Interpretations of the Derivative
Before moving on to the section where we learn how to
compute derivatives by avoiding the limits we were evaluating in the previous
section we need to take a quick look at some of the interpretations of the
derivative. All of these interpretations arise from recalling how our
definition of the derivative came about. The definition came about by noticing
that all the problems that we worked in the first
section in the chapter on limits required us to evaluate the same limit.
Rate of Change
The first interpretation of a derivative is rate of
change. This was not the first problem that we looked at in the limit chapter,
but it is the most important interpretation of the derivative. If
 represents
a quantity at any x then the derivative
 represents
the instantaneous rate of change of
 at
 .
Example
Suppose that the amount of water in a holding tank at t minutes
is given by
 .
Determine each of the following.
(a)
Is the volume of water in the tank increasing or decreasing at
 minute?
(b)
Is the volume of water in the tank increasing or decreasing at
 minutes?
(c)
Is the volume of water in the tank changing faster at
 or
 minutes?
(d)
Is the volume of water in the tank ever not changing? If so, when?
Solution
In the solution to this example we will use both notations
for the derivative just to get you familiar with the different notations.
We are going to need the rate of change of the volume to
answer these questions. This means that we will need the derivative of this
function since that will give us a formula for the rate of change at any time
t. Now, notice that the function giving the volume of water in the tank is
the same function that we saw in Example 1 in the last
section except the letters have changed. The change in letters between the
function in this example versus the function in the example from the last
section won�t affect the work and so we can just use the answer from that
example with an appropriate change in letters.
The derivative is.
Recall from our work in the first limits section that we
determined that if the rate of change was positive then the quantity was
increasing and if the rate of change was negative then the quantity was
decreasing.
We can now work the problem.
(a) Is the volume of water in the tank increasing
or decreasing at
 minute?
In this case all that we need is the rate of change of the
volume at
 or,
So, at
 the
rate of change is negative and so the volume must be decreasing at this time.
(b) Is the volume of water in the tank increasing
or decreasing at
 minutes?
Again, we will need the rate of change at
 .
In this case the rate of change is positive and so the
volume must be increasing at
 .
(c) Is the volume of water in the tank changing
faster at
 or
 minutes?
To answer this question all that we look at is the size of
the rate of change and we don�t worry about the sign of the rate of change. All
that we need to know here is that the larger the number the faster the rate of
change. So, in this case the volume is changing faster at
 than
at
 .
(d) Is the volume of water in the tank ever not
changing? If so, when?
The volume will not be changing if it has a rate of change
of zero. In order to have a rate of change of zero this means that the
derivative must be zero. So, to answer this question we will then need to solve
This is easy enough to do.
So at
 the
volume isn�t changing. Note that all this is saying is that for a brief instant
the volume isn�t changing. It doesn�t say that at this point the volume will
quit changing permanently.
If we go back to our answers from parts (a) and (b) we can
get an idea about what is going on. At
 the
volume is decreasing and at
 the
volume is increasing. So at some point in time the volume needs to switch from
decreasing to increasing. That time is
 .
This is the time in which the volume goes from decreasing
to increasing and so for the briefest instant in time the volume will quit
changing as it changes from decreasing to increasing.
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