Download gcua11e_ppt_1_1

Chapter 1
Functions
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115
Chapter Outline

The Slope of a Straight Line

The Slope of a Curve at a Point

The Derivative

Limits and the Derivative

Differentiability and Continuity

Some Rules for Differentiation

More About Derivatives

The Derivative as a Rate of Change
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 115
§ 1.1
The Slope of a Straight Line
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 115
Section Outline

Nonvertical Lines

Positive and Negative Slopes of Lines

Interpretation of a Graph

Properties of the Slope of a Nonvertical Line

Finding the Slope and y-Intercept of a Line

Sketching the Graph of a Line

Making Equations of Lines

Slope as a Rate of Change
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 115
Nonvertical Lines
Definition
Equations of Nonvertical
Lines: A nonvertical line L
has an equation of the form
y  mx  b.
The number m is called the
slope of L and the point (0, b)
is called the y-intercept. The
equation above is called the
slope-intercept equation of L.
Example
y  3x  4
For this line, m = 3 and
b = -4.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 115
Lines – Positive Slope
EXAMPLE
The following are graphs of equations of lines that have positive slopes.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 115
Lines – Negative Slope
EXAMPLE
The following are graphs of equations of lines that have negative slopes.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 115
Interpretation of a Graph
EXAMPLE
A salesperson’s weekly pay depends on the volume of sales. If she sells x units of goods, then
her pay is y = 5x + 60 dollars. Give an interpretation of the slope and the y-intercept of this
straight line.
SOLUTION
First, let’s graph the line to help us understand the exercise.
460
(80, 460)
pay
360
260
160
60
-20-40
30
80
130
# of sales
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 115
Interpretation of a Graph
CONTINUED
The slope is 5, or 5/1. Since the numerator of this fraction represents the amount of change in
her pay relative to the amount of change in her sales, the denominator, for every 1 sale that she
makes, her pay increases by 5 dollars.
The y-intercept is 60 and occurs on the graph at the point (0, 60). This point suggests that when
she has executed 0 sales, her pay is 60 dollars. This $60 could be referred to as her base pay.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 115
Properties of the Slope of a Nonvertical Line
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 115
Properties of the Slope of a Line
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 115
Finding Slope and y-intercept of a Line
EXAMPLE
Find the slope and y-intercept of the line y 
x 1
.
3
SOLUTION
First, we write the equation in slope-intercept form.
y
x 1
3
This is the given equation.
y
x 1

3 3
Divide both terms of the numerator of
the right side by 3.
y
1
1
x
3
3
Rewrite
x
1
as x .
3
3
Since the number being multiplied by x is 1/3, 1/3 is the slope of the line. Since the other 1/3 is
the number being added to the term containing x, 1/3, or (0, 1/3), is the y-intercept.
Incidentally, it was a complete coincidence that the slope and y-intercept were the same number.
This does not normally occur.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 115
Sketching Graphs of Lines
EXAMPLE
Sketch the graph of the line passing through (-1, 1) with slope ½.
SOLUTION
We use Slope Property 1. We begin at the given point (-1, 1) and from there, move up one unit and
to the right two units to find another point on the line.
(-1, 1)
-5
-3
5
5
3
3
(-1, 1)
1
-1 -1
1
3
5
-5
-3
1
-1 -1
-3
-3
-5
-5
1
3
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 115
5
Sketching Graphs of Lines
CONTINUED
Now we connect the two points that have already been determined, since two points determine a
straight line.
5
3
1
-5
-3
-1 -1
1
3
5
-3
-5
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 115
Making Equations of Lines
EXAMPLE
Find an equation of the line that passes through the points (-1/2, 0) and (1, 2).
SOLUTION
To find an equation of the line that passes through those two points, we need a point (we already
have two) and a slope. We do not yet have the slope so we must find it. Using the two points we
will determine the slope by using Slope Property 2.
m
20
2
2 4

 2 
3
3 3
1  1
2
2
 
We now use Slope Property 3 to find an equation of the line. To use this property we need the
slope and a point. We can use either of the two points that were initially provided. We’ll use the
second (the first would work just as well).
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 15 of 115
Making Equations of Lines
CONTINUED
y  y1  mx  x1 
This is the equation from Property 3.
y  2  4 3 x 1
(x1, y1) = (1, 2) and m = 4/3.
y 2  4 3x 4 3
Distribute.
y  4 3x  2 3
Add 2 to both sides of the equation.
NOTE: Technically, we could have stopped when the equation looked like y  2  4 3 x 1 since
it is an equation that represents the line and is equivalent to our final equation.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 16 of 115
Making Equations of Lines
EXAMPLE
Find an equation of the line that passes through the point (2, 0) and is perpendicular to the line
y = 2x.
SOLUTION
To find an equation of the line, we need a point (we already have one) and a slope. We do not yet
have the slope so we must find it. We know that the line we desire is perpendicular to the line
y
= 2x. Using Slope Property 5, we know that the product of the slope of the line desired and the
slope of the line y = 2x is -1. We recognize that the line y = 2x is in slope-intercept form and
therefore the slope of the line is 2. We can now find the slope of the line that we desire. Let the
slope of the new line be m.
(slope of a line)(slope of a new line) = -1
2m = -1
m = -0.5
This is Property 5.
The slope of one line is 2 and the
slope of the desired line is denoted by
m.
Divide.
Now we can find the equation of the desired line using Property 3.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 17 of 115
Making Equations of Lines
CONTINUED
y  y1  mx  x1 
y  0  0.5x  2
y  0 . 5 x  1
This is the equation from Property 3.
(x1, y1) = (2, 0) and m = -0.5.
Distribute.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 18 of 115
Slope as a Rate of Change
EXAMPLE
Compute the rate of change of the function over the given intervals.
y  2x  27
0, 1, 0, 0.5
SOLUTION
We first get y by itself in the given equation.
y  2 x  27
y  2 x  27
This is the given equation.
Subtract 2x from both sides.
Since this is clearly a linear function (since it’s now in slope-intercept form) it has constant slope,
namely -2. Therefore, by definition, it also has a constant rate of change, -2. Therefore, no matter
what interval is considered for this function, the rate of change will be -2. Therefore the answer,
for both intervals, is -2.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 19 of 115