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3.5 Graphing Linear Equations in 3 Variables PowerPoint PPT Presentation


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In a three dimensional system, how many “areas” do you have? Where are the axes in a three dimensional system? How do you write a linear equation in x, y, and z as a function of two variables?. 3.5 Graphing Linear Equations in 3 Variables. Graphing in Three Dimensions.

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3.5 Graphing Linear Equations in 3 Variables

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3 5 graphing linear equations in 3 variables l.jpg

In a three dimensional system, how many “areas” do you have?

Where are the axes in a three dimensional system?

How do you write a linear equation in x, y, and z as a function of two variables?

3.5 Graphing Linear Equations in 3 Variables


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Graphing in Three Dimensions

Solutions of equations in three variables can be pictured with a three-dimensional coordinate system.

To construct such a system, begin with the xy-coordinate plane in a horizontal position.

Then draw the z-axis as a vertical line through the origin.

In much the same way that points in a two dimensional coordinate system are represented by ordered pairs, each point in space can be represented by an ordered triple (x, y, z).

Drawing the point represented by an ordered triple is called plotting the point.

The three axes, taken two at a time, determine three coordinate planes that divide space into eight octants.

The first octant is one for which all three coordinates are positive.


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Plotting Points in Three Dimensions

Plot the ordered triple in a three-dimensional coordinate system.

(–5, 3, 4)

SOLUTION

To plot (–5, 3, 4), it helps to first find the point (–5, 3) in the xy-plane.

The point (–5, 3, 4), lies four units above.


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Plotting Points in Three Dimensions

Plot the ordered triple in a three-dimensional coordinate system.

(–5, 3, 4)

(3, – 4, –2)

SOLUTION

SOLUTION

To plot (–5, 3, 4), it helps to first find the point (–5, 3) in the xy-plane.

To plot (3, – 4, –2) it helps to first find the point (3, – 4) in the xy-plane.

The point (3, – 4, –2) lies two units below.

The point (–5, 3, 4), lies four units above.


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Graphing in Three Dimensions

A linear equation in three variablesx, y, z is an equation in the form

ax + by + cz = d

where a, b, andc are not all 0.

An ordered triple (x, y, z) is a solution of this equation if the equation is true when the values of x, y, and z are substituted into the equation.

The graph of an equation in three variables is the graph of all its solutions.

The graph of a linear equation in three variables is a plane.

A linear equation in x, y, and z can be written as a function of two variables.

To do this, solve the equation for z. Then replace z with f(x,y).


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Graphing a Linear Equation in Three Variables

Sketch the graph of 3x + 2y + 4z = 12.

SOLUTION

Begin by finding the points at which the graph intersects the axes.

Let x = 0, and y = 0, and solve for z to get

z = 3.

This tells you that the z-intercept is 3, so plot the point (0, 0, 3).

In a similar way, you can find the x-intercept is 4 and the y-intercept is 6.

After plotting (0, 0, 3), (4, 0, 0) and (0, 6, 0), you can connect these points with lines to form the triangular region of the plane that lies in the first octant.


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Evaluating Functions of Two Variables

3

4

3

4

1

4

=

(12 – 3x – 2y)

z =

1

4

1

4

f(1, 3) =

f(x, y) =

(12 – 3x – 2y)

(12 – 3(1) – 2(3))

(

)

1, 3, .

This tells you that the graph of f contains the point

Write the linear equation 3x + 2y + 4z = 12 as a function of x and y.

Evaluate the function when x = 1 and y = 3. Interpret the result geometrically.

SOLUTION

3x + 2y + 4z = 12

Write original function.

4z = 12 –3x –2y

Isolate z-term.

Solve for z.

Replace z with f(x, y).

Evaluate when x = 1 and y = 3.


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Using Functions of Two Variables in Real Life

Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35.

Write a model for the total amount you will spend as a function of the number of pounds of bluegrass and rye.

SOLUTION

Your total cost involves two variable costs (for the two types of seed) and one fixed cost (for the spreader).

Verbal Model

Total cost

Bluegrass cost

Bluegrass amount

Rye cost

Rye amount

Spreader cost

=

+

+


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Modeling a Real-Life Situation

Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35.

Write a model for the total amount you will spend as a function of the number of pounds of bluegrass and rye.

Total cost = C

(dollars)

Labels

Bluegrass cost = 2

(dollars per pound)

Bluegrass amount = x

(pounds)

Rye cost = 1.5

(dollars per pound)

Rye amount = y

(pounds)

Spreader cost = 35

(dollars)


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Modeling a Real-Life Situation

Total cost

Bluegrass cost

Bluegrass amount

Rye cost

Rye amount

Spreader cost

=

+

+

Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35.

Write a model for the total amount you will spend as a function of the number of pounds of bluegrass and rye.

Algebraic

Model

C = 2x + 1.5y + 35

Evaluate the model for several different amounts of bluegrass and rye, and organize your results in a table.

To evaluate the function of two variables, substitute values of x and y into the function.


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Modeling a Real-Life Situation

Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35.

Evaluate the model for several different amounts of bluegrass and rye, and organize your results in a table.

For instance, when x= 10 and y=20, then the total cost is:

C = 2 x + 1.5y+ 35

Write original function.

C = 2 (10) + 1.5(20)+ 35

Substitute for x and y.

= 85

Simplify.


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Modeling a Real-Life Situation

y Rye (lb)

0

10

20

30

40

10

$70

$85

$100

$115

xBluegrass (lb)

20

$90

$105

$120

$135

30

$110

$125

$140

$155

40

$130

$145

$160

$175

Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35.

Evaluate the model for several different amounts of bluegrass and rye, and organize your results in a table.

The table shows the total cost for several different values of x and y.


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In a three dimensional system, how many “areas” do you have?

There are eight octants.

Where are the axes in a three dimensional system?

X-axis goes towards you and from you, y-axis goes right and left and z-axis goes up and down.

How do you write a linear equation in x, y, and z as a function of two variables?

f(x,y)


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Page 173, 19-35 odd,

39-45 odd

Assignment 3.5


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