Section 1
Average Rates of Change
Nothing endures but change.
Heraclitus
calculus is the study of rates of change. “Velocity”
D
means the same thing as “rate of change”, although we usually use
velocity to mean the rate of change of distance with respect to time. The
IFFERENTIAL
average velocity of an object is the distance it travels divided by how long
it takes to travel that distance. If a person walks 200 yards in 25 seconds,
their average velocity will be 200 yards/25 seconds = 8 yards/second.
Note that speed is not quite the same as velocity. “Speed” indicates
how fast an object is traveling; “velocity” indicates how fast and in which
direction. If an object is moving on a line, we will indicate one direction on the line as positive and one direction as negative. If there is a
horizontal line in front of us, we will normally take the right to be the
positive direction and left to be the negative. If an object has speed 5
meters/second, it could be moving to the left or the right. If an object has
velocity 5 meters/second, it will be moving to the right. An object with
velocity -5 meters/second will be moving to the left. In higher dimensions,
describing velocities involves more than a single number; in this class, we
will only deal with motion along a line.
To find the average velocity of an object we typically have to find out
how far the object travels (its change in position) and and the length of
time it takes (the change in time). So let’s begin by discussing change.
1
We’ll start with some notation. The symbol ∆ (the Greek letter Delta,
corresponding to the Latin letter “D”, as in Difference) is used to indicate
how much a quantity changes, and so “∆x” represents the amount the
variable x changes and can be read “the change in x”.
Definition
If a quantity x changes from the value a to the value b, then
∆x = b − a.
Example. If the quantity x increases from 3 to 7, then
∆x = 7 − 3 = 4.
Of course, there is nothing special about the letter x.
2
Example. If the quantity t changes from 2 to 9, then
∆t = 9 − 2 = 7.
2
There is no reason why b has to be larger than a in the above definition;
the quantity does not have to increase.
If the value of y = f ( x ) depends on x, then a change in the value of
x will cause the value of y to change also, so the change in these two
quantities will be related.
Example. Let y = x2 . If x changes from x = 1 to x = 3, then y will
change from y = 12 = 1 to y = 32 = 9. We will have
∆x = 3 − 1 = 2
∆y = 9 − 1 = 8.
2
The average rate of change of one quantity with respect to another is
the ratio of the change in the first quantity changes to that of the second.
Definition
The average rate of change of y = f ( x ) (with respect to x) from x = a to
x = b is
∆y
f (b) − f ( a)
=
.
∆x
b−a
2
Example. Let y = f ( x) = 2x2 . Find the average rate of change of y
with respect to x from x = 2 to x = 5.
We first of all have ∆x = 5 − 2 = 3. Next, as x changes from x = 2 to x = 5,
y = f ( x ) will change from y = f (2) = 2 · 22 = 8 to y = f (5) = 2 · 52 = 50,
so ∆y = 50 − 8 = 42. Putting it together, the average rate of change of y
from x = 2 to x = 5 is
∆y
42
=
= 14
2
∆x
3
Example. Let y = f ( x) = 1/x. Find the average rate of change of y
with respect to x from x = 1 to x = 3.
Since ∆x = 3 − 1 = 2 and ∆y = f (3) − f (1) = 1/3 − 1/1 = −2/3, the
average rate of change of y from x = 1 to x = 3 is
−2/3
2
∆y
=
=− .
∆x
1
3
2
The rate of change of y with respect to x tells us how fast y changes
relative to x. If ∆y/∆x = 2, for example, then y is changes twice as much
as x over the given interval, and so on the average y is changing twice as
fast over the interval. The larger the rate of change, the faster y is changing
compared to x. Note that in the second example above, the rate of change
is negative; that means that y is changing in the opposite direction as x —
the quantity y is getting smaller as x gets larger.
Example. A ball is dropped from the top of a tall building, so that in t
seconds it will have fallen y = f (t ) = 16t 2 feet. What is the average
velocity of the ball from t = 1 to t = 3 seconds?
Since
∆t = 3 − 1 = 2 seconds
and
∆y = f (3) − f (1) = 144 − 16 = 128 feet,
the average velocity of the ball from time t = 1 second to t = 3 seconds is
∆y
128
=
= 64 feet/sec
∆t
2
2
If we are given the velocity, then we can use that to find the distance
travelled.
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Example. Suppose a bike travels for 3 hours at 25 miles/hour. How far
does it travel?
The bike will travel a distance of 3 hours times 25 miles/hour for a total
of 75 miles.
2
Example. Suppose y = f ( x) and the average rate of change of y with
respect to x from x = 1 to x = 5 is ∆y/∆x = 7. Find ∆y.
Since ∆x = 5 − 1 = 4 and ∆y/∆x = 7, we get ∆y = (∆y/∆x ) ∆x = 7 · 4 =
28.
2
If we know how far an object moves and its beginning position, we
can figure out its ending position.
Example. A car travels directly away from us for 5 hours, at an average
velocity of 40 miles/hour. If it starts out 120 miles away, how far away
is it after the 5 hours?
The car travels away from us a distance of 5 · 40 = 200 miles. Since it
starts off 120 miles away, it ends up a distance of 120 + 200 miles.
2
Example. Suppose y = f ( x). If f (2) = 12 and, as x changes from x = 2
to x = 4, the average rate of change of y with respect to x is ∆y/∆x = 5.
What is f (4)?
As x changes from x = 2 to x = 4, the amount that y = f ( x ) changes is
∆y = (∆y/∆x ) ∆x = 5 · 2 = 10. Since at x = 2, y begins with the value
f (2) = 12, y ends up with the value f (4) = 12 + 10 = 22.
2
Geometrically, the average rate of change of a function can be interpreted as the slope of a secant line. A secant line is a line through two
points of a specified curve. We will usually want to specify the points.
Definition
The secant line to the graph y = f ( x ) from x = a and x = b is the line
through ( a, f ( a)) and (b, f (b)). (See figure 1.1.)
The slope of the secant line is ( f (b) − f ( a))/(b − a) = ∆y/∆x, which
is exactly the average rate of change of f from x = a to x = b.
Differential calculus, unfortunately, is not the study of average velocities, but rather the study of instantaneous velocities — how fast something
is changing at an instant or point. But now we’re faced with a problem: in
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y = f (x)
m=
f (b)− f ( a)
b− a
(b, f (b))
f (b) − f ( a)
( a, f ( a))
b−a
a
b
Figure 1.1: A secant line.
an instant, the dependent variable doesn’t change. In the formula for rate
of change,
∆y
,
∆x
the denominator ∆x will be 0, and division by 0 is forbidden. So we need
a different, but not too different, definition for velocity at a point. Any
reasonable definition of velocity at a point should give values close to the
velocities near the point.
Let’s look at a concrete situation. In the example with the falling ball,
the ball will have fallen f (t) = 16t2 feet in t seconds. How fast is it going
in exactly 3 seconds? Let’s look at the rate of change near t = 3 seconds.
In other words, let’s look at the rate of change from 3 seconds to t seconds
for some values of t close to 3.
• From t = 3 to t = 3.1 seconds, the ball’s average velocity will be
f (3.1) − f (3)
153.76 − 144
=
= 97.6 feet/sec.
3.1 − 3
.1
• From t = 3 seconds to t = 3.01 seconds, the average velocity is
f (3.01) − f (3)
144.9616 − 144
=
= 96.16 feet/sec
3.01 − 3
.1
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• From t = 3 seconds to t = 3.001 seconds, the average velocity is
144.096016 − 144
f (3.001) − f (3)
=
= 96.016 feet/sec.
3.001 − 3
.001
The velocity of the ball at t = 3 seconds should be close to these average
velocities, and should probably be closer to the average velocities over the
smaller intervals. It looks as if these average velocities are approaching
96 feet/sec, and indeed they are. We wouldn’t be too far off if we said
that the velocity at a point is simply the average velocity over a small
interval near the point. In fact, that is a pretty good way to think of an
instantaneous rate of change.
Informal Definition. The instantaneous rate of change of y = f ( x ) (with
respect to x) at x = a is the average rate of change of y = f ( x ) over a
small interval near x = a.
2
Example. Going back to our previous example with the falling ball,
where the ball will have fallen f (t) = 16t2 feet in t seconds, we could (not
quite accurately) say that the ball’s velocity in exactly 3 seconds is the
average velocity over the interval from t = 3 to t = 3.001 seconds; namely
the 96.016feet/sec.
This isn’t quite the exact answer; what’s more, it depends on the small
interval we use. The average velocity over the interval from t = 3 to
t = 3.0005 seconds, for example, is 96.008feet/sec, which is a different
value than above. Nevertheless, this is a useful way to think about the
instantaneous rate of change.
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