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As mentioned earlier in the section, the concepts of slope and the more general rates of change are important concepts not
just in Mathematics, but also in other fields. Many important phenomena are modeled using non-linear functions, and while
the rates of change of these functions are not constant, we can sample the function at two points and compute what is known
as an average rate of change between them to give some sense as to the function’s behavior over that interval. (We are basically pretending that the function is linear on a short interval to see what we can say about its
behavior.)
Let \(f\) be a function defined on the interval \([a,b]\).
The average rate of change of \(f\) over \([a,b]\) is defined as:
Geometrically, the average rate of change is the slope of the line (This line is called a secant line.) containing \((a, f(a))\) and
\((b, f(b))\).
As with Definitions absmaxmindefn and incdeccnstdefn, the wording in Definition arc, while referring to the function \(f\), is really making a statement about its
outputs \(f(x)\).
If \(f\) is increasing over \([a,b]\), then the average rate of change will be positive. Likewise, if \(f\) is decreasing or constant, the average rate
of change will be negative or \(0\), respectively. (Think about this for a moment.) However, as the next example demonstrates, the
converses of these statements aren’t always true. (For example, the average rate of change over an interval
could be positive yet the function could decrease over part of that interval and then increase on a different
part.)
The formula \(s(t) = -5t^{\,2} + 100t\) for \(0 \leq t \leq 20\) gives the height, \(s(t)\), measured in feet, of a model rocket above the Moon’s surface as a function of the time \(t\),
in seconds after lift-off.
Find \(s(0)\), \(s(5)\), \(s(10)\), \(s(15)\) and \(s(20)\) and use these along with a graphing utility to graph \(y = s(t)\).
State the range of \(s\) and interpret the extrema, if any exist.
Find and interpret the \(t\)- and \(y\)-intercepts.
Find and interpret the interval(s) over which \(s\) is increasing, decreasing or constant.
Find and interpret the average rate of change of \(s\) over the intervals \([0,5]\), \([5,10]\), \([10, 20]\) and \([5, 15]\).
To find \(s(0)\), we substitute \(t=0\) into the formula for \(s(t)\): \(s(0) = -5(0)^2+100(0) = 0\). Similarly, \(s(5) = -5(5)^2+100(5) = -5(25)+500 = -125+500 = 375\). Continuing, we obtain: \(s(10) = 500\), \(s(15) = 375\) and \(s(20) = 0\). We construct a table of
values and with a graphing utility we obtain:
Projecting the graph to the \(y\)-axis, we see that the range of \(s\) is \([0, 500]\) so the minimum of \(s\) is \(0\) and the maximum is \(500\). This means
that the rocket at some point is on the surface of the Moon and reaches its highest altitude of \(500\) feet above the lunar
surface.
The first intercept we see is \((0, 0)\) which is both a \(t\)- and a \(y\)-intercept. Since \(t\) is the time after lift-off and \(y = s(t)\) is
the height above the Moon’s surface, the point \((0, 0)\) means that the model rocket was launched (\(t=0\)) from the
Moon’s surface (\(s(t) = 0\)). The remaining intercept, \((20, 0)\), is another \(t\)-intercept. This means that \(20\) seconds after lift-off (\(t=20\)),
the model rocket returns to the Moon’s surface (\(s(t) = 0\)). That is, the ‘time of flight’ of the model rocket is 20
seconds.
Referring to Definition incdeccnstdefn, \(s\) increases over the interval \([0, 10]\), since for those values of \(t\), as we read from left to right, the graph of
the function is rising meaning the \(y\) values (hence \(s(t)\) values) are getting larger. Thus the model rocket is heading upwards
for the first \(10\) seconds of its flight. We find that \(s\) decreases over the interval \([10, 20]\), indicating once it has reached its highest
altitude of \(500\) feet \(10\) seconds into the flight, the rocket begins to fall back to the surface of the Moon, landing \(20\) seconds after
lift-off.
To find the average rate of change of \(s\) over the interval \([0, 5]\) we compute
In other words, the height is increasing at an
average rate of \(75\) feet per second during the first 5 seconds of flight. The rate here is called the average velocity of the
rocket over this interval. Velocity differs from speed in that velocity comes with a direction. In this case, a positive
velocity indicates that the rocket is traveling upwards, since when \(s\) is increasing, the model rocket is climbing
higher.
Similarly, the average rate of change of \(s\) over the interval \([5, 10]\) works out to be \(25\). This means that the average velocity over
the next \(5\) seconds of the flight has slowed to \(25\) feet per second. The model rocket is still, on average, traveling upwards,
albeit more slowly than before.
Over the interval \([10, 20]\), the average rate of change of \(s\) works out to be \(-50\). This means that, on average, the rocket is falling at a
rate of \(50\) feet per second. The rocket has managed to fall from its highest point \(500\) feet above the surface of the Moon back
to the Moon’s surface in \(10\) seconds so this makes sense. Finally, the average rate of change of \(s\) over \([5, 15]\) is
\(0\). This means that the model is the same height above the ground after \(5\) seconds (\(375\) feet) as it is after \(15\)
seconds.
Geometrically, the average rate of change of a function over an interval can be interpreted as the slope of a secant line.
Below on the left is a dotted line containing \((0, 0)\) and \((5, 375)\) (which has slope \(75\)) along with a dotted line containing the points \((5, 375)\) and \((10, 500)\)
(which has slope \(25\)). Visually, the lines help demonstrate that, while \(s\) is increasing over \([0, 10]\), the rate of increase is slowing
down as \(t\) nears \(10\).
The graph above on the right depicts a dotted line through \((10, 500)\) and \((20,0)\) indicating a net decrease over that interval. We also
have a horizontal line (\(0\) slope) containing the points \((5, 375)\) and \((15, 375)\), which shows no net change between those two points,
despite the fact that the rocket rose to its maximum height then began its descent during the interval
\([5, 15]\).
An important lesson from the last example is that average rates of change give us a snapshot of what is happening at the
endpoints of an interval, but not necessarily what happens over the course of the interval. Calculus gives us tools to
compute slopes at points which correspond to instantaneous rates of changes. While we don’t quite have
the machinery to properly express these ideas, we can hint at them in the Exercises. Speaking of exercises
…
PreCalculus