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RATE of CHANGE
SLOPE 12
12
xx
yym
orx
ym
m1
m2
m1 = m2
Slope of a line is constant
Slope is known as the change in ‘y’ with respect to ‘x’
m is - m is +
m is 0
Slope of a curve varies
Slope of the tangent line = slope of the curve at the point of contact
One application of slope is velocity
velocity is the change in position with respect to time.
Two aspects of velocity
I. Average velocity
Time (h) Distance (km)
0 0
1 105
2 197
3 315
4 402
5 500
What is the average velocity between the 2nd and 4th hours?
(2, 197) ; (4, 402)
24
197402
velocity
velocity is 102.5 km/h
One application of slope is velocity
velocity is the change in position with respect to time.
Two aspects of velocity
I. Average velocity
Time (h) Distance (km)
0 0
1 105
2 197
3 315
4 402
5 500
What is the average velocity for the first 3 hours of the trip?
(0, 0) ; (3, 315)
velocity is 105 km/h
03
0315
velocity
Distance (km)
Time (minutes)
x
y
Tangent of the curve
Instantaneous rate of change at the given moment in time.
Kim and Duncan took a trip and recorded information about time and distance travelled.
0 0
10 15
20 30
30 45
40 60
50 80
60 100
70 105
80 110
90 135
100 150
110 155
Time (minutes)
Number of kilometres traveled
Distance (km)
Time (minutes)
0 10 20 30 40 50 60 70 80 90 100 110 120 130 10
20
30
40
50
60
70
80
90
100
110
120
130
150
140
160
x
y
Kim and Duncan took a trip and recorded information about time and distance travelled.
0 0
10 15
20 30
30 45
40 60
50 80
60 100
70 105
80 110
90 135
100 150
110 155
Time (minutes)
Number of kilometres traveled
x
y
Distance (km)
Time (minutes)
0 10 20 30 40 50 60 70 80 90 100 110 120 130 10
20
30
40
50
60
70
80
90
100
110
120
130
150
140
160
x
y
0 10 20 30 40 50 60 70 80 90 100 110 120 minutes
10
150
140
130
120
110
100
30
20
50
40
70
60
90
80
160km
t
dr
Find their average speed (velocity) over the first 50 min of the trip. Express your answer in km/hour
Starting values?
Ending values?
(0 min, 0 km)
(50 min, 80 km)
12
12
tt
ddr
050
080
r
1.6km/min x 60 min = 96km/h
min/6.1min50
80km
kmr
t
dr
12
12
tt
ddr
y2 y1
x2 x2
distance
time m =
slope
Rate of change is really a calculation of the slope of a line between any two given points.
x
y
0 10 20 30 40 50 60 70 80 90 100 110 120 minutes
10
150
140
130
120
110
100
30
20
50
40
70
60
90
80
160km
t
dr
Find their average velocity between 40 and 80 min into the trip. Express your answer in km/hour
Starting values?
Ending values?
(40 min, 60 km)
(80 min, 110 km)
12
12
tt
ddr
1km/min x 60 min = 75km/h
min/25.1min40
50km
kmr
4080
60110
r
x
y
0 10 20 30 40 50 60 70 80 90 100 110 120 minutes
10
150
140
130
120
110
100
30
20
50
40
70
60
90
80
160km
t
dr
Find their average velocity between 30 and 90 min into the trip. Express your answer in km/hour
Starting values?
Ending values?
(30 min, 45 km)
(90 min, 135 km)
12
12
tt
ddr
1.5km/min x 60 min = 90km/h
3090
45135
r
min/5.1min60
90km
kmr
x
y
0 10 20 30 40 50 60 70 80 90 100 110 120 minutes
10
150
140
130
120
110
100
30
20
50
40
70
60
90
80
160km
If the average velocity the entire trip was 80km/h what would the graph look like
60 min – 80 km
120 min – 160 km
min km
5 5
10 10
20 20
40 30
60 40
80 50
90 60
100 80
x
y
Distance (km)
20
40
60
80
100
120
0 10 20 30 40 50 60 70 80 90 100
Time (min)
CAR TRIP
Distance (km)
x
y
min km
5 5
10 10
20 20
40 30
60 40
80 50
90 60
100 80
A
B
C
D
0 10 20 30 40 50 60 70 80 90 100 110
Time (min)
20
40
60
80
100
120
Calculate the average velocity from
A to B B to C
C to D A to D
A
B
C
D
60km/h 30km/h
90km/h 47.5km/h
A toy rocket is launched from the ground and its flight path is described as y = -x2 + 13 x – 30.
h = - (t)2 + 13t - 30
h represents th
e
height (in
meters) of th
e
rocket‘t’
represents the tim
e
(in seconds)of th
e rocket
flight
h = -t2 t
h = - (t)2 + 13t - 30
x
y
10
5
15
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 seconds
metres
Fuse is lit but nothing happens for 3 seconds
It strikes the ground 10 seconds after the fuse is lit
Vertex tells how high the rocket reaches
h = - (t)2 + 13t - 30
x
y
10
5
15
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 seconds
metres
Velocity or speed of the rocket can now be determine
Distance (km)
Time (minutes)
0 10 20 30 40 50 60 70 80 90 100 110 120 130 10
20
30
40
50
60
70
80
90
100
110
120
130
150
140
160
x
y
Along a curve the rate of change is continually shifting
To find the average rate of change between two points we determine a straight line distance between them
x
y
A constant positive rate of
change (acceleration)
A constant negative rate of change
(negative acceleration)
deceleration
x
y
(6s, 9m)
(10s, 17m)
12
12
tt
ddr
ss
mmr
610
917
s
mr
4
8 smr /2
x
y
(18s, 9m)
(16s, 14m)
12
12
tt
ddr
ss
mmr
1618
149
s
mr
2
5 smr /5.2
ab
afbfr
)()(
Formula for rate of change
a represents the initial time
b represents the ending time
f(a) represents the height calculated at time ‘a’
f(b) represents the height calculated at time ‘b’
x
y
ab
afbfr
)()(
Formula for rate of change
x
y
h = - (t)2 + 13t - 30
(a)f(a)
h = - (t)2 + 13t - 30
(b)f(b)
ab
afbfr
)()(Formula for rate of change
a = 1s
b = 3s
f(a) = - 4.9(1)2 + 15(1) + 10
A projectile is traveling along a path determined by h = - 4.9x2 + 15x + 10.
What is its average rate of speed between 1 and 3 seconds”
f(b) = - 4.9(3)2 + 15(3) + 10
f(a) = - 4.9 + 25
f(a) = 20.5m
f(b) = - 44.1 + 55
f(b) = 10.9m
ab
afbfr
)()(Formula for rate of change
a = 1s
b = 3s
f(a) = - 4.9(1)2 + 15(1) + 10
f(b) = - 4.9(3)2 + 15(3) + 10
f(a) = - 4.9 + 25
f(a) = 20.1m
f(b) = - 44.1 + 55
f(b) = 10.9m
13
1.209.10
rab
afbfr
)()(
2
2.9r smr /6.4
Falling at an average sped of 4.8m/s
Given the function h = – 4.9t2 + 20t + 15
What was the initial height of the projectile?
When did the projectile reach maximum height?
15m
Find ‘t’ sa
b04.2
8.9
20
)9.4(2
20
2
h = – 4.9t2 + 20t + 15
h = - 4.9(2.04)2 + 20(2.04) + 15
h = 35.41m
Maximum height
( 2.04s , 35.41m)
2.04s
35.41m
Distance (km)
Time (minutes)
x
y
Tangent of the curve
Instantaneous rate of change at the given moment in time.
I. Instantaneous velocity – how far am I traveling at one specific moment?
Need the slope of the tangent at the specific point.
I. Instantaneous velocity – how far am I traveling at one specific moment?
Need the slope of the tangent at the specific point.
(15.01s, ?m)
(14.99s, ?m)
x
y
● (15s, 16m)
ab
afbfr
)()(
Formula for rate of change
a represents the initial time
b represents the ending time
f(a) represents the height calculated at time ‘a’
f(b) represents the height calculated at time ‘b’
x
y
(15.01s, ?m)
(14.99s, ?m)
Two people have been stranded at sea in a life raft. Their only chance of being rescued is if somebody
sees the flare shot from their flare gun. Assume the flare follows the path given by h = -t2 + 10t + 25 (h
= height in metres and t = time in seconds)
What is the maximum height the flare will reach?
How long will it be before the flare falls back into the sea??
How high is the flare after 3 seconds?
What is the average velocity of the flare between 2 and 3 seconds?
At 6 seconds what is the velocity of the flare?
12.07s
50m
46m
5m/s
-2m/s
A giant yo-yo is flung electronically and it follows a path h = t2 –10t +24
(h = height in metres and t = time in seconds)
What is the minimum depth the yo-yo will reach?
What was the height of the yo-yo when it was launched??
How high is the yo-yo after 9 seconds?
What is the average velocity of the yo-yo between 8 and 10 seconds?
At 7 seconds what is the velocity of the yo-yo?
24m
-1m
15m
8m/s
3m/s
Little Jimmy is strolling around the pool when he drops his favorite toy into the deep end!! The deep end is 8 meters deep. His babysitter jumps into the pool and swims along the path d = x2 – 10x + 15. (h = height in metres and t = time in seconds)
How deep can the babysitter swim?
Will she succeed in reaching the toy?
How long before she returns to the surface?
What is the average velocity of the swimmer between 2 and 5 seconds?
At 7 seconds what is the velocity of the swimmer?
yes
10m
8.16s
3.7m/s
4m/s
The height of a football after is is thrown can be determined by the function h = - 4.9t2 + 28t + 2. (h = height in metres and t = time in seconds)
What is the maximum height the ball will reach
After how many seconds does it reach the max height?
How long before the ball returns to the ground?
What is the average velocity of the football between 1 and 3 seconds?
At precisely 4 seconds what is the velocity of the football?
2.9s
42m
5.78s
8.4m/s
-11.2m/s
Distance (km)
Time (minutes)
0 10 20 30 40 50 60 70 80 90 100 110 120 130 10
20
30
40
50
60
70
80
90
100
110
120
130
150
140
160
x
y
