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Calculus 1 Problems & Solutions – Chapter 2 – Section 2.1.1 |
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2.1.1
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1. Average Rates Of Change |
Consider a car moving on a straight even road in the same
direction. Suppose it travels 300 km in 3 hours. If the velocity
of the car is constant, then it is 300/3 = 100 km/h. This is also the average
velocity. Suppose the velocity changes several
times. It's about 100 km/h much of the time, but sometimes it's around 90 km/h,
and sometimes it's around 110 km/h.
The average velocity of the car during that time period is still 300/3 = 100
km/h. The average velocity is the velocity that
the car would have if it travelled the same distance in the same amount of time
at a constant velocity. Velocity is the rate
of change of distance travelled with respect to time: velocity = distance
travelled / time elapsed. Distance travelled is a
function of time. Recall that the rate of change of a function f(x) is the
change of the function f(x) generated
by a 1-unit
change of the variable x.
Note:
More precise definitions are as follows. The position of an object is the
signed displacement from a fixed point
called the origin to the object. It's a function of time elapsed. Velocity of
the object is the rate of change of its position
with respect to time. Let s(t) be the
position of the object at time t. The average velocity of the object over [t1, t2],
where
t1 and t2
are 2 points in time and t1 < t2,
is (change of position)/(change of time) = (s(t2)
– s(t1))/(t2
– t1). It may be that
s(t2) > s(t1)
or s(t2) < s(t1)
or s(t2) = s(t1).
So velocity is a signed quantity. The car example above is ok because the car
moves in the same direction, so that the distance travelled is the same as the
change of position. If the car may change
direction then the distance travelled may differ from the change of position
(eg, if t1 < t2
< t3 but s(t1)
< s(t3) < s(t2)
then
distance travelled = |s(t2) – s(t1)| + |s(t3) – s(t2)| and change of position = s(t3) – s(t1);
clearly distance travelled >
change of position) and thus the more precise definitions are required.
Let the domain of the function f(x) contain
points a
and b.
See Figs. 1.1 and 1.2. When x
changes from a
to b,
f(x)
changes from f(a) to f(b). During
the change of x
from a
to b,
how much does f(x) change
when x
changes by 1 unit?
The answer is the total change of f(x) divided by the total change of x, which is (
f(b) – f(a))/(b – a). This
quotient gives
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Fig. 1.1
Average rate of change of f over [a, b] is |
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Fig. 1.2
Average rate of change of f over [a, b] is |
the change of f(x) per 1-unit change of x over [a, b], and is the average rate of change of f(x) over [a, b].
If f(x) changes at a constant rate, then
the graph of f
over [a,
b]
is a straight-line segment, as in Fig. 1.1. Every 1-unit
change of x
contributes an equal amount of change of f(x).The average rate of change
of f(x) over [a, b] is
( f(b) – f(a))/(b – a), which is
the slope of the line segment. The rate of change of f(x) at any
point or instant x
in [a,
b]
is equal to the average rate of change ( f(b) – f(a))/(b – a).
Now, suppose f(x) changes at
a changing rate. See Fig. 1.2. The graph of f is a curve. The colored regions
show two
1-unit changes of x
for which f(x) changes by
different amounts, v
and w.
This confirms that the rate of change of f(x)
over [a,
b]
changes. Not every 1-unit change of x contributes an equal share of the
change of f(x). Thus, the
quotient
( f(b) – f(a))/(b – a), which is
a constant (because a,
b, f(a), and f(b) are all
constants), doesn't give the actual rate of
change of f(x). Instead,
it gives the rate of change that f(x) would have if f(x) changed by the same amount f(b) – f(a)
at a constant rate of ( f(b) – f(a))/(b – a)). It's
called the average rate of change of f(x) over [a, b].
Definition 1.1
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The average rate of change of f over [a, b] is the quantity:
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Fig. 1.3
Average rate of change of f corresponding to a change of x |
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Fig. 1.4
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Example 1.1
Solution
The required average velocity is:
EOS
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2. Instantaneous Rates Of Change |
For
a car travelling on a road between time t1
and time t2, at any point of time, or instant, between t1
and t2 it has a
velocity, called instantaneous velocity. This is true whether the velocity is
constant or not, as evidenced by the
speedometer. If the velocity is constant, then the instantaneous velocity at
any point between t1 and t2
equals the average
velocity over [t1, t2].
Otherwise, the two types of velocity generally are not equal. In this case, how
can we compute the
instantaneous velocity at a point in [t1,
t2]? The answer is to use the concept of limit, as we're
about to see in the general
case of instantaneous rate of change, as follows.
Let a be a point in dom(f
). See Figs. 2.1 and 2.2. The rate of change of f(x)
at point a is called the instantaneous
rate of change, or simply the rate of change, of f(x)
at a.
Yes, the adjective instantaneous
isn't necessary when we talk about the rate of change at a point or instant.
What else
could the rate be? Remember, we talk
about the rate of change at a point or instant.
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Fig. 2.1
Instantaneous rate of
change of f at a equals average rate of |
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Fig. 2.2
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If the instantaneous rate of
change of f(x) at x = a exists,
then it's clear that it must be unique. So we define it by using
the (two-sided) limit, not one-sided limits, which may be different when they
exist.
Definition 2.1
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The instantaneous rate
of change, or simply the rate of change, of f(x)
at the point x = a is this limiting
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Of
course if this limit at point a doesn't exist, then the instantaneous rate of change
of f(x) at a doesn't exist. Note that
this limit is a two-sided one. Thus the values of h consist of negative values as well as positive ones.
Example 2.1
Solution
The required instantaneous velocity is:
EOS
Problems & Solutions |
1. Let f(x) = 1/(x – 1). Find the average rate of change of f over the interval [2, 5].
Solution
The required average rate of change is:
2. Let f(x) = x2 – 3x. Calculate the average rate of
change of f
over the interval [0, 2], its instantaneous rate of change
at the midpoint of that interval, and the absolute
value of the difference between the two rates of change.
Solution
Average rate of change:
Absolute value of the difference:
|(–1) – (–1)| = 0.
a. Calculate the average velocity of the object during the 2-hour
time interval from t
= 1 to t
= 3.
b. Calculate the average velocity of the object during the 1-hour
time interval from t
= 1 to t
= 2.
c. Calculate the average velocity of the object during the
half-hour time interval from t
= 1 to t
= 1.5.
d. Calculate the average velocity of the object during the
one-tenth-of-an-hour time interval from t = 1 to t = 1.1.
e. On the basis of parts a–d, guess the instantaneous
velocity of the object at time t = 1.
f. Calculate the exact instantaneous velocity of the object
at time t
= 1.
Solution
Let f(t) = s = 3t2.
4. Two positive
electrical charges start 5 cm apart at time t = 0 sec, and the distance between
them increases at a
constant
rate of 1 cm/sec. The force of repulsion between them is given by F = 10/(t + 5)2 dynes.
a. Find the average rate of change of the force of repulsion from
time t
= 0 sec to time t
= 5 sec.
b. Find the instantaneous rate of change of the force of repulsion
at time t
= 5 sec.
Solution
Solution
Let x be an arbitrary point of dom(f ). The instantaneous rate of change of f at x is:
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