tom.h.wilson Dept. Geology and Geography West Virginia University

Transcription

1 tom.h.wilson Dept. Geology and Geography West Virginia University

2 Objectives for the day 8.13 and 8.14 due next time In-class digital approach to differentiation including simple in-class exercise. Excel extensions of problems 8.13 and 8.14 starting with basic layout. More practice with derivatives see review worksheet

3 Some dates to highlight 8.13 and 8.14 due next time Next class address questions concerning the Excel version of problems 8.13 and Excel problems due next Thursday. Look over problems 8.16, 8.17 and 8.18 for discussion/assignment Next week we will have some in-class integration efforts to work through. Begin reading Chapter 9. We should wrap up derivatives next week.

4 Text problems 8.13 and 8.14 Last opportunity for questions Do problems as noted in text and provide additional detail as noted in handout.

5 Things to focus on for txt problems 8.13 see handout from last time As you answer these questions go beyond answers in the text as noted at right.

6 Provide details of your analysis as noted below and in class Any final questions? As you answer these questions provide details in solutions for (i), x at X/2 (ii), the derivative (iii) and salinity estimation at points on either side of the basin center. Due next Tuesday

7 Refer back to slides from Tuesday for additional perspectives These and more are there for your review.

8 Today we will go through an Excel exercise designed to provide graphical illustration of the exponential differentiation rule. dae dx cx de x e x dx cx d( cx) Ae dx cae cx or alternatively d e o dz z doe dz e The second equations above illustrate combined use of the rule for differentiating natural exponential functions and the chain rule. o z z o e z d( z ) 1 oe dz z

9 Computer exercise: Using the computer to calculate the derivative of a natural exponential function slope = lim z 0 e z z e e z z z d e z dz z z

10 Let s go over to pages 122 and 123 and calculate the derivative in discrete form Bring up Excel Your name in the title These slides are for general reference and we may do things a little differently in lab

11 We could make the problem a little more complex by introducing a o and a compaction coefficient c or d c0e dz cz

12 d coe dx cz Evaluate explicitly for 0 =0.5 and c = 0.5km -1 d dx

13 Note that the discrete approximation and derivative are very close in value have small error The error maintains a constant fractional or percent difference

14 Hand in a plot before leaving for 5 points put your name in the title

15 Follow up on derivatives of logs. The slope of the natural log. Slides concept review for your reference Note that the slopes (derivative) near zero is positive and very large. Red dashed line drawn off to the side indicates slope is positive and nearly infinite.

16 Think about the relationship between the two Again near zero the derivative (slope) has a very high positive value. With the slope decreasing very quickly over the range 0 to 1 and is always positive. Lastly note that this has the form 1/x

17 1/x plotted on a lgorithmic scale 1 ln(1/ x) ln( x ) ln( x) Not that in taking the ln(1/x) we simply get the negative of the ln(x) ln( x)

18 d(lny)/dx=(1/y)dy/dx (using the chain rule) y ln( x 2 ) dy dx d(ln x) 1 dx x dy dh. dh dx How about d 2 cx (ln( x e )) dx Take a couple minutes and work through this.

19 Other bases The derivative of logarithmic functions Given > y ln(x) dy 1 dx x For the more general case where the base is unequal to e y log a ( x) d log a( x) log a( e) dx x The power you raise the base 10 to to get 10 is log 10 (e)=0.434

20 Another way to look at this is to recall the earlier definition for logs with arbitrary base Recall that for arbitrary base, we have log x 10 log b x or equivalently log10 b ln x ln b dlog b x d(ln x / ln b) 1 d ln x so dx dx ln b dx for b=10, ln10 x x x So we treat derivatives of log 10 the same way we do the ln, but have to multiply by the factor

21 Log and ln 4 d log 10( x) 0.43 dx x d ln( x) 1 dx x y= log(x) or ln (x) log(x) ln(x) x The factor 0.43 (log 10 e) suggests the slope on the log 10 curve will be less than that of the ln curve, which as the graph indicates it should be

22 Derivatives of exponential functions with base unequal to e. Slides for quick review The derivative of exponential functions In general for ax y e dy d( ax ) dx dx ax e ax ae x y a If express a as e n n nx so that y e x e then dy dx d dx e nx ne nx

23 dy dx d dx e nx ne nx Since in general x nx a e and n ln(a) dy dx ln( a). x a a can be thought of as a general base. It could be 10 or 2, etc.

24 Locating the maxima and minima of a function using derivatives We ll discuss a problem in the text in more detail next time, but for now, recall from graphical considerations alone, that the slope or derivative will be zero when a function is at a minimum. The tangent, derivative, slope at a point, in this case, the maximum, 0 slope is zero. Note that the slope coming from negative to positive x is continually decreasing in value. The variation in the slope is the second derivative and it is negative across the maximum in a function.

25 Locating the maxima and minima of a function using derivatives The function slopes decrease across a maximum. In other words, the slope of a function defining the slope will be negative. zero Take the derivative and second derivative of y=-2x 2 +2x+20 y =-4x+2 = 0 at x=0.5 y =-4 Thus, the second derivative is negative and a negative second derivative indicates the function has a maximum at that point the point of zero slope.

26 At a function minimum the second derivative is positive The function slopes increase across a minimum. In other words, the slope of a function defining the slope will be positive. zero Take the derivative and second derivative of y=2x 2 +2x+20 y =4x+2 = 0 at x=-0.5 y =4 The second derivative is positive and indicates this is a minimum

27 Application - consider the energy expended during normal faulting presented by Waltham See pages of Waltham e t m L We assume that the energy expended to produce normal fault displacement is proportional to the length of the fault surface (L) and fault displacement (m) Such that w=lm.

28 Use trig to Express length of the fault cut (L) and displacement (m) in terms of layer thickness (t) and fault gap (e), respectively. t m m e cos e cos t sin L e m L t L t sin

29 Energy Lm Look over this problem in the text. We will discuss further next time w Lm et cossin Work should be proportional to the displacement m and length of cut L Is a proportionality constant When the layer is placed in tension it will break in response to a certain minimum amount of work. The question is what is the angle required to minimize the energy expended to fracture the layer and produce the observed displacement. Finding the minimum requires taking 1 st and 2 nd derivatives of the above expression 2 dw & d w 2 d d Go to setup for Excel problems

30 Review relationships of function minima and maxima to 1 st and 2 nd derivatives The function y

31 The form of the expression suggests use of the quotient rule / For the first derivative d f g given w dw d let f=et and g = cossin. To organize our computation define the various terms required to use the quotient rule f et g cossin et cossin d df 0 d dg d(cossin ) d d The product rule is required to solve dg/d.

32 Application of the product rule dg d sin d cos cos sin d d d 2 2 cos sin Substitute into the formula for the quotient rule and solve d f df dg g g f d d 2 d g to get dw d et sin cos cos sin 2 2 For this to be 0, sin cos must be equal to 0, which implies that =45 o.

33 Is the point where the slope is 0 a maximum or minimum? derivative((sin(x)^2-cos(x)^2)/((cos(x)^2)(sin(x)^2))) Where is 45 degrees?

34 Is the point where the slope is 0 a maximum or minimum? From the text d w d 2 et(sin cos ) cos sin Evaluated at 45 o yields 2 d w 2 d 2 et( ) et( 1 ) =8et The result is positive since, e and t are positive.

35 The second derivative is positive so the work is a minumum d w d 2 et(sin cos ) cos sin =8 45 o

36 The second derivative is positive so the work is a minumum While this is a good example, the fault surface generally forms an angle of about 30 o with S v. =8 45 o

37 A normal fault forms in response to stress differences such that S v >S Hmax >S hmin S v

38 Failure along pre-existing fault and fracture surfaces Green and blue faults and fractures are stable

39 Faults are steeper than 45 o and the angle with S v, ~30 o S v e t m L As we get into integral calculus think about how you could calculate S v

40 We ll finish today with a quick look at Excel versions of problems 8.13 and 8.14 and basic setup Only one value of Smax used in this exercise Setup problem as illustrated in your handout. Only one value of Smax used! Make that change in your handout

41 Problem 8.14 setup

42 See handout The setup is laid out for you in the handout and in the above. Take a few minutes to set up the problem in your excel file.

43 Derivative review worksheet look over for next time

44 Evaluate the derivative see today s handout

45 These and more see today s in-class worksheet See question 8.8 in the text (page 148)

46 Read Chapter 9 Integral calculus derivatives in reverse coming soon Integral calculus d

47 Some dates to highlight Text problems 8.13 and 8.14 due next time Due date for Excel problems next Thursday. Bring questions to class about problems 8.16, 8.17 and 8.18 next Tuesday Begin reading Chapter 9. We should wrap up derivatives and begin discussion of integrals next week.