module - introduction to algebra - part...

LAST REVISED April, 2009

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Calculus Module C09

Delta Process

Module C09 − Delta Process

1

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. Find the average rate of change in a function from first principles. 2. Represent average rates of change graphically. 3. Find the average rate of change of a function by using the delta process. 4. 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.

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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 2( )f x x x= + , then as an example:

2(2) 2 2 4 2 6f = + = + = and 2( 1) ( 1) ( 1) 0f − = − + − = In calculus a change or increment of a variable x is written xΔ and is read "delta x". This

xΔ indicates a change in x, sΔ a change in distance s, tΔ a change in time t, and so on for other variables. For convenience, the quantities 2)( xΔ , 3)( xΔ , are written 2xΔ ,

3xΔ , where it is understood that the xΔ portion is a single unit and is being squared, cubed, and so on. So, for ( )f x = 2x x+ let’s replace x with x x+ Δ to get:

2 2 2( ) ( ) ( ) ( 2 ) ( )f x x x x x x x x x x x x+ Δ = + Δ + + Δ = + ⋅Δ + Δ + + Δ2 22

x x x x x x= + ⋅Δ + Δ + + Δ

2 22

x x x x x x= + + ⋅Δ + Δ + Δ , ← place the higher powers of xΔ at the end of the expression.

EXAMPLE 1 If 2( )f x x= + x , find (2 )f x+ Δ . SOLUTION:

2(2 ) (2 ) (2 )f x x+ Δ = + Δ + + Δx 24 4 2x x x= + Δ + Δ + + Δ 26 5 x x= + Δ + Δ

NOTE: If xΔ is a small change in x, i.e. numerically less than 1, then 2xΔ is even smaller. The increment xΔ of a variable may be determined by taking the difference in x as it increases or decreases from one value of 1x x= to another value of 2x x= . Algebraically the increment in x is represented as 2 1x x xΔ = − and the increment in as . y 2 1y y y−Δ =

2


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 1x x= to another 2x x= . This average rate of change in the function with respect to x is given by the quotient:

yx

ΔΔ

= 2 1

2 1

y ychange in ychange in x x x

−=

−

EXAMPLE 1 From first principles, find the average rate of change in the function 2y x= as x changes from 1x = to .

3x =

SOLUTION: From first principles, at and at 21 ( ) (1) 1x y f x f= ; = = = = ,1 923 (3) 3x y f= ; = = = .

From yx

ΔΔ

= average rate of change in y with respect to x

and yx

ΔΔ

= 2

2 1

y y1

x x−−

, we have:

9 1 8 43 1 2

yx

Δ −= = =

Δ −

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


Experiential Activity One 1. If , find 2( ) 1f x x= − (1 )f x+ Δ

2. If , find 2( ) 3 2 7g x x x= + + (2 )g x− Δ

3. Find and yΔ yx

ΔΔ

from first principles, given:

a) 2 3y x= − and x changes from 3.3 to 3.5 b) and x changes from 02 4y x x= + 7. to 0 85.

4. From first principles, find the average rate of change in the function as x changes from to

3 1y x= +3x = 6x = .

5. From first principles, find the average rate of change in the function

as x changes from

2 3 5y x x= − +3 03. to 3 02. .

Experiential Activity One Answers 1. 22 x xΔ + Δ 2. 223 14 3x x− Δ + Δ

3. (a) 0 4 2yy xΔΔ = . , =Δ (b) 0 8325 5 55yy x

ΔΔ = . , = .Δ

4. 3yx

Δ =Δ

5. 0 0305 3 050 01yx

Δ − .= =Δ − . .

4


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 2y x= . It was found that as x changed from 1 to 3 that changed from 1 to 9. Thus the average rate of change of

wrt

y y

x , yx

ΔΔ , was 4 to 1.

This is shown graphically below:

5

FIGURE 1 From Figure 1 it is seen that the average rate of change of the function with respect to x

over the interval is 41

yx

Δ =Δ . This also represents the slope of the secant line through

points (1,2) and (3,9). Remember: A secant line to a curve cuts the curve in two points.

x

O 1

y = x2

1234

y

56789

10

Δy 9 −1 Δx 3 −1 = =

(2,4)

(1,1)

(3,9)

Δy

Δx

2 3 4−1−2−3−4


EXAMPLE 1 Find the average rate of change in the function 4 1y x= − as x changes from 1x = to

. Interpret the results graphically. 2x = SOLUTION:

From first principles, given 14 1 x 1

6

y x= − , = and 2 2x = : at and at

1 11 4(1) 1x y= , = − = 3

2 22 4(2) 1 7x y= , = − =

2 12 1

y yyx xx−Δ = −ΔFrom we have

7 32 1

yx

Δ −=Δ −

41

yx

Δ =Δ

4=

Graphically we get:

xO 1

y = 4x − 1

1 2 3 4

y

5 6 7 8

(1,3)

(2,7)

Δy

Δx

2 3

Δy 7 – 3 4Δx 2 – 1 1 = =


EXAMPLE 2

12y x= as x changes from 1x = to : 6x =Find the average rate of change in the function

1. from first principles, then 2. interpret the results graphically

SOLUTION: 1. From first principles, given 12 121y = = , 1x = 1 and 2 6x = :

At 1 1

121 11x y= , = = 2

at 2 2126 26x y= , = =

From yx

ΔΔ = average rate of change in y with respect to x, and

2 12 1

y yyx xx−Δ = −Δ , we have

102 12 256 1

yx

Δ −−= = =Δ − −

2yx

Δ = −Δ

2. Interpreting graphically, we have:

−8

y

7

NOTE: yx

ΔΔ represents the slope of the secant line through the points (1 12), and (

on the curve of the function. From Figure 2, the ratio

6 2),

yx

ΔΔ gives the average rate of

change of y with respect to x.

x

(1,12)

Δy Δx

16

12

8

4

4

−8

−12

−6 −4 −2 2 4 6 8

Δy 2 −12 −10 Δx 6 – 1 5 = = = −2

(6,2)

12 xy =


Experiential Activity Two 1. Sketch . Find 2 3y x= + yΔ as x changes from 1 to 1.3. 2. Sketch . Find 3 2y x= + yΔ as x changes from 2 to 2.5. 3. Find the average rate of change in the function 2 3y x x 1= + + as x changes

from 3 to 4. 4. Find the average rate of change in the function 24

3y x= + as x changes

from 1 to 5. Interpret the results graphically.

Experiential Activity Two Answers Left to student to complete.

8


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

9

Let us now develop the general formula for yΔ 2y x given xΔ = . We change x by some

increment. We use the general increment xΔ . The new value of x is then x x+ Δ

y y+ Δ 2y x=2( )y y x+ Δ

2 22y y x x x x+ Δ = + ⋅Δ + Δ

2x y

. Using this new value for x in the original equation, we get a new value for y which can be represented by . From the new equation becomes: x= + Δ Expanding:

Subtracting the original function, , to find the expression for y = , we get Δ

2 2

2

2y y x x x+ Δ = + Δ + Δx

( )y x− =22y x x xΔ = ⋅Δ + Δ

yx

ΔΔ x which we are after, divide both sides of the equation by To obtain the ratio Δ :

22y x x xx x

+ΔΔ ΔΔ ⋅Δ=

which reduces to

2y x xxΔ = + ΔΔ , which is the desired result.

This general formula 2yx

ΔΔ x x= + Δ 2y x= , the ratio says that for the specific function,

yx

ΔΔ for the average rate of change will always be equal to “twice the initial value of x

plus the increment in x ". NOTE: In TML answer format always put xΔ raised to a power greater than one at the end of the expression, i.e. treat 2 3x xΔ , Δ etc. the same as in your text.


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 x+ Δ and replace with y y y+ Δ in the original function, and

expand if necessary. 2. Subtract the original function. Divide both sides of the result by xΔ . Now we have

the ratio yx

ΔΔ .

3. The expression shows the rule for finding yx

ΔΔ of the particular function involved.

Observe that the average rate of change, yx

ΔΔ , can also be written in general form as:

( ) ( ) ( )y f x f x x f xx x x

Δ Δ +Δ −= =Δ Δ Δ ,

which is a mathematical summary of the delta process where:

• STEP #1 is represented by ( )f x x+ Δ ,

• STEP #2 is represented by ( ) ( )f x x f x+ Δ − , and

• STEP #3 is represented by ( ) ( )f x x f xx

+Δ −Δ

EXAMPLE 1

Find the average rate of change in the function 12y x= as x changes from 1x = to 6x = from the general formula using the delta process. SOLUTION:

From the delta process given 12yx

= :

STEP 1: (substitution): 12y y x x+ Δ = +Δ

10


STEP 2: (subtracting the original function):

12

12( )

y yx x

yx

+ Δ =+ Δ

− =

12 12yx x x

Δ = −+ Δ

and combining terms with a common denominator,

12( ) 12( )( )

x x xy x x x− +ΔΔ = +Δ

and simplifying,

yΔ = 12 12 12( )

x x xx x x− − ⋅Δ

+Δ

OR yΔ = 12

( )x

x x x− ⋅Δ+Δ

STEP 3: (dividing both sides by xΔ ):

12 1( )

y xx xx x x

Δ − ⋅Δ= ⋅Δ Δ+Δ

we obtain

12( )

yx x x x

Δ −=Δ +Δ

Now evaluating the formula 12(

yx )x x x

Δ −=Δ +Δ over the interval where 1x = and

, we have:

6 1 5xΔ = − =

12 1261(1 5)

yx

Δ − −= =Δ +

therefore 2yx

Δ = −Δ

11


Experiential Activity Three

1. Using the delta process find yΔ and yx

ΔΔ for:

a. and 2y x= − 3 x changes from 3.3 to 3.5

b. and 2 4y x x= + x changes from 0.7 to 0.85

2. Since the average velocity, sv tΔ= Δ , find v given:

a. and t changes from 2 s to 3 s. 23s t= + 5

2b. 22 5s t t= + − and t changes from 2 s to 4 s.

3. Determine the general formula of yx

ΔΔ by the delta process for each of:

a. 2 6 7y x x= − +

b. 2 4 3y x x= + −

4. Find the average rate of change in the function 243y x= + using the delta process.

5. Find yx

ΔΔ by using the delta process for the function 3 3y x x= − and x changes from

6 to 7.

Experiential Activity Three Answers 1. a) 0.4; 2

b) 0.8325; 5.55 2. a) 15

b) 17 3. a) 2 6x x+ Δ −

b) 2 4x x+ Δ +

4. 24( 3)( 3 )x x x

−+ + +Δ

5. 124

12


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.

The notation for average rate of change yx

ΔΔ becomes a means of determining

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= .

13

The average rate of change of the function in the interval between P and Q is yx

ΔΔ which

represents the slope of the secant line through the points P and Q.

y

x

Δy

Δx

P

Q

y = f (x)

Figure 1


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 are given as , and . Secant lines through each of , and joining Q are shown.

P Q→

1P 2P 3P 1P 2P 3P

14

As note that . See figure 2. Note further that as ( approaches

as a limit), the secant lines , , and approach a tangent line to the curve at point Q . At the instant the secant line “becomes” a tangent line (at point Q ), we no longer are measuring the

P Q→ 0xΔ → P Q→ PQ PQ 1PQ 2P Q 3PQ

average rate of change in the function, but now have the instantaneous rate of change in the function at point on the curve. This instantaneous rate of change may be determined bu calculating the slope of the tangent line to the curve at . This slope becomes the limit of the slopes of the secant lines as .

Q

Q P Q→

Where m represents the slope of the tangent line at Q and yx

ΔΔ represents the slope of the

secant line , using limit notation we write: PQ

0limx

ymxΔ →

Δ=

Δ

This limit in calculus is called the derivative and is denoted by dydx and is read the

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:

( )x

dy df x f x D y ydx dx

′ ′, , , ,

dydx and are the most popular in this text. y′

The notation ( )f x′ is useful in evaluating derivatives.

y

x

P

Q

y = f (x)

Figure 2

P1

P2

P3


15

Definition: The derivative of a function ( )f x with respect to the independent variable x is the limit approached by the ratio of ( )f xΔ to xΔ as 0xΔ → . Remember: ( )f xΔ is the same as . yΔ

In symbols,

0limx

dy ydx xΔ →

Δ=

Δ

or

0

( )limx

dy f xdx xΔ →

Δ=

Δ

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

of evaluating the limit of the general formula for yx

ΔΔ as 0xΔ → through substitution

(see module 6). Figure 3 below illustrates the relationship between xΔ and , and between dx yΔ and . Observe

dyx dxΔ = , and . The line PD is a tangent line to the point P. y dyΔ ≠

D

x

y

P(x,y)

dx = Δ x

y = f (x)

C

dy

Δ y

Q(x +Δ x, y + Δy)

θ

Figure 3


3

EXAMPLE 1 Determine the derivative of y with respect to x when 22y x= + using the 4-step delta process.

Remember: We wish to determine the ratio yx

ΔΔ , then take the limit of this expression as

. 0xΔ → SOLUTION:

STEP 1: Replace x by x x+ Δ and by y y y+ Δ in the original function (and expand if necessary).

2

2 2

2( ) 32 4 2

y y x xy y x x x x+ Δ = + Δ +

+ Δ = + ⋅Δ + ⋅Δ + 3

33)

STEP 2: Subtract the original function from the result of STEP 1.

2 2

2

2 4 2( 2y y x x x xy x+ Δ = + ⋅Δ + ⋅Δ +

− = +

24 2y x x xΔ = ⋅Δ + ⋅Δ STEP 3: Divide both sides by xΔ .

24 2

4 2

y x x xx xy x xx

Δ ⋅Δ + ⋅Δ=Δ ΔΔ = + ⋅ΔΔ

STEP 4: Evaluate the limit of both sides of the equation as 0xΔ → (LHS by definition, RHS by substitution).

dydx = 0

limxΔ →

yx

ΔΔ

= 0lim (4 2 )x

x xΔ →

+ ⋅Δ

dydx = 4 0x +

dydx = 4x

Question: What does 4dy xdx = represent geometrically?

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 22 3y x= + .

16


EXAMPLE 2

For , find 3 4y x= + dydx .

SOLUTION: STEP 1: Substitution.

y + Δyy 4

44)

= 3( ) 4x x+ Δ + y + Δ = 3 3x x+ Δ +

STEP 2: Subtraction.

3 3

( 3y y x xy x+ Δ = + Δ +

− = +

3y xΔ = Δ STEP 3: Division.

yx

ΔΔ = 3 x

xΔΔ

yx

ΔΔ = 3

STEP 4: Limit as . 0xΔ →

0 0lim lim 3 3x x

yyx xΔ → Δ →

ΔΔ = =Δ Δ=

NOTE: The function represents a straight line. The slope of a straight line is constant.

3 4y x= +

17


EXAMPLE 3

Find the derivative of 31y x= using the delta process.

SOLUTION: STEP 1: Replace y by y y+ Δ and x by x x+ Δ in the original function. 3

1(y y )x x+ Δ = +Δ (2)

STEP 2: Subtract the original function from (2).

3

3

1( )1( )

y y x x

y x

+ Δ = +Δ

− =

3 31 1y x x xΔ = −+Δ

STEP 3: Combine the RHS of equation (3) with a LCD of 3 ( 3)x x x+ Δ to get:

yΔ = 3 3

3 3( )

( )x x xx x x− +Δ

+Δ

yΔ = 3 3 2 2 3

3 33 3

( )x x x x x x x

x x x− − Δ − Δ −Δ

+Δ

STEP 4: Divide through by xΔ .

2 23 3

2 23 3

3 3( )

3 3( )

y 3x x x x xx x x x x

x x x xx x x

Δ − Δ − Δ −Δ=Δ Δ ⋅ +Δ− − Δ −Δ= +Δ

STEP 5: Take limits as 0xΔ → .

26

4

3

3

dy xxdx

x

−=

−=

Thus the first derivative of 3

1y x= is 43x− .

18


Experiential Activity Four

Using the 4-step delta process, find the derivative of the following: PART A 1. 22y x= 5. 2 2 3y x x= + −

2. 23 4y x= + 6. 21y x=

3. 6 8y x= + 7. y x=

4. 24 3y x= − − x

PART B

8. 122y x

−= + 11. 24A rπ=

9. 3 2xy x−= 12. 2

16y x=

10. 2A rπ= 13. 343V rπ=

Experiential Activity Four Answers

PART A

1. 4dy xdx = 5. 2 2dy xdx = +

2. 6dy xdx = 6. 32dyxdx = −

3. 6dydx = 7. 1dy

dx =

4. 3 2dy xdx = − −

PART B

8. 212

( 2)dydx x= + 11. 8dA rdr π=

9. 23dy

xdx−= 12. 3

32dydx x

−=

10. 2dA rdr

π= 13. 24dV rdr π=

19


REVIEW

1. The ratio dydx gives and average rate of change in y with respect to x.

2. The limit of the ratio yx

ΔΔ as 0xΔ → gives an instantaneous rate of change of

y with respect to x.

3. The notation “0

limx

yxΔ →

ΔΔ

” is a unitized package. Taken as a whole it gives us the

derivative of y with respect to x, represented by the symbol dydx

.

4. The derivative is also the instantaneous rate of change. 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 then as well. We are concerned with the limit of the ratio 0xΔ → 0yΔ → yx

ΔΔ as

, NOT when 0Δ →x 0xΔ = .

8. The geometric interpretation of the ratio yx

ΔΔ is “the equation for the slope of the

secant line joining two points on a curve”. 9. The “dependent variable” is always differentiated with respect to the “independent

variable”. For example, in the formula 2A rπ= , the derivative notation would be dAdr , since the numerator is always the dependent variable and the denominator is

always the independent variable.

20


REVIEW EXERCISE

21

21. Given , find: 22 3y x x= + −a) from first principles as x changes from 1 to 1.50. yΔ

b) yx

ΔΔ from first principles as x changes for 1 to 1.50.

c) dydx using the 4-step delta process.

2. Given 2 4 1s t t= − − + , find: a) sΔ from first principles as t changes from 2 to 2.1. b) s

tΔΔ from first principles as t changes from 2 to 2.1.

c) dsdt using the 4-step delta process.

Answers for Review Exercise 1. a) 4 2. a) 0 81− .

b) 8 b) 8 1− .

c) 4 3x + c) 2 4t− −

Module C09 – Delta Process 22

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.


List for Module C09 1. 2 26 6 6 3 4 2x x x x x x+ Δ + + Δ − + Δ 2. 2 212 12 6 4x x x x+ Δ − + Δ 3. 2 212 12 6 4x x x x− − Δ − − Δ 4. 2 23 3 6 3 4x x x x x x+ Δ + + Δ + + Δ 5. 2 26 6 6 3 4 2x x x x x x+ Δ − − Δ + + Δ 6. 2 29 9 3x x x x+ Δ + Δ 7. 2 23 3 8 4x x x x x x+ Δ − − Δ + Δ 8. 2 29 9 4 8 4 3x x x x x x+ Δ + Δ + + + Δ 9. 2 212 12 24 12 4x x x x x x+ Δ − − Δ + Δ 10. 2 212 12 4 2 4x x x x x x− − Δ + + Δ − Δ 11. 2 23 24 3 12x x x x x x+ + Δ + Δ + Δ 12. 2 23 24 3 12x x x x x x− + Δ − Δ + Δ 13. 2 212 12 6 4x x x x− + Δ + + Δ 14. 2 29 9 5x x x x+ Δ + Δ 15. 2 212 12 6 4x x x x− Δ + − Δ

23

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