AMATYC

November 11th, 2011

“Dynamic Solution Exercises”“Dynamic Solution Exercises”

Dr. Elliott Ostler –

UNO Mathematics Education Professor

Michael Flesch –

Metropolitan Community College Math Instructor

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Welcome

• Introductions

• History of the Project

• Contents of your Packet• Contents of your Packet

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Website for Materials

• All contents of your packet can be found at: http://resource.mccneb.edu/math

• You will need to locate the class from the pull down menu on the left side. pull down menu on the left side.

• Math 0910 for Basic Math materials

• Math 0960 for Beginning Alg. materials

• Math 1310 for Intermediate Alg. materials

• Math 1420 for College Alg. materials

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Email Addresses

• Michael Flesch - mflesch@mccneb.edu

• Dr. Elliott Ostler – eliottostler@gmail.com

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Suggestions for Making

Vertical Alignment Between

Courses Tangible throughout

the Mathematics Curriculum

Using:

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Using:

Dynamic Solution Exercises

(DSE’s)

Overview of Session

• Our presentation justifies the need for, and

offers some suggestions on the selection and

implementation of mathematical problems

known as Dynamic Solution Exercises.

• Our intent is to help provide insight into • Our intent is to help provide insight into

how mathematics teachers can go about

making vertical articulation a cooperative and

tangible part of the mathematics curriculum.

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Overview Continued

• Some sample Dynamic Solution Exercises are provided based on research at Metropolitan Community College in Omaha NE. Omaha NE.

• Some strategies for selecting and building a DSE instructional environment will be discussed.

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Dr. Elliott Ostler

• Filmed at the University of NE at Omaha

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DSE Example - Mixture Problem

• You have been asked to mix two different strength weed killers together to obtain 10 liters of weed killer that has a 61% strength. On hand you have one barrel of strength. On hand you have one barrel of weed killer that is 40% strength and another and another barrel that is 70%. How many liters of each should be mixed to obtain 10 liters that is 61% weed killer?

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Basic Mathematics

• This problem could be done by having the students investigate the different possibilities that will occur.

• The students will perform the calculations • The students will perform the calculations fill in a table, and then make conclusions based on the results.

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Questions

• At what combination of Amount 1 and Amount 2 was the percent of weed killer 61%?

• At what combination of Amount 1 and • At what combination of Amount 1 and Amount 2 was the percent of weed killer the least? The most?

• Based on amounts used and the percent of weed killer in each, what generalizations might you make?

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Beginning Algebra

• We now use the variable (x) to represent Amount 1 and (10 – x) to represent Amount 2. This allows us to write an equation in terms of x.

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equation in terms of x.

• We can see how we might write an algebraic equation to solve with any desired percent in the final solution.

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2nd Part of Question

• 70%*x + 40%*(10-x) = 6.1 liters61%*10 Liters = 6.1 Liters

• You now want 10 liters of a mixture of the two weed killers with a final strength of two weed killers with a final strength of 49%. Without having to construct another table, write one equation in terms of x using the model from above that will allow you to find the correct combinations of the two liquids?

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Solution Is An Equation

• 70%*x + 40%*(10-x) = 4.9 liters49%*10 Liters = 4.9 Liters

• Solve this equation to get x = 3 and y = 7

• We can now found the correct combination • We can now found the correct combination for any percent mix we are looking for algebraically.

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Intermediate Algebra

• Introduces a second variable allowing us to set up a system of two equations and unknowns and find a solution by both an algebraic or geometric process.

• It allows us to write the two constraints as linear equations and find the solutions by graphing the two constraints and finding the intersection.

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Algebraic Solution

• We have two equations and two unknowns to describe the system:

• x + y = 10

• 0.70*x + 0.40*y = 0.61*10 = 6.1• 0.70*x + 0.40*y = 0.61*10 = 6.1

• Solve this system using the substitution or elimination method.

• Solution will again be x = 7 and y = 3

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Geometric Solution

• Now solve the system by graphing.

• 0.7x + 0.4y = 6.1 and x + y = 10

• y = -1.75x + 15.25 and y = -x + 10 can • y = -1.75x + 15.25 and y = -x + 10 can be graphed to find the solution.

• Plot these equations and find the intersection.

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Part 2

• You have been directed to put together 10 liters of a mixture of the two different weed killers that will have a final strength of 52%. Can you construct a system of two 52%. Can you construct a system of two equations and two unknowns using the model from above that will allow you to find the correct combinations of the two liquids? Solve the system algebraically and geometrically by graphing.

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Algebraic Solution

• The two equations are: x+y = 10 & 0.70x + 0.40y = 0.52*10 = 5.2

• The algebraic solution can be accomplished with the substitution ofaccomplished with the substitution ofy = -x+10 into the other equation to get:

0.70x + 0.40(-x + 10) = 5.2

• The solution is x = 4 and y = 10-x = 6

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Geometric Solution

• The geometric solution can be found by solving each equation for y and graphing.

• y = -x + 10 and 0.70x + 0.40y = 5.2• y = -x + 10 and 0.70x + 0.40y = 5.2

so y = -7/4*x + 13

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College Algebra

• Algebraic and geometric interpretation of of an inconsistent system . This example will have a solution, but it will fall outside the allowable physical constraints so thus there will be no solution.

• The algebraic and geometric process will yield a solution, but not one that is within the defined possibilities for x and y. The student needs to recognize and interpret their results to the algebraic and geometric solutions.

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Statement of the Problem

• You have been asked to mix two different strength weed killers together to obtain 10 liters of weed killer that has a 36% strength. On hand you have one barrel of strength. On hand you have one barrel of weed killer that is 80% strength and another and another barrel that is 40%. How many liters of each should be mixed to obtain 10 liters that is 36% weed killer?

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Algebraic Solution

• Through substitution, we have:

• 0.80*x + 0.40(10-x) = 3.6

• x= -1 and since x + y = 10, y = 11 so our ordered pair is (-1, 11)ordered pair is (-1, 11)

• Our values for x and y are outside the physical constraints of allowable values for x and y which must be between 0 and 10.

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Geometric Solution

• Solve for y and graph each line.

• y = -x + 10 and 0.80x + 0.40y = 3.6

• y = -x + 10 and y = -2x + 9• y = -x + 10 and y = -2x + 9

• We see how the intersection is outside the allowable physical constraints of x and y.

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Some DSE’s for Basic Math• Concept #1 – Division by Zero

• Concept #2 – Comparing additionverses multiplication

• Concept #3 - Location of Negative Signin Fractions

• Concept #4 – The Distributive Property

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Some DSE’s for Basic Math

• Concept #5 - Distributive Property –Factoring

• Concept #6 - Reducing Fractionalexpressions to lowestexpressions to lowestterms by factoring

• Concept #7 - Multiplication – Rewritingto use Distributive Prop.

• Concept #8 - Quantities Equal to -1

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Some DSE’s for

Elementary Algebra• Concept #1 – Comparing Addition

verses Multiplication

• Concept #2 - Location of Negative Signin a Fraction in a Fraction

• Concept #3 – Division by Zero

• Concept #4 – The Distributive Property

• Concept #5 - Distributive Property inReverse – Factoring

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Some DSE’s for

Elementary Algebra• Concept #6 - Distributing -1 to a

Quantity

• Concept #7 – Interpreting fractionalCoefficient - Variable TermCoefficient - Variable Term

• Concept #8 - Clearing Fractions in anEquation

• Concept #9 - Multiplication – Rewritingusing Distributive Prop.

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Some DSE’s for

Elementary Algebra• Concept #10 - Simplifying Rational

Expressions – FindRestricted Values

• Concept #11 – Simplifying Rational• Concept #11 – Simplifying RationalExpressions withNumerical Values

• Concept #12 - Simplifying RationalExpressions Equal to 1

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Some DSE’s for

Elementary Algebra• Concept #13 – Simplifying Rational

Expressions withVariable Terms

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Some DSE’s for

Intermediate Algebra• Concept #1 – The Distributive Property

• Concept #2 – Comparing Additionverses Multiplication

• Concept #3 - Distributing -1 to Quantity• Concept #3 - Distributing -1 to Quantity

• Concept #4 - Multiplication – Rewritingto use Distributive Prop,

• Concept #5 - Using Distributive Prop. InReverse – Factoring

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Some DSE’s for

Intermediate Algebra• Concept #6 – Division by Zero

• Concept #7 - Simplifying RationalExpressions & Functions,Identify Restricted ValuesIdentify Restricted Values

• Concept #8 – Simplifying RationalExpressions withNumerical Values

• Concept #9 - Simplifying RationalExpressions Equal to One

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Some DSE’s for

Intermediate Algebra• Concept #10 – Simplifying Rational

Expressions withVariable Terms

• Concept #11 - Location of Negative• Concept #11 - Location of NegativeSign in a Fraction

• Concept #12 - Clearing Fractions in anEquation

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Some DSE’s for

Intermediate Algebra• Concept #13 – Adding Rational

Expressions withVariable Terms

• Concept #14 – Solving Rational• Concept #14 – Solving RationalEquations

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DSE’s in Detail for

Basic Mathematics• We will take a look at some examples of

what was developed in your packet of materials concerning the DSE concepts for a Basic Mathematics Course.a Basic Mathematics Course.

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DSE’s in Detail for

Intermediate Algebra• We will take a look at some examples of

what was developed in your packet of materials concerning the DSE concepts for an Intermediate Algebra Course.an Intermediate Algebra Course.

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Key Ideas for a DSE:Key Ideas for a DSE:

• Use what best supports Primary NeedStandards

• Large in Scope and often evaluated by checklist

• Repeated through successive courses/years

• Focus on Relationships not Process

• Forces Synthesis of mathematical ideas

Discussion & Final Comments

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Good DSE’s:Good DSE’s:• Sundial: On a rectangular, horizontal sundial, mark

the Dial Plate.

• Evaporating Solution: We want to keep the amount ofan antibacterial solution in a swimming pool at a constant

level. Show this using multiple iterations of a linear function.

• Optimize the volume: Removing a wedge from a circlewould create a cone. What is the best size piece to

remove that would allow for the maximum cone value?

• Find the “Shortest” Path: An insect travels at 8ft./sec overdry ground and 5ft./sec over water. If he sees food 28 feetdown the shore and 18 feet from the shore. How fast can

it get the food?

Adding to Forms of OneAdding to Forms of One

or: 13.07. =++

How about:

3.7.0+

0.1

or: Does the sum of and equal ? 7. 3. 1or: Does the sum of and equal ? 7. 3. 1

or: xxx 17.3. =+

or: 1cossin22

=+ xx

or: 13

5

3

22

2

limlim =+

+

−

∞→∞→ xx

x

xx

Simple to complex:Simple to complex:

Linear Equations Arithmetic Sequences

Key ideas:

• y-axis intercept: b

• Slope: m

• Points on a line: (x,y)

Key ideas:

• First term of the sequence: a1

• Common Difference: d

• General term: an at the nth term

Formula:

y = mx + b

Formula:

an = a1 + (n-1)d

Statement of the problem

• You are working for a lawn service and have been asked to mix two different strength weed killers together to obtain 10 liters of weed killer that has a 36% liters of weed killer that has a 36% strength. On hand you have one barrel of weed killer that is 80% strength and another and another barrel that is 40%. How many liters of each should be mixed to obtain 10 liters that is 36% weed killer?

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A Key for ranking MathematicalA Key for ranking Mathematical

Tasks:Tasks:

Simple Complex

Task

Algorithmic Interpretive Constructive

Concrete Abstract Concrete Abstract

Single Multiple Single Multiple Single Multiple Single Multiple