Transcript Find Slope and Rate of Change

Find Slope and Rate of
Change
Chapter 2.2
How Fast is He Walking?
How Fast is He Walking?
• The man moves a distance of 3 feet for every second that passes
• What this shows is a relation between the distance the man has walked
(in feet) and the time that has passed (in seconds)
• We could express this relation (which is also a function) as
𝑑 = 3𝑡
• In the above function, d is the distance the man has traveled and t is
the time in seconds
• Note that this has the same pattern as 𝑦 = 𝑚𝑥 + 𝑏, where m is three
and b is zero
• Now compare this to a slightly different picture
How Fast is He Walking?
Rate of Change
• In this second animation, the man’s speed is itself changing after every
second passes
• The first animation is an example of a constant rate of change
• The second animation is an example of a variable rate of change
(which means that the rate itself changes)
• The graph of two related quantities (like distance and time when
something is moving) will be a line if the rate of change is constant
• This rate of change is commonly called the slope
Rate of Change
• By comparison, when the rate of change itself changes, the graph
showing the relation between two quantities is a curve
• In the previous animation, the equation showing the distance traveled
from the start at a given time is 𝑑 = 𝑡 2
• Note that this does not have the form 𝑦 = 𝑚𝑥 + 𝑏; its form is 𝑦 = 𝑥 2
• You will study functions where the independent variable (x) is squared
in a later chapter
Slope of a Line
We define the slope of a line as follows:
Suppose that a line passes through two points represented by
(𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ). Then the slope of the line passing through these
points is
𝑦2 − 𝑦1 𝑦1 − 𝑦2
𝑚=
=
𝑥2 − 𝑥1 𝑥1 − 𝑥2
Slope of a Line
Slope of a Line
• We sometimes say that the slope of a line is the measure of its rise
divided by the measure of its run
• That is
𝑦2 − 𝑦1 rise
=
𝑥2 − 𝑥1 run
• If you use this way of thinking about slope, remember that rise can be
either positive or negative, and the same goes for run
Example
What is the slope of the line passing through the points (−2,1) and
(3,5)?
• Let 𝑥1 = −2, 𝑦1 = 1, 𝑥2 = 3, 𝑦2 = 5
• Use the formula
𝑦2 − 𝑦1
5−1
4
𝑚=
=
=
𝑥2 − 𝑥1 3 − −2
5
4
.
5
• The slope is This means that, for every five spaces to the right, the
line rises by 4 spaces
Guided Practice
Find the slope of the line passing through the given points.
a) (0,3) and (4,8)
b) (−5,1) and (5, −4)
c) (−3, −2) and (6,1)
d) (7,3) and (−1,7)
Guided Practice
a) Let 𝑥1 = 0, 𝑦1 = 3, 𝑥2 = 4, 𝑦2 = 8
8−3 5
𝑚=
=
4−1 4
c) Let 𝑥1 = −3, 𝑦1 = −2, 𝑥2 = 6, 𝑦2 =
1
1 − −2
3 1
𝑚=
= =
6 − −3
9 3
b) Let 𝑥1 = −5, 𝑦1 = 1, 𝑥2 = 5, 𝑦2 =
−4
−4 − 1
5
1
𝑚=
=−
=−
5 − −5
10
2
d) Let 𝑥1 = 7, 𝑦1 = 3, 𝑥2 = −1, 𝑦2 = 7
7−3
4
1
𝑚=
=
=−
−1 − 7
−8
2
Classifying Lines by Slope
• You should be able to tell by looking at a line whether its slope is
positive, negative, zero, or undefined
• Vertical lines have undefined slopes because, using the slope formula,
the denominator yields zero and division by zero is not defined
• Horizontal lines have zero slopes because, using the slope formula, the
numerator yields zero and every non-zero number divided by zero is
zero
• For the last two cases, imagine that you are walking on a line from left
to right in the coordinate plane
Negative Slope
Negative Slope
Positive Slope
Positive Slope
Classification of Line by Slope
• A vertical line has an undefined slope
• A horizontal line has a slope of zero
• A line that falls from left to right has a negative slope
• A line that rises from left to right has a positive slope
Example
Without graphing, tell whether the line through the given points rises,
falls, is horizontal, or is vertical.
a)
b)
c)
d)
1−1
0
(−5,1) and (3,1): 𝑚 =
=
=0
−5−3
−8
−4−0
4
(−6,0) and (2, −4): 𝑚 =
=−
2− −6
8
8−3
5
(−1,3) and (5,8): 𝑚 =
=
5− −1
6
−1−6
7
(4,6) and (4, −1): 𝑚 =
=−
4−4
0
=
1
−
2
Guided Practice
Without graphing, tell whether the line through the given points rises,
falls, is horizontal, or is vertical.
a) (−4,3) and (2, −6)
b) (7,1) and (7, −1)
c) (3, −2) and (5, −2)
d) (5,6) and (1, −4)
Guided Practice
Without graphing, tell whether the line through the given points rises,
falls, is horizontal, or is vertical.
a) (−4,3) and (2, −6): the line falls from left to right
b) (7,1) and (7, −1): the line is vertical
c) (3, −2) and (5, −2): the line is horizontal
d) (5,6) and (1, −4): the line rises from left to right
Parallel & Perpendicular Lines
• Recall from geometry that
• Two lines are parallel if they never intersect
• Two lines are perpendicular if they intersect at right angles (90˚)
• It is possible to show how the slopes of lines that are parallel or lines
that are perpendicular are related, but this is lengthy so we will just
remember the relationship
Parallel & Perpendicular Lines
• Two non-vertical lines are parallel if and only if their slopes are equal
• The phrase “if and only if” means two things:
• Two lines are parallel if their slopes are equal
• Two lines with equal slopes are parallel
• Two non-vertical lines are perpendicular if and only if their slopes are
opposite reciprocals
• We can also say, if two lines have slopes 𝑚1 and 𝑚2 , then they are
perpendicular if and only if 𝑚1 𝑚2 = −1
• The phrase “if and only if” means two things:
• Two lines are perpendicular if their slopes are opposite reciprocals
• Two line with slopes that are opposite reciprocals are perpendicular
Example
Tell whether the lines are parallel, perpendicular, or neither.
a) Line 1: through −2,2 and 0, −1 ; Line 2: through −4, −1 and
2,3
b) Line 1: through 1,2 and 4, −3 ; Line 2: through −4,3 and
−1, −2
Example
Tell whether the lines are parallel, perpendicular, or neither.
a) Line 1: through −2,2 and 0, −1 ; Line 2: through −4, −1 and
2,3
Slope of Line 1: 𝑚1 =
Slope of Line 2: 𝑚2 =
−1−2
0− −2
3− −1
2− −4
=
=
3
−
2
4
2
=
6
3
The slopes are opposite in sign and reciprocals, so the lines are
perpendicular
We can also multiply the slopes:
3 2
− ⋅
2 3
=
6
−
6
= −1
Example
Tell whether the lines are parallel, perpendicular, or neither.
b) Line 1: through 1,2 and 4, −3 ; Line 2: through −4,3 and
−1, −2
Slope of Line 1:
Slope of Line 2:
−3−2
5
=−
4−1
3
−2−3
5
=−
−1−(−4)
3
The slopes are equal, so the lines are parallel
Guided Practice
Tell whether the lines are parallel, perpendicular, or neither.
a) Line 1: through (−2,8) and (2, −4); Line 2: through (−5,1) and
(−2,2)
b) Line 1: through (−4, −2) and (1,7); Line 2: through (−1, −4) and
(3,5)
Guided Practice
Tell whether the lines are parallel, perpendicular, or neither.
a) Line 1: through (−2,8) and (2, −4); Line 2: through (−5,1) and
(−2,2)
Slope of Line 1:
Slope of Line 2:
−4−8
12
=−
2− −2
4
2−1
1
=
−2− −5
3
= −3
Since the slopes are opposite in sign and reciprocals, the lines are
1
3
perpendicular. Also, −3 ⋅ = − = −1
3
3
Guided Practice
Tell whether the lines are parallel, perpendicular, or neither.
b) Line 1: through (−4, −2) and (1,7); Line 2: through (−1, −4) and
(3,5)
Slope of Line 1:
Slope of Line 2:
7−
1−
5−
3−
−2
−4
−4
−1
=
=
9
5
9
4
The slopes are neither equal nor opposite reciprocals, so the lines are
neither parallel nor perpendicular
Rate of Change
• At the beginning of this presentation, we considered slope as the ratio
of the change in the distance a man walked compared to the time that
passed
• The slope of a line is always the ratio of the change in one quantity
compared to the change in another
• That is, slope is an average rate of change
Example
A Giant Sequoia tree had a diameter of 137 inches in 1965 and a
diameter of 141 inches in 2005. What is the average rate of change in
the tree’s diameter over this period of time?
Using units will help us determine how to proceed. The average rate of change in
diameter (in inches) over time (in years) has units of inches per year. As a fraction,
inches
this would be written as
. We can think of the values given in the problem as
year
two points in the coordinate plan, (1965, 137) and (2005, 141). The slope (or
average rate of change) is
141 − 137
4
1
=
=
2005 − 1965 40 10
So the average rate of change (or growth) is 1/10 inch per year.
Guided Practice
A Giant Sequoia tree has a diameter of 248 inches in 1965 and a
diameter of 251 inches in 2005. Find the average rate of change in the
diameter. Include units in your answer.
Guided Practice
A Giant Sequoia tree has a diameter of 248 inches in 1965 and a
diameter of 251 inches in 2005. Find the average rate of change in the
diameter. Include units in your answer.
We take the related values as ordered pairs: (1965, 248) and (2005, 251).
Now calculate the average rate of change (slope):
251 − 248
3
=
= 0.075
2005 − 1965 40
The average change in diameter over this period of time is 0.075 inches per year.
Exercise 2.2
• Page 86, #3-39 odds