The purpose of this section is to remind us of one of the
more important applications of derivatives.
That is the fact that represents the rate of change of . This is an application that we repeatedly saw
in the previous chapter. Almost every
section in the previous chapter contained at least one problem dealing with
this application of derivatives. While this
application will arise occasionally in this chapter we are going to focus more
on other applications in this chapter.
So, to make sure that we don’t forget about this application
here is a brief set of examples concentrating on the rate of change application
of derivatives. Note that the point of
these examples is to remind you of material covered in the previous chapter and
not to teach you how to do these kinds of problems. If you don’t recall how to do these kinds of
examples you’ll need to go back and review the previous chapter.
Example 1 Determine
all the points where the following function is not changing.
First we’ll need to take the derivative of the function.
function will not be changing if the rate of change is zero and so to answer
this question we need to determine where the derivative is zero. So, let’s set this equal to zero and solve.
The solution to
this is then,
If you don’t
recall how to solve trig equations check out the Solving Trig Equations sections in the
Example 2 Determine
where the following function is increasing and decreasing.
As with the first problem we first need to take the derivative
of the function.
Next, we need
to determine where the function isn’t changing. This is at,
function is not changing at three values of t. Finally, to determine
where the function is increasing or decreasing we need to determine where the
derivative is positive or negative.
Recall that if the derivative is positive then the function must be
increasing and if the derivative is negative then the function must be
decreasing. The following number line
gives this information.
So, from this
number line we can see that we have the following increasing and decreasing
If you don’t remember how to solve polynomial and rational
inequalities then you should check out the appropriate sections in the Review
Finally, we can’t forget about Related
Example 3 Two
cars start out 500 miles apart. Car A
is to the west of Car B and starts driving to the east (i.e. towards Car B) at 35 mph and at the same time Car B starts
driving south at 50 mph. After 3 hours
of driving at what rate is the distance between the two cars changing? Is it increasing or decreasing?
The first thing to do here is to get sketch a figure
showing the situation.
In this figure y represents the distance driven by
Car B and x represents the distance
separating Car A from Car B’s initial position and z represents the distance separating the two cars. After 3 hours driving time with have the
following values of x and y.
We can use the
Pythagorean theorem to find z at
this time as follows,
Now, to answer
this question we will need to determine given that and . Do you agree with the signs on the two
given rates? Remember that a rate is
negative if the quantity is decreasing and positive if the quantity is
We can again
use the Pythagorean theorem here.
First, write it down and the remember that x, y, and z are all changing with time and so
differentiate the equation using Implicit
Finally, all we
need to do is cancel a two from everything, plug in for the known quantities
and solve for .
So, after three
hours the distance between them is decreasing at a rate of 14.9696 mph.
So, in this section we covered three “standard” problems
using the idea that the derivative of a function gives its rate of change. As mentioned earlier, this chapter will be
focusing more on other applications than the idea of rate of change, however,
we can’t forget this application as it is a very important one.