*}

b) The number of fish in generation 10 is \begin{align*} y = \frac{2023.6} {1 + 1706.3(2.71)^{-.484(10)}} = 139.6\end{align*} thousand fish

c) The number of fish in generation 25 is \begin{align*} y = \frac{2023.6} {1 + 1706.3(2.71)^{-.484(25)}} = 2005\end{align*} thousand fish

Step 6 Check

To check the validity of the solutions, let's plot the answers to b) and c) on the scatter plot. We see that the answer to both b) and c) are close to the rest of the data.

We conclude that a logistic function represents the situation more accurately than an exponential function. However, for small populations the exponential function is an equally good representation, and it is much easier to use in most cases.

## Review Questions

1. In Example 1, evaluate the length of the spring for weight \begin{align*}= 3 \ ounces\end{align*} by
1. Using the linear function
2. Using the cubic function
3. Figuring out which function is best to use in this situation.
2. In Example 1, evaluate the length of the spring for weight \begin{align*}= 15 \ ounces\end{align*} by
1. Using the linear function
2. Using the cubic function
3. Figuring out which function is best to use in this situation.
3. In Example 2, evaluate the height of the water in the cylinder when \begin{align*}t = 4.2 \ seconds\end{align*} by
1. Using the linear function
2. Using the quadratic function
3. Figuring out which function is best to use in this situation.
4. In Example 2, evaluate the height of the water in the cylinder when \begin{align*}t = 19 \ seconds\end{align*} by
1. Using the linear function
2. Using the quadratic function
3. Figuring out which function is best to use in this situation.
5. In Example 3, evaluate the height of the ball when \begin{align*}t = 5.2 \ seconds\end{align*}. Find when the ball is at its highest point.
6. In Example 4, evaluate the number of fish in generation 8 by
1. Using the exponential function
2. Using the logistic function
3. Figuring out which function is best to use in this situation.
7. In Example 4, evaluate the number of fish in generation 18 by
1. Using the exponential function
2. Using the logistic function
3. Figuring out which function is best to use in this situation.

## Review Answers

1. 2.6 inches
2. 2.6 inches
3. Both functions give the same result. The linear function is best because it is easier to use.
1. 5 inches
2. 4.5 inches
3. The two functions give different answers. The cubic function is better because it gives a more accurate answer.
1. 34.96 cm
2. 35.07 cm
3. The results from both functions are almost the same. The linear function is best because it is easier to use.
1. -18.02 cm
2. 3.5 cm
3. The two function give different results. The quadratic function is better because it gives a more accurate answer.
1. 48.6 meters
2. 3.7 seconds
1. 51,000
2. 55,000
3. The results from both functions are almost the same. The linear function is best because it is easier to use.
1. 2,956,000
2. 1,571,000
3. the two functions give different results; the logistic function is better because it gives a more accurate answer.

## Texas Instruments Resources

In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9620.

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