Geology Geomathematics. An introduction to differential calculus. tom.h.wilson

Transcription

1 Geology Geomathematics An introduction to differential calculus tom.h.wilson tom.wilson@mail.wvu.edu Dept. Geology and Geography West Virginia University

2 Developing basic concepts and learning some differentiation rules Developing the relationship of the slope of a function to its derivative illustrated using exponential, trig and polynomial functions. Basic rules: Power rule Sum rule Chain rule Product and quotient rules Exponential rules Some in-class example problems

3 Problems handed out last time - Turn in porosity problem before leaving In the porosity density relationship = 0 e (-z/), assume 0 is 0.4 and is 1.5. What is the slope (or porosity gradient) between depths of 1 and km in this area? What is the slope between 1.4 and 1.6 kilometers? Calculate these slopes out explicitly and sketch your results on the graph of this function on the back of this page. The derivative is just a slope: the slope at a point on the curve or the tangent.

4 Between 1 and km depths, slope =-0.099/km Between 1.4 and 1.6 km depths, slope =-0.098/km z Slope = /z Slope at this point is the derivative The derivative or slope of the tangent line at the depth 1.5 km = /km

5 At.5km, the slope is at 3.5 km the slope is at 0.5, Porosity depth relationship =0.4e -z/ PHI Z

6 Slopes evaluated over z of 0.km about a given depth 0.5 ~ 0.19 z 1.5 ~ 0.1 z 3 ~ 0.04 z Check in-class work You ll find the derivative d o e dz z Evaluation of the derivative at each of these points yields d z o e e dz e e The individual slope calculations provide approximate estimates of the slope at the midpoint. The derivative is exact and also provides an analytical expression that is easily plotted.

7 Slopes of the cosine are a little more obvious

8 0, (at 0.785), -1 (at 1.571), (at.356) 1.0 Cosine function 0.5 -sin() cos () Calculation of these slopes, point-bypoint, outline the negative sine function (radians)

9 For y=x 40 y=x y x

10 Derivative of a constant: i.e. y-a y y=a What s the slope? So derivative of a constant is 0 The y is zero so y/x and / are always 0. x

11 The book works through the differentiation of y = x, so let s try y =x 4. y ( x ) 4 y multiplying that out -- you get... x 4 3 4x 6x ( ) 4x( ) ( ) 3 4

12 y x 4 3 4x 6x ( ) 4x( ) ( ) 3 4

13 y x 4 3 4x 6x ( ) 4x( ) ( ) 3 4 Remember the idea of the and is that they represent differential changes that are infinitesimal - very small. So if is (that s 1x10-4 ) then () = (or 1x10-8 ) () 3 = 1x10-1 and () 4 = 1x So even though is very small, () is orders of magnitude smaller

14 y x 4 3 4x 6x ( ) 4x( ) ( ) 3 4 so that we can just ignore all those terms with () n where n is greater than 1. Our equation gets simple fast y x 4 4x 3 Also, since y =x 4, we have y y 4x 3 and then - 4x 3

15 4x 3 Divide both sides of this equation by to get 3 4x This is just another illustration of what you alrea know as the power rule,

16 which - in general for n y ax is n nax Just as a footnote, remember that the constant factors in an expression carry through the differentiation. This is obvious when we consider the derivative - 1 The shift does not change the slope (derivative). y ax b + =

17 Examining the effects of differential increments in y and x we get the following y y a( x ) b a( x x ) y ( ax b) ax y y ax Where y=ax +b b a( x)

18 Alternatively d ax b dax db = dax = =ax ( ) Distribute the d/ operator =0 since b is a constant. Line with slope a and 0 intercept.

19 Don t let negative exponents fool you. If n is -1, for example, we still have n nax 1 or just ax

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21 Given the function - what is y( x) f ( x) g( x) We just differentiate f and g individually and take their sum, so that? df dg Distribute the d/ operator

22 Take the simple example y ( x c) ( ax 4 b) - what is? We can rewrite y x c ax 4 b Then just think of the derivative operator as being a distributive operator that acts on each term in the sum.

23 Where y x c ax 4 b then - d x c ax b 4 ( ) and On the first term apply the power rule 4 dc dax db What happens to dc? What is the derivative of a flat line?

24 Successive differentiations yield 4 dc dax db 3 x 0 4ax 0 Thus - x 4ax 3

25 Differences are treated just like sums so that d ( x c) ( ax 4 b) is just x 4ax 3

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27 Differentiating functions of functions - Given a function y ( x 1) ( x 1) h( x) write h we consider d h dh dh y compute h dh d Then compute x 1 x take the product of the two, yielding and dh. dh

28 dh. dh ( x 1).x 4x( x 1) We can also think of the application of the chain rule especially when powers are involved as working form the outside to inside of a function y ( x 1)

29 Where y ( x 1) ( x 1) 1.x Derivative of the quantity squared viewed from the outside. Again use power rule to differentiate the inside term(s)

30 Using a trig function such as y sin( ax) (the angle is another function ax) let then Which reduces to h ax dh. dh cos(ax).a or just acos(ax)

31 A brief look at derivatives of trig functions. Consider dsin()/d. Start with the following - sin( ) sin( ) identities sin( ) sin cos cos sin sin( ) sin cos cos sin cos( ) cos cos sin sin cos( ) cos cos sin sin Take notes as we go through this and the derivative of the cosine in class.

32 We end up with cos lim sin sin... 3! 5! 7! When is small (such as in ), sin~ We can also see this graphically using arc length relationships

33 In general if y f ( g( h( i(...( q( x))...)))) then df. df dg. dg dh dh dq.... di

34 How do you handle derivatives of functions like ) ( ) ( ) ( x g x f x y? or ) ( ) ( ) ( x g x f x y The products and quotients of other functions

35 Removing explicit reference to the independent variable x, we have y fg Going back to first principles, we have y ( f df )( g dg) Evaluating this yields y fg gdf fdg dfdg Since dfdg is very small we let it equal zero; and since y=fg, the above becomes -

36 Product rule applied to straight line formula with its constants y y=ax+b b is the intercept The slope a=y/x is a constant Product rule x b, the intercept is a constant that just gets added to the ax and shifts it up or down. The slope does not change. /=xda/+a/ +db/ = a

37 gdf fdg & df dg g f Which is a general statement of the rule used to evaluate the derivative of a product of functions. The quotient rule is just a variant of the product rule, which is used to differentiate functions like f y g

38 The quotient rule states that d f g g df g f dg The proof of this relationship can be tedious, but I think you can get it much easier using the power rule Rewrite the quotient as a product and apply the product rule to y as shown below f 1 y fg g

39 We could let h=g -1 and then rewrite y as y fh Its derivative using the product rule is just df h f dh dh = -g - dg and substitution yields df g f dg g

40 Multiply the first term in the sum by g/g (i.e. 1) to get > g g df g f dg g Which reduces to g df g f dg the quotient rule

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42 Next time we ll use Excel to demonstrate that the rules noted below accurately characterize slope variations. x de e x Basically indestructible in this form cx For a function like Ae, this is not the case. Calculating the derivative becomes a little more complex. Rewrite the function Take derivative of the exponent cx dae cx d( cx) cx Ae cae This is an application of the rule for differentiating exponents and the chain rule de h( x) e h( x) dh

43 Basic rule for differentiating exponential functions de x e x dae cx Ae cx ( ) Sketch and discuss d cx cae cx Rewrite the exponential function and multiply it by the derivative of the exponent a two-step process.

44 Second derivative? cx d Ae dae cx cae cx and cx d Ae dcae cx

45 Follow up on carrying the constants through Use the product rule to differentiate a simple function like y ax dg df f g.

46 If we finish these today hand in otherwise bring in for discussion next time

47 Next time we ll put the rule to the test using Excel d dz cz c0e In the lab exercise c = 1. derivative

48 Look over problems 8.13 and 8.14 Bring questions to class next time No due date set at present for these two problems.

49 Next time we ll continue with exponentials and logs, but also have a look at question 8.8 in Waltham (see page 148). Find the derivatives of x ( i ) x. e ( ii) ( iii) z y 3.sin( ) x.cos(x) x.tan( x) 4 ( iv) B 3.ln( ) 17

50 Before leaving Hand in analysis of porosity/depth relationship Depending on progress in the class today I may pick up the in-class differentiation problems.

51 Looking ahead continue reading Chapter 8 Differential Calculus Look over problems 8.13 and 8.14