Two kinds of rate of change
Q: A car travels 110 miles in 2 hours. What’s its average rate of change
(speed)?
A: 110/2 = 55 mi/hr. That is, if we drive 55 miles in an hour, then in 2
hours, we will have driven 110 miles.
Q: If you are driving and suddenly look at your odometer, which says 60
mi/hr, what kind of rate of change is that?
A: Instantaneous R.O.C. That is, the rate at that particular time instance.
Average R.O.C. is over a period of time
Instantaneous R.O.C. is at a given point of time.
Average rate of change
A rocket is shot straight up, given by
the function f(x) = -16x^2+128x, where x =
time, f(x) = height at time x.
P
Height (feet)
0
Q: What is the average speed between
Q P (at x = 4) and Q (at x = 7)?
1. Find the slope of the secant line between
P and Q.
1
4
5
Time (seconds)
7 8
2. Find the coordinates of P and Q. For P,
f(4) = -16(4)^2+128(4) = 256, P(4,256).
For Q, f(7) = -16(7)^2+128(7) = 112, so
Q(7,112).
3. Slope = (112-256)/(7-4) = -48 ft/sec
Instantaneous rate of change
How fast is the rocket moving at precisely
X = 4 seconds?
P
Height (feet)
Q: What is the average R.O.C as x changes
from 4 to 6?
0
f (6)  f ( 4)
192  256

 32
64
2
Q: What is the average R.O.C as x changes
from 4 to 5?
f (5)  f ( 4)
240  256

 16
54
1
1
4
5
6
Time (seconds)
Q: What is the average R.O.C as x changes
8 from 4 to 4.001?
f (4.001)  f (4) 255.999984  256

 0.016
4.001  4
0.001
What do you think the instantaneous R.O.C would be at x = 4?
Definition
Average Rate of Change of a
quantity over a period of
time is the amount of change
divided by the time required
for the change.
Example
f(x) = 0.1×x2
12
10
8
250
6
4
2
0
5
10
Average
speed is
35.7 mph
measured
over 7
hours
7
time in hours
15
Important Idea
The average rate of change
(speed) over a time period is
the slope of the secant line
connecting the beginning
and end of the time period.
Average y
y2  y1

Rate of 

t
t

t
2
1
Change
Try
This
Describe
in words
how you
could find
the speed
at exactly
the 5th
hour.
f(x) = 0.1×x2
12
10
8
6
4
2
0
5
5
10
15
time in hours
Solution
The instantaneous
th
velocity at exactly the 5
hour is the slope of the
line tangent to the
velocity function at t=5.
Important Idea
The instantaneous
velocity at a point, or any
other rate of change, is
the slope of the tangent
line at the point
The derivative can be used to determine the rate of change of one variable with respect
to another.
Ex: Population growth, production rates, rate of water flow, velocity and acceleration.
Ex: Free fall Position function. A function, s, that gives position (relative to the origin) of an
object as a function of time.
1 2
s  gt  vot  so
2
A ball dropped from a 160 foot building:
Therefore, the
average velocity is
change in dis tan ce y

change in time
x
Find average velocity over each time interval
[1, 2]
[1,1.5]
[1,1.1]
Negative velocity indicates object is falling
Generally if, s = s(t) is the position for an object moving in a
straight line, then the velocity of the object at time t is:
s (t  t )  s (t )
v(t )  lim
 s '(t )
t 0
t
Find instantaneous velocity when t=1.1 sec
Position Function
s (t )  16t  vot  so
2
Velocity Function
s '(t )  v(t )  32t  vo
Acceleration Function
s ''(t )  v '(t )  a(t )  32