Instantaneous Rate of Change:
Last section we discovered that the average rate of change in F(x) can also be interpreted
as the slope of a scant line. The average rate of change involves the change in F(x) over a
designated interval [x1, x2] or between the two points (x1, F(x1)) and (x2, F(x2)). The
secant line passes through these two points.
A logical extrapolation would indicate that the instantaneous rate of change in F(x) at a
point x would be the same as the slope of a tangent line touching the graph of F(x) at that
same point (x, F(x)). In the case of average rate of change you can find the slope of the
secant line easily because you have two points to work with. But in the case of the
instantaneous rate of change there is only one point and therefore it is not possible to
directly calculate the slope of the tangent line. Is it possible to find this value that will be
both the slope of the tangent line and the instantaneous rate of change? We will now
define a process that will help us do just that.
Finding Instantaneous Rate of Change and the Slope of the Tangent Line.
We will again consider the following profit function from the last section:
P(q) = -2q2 + 400q – 5000 where q represents quantity of items sold.
P
Profit and slope of tangent line
13900
11900
9900
7900
5900
3900
1900
-100
Profit
Tangent line
0 20 40 60 80 10 12 14 16 18 20 22
0 0 0 0 0 0 0
x
Figure 1
We would like the find the instantaneous rate of change in profit for q = 40. We will find
the instantaneous rate of change numerically. This method proceeds as follows:
Since we can not find the slope of the tangent line touching the graph at q = 40 by direct
calculation we will try to guess its value as precisely as possible. In order to do this we
must find slopes of secant lines that are very close to the tangent line at the point
(40, 7800). First we will find the slope of secant line through the points ( 40, 7800) and
(41, 8030). Then we will repeat the process for points ( 40, 7800) and ( 40.5, 7919.5). We
then will repeat again for the points ( 40, 7800), (40.05, 7811.995) and so on. In order
to be very precise we will also need to consider points on the left side of q = 40 as well
such as ( 39.99, 7797.599) and ( 40, 7800). The process is better illustrated in the table
below.
using points from the right side of q = 40
x1 x2
P(x1)
P(x2)
40
41
7800
8038
40
40.5
7800
7919.5
40
40.05
7800
7811.995
40
40.005
7800
7801.2
40
40.0005
7800
7800.12
[P(x2) - P(x1)] / [x2 - x1]
238
239
239.9
239.99
239.999
using points from the left side of q = 40
x1 x2
P(x1)
P(x2)
40
39
7800
7558
40
39.9
7800
7775.98
40
39.99
7800
7797.6
40
39.999
7800
7799.76
40
39.9999
7800
7799.976
[P(x2) - P(x1)] / [x2 - x1]
242
240.2
240.02
240.002
240.0002
Figure 2
As we look at the slopes of the many secant lines in the chart above we notice that as the
secant lines pass through points that are closer and closer together their slopes seem to be
closer and closer to the same value both on the right and left side of the point (40, 7800).
Therefore we will make an estimate (or a good guess) that the slope of the tangent line is
m = 240. The closer the points involved in calculations the better the estimate.
In the process above we found the slope of the tangent line numerically. We will also
learn to find it algebraically.
In order to develop the algebraic approach we need to go back to idea of a scant line. If
we have a function f(x) a secant line intersect the graph of this function in two places.
Therefore it shares two points with f(x). We will call those two points (x, f(x)) and
(x + h, f(x + h)). In order to find the slope of this second line we need to find the change
in the y-values divided by the change in the x-values. This will give us the secant line
f ( x + h) − f ( x ) f ( x + h) − f ( x )
=
. This is a generalization of the slope of
slope m =
( x + h) − x
h
any secant line.
If we now think back to the numerical process above, you will remember that we found
the slopes of several secant lines. Each secant line was closer to the tangent line of
interest than the secant line before it and the slope values were getting closer to the slope
value of the tangent line. For each succeeding secant line the two points used were closer
together than before. One way to express this change in points is to use the notation
“ Limit ”. This notation means that the distance “h” between the x values is getting
h →0
smaller and smaller. The distance h is approaching the value zero. Therefore the limit of
the slope values of the secant lines will approach the slope value of the tangent. The
symbolic notation that describes this process is given as follows.
dy
f ( x + h) − f ( x )
∆y
= lim it
= lim it
h →0
∆x → 0 ∆x
dx
h
dy
∆y
is notation for the slope of the tangent line and
is notation for the slope of the
dx
∆x
secant line. This formula is also called the Derivative (to be discussed in a later section)
(
Now we will look at an example how to find the slope of a tangent line touching the
graph of f(x) at some value x = a.
Example 1
Find the slope of the tangent line touching the graph f ( x) = x 2 + 3 at x = 5.
Solution:
Before we start the problem we will review notation: Review the following evaluations:
f(x) = x2 + 3
f(4) = (4)2 + 3 = 19
f(7) = (7)2 + 3 = 52
f(m) = m2 + 3
f(x + h) = (x + h)2 + 3
Now we will now proceed to solve the example problem.
dy
f ( x + h) − f ( x )
The slope of the tangent line is given to be
= lim it
so we need to
h →0
dx
h
make the appropriate substitution as follows:
dy
[( x + h) 2 + 3] − [ x 2 + 3]
x 2 + 2 xh + h 2 + 3 − x 2 − 3
h(2 x + h)
= lim it
= lim it
= Limit
=
h →0
h →0
h →0
dx
h
h
h
lim it (2 x + h) = 2 x = 2(5) = 10 . Therefore the slope of the tangent line touching the graph
h →0
at x = 5 is m = 10. See the graph below.
F(x) and tangent line
45
35
25
15
5
-5
-5 -4 -3 -2-15
-1 0 1 2 3 4 5 6 7 8 9 10
-25
f(x)
T
x
Figure 3
In order to verify that this is the correct Tangent line we will find the equation of the
tangent line that touches the graph of f ( x) = x 2 + 3 at x = 5.
Solution:
In order to find the equation of any line you need the following:
1) a point
2) a slope
3) the point slope form of the equation y − y1 = m( x − x1 )
Since we are interested in the tangent line that touches the graph of f ( x) = x 2 + 3 at
4) x = 5, we need to find that the y –value f(5) = 28. So now we have the point (5,
28). Next we find the slope to the tangent line at x = 5. Since the derivative of
dy
= 2 x is the general slope formula for any tangent line touching f (x) we found
dx
dy
= 10 . Finally we use
above have that the slope of the tangent line at x = 5 is
dx
the point and slope information to create the equation y − y1 = m( x − x1 ) to get
y − 28 = 10( x − 5) . Now simplify to get y = 10x – 22 which is the tangent line
equation pictured in Figure 3 above.
Example 2
Find the instantaneous rate of change in Revenue R = −20q 2 + 4q at q = 30 where q is
the quantity of items sold.
a) Using the table in Figure 2 as a guide, find the instantaneous rate of change
numerically from the left side. Calculate the slope of secants lines through the
following points where
3) q1=30 and q2= 29.99
1) q1=30 and q2= 29.5
2) q1=30 and q2= 29.9
4) q1=30 and q2= 29.999
b) Now find the instantaneous rate of change numerically from the right side.
Calculate the slope of secants lines through the following points where
3) q1=30 and q2= 30.001
1) q1=30 and q2= 30.1
2) q1=30 and q2= 30.01
4) q1=30 and q2= 30.0001
c) Use your calculations from a) and b) to estimate the slope of the tangent line
at q = 30 which will also the instantaneous rate of change in revenue at q = 30.
Now we will do the problem again algebraically:
To find the instantaneous rate of change at q = 30, we need to find
dR
R ( x + h) − f ( x )
(−20(q + h) 2 + 4(q + h)) − (−20q 2 + 4q )
= lim it
= Limit
h →0
h →0
dq
h
h
= Limit
h →0
−20q 2 − 40qh − 20h 2 − 20h 2 + 4q + 4h + 20q 2 − 4q
h(−40q − 20h + 4)
= Limit
h →o
h
h
= -40q + 4 = -40(30) + 4 = -1196
The Marginal Concept Revisited: In the section on average rate of change we defined
the term marginal as the average rate change in the dependent variable over a one unit
change in the independent variable. When considering instantaneous rate of change you
are considering a very small change in the independent variable. However when you are
considering a change in quantity of items the smallest change will be a unit of one. ( you
would not make a fraction of an item to sell) Therefore the instantaneous rate of change
when quantity is involved will be equivalent to the definition of marginal.