Calculus Module C09. Delta Process. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

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1 Calculus Module C09 Delta Process Copyright This publication The Northern Alberta Institute of Technology 00. All Rights Reserved. LAST REVISED April, 009

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3 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 Eplorer 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. 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. 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 eplain how this is accomplished. Module C09 Delta Process

4 OBJECTIVE ONE When you complete this objective you will be able to Find the average rate of change in a function from first principles. Eploration Activity REVIEW OF FUNCTIONAL NOTATION In defining the derivative, the functional notation of Module is convenient. Briefly in review, if f ( ) = +, then as an eample: f () = + = 4 + = 6 and f ( ) = ( ) + ( ) = 0 In calculus a change or increment of a variable is written Δ and is read "delta ". This Δ indicates a change in, Δ s a change in distance s, Δ t a change in time t, and so on for other variables. For convenience, the quantities ( Δ ), ( Δ ), are written Δ, Δ, where it is understood that the Δ portion is a single unit and is being squared, cubed, and so on. So, for f ( ) = + let s replace with + Δ to get: f ( +Δ ) = ( +Δ ) + ( +Δ ) = ( + Δ +Δ ) + ( +) = + Δ +Δ + + = + + Δ +Δ +Δ, place the higher powers of Δ at the end of the epression. EXAMPLE If f ( ) SOLUTION: = +, find ( ) f + Δ. f ( +Δ ) = ( +Δ ) + ( + ) = 4+ 4Δ +Δ + +Δ = 6+ 5Δ +Δ NOTE: If smaller. Δ is a small change in, i.e. numerically less than, then Δ is even The increment Δ of a variable may be determined by taking the difference in as it increases or decreases from one value of = to another value of =. Algebraically the increment in is represented as Δ = and the increment in y as Δ y = y y. Module C09 Delta Process

5 AVERAGE RATE OF CHANGE OF A FUNCTION It is possible to use these increments in and y to determine the average rate of change in the function f ( ) as changes from = to another =. This average rate of change in the function with respect to is given by the quotient: Δ = change in y y y = change in EXAMPLE From first principles, find the average rate of change in the function from = to =. y = as changes SOLUTION: From first principles, at = ; y = f( ) = f() = =, and at = ; y = f() = = 9. From Δ = average rate of change in y with respect to and Δ = y y, we have: 9 8 = = = 4 We conclude that as changes by unit, the function changes by 4; or the average rate of change in the function over the interval from = to = is 4:. Module C09 Delta Process

6 Eperiential Activity One. If f( ) =, find f ( + Δ ). If g ( ) = + + 7, find g( Δ ). Find and from first principles, given: a) y = and changes from. to.5 b) y = + 4 and changes from 0. 7 to From first principles, find the average rate of change in the function y = + as changes from = to = From first principles, find the average rate of change in the function as changes from. 0 to. 0. y = + 5 Eperiential Activity One Answers.. Δ +Δ 4Δ + Δ. (a) Δ y = 04., = (b) Δ y = 0. 85, = Δ y 4. = 5. = = Module C09 Delta Process

7 OBJECTIVE TWO When you complete this objective you will be able to Represent average rates of change graphically. Eploration Activity From Objective, Eample we have the function y =. It was found that as changed from to that y changed from to 9. Thus the average rate of change of wrt,, was 4 to. Δ y This is shown graphically below: y y = (,) (,4) (,9) 9 = = 4 O 4 FIGURE From Figure it is seen that the average rate of change of the function with respect to Δ y over the interval is = 4. This also represents the slope of the secant line through points (,) and (,9). Remember: A secant line to a curve cuts the curve in two points. Module C09 Delta Process 5

8 EXAMPLE Find the average rate of change in the function y = 4 as changes from = to =. Interpret the results graphically. SOLUTION: From first principles, given y= 4, = and = : at and at =, y = 4() = 7 =, y = 4() = From Δ y y y = = 7 Δ y = 4 = 4 we have Graphically we get: y (,) (,7) y = = = O 6 Module C09 Delta Process

9 EXAMPLE Find the average rate of change in the function y = as changes from = to = 6 :. from first principles, then. interpret the results graphically SOLUTION:. From first principles, given y = =, = and = 6 : At =, y = = at = 6, y = = 6 From = average rate of change in y with respect to, and Δ Δ y y y =, we have Δ = = 0 = 6 5 Δ y =. Interpreting graphically, we have: y 6 (,) 0 = = = (6,) 8 y = NOTE: represents the slope of the secant line through the points ( ), and ( 6, ) on the curve of the function. From Figure, the ratio gives the average rate of change of y with respect to. Module C09 Delta Process 7

10 Eperiential Activity Two. Sketch y = +. Find Δ y as changes from to... Sketch y = +. Find Δ y as changes from to.5.. Find the average rate of change in the function from to 4. y = + + as changes 4. Find the average rate of change in the function y = 4 as changes + from to 5. Interpret the results graphically. Eperiential Activity Two Answers Left to student to complete. 8 Module C09 Delta Process

11 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. Eploration Activity THE DELTA PROCESS Let us now develop the general formula for given y =. We change by some increment. We use the general increment Δ. The new value of is then +. Using this new value for in the original equation, we get a new value for y which can be represented by y+. From y = the new equation becomes: y+ = ( +) Epanding: y+δ y= + Δ + Subtracting the original function, y = y+δ y = + Δ + ( y = ) Δ y= Δ +Δ, to find the epression for, we get To obtain the ratio which we are after, divide both sides of the equation by Δ : = Δ +Δ which reduces to Δ y = +Δ, which is the desired result. This general formula = + says that for the specific function, y =, the ratio for the average rate of change will always be equal to twice the initial value of plus the increment in ". NOTE: In TML answer format always put Δ raised to a power greater than one at the end of the epression, i.e. treat Δ,Δ etc. the same as in your tet. Module C09 Delta Process 9

12 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:. Replace with +Δ and replace y with y+ Δ y in the original function, and epand if necessary.. Subtract the original function. Divide both sides of the result by Δ. Now we have the ratio Δ.. The epression shows the rule for finding of the particular function involved. Observe that the average rate of change,, can also be written in general form as: Δ Δ f( ) f( +) f( ) = =, which is a mathematical summary of the delta process where: STEP # is represented by f ( + Δ ), STEP # is represented by f ( + ) f( ), and STEP # is represented by f ( +) f( ) EXAMPLE Find the average rate of change in the function y = as changes from = to = 6 from the general formula using the delta process. SOLUTION: From the delta process given y = : STEP : (substitution): y +Δ y = +Δ 0 Module C09 Delta Process

13 STEP : (subtracting the original function): y+δ y = +Δ ( y = ) Δ y = + and combining terms with a common denominator, ( ) ( +) Δ y = ( +) and simplifying, Δ y = ( +) OR Δ y = ( +) STEP : (dividing both sides by we obtain = ( +) Δ ): = Δ ( +) Now evaluating the formula = Δ ( +) Δ = 6 = 5, we have: over the interval where = and = = Δ (+ 5) 6 Δ y therefore = Module C09 Delta Process

14 Eperiential Activity Three. Using the delta process find Δ y and for: a. y = and changes from. to.5 b. y = + 4 and changes from 0.7 to Since the average velocity, v = Δs, find v given: Δ t a. s= t +5 and t changes from s to s. b. s = t + 5t and t changes from s to 4 s.. Determine the general formula of a. y = by the delta process for each of: b. y = Find the average rate of change in the function y = 4 using the delta process Find by using the delta process for the function y = and changes from 6 to 7. Eperiential Activity Three Answers. a) 0.4; b) 0.85; a) 5 b) 7. a) + 6 b) +Δ ( + )( + +) 5. 4 Module C09 Delta Process

15 OBJECTIVE FOUR When you complete this objective you will be able to Use the delta method to find derivatives. Eploration 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 becomes a means of determining instantaneous rate of change. We proceed with Figure below in which two points P and Q are located on the graph of the function y = f( ). y y = f () P Q Figure The average rate of change of the function in the interval between P and Q is represents the slope of the secant line through the points P and Q. which Module C09 Delta Process

16 We now fi point Q on the curve and allow point P to move along the curve and approach Q. This is illustrated in Figure. Successive locations of P as P Q are given as, and. Secant lines through each of, and P joining Q are shown. P P P P P y y = f () P P P Q P Figure As P Q note that 0. See figure. Note further that as P Q ( P approaches Q as a limit), the secant lines PQ, PQ, PQ and PQ 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 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. Where m represents the slope of the tangent line at Q and represents the slope of the secant line PQ, using limit notation we write: m = lim Δ 0 This limit in calculus is called the derivative and is denoted by and is read the d derivative of y with respect to. Any one of the following symbols is used to indicated the derivative of y with respect to : df ( ),, f, D y, y d d and y are the most popular in this tet. d The notation f ( ) is useful in evaluating derivatives. 4 Module C09 Delta Process

17 Definition: The derivative of a function f ( ) with respect to the independent variable is the limit approached by the ratio of Δ f ( ) to Δ as 0. Remember: Δ f ( ) is the same as. In symbols, = lim d 0 or Δf ( ) = lim d 0 In less technical language, the derivative of y with respect to is the instantaneous rate of change of y with respect to at the instant. 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 as Δ 0 through substitution (see module 6). Figure below illustrates the relationship between Δ and d, and between Observe Δ = d, and. The line PD is a tangent line to the point P. Δ y and. y Q( +Δ, y + ) y = f () P(,y) Figure θ d = Δ D C Δ y Module C09 Delta Process 5

18 EXAMPLE Determine the derivative of y with respect to when y = + using the 4-step delta process. Remember: We wish to determine the ratio, then take the limit of this epression as Δ 0. SOLUTION: STEP : Replace by + Δ and y by y+ Δ y in the original function (and epand if necessary). y+δ y= ( +Δ ) + y+δ y= + 4 Δ + Δ + STEP : Subtract the original function from the result of STEP. y+δ y = + 4 Δ + Δ + ( y = + ) Δ y = 4 Δ + Δ STEP : Divide both sides by = 4 Δ + Δ y = 4 + Δ Δ. STEP 4: Evaluate the limit of both sides of the equation as 0 (LHS by definition, RHS by substitution). d = lim 0 d d lim (4 + Δ ) = 0 = = 4 Question: What does = 4 represent geometrically? d 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 = +. 6 Module C09 Delta Process

19 EXAMPLE For y = + 4, find SOLUTION: STEP : Substitution. d. y+ = ( +Δ ) + 4 y+δ y = + Δ + 4 STEP : Subtraction. y+δ y= + Δ + 4 ( y = + 4) Δ y= Δ STEP : Division. = Δ = STEP 4: Limit as 0. Δ y = lim = lim = 0 0 NOTE: The function constant. y = + 4 represents a straight line. The slope of a straight line is Module C09 Delta Process 7

20 EXAMPLE Find the derivative of y = using the delta process. SOLUTION: STEP : Replace y by y+ Δ y and by + Δ in the original function. y+δ y= ( +) () STEP : Subtract the original function from (). y+δ y= ( +) ( y = ) Δ y = + STEP : Combine the RHS of equation () with a LCD of ( ) +Δ to get: Δ y = Δ y = ( +) ( +) ( +) STEP 4: Divide through by Δ. Δ Δ Δ Δ Δ Δ +Δ = Δ Δ ( + ) y = ( ) STEP 5: Take limits as 0. = 6 d = 4 Thus the first derivative of y = is. 4 8 Module C09 Delta Process

21 Eperiential Activity Four Using the 4-step delta process, find the derivative of the following: PART A. y = 5. y = +. y = y =. y = y = 4. y = 4 PART B 8. y = +. A= 4π r 9. y =. y = 6 0. A= π r. V = 4 π r Eperiential Activity Four Answers PART A. 4 d = 5. d = +. 6 d = 6. d =. 6 d = 7. d = 4. d = PART B 8. = d ( + ). da = 8π r dr 9. = d. = d 0. da = π r. dv = 4π r dr dr Module C09 Delta Process 9

22 REVIEW. The ratio d gives and average rate of change in y with respect to.. The limit of the ratio y with respect to. as Δ 0 gives an instantaneous rate of change of. The notation lim is a unitized package. Taken as a whole it gives us the 0 Δ derivative of y with respect to, represented by the symbol d. 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 0 then 0 as well. We are concerned with the limit of the ratio 0, NOT when Δ = 0. as 8. The geometric interpretation of the ratio secant line joining two points on a curve. is the equation for the slope of the 9. The dependent variable is always differentiated with respect to the independent variable. For eample, in the formula A= π r, the derivative notation would be da, since the numerator is always the dependent variable and the denominator is dr always the independent variable. 0 Module C09 Delta Process

23 REVIEW EXERCISE. Given y = +, find: a) from first principles as changes from to.50. b) from first principles as changes for to.50. c) using the 4-step delta process. d. Given s = t 4t+, find: a) Δ s from first principles as t changes from to.. b) Δs from first principles as t changes from to.. Δt c) ds dt using the 4-step delta process. Answers for Review Eercise. a) 4. a). 08 b) 8 b). 8 c) 4 + c) t 4 Module C09 Delta Process

24 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. Module C09 Delta Process

25 List for Module C Δ Δ Δ 6 4Δ + Δ + 6+ Δ + 4+Δ Δ + 4+ Δ 9 + 9Δ + Δ + 8 4Δ +Δ 9 + 9Δ + 4Δ Δ + 4 Δ + 4Δ Δ Δ + 4+ Δ + Δ +Δ 4+ Δ +Δ + Δ Δ 9 + 9Δ + 5Δ Δ + 6 4Δ Module C09 Delta Process

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