Calculus
Module C09
Delta Process
Copyright
This publication © The Northern
Alberta Institute of Technology
2002. All Rights Reserved.
LAST REVISED April, 2009
Delta Process
Statement of Prerequisite Skills
Complete all previous TLM modules before beginning this module.
Required Supporting Materials
Access to the World Wide Web.
Internet Explorer 5.5 or greater.
Macromedia Flash Player.
Rationale
Why is it important for you to learn this material?
This unit provides the fundamental theory behind what calculus is and how it was
developed.
Learning Outcome
When you complete this module you will be able to…
Learning Objectives
1.
2.
3.
4.
Find the average rate of change in a function from first principles.
Represent average rates of change graphically.
Find the average rate of change of a function by using the delta process.
Use the delta method to find the derivative of a function.
Connection Activity
Assume you throw a ball to a friend. The path that it follows is a parabola and can be
described mathematically by a quadratic function. Using the function you can determine
the vertical and horizontal distance at any point in the path. However, if you’re interested
in determining the instantaneous rate of change of the ball at any point of the path you
would need calculus. The delta process will help explain how this is accomplished.
1
Module C09 − Delta Process
OBJECTIVE ONE
When you complete this objective you will be able to…
Find the average rate of change in a function from first principles.
Exploration Activity
REVIEW OF FUNCTIONAL NOTATION
In defining the derivative, the functional notation of Module 3 is convenient. Briefly in
review, if f ( x) = x 2 + x, then as an example:
f (2) = 22 + 2 = 4 + 2 = 6
and f ( −1) = (−1) 2 + ( −1) = 0
In calculus a change or increment of a variable x is written Δx and is read "delta x". This
Δx indicates a change in x, Δs a change in distance s, Δt a change in time t, and so on
for other variables. For convenience, the quantities (Δx) 2 , (Δx)3 , " are written Δx 2 ,
Δx 3 , " where it is understood that the Δx portion is a single unit and is being squared,
cubed, and so on.
So, for f ( x) = x 2 + x let’s replace x with x + Δx to get:
f ( x + Δx) = ( x + Δx) 2 + ( x + Δx) = ( x 2 + 2 x ⋅ Δx + Δx 2 ) + ( x + Δx)
= x 2 + 2 x ⋅ Δx + Δx 2 + x + Δx
= x 2 + x + 2 x ⋅ Δx + Δx + Δx 2 , ← place the higher powers of Δx at the end
of the expression.
EXAMPLE 1
If f ( x) = x 2 + x , find f (2 + Δx).
SOLUTION:
f (2 + Δx) = (2 + Δx) 2 + (2 + Δx)
= 4 + 4Δx + Δx 2 + 2 + Δx
= 6 + 5Δx + Δx 2
NOTE: If Δx is a small change in x, i.e. numerically less than 1, then Δx 2 is even
smaller.
The increment Δx of a variable may be determined by taking the difference in x as it
increases or decreases from one value of x = x1 to another value of x = x2 . Algebraically
the increment in x is represented as Δx = x2 − x1 and the increment in y as Δy = y2 − y1 .
2
Module C09 – Delta Process
AVERAGE RATE OF CHANGE OF A FUNCTION
It is possible to use these increments in x and y to determine the average rate of change in
the function f ( x) as x changes from x = x1 to another x = x2 . This average rate of
change in the function with respect to x is given by the quotient:
Δy
change in y y2 − y1
=
=
Δx
change in x x2 − x1
EXAMPLE 1
From first principles, find the average rate of change in the function y = x 2 as x changes
from x = 1 to x = 3 .
SOLUTION: From first principles,
at x = 1;
From
and
y = f ( x) = f (1) = 12 = 1, and at x = 3;
y = f (3) = 32 = 9 .
Δy
= average rate of change in y with respect to x
Δx
y − y1
Δy
= 2
, we have:
Δx
x2 − x1
Δy 9 − 1 8
=
= =4
Δx 3 − 1 2
We conclude that as x changes by 1 unit, the function changes by 4; or the average rate
of change in the function over the interval from x = 1 to x = 3 is 4:1.
3
Module C09 − Delta Process
Experiential Activity One
1. If f ( x) = x 2 − 1 , find f (1 + Δx)
2. If g ( x) = 3 x 2 + 2 x + 7 , find g (2 − Δx)
3. Find Δy and
Δy
from first principles, given:
Δx
a) y = 2 x − 3 and x changes from 3.3 to 3.5
b) y = x 2 + 4 x and x changes from 0.7 to 0.85
4. From first principles, find the average rate of change in the function y = 3x + 1 as x
changes from x = 3 to x = 6.
5. From first principles, find the average rate of change in the function y = x 2 − 3x + 5
as x changes from 3.03 to 3.02 .
Experiential Activity One Answers
1. 2Δx + Δx 2
2. 23 − 14Δx + 3Δx 2
3. (a) Δy = 0.4,
Δy
Δy
= 2 (b) Δy = 0.8325,
= 5.55
Δx
Δx
4.
Δy
=3
Δx
5.
Δy −0.0305
=
= 3.05
Δx
−0.01
4
Module C09 – Delta Process
OBJECTIVE TWO
When you complete this objective you will be able to…
Represent average rates of change graphically.
Exploration Activity
From Objective 1, Example 2 we have the function y = x 2 . It was found that as x
changed from 1 to 3 that y changed from 1 to 9. Thus the average rate of change of y
Δy
wrt x ,
, was 4 to 1.
Δx
This is shown graphically below:
y
10
8
7
6
5
4
3
2
1
−4
−3
−2
(3,9)
9
y = x2
−1
O
Δy
9 −1
=
Δx = 3 −1
Δy
(2,4)
(1,1)
Δx
x
1
2
3
4
FIGURE 1
From Figure 1 it is seen that the average rate of change of the function with respect to x
Δy 4
over the interval is
= . This also represents the slope of the secant line through
Δx 1
points (1,2) and (3,9).
Remember: A secant line to a curve cuts the curve in two points.
5
Module C09 − Delta Process
EXAMPLE 1
Find the average rate of change in the function y = 4 x − 1 as x changes from x = 1 to
x = 2 . Interpret the results graphically.
SOLUTION:
From first principles, given y = 4 x − 1,
and at x2 = 2, y2 = 4(2) − 1 = 7
From
Δy
=
Δx
Δy
Δx
Δy
Δx
x1 = 1 and x2 = 2 : at x1 = 1,
y2 − y1
x2 − x1 we have
= 7−3
2−1
4
=
1
=4
Graphically we get:
y
8
7
6
5
4
3
2
1
O
(2,7)
y = 4x − 1
Δy
(1,3)
Δy
7–3 4
=
Δx = 2 – 1 1
Δx
1
2
3
x
6
Module C09 – Delta Process
y1 = 4(1) − 1 = 3
EXAMPLE 2
Find the average rate of change in the function y = 12
x as x changes from x = 1 to x = 6 :
1. from first principles, then
2. interpret the results graphically
SOLUTION:
1. From first principles, given y = 12 = 12 , x1 = 1 and x2 = 6 :
1
At x1 = 1,
at x2 = 6,
y1 = 12 = 12
1
y2 = 12 = 2
6
Δy
= average rate of change in y with respect to x, and
Δx
From
Δy y2 − y1
=
, we have
Δx x2 − x1
Δy 2−12 −10
=
=
= −2
Δx 6−1
5
Δy
= −2
Δx
2. Interpreting graphically, we have:
y
Δy
2 −12 −10
=
= −2
Δx = 6 – 1
5
16
12
(1,12)
8
4
−8
−6
−4
−2
Δy
(6,2)
Δx
2
4
6
8
x
4
y=
12
x
−8
−12
Δy
represents the slope of the secant line through the points (1,12) and (6, 2)
Δx
Δy
gives the average rate of
on the curve of the function. From Figure 2, the ratio
Δx
change of y with respect to x.
NOTE:
7
Module C09 − Delta Process
Experiential Activity Two
1. Sketch y = 2 x + 3 . Find Δy as x changes from 1 to 1.3.
2. Sketch y = x 3 + 2 . Find Δy as x changes from 2 to 2.5.
3. Find the average rate of change in the function y = x 2 + 3x + 1 as x changes
from 3 to 4.
4. Find the average rate of change in the function y = 24 as x changes
x +3
from 1 to 5. Interpret the results graphically.
Experiential Activity Two Answers
Left to student to complete.
8
Module C09 – Delta Process
OBJECTIVE THREE
When you complete this objective you will be able to…
Find the average rate of change of a function by using the delta process.
Exploration Activity
THE DELTA PROCESS
Δy
given y = x 2 . We change x by some
Δx
increment. We use the general increment Δx . The new value of x is then x + Δx . Using
this new value for x in the original equation, we get a new value for y which can be
represented by y + Δy . From y = x 2 the new equation becomes:
Let us now develop the general formula for
y + Δy = ( x + Δx) 2
Expanding:
y + Δy = x 2 + 2 x ⋅ Δx + Δx 2
Subtracting the original function, y = x 2 , to find the expression for Δy , we get
y + Δy = x 2 + 2 xΔx + Δx 2
− (y
= x2 )
Δy = 2 x ⋅ Δx + Δx 2
Δy
which we are after, divide both sides of the equation by Δx :
Δx
Δy 2 x⋅Δx +Δx2
=
Δx
Δx
To obtain the ratio
which reduces to
Δy
= 2 x + Δx , which is the desired result.
Δx
This general formula
Δy
= 2 x + Δx says that for the specific function, y = x 2 , the ratio
Δx
Δy
for the average rate of change will always be equal to “twice the initial value of x
Δx
plus the increment in x ".
NOTE: In TML answer format always put Δx raised to a power greater than one at the
end of the expression, i.e. treat Δx 2 , Δx 3 etc. the same as in your text.
9
Module C09 − Delta Process
This method of deriving the equation that shows the rule for finding the average rate of
change in a function, is called the delta process, and may be summarized into three steps
as follows:
1. Replace x with x + Δx and replace y with y + Δy in the original function, and
expand if necessary.
2. Subtract the original function. Divide both sides of the result by Δx . Now we have
Δy
.
the ratio
Δx
Δy
of the particular function involved.
3. The expression shows the rule for finding
Δx
Observe that the average rate of change,
Δy
, can also be written in general form as:
Δx
Δy Δf ( x) f ( x +Δx)− f ( x)
=
=
,
Δx
Δx
Δx
which is a mathematical summary of the delta process where:
•
STEP #1 is represented by f ( x + Δx ) ,
•
STEP #2 is represented by f ( x + Δx) − f ( x) , and
•
STEP #3 is represented by
f ( x +Δx)− f ( x)
Δx
EXAMPLE 1
Find the average rate of change in the function y = 12
x as x changes from x = 1 to x = 6
from the general formula using the delta process.
SOLUTION:
From the delta process given y =
12
:
x
STEP 1: (substitution):
y + Δy = 12
x+Δx
10
Module C09 – Delta Process
STEP 2: (subtracting the original function):
12
x + Δx
12
−( y
= )
x
12
12
Δy =
−
x + Δx x
y + Δy =
and combining terms with a common denominator,
Δy =
12( x)−12( x +Δx)
x( x +Δx)
and simplifying,
Δy
=
Δy
OR
=
12 x−12 x −12⋅Δx
x( x+Δx)
−12⋅Δx
x( x +Δx)
STEP 3: (dividing both sides by Δx ):
Δy −12⋅Δx 1
=
⋅
Δx x( x+Δx) Δx
we obtain
Δy
= −12
Δx x( x+Δx)
Now evaluating the formula
Δy
= −12 over the interval where x = 1 and
Δx x( x +Δx)
Δx = 6 − 1 = 5 , we have:
Δy
= −12 = −12
Δx 1(1+5)
6
therefore
Δy
= −2
Δx
11
Module C09 − Delta Process
Experiential Activity Three
1. Using the delta process find Δy and
Δy
for:
Δx
a.
y = 2 x − 3 and x changes from 3.3 to 3.5
b.
y = x 2 + 4 x and x changes from 0.7 to 0.85
2. Since the average velocity, v = Δs , find v given:
Δt
a.
s = 3t 2 + 5 and t changes from 2 s to 3 s.
b.
s = 2t 2 + 5t − 2 and t changes from 2 s to 4 s.
3. Determine the general formula of
a.
y = x2 − 6 x + 7
b.
y = x2 + 4 x − 3
Δy
by the delta process for each of:
Δx
4. Find the average rate of change in the function y = 24 using the delta process.
x +3
5. Find
Δy
by using the delta process for the function y = x 3 − 3 x and x changes from
Δx
6 to 7.
Experiential Activity Three Answers
1. a)
b)
2. a)
b)
3. a)
b)
0.4; 2
0.8325; 5.55
15
17
2 x + Δx − 6
2 x + Δx + 4
−24
4.
( x+3)( x+3+Δx)
5. 124
12
Module C09 – Delta Process
OBJECTIVE FOUR
When you complete this objective you will be able to…
Use the delta method to find derivatives.
Exploration Activity
THE DERIVATIVE BY THE DELTA PROCESS
At this point it should be apparent what is meant by an average rate of change and how to
find it. In many situations, however, an average rate of change is insufficient. Often it is
necessary to determine the rate of change at an instant.
Δy
becomes a means of determining
The notation for average rate of change
Δx
instantaneous rate of change. We proceed with Figure 1 below in which two points P and
Q are located on the graph of the function y = f ( x) .
y = f (x)
y
P
Δy
Δx
Q
x
Figure 1
The average rate of change of the function in the interval between P and Q is
Δy
which
Δx
represents the slope of the secant line through the points P and Q.
13
Module C09 − Delta Process
We now fix point Q on the curve and allow point P to move along the curve and
approach Q. This is illustrated in Figure 2. Successive locations of P as P → Q are given
as P1 , P2 and P3 . Secant lines through each of P1 , P2 and P3 joining Q are shown.
y = f (x)
y
P
P1
P2
P3
Q
x
Figure 2
As P → Q note that Δx → 0 . See figure 2. Note further that as P → Q ( P approaches
Q as a limit), the secant lines PQ , PQ
, P2 Q and P3Q approach a tangent line to the
1
curve at point Q . At the instant the secant line “becomes” a tangent line (at point Q ), we
no longer are measuring the average rate of change in the function, but now have the
instantaneous rate of change in the function at point Q on the curve. This instantaneous
rate of change may be determined bu calculating the slope of the tangent line to the curve
at Q . This slope becomes the limit of the slopes of the secant lines as P → Q .
Δy
represents the slope of the
Where m represents the slope of the tangent line at Q and
Δx
secant line PQ , using limit notation we write:
Δy
m = lim
Δx → 0 Δx
dy
This limit in calculus is called the derivative and is denoted by
and is read the
dx
derivative of y with respect to x. Any one of the following symbols is used to indicated
the derivative of y with respect to x:
dy df ( x)
,
, f ′x, Dx y, y ′
dx
dx
dy
"
and y ′ are the most popular in this text.
dx
The notation f ′( x) is useful in evaluating derivatives.
14
Module C09 – Delta Process
Definition: The derivative of a function f ( x) with respect to the independent
variable x is the limit approached by the ratio of Δf ( x) to Δx as Δx → 0 .
Remember: Δf ( x) is the same as Δy .
In symbols,
dy
Δy
= lim
dx Δx →0 Δx
or
dy
Δf ( x)
= lim
Δ
x
→
0
dx
Δx
In less technical language, the derivative of y with respect to x is the instantaneous rate of
change of y with respect to x at the instant x.
The derivative of a function is found using the delta process, and adding the fourth step
Δy
of evaluating the limit of the general formula for
as Δx → 0 through substitution
Δx
(see module 6).
Figure 3 below illustrates the relationship between Δx and dx , and between Δy and dy .
Observe Δx = dx , and Δy ≠ dy . The line PD is a tangent line to the point P.
y
Q(x +Δ x, y + Δy)
y = f (x)
Δy
D
P(x,y)
dy
θ
dx = Δ x
C
x
Figure 3
15
Module C09 − Delta Process
EXAMPLE 1
Determine the derivative of y with respect to x when y = 2 x 2 + 3 using the 4-step delta
process.
Δy
, then take the limit of this expression as
Remember: We wish to determine the ratio
Δx
Δx → 0 .
SOLUTION:
STEP 1: Replace x by x + Δx and y by y + Δy in the original function (and expand if
necessary).
y + Δy = 2( x + Δx) 2 + 3
y + Δy = 2 x 2 + 4 x ⋅ Δx + 2 ⋅ Δx 2 + 3
STEP 2: Subtract the original function from the result of STEP 1.
y + Δ y = 2 x 2 + 4 x ⋅ Δ x + 2 ⋅ Δx 2 + 3
−( y
= 2 x2
+ 3)
Δy = 4 x ⋅ Δx + 2 ⋅ Δx 2
STEP 3: Divide both sides by Δx .
Δy 4 x⋅Δx + 2⋅Δx2
=
Δx
Δx
Δy
= 4 x + 2 ⋅ Δx
Δx
STEP 4: Evaluate the limit of both sides of the equation as Δx → 0 (LHS by definition,
RHS by substitution).
dy
lim Δy
dx = Δx →0 Δx
dy
dx
dy
dx
Question: What does
(4 x + 2 ⋅ Δx)
= Δlim
x →0
= 4x + 0
= 4x
dy
= 4 x represent geometrically?
dx
Answer: It represents the equation of the slope of the tangent line to the curve of at any
point on the curve of the graph of y = 2 x 2 + 3 .
16
Module C09 – Delta Process
EXAMPLE 2
For y = 3 x + 4 , find
dy
.
dx
SOLUTION:
STEP 1: Substitution.
y + Δy = 3( x + Δx) + 4
y + Δy = 3 x + 3Δx + 4
STEP 2: Subtraction.
y + Δy = 3x + 3Δx + 4
−( y
= 3x
+ 4)
Δy = 3Δx
STEP 3: Division.
Δy
Δx
Δy
Δx
= 3ΔΔxx
= 3
STEP 4: Limit as Δx → 0 .
Δy
Δy
= lim
= lim 3 = 3
Δx Δx →0 Δx Δx →0
NOTE: The function y = 3 x + 4 represents a straight line. The slope of a straight line is
constant.
17
Module C09 − Delta Process
EXAMPLE 3
Find the derivative of y = 13 using the delta process.
x
SOLUTION:
STEP 1: Replace y by y + Δy and x by x + Δx in the original function.
y + Δy =
1
(2)
( x +Δx)3
STEP 2: Subtract the original function from (2).
1
( x +Δx)3
−( y
= 13 )
x
1
Δy =
− 1
x +Δx3 x3
y + Δy =
STEP 3: Combine the RHS of equation (3) with a LCD of x 3 ( x + Δx)3 to get:
Δy =
x3 −( x +Δx)3
x3 ( x +Δx)3
3
3
2
x2 −Δx3
Δy = x − x −3 x3 Δx −3 xΔ
x ( x +Δx)3
STEP 4: Divide through by Δx .
Δy −3 x2Δx −3 xΔx2 −Δx3
=
Δx
Δx⋅x3 ( x +Δx)3
2
x2
= −3 x 3−3 xΔx −Δ
3
x ( x +Δx)
STEP 5: Take limits as Δx → 0 .
dy −3 x2
= 6
x
dx
= −34
x
Thus the first derivative of y = 13 is − 34 .
x
x
18
Module C09 – Delta Process
Experiential Activity Four
Using the 4-step delta process, find the derivative of the following:
PART A
1. y = 2 x 2
5. y = x 2 + 2 x − 3
2. y = 3 x 2 + 4
6. y = x12
3. y = 6 x + 8
7. y = x
4. y = 4 − 3 x − x 2
PART B
8.
y = −12
x+2
11.
A = 4π r 2
9.
y = 3−x2 x
12.
y = 162
x
10.
A = π r2
3
13. V = 43 π r
Experiential Activity Four Answers
PART A
1.
2.
3.
4.
dy
dx
dy
dx
dy
dx
dy
dx
= 4x
5.
= 6x
6.
=6
7.
dy
= 2x + 2
dx
dy
= − 23
x
dx
dy
=1
dx
= −3 − 2 x
PART B
8.
9.
10.
dy
= 12
dx ( x + 2)2
dy −3
=
dx x2
dA
= 2π r
dr
= 8π r
11. dA
dr
dy −32
= 3
12.
dx
x
= 4π r 2
13. dV
dr
19
Module C09 − Delta Process
REVIEW
1. The ratio
dy
gives and average rate of change in y with respect to x.
dx
2. The limit of the ratio
Δy
as Δx → 0 gives an instantaneous rate of change of
Δx
y with respect to x.
Δy
” is a unitized package. Taken as a whole it gives us the
Δx
dy
.
derivative of y with respect to x, represented by the symbol
dx
4. The derivative is also the instantaneous rate of change.
3. The notation “ lim
Δx → 0
5. The 4-step delta process provides an algebraic means of determining the derivative of
a function.
6. Basic to the definition of the derivative and to the delta process is the concept of the
limit.
7. As Δx → 0 then Δy → 0 as well. We are concerned with the limit of the ratio
Δx → 0 , NOT when Δx = 0 .
Δy
as
Δx
Δy
is “the equation for the slope of the
Δx
secant line joining two points on a curve”.
8. The geometric interpretation of the ratio
9. The “dependent variable” is always differentiated with respect to the “independent
variable”. For example, in the formula A = π r 2 , the derivative notation would be
dA , since the numerator is always the dependent variable and the denominator is
dr
always the independent variable.
20
Module C09 – Delta Process
REVIEW EXERCISE
1. Given y = 2 x 2 + 3 x − 2 , find:
a) Δy from first principles as x changes from 1 to 1.50.
Δy
b)
from first principles as x changes for 1 to 1.50.
Δx
dy
c)
using the 4-step delta process.
dx
2. Given s = −t 2 − 4t + 1 , find:
a) Δs from first principles as t changes from 2 to 2.1.
b) Δs from first principles as t changes from 2 to 2.1.
Δt
ds
using the 4-step delta process.
c)
dt
Answers for Review Exercise
1.
a) 4
2.
a)
−0.81
b) 8
b) −8.1
c) 4 x + 3
c)
−2t − 4
21
Module C09 − Delta Process
Practical Application Activity
Complete the Delta Process assignment in TLM.
Summary
This unit introduces you to the fundamental process for calculus and introduced you to
the concept of the derivative.
In later modules you will learn how to find the derivative of a function by way of
formulae.
22
Module C09 – Delta Process
List for Module C09
1.
6 x 2 + 6 xΔx + 6 x + 3Δx − 4 + 2Δx 2
2.
12 x 2 + 12 xΔx − 6 + 4Δx 2
3.
−12 x 2 − 12 xΔx − 6 − 4Δx 2
4.
3x 2 + 3 xΔx + 6 x + 3Δx + 4 + Δx 2
5.
6 x 2 + 6 xΔx − 6 x − 3Δx + 4 + 2Δx 2
6.
9 x 2 + 9 xΔx + 3Δx 2
7.
3x 2 + 3 xΔx − 8 x − 4Δx + Δx 2
8.
9 x 2 + 9 xΔx + 4Δx + 8 x + 4 + 3Δx 2
9.
12 x 2 + 12 xΔx − 24 x − 12Δx + 4Δx 2
10.
−12 x 2 − 12 xΔx + 4 x + 2Δx − 4Δx 2
11.
3x 2 + 24 x + 3xΔx + 12Δx + Δx 2
12.
3x 2 − 24 x + 3xΔx − 12Δx + Δx 2
13.
−12 x 2 + 12 xΔx + 6 + 4Δx 2
14.
9 x 2 + 9 xΔx + 5Δx 2
15.
12 x 2 − 12 xΔx + 6 − 4Δx 2
23
Module C09 − Delta Process