geology 351 - geomathematics an introduction to...

Geology 351 - Geomathematics

Tom Wilson, Department of Geology and Geography

[email protected]

Dept. Geology and GeographyWest Virginia University

An introduction to differential calculus

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

Problems handed out last time -

Turn in porosity problem before leaving


In the porosity density relationship = 0e(-z/), assume 0 is 0.4 and is 1.5. What is the

slope (or porosity gradient) between depths of 1 and 2 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.

Between 1 and 2 km depths, slope =-0.099/km

Between 1.4 and 1.6 km depths, slope =-0.0982/km


The derivative or slope of the tangent line at the depth 1.5 km =-0.0981/km

z

Slope = /z

Slope at this point is the derivative

At 2.5km, the slope is -0.05

at 3.5 km the slope is -0.0259

at 0.5, -0.191


Z

0 1 2 3 4 5

PH

I

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4Porosity depth relationship

=0.4e-z/1.5

Check in-class work


0.5~ 0.19

1.5~ 0.1

3~ 0.04

z

z

z

Slopes evaluated over z of

0.2km about a given depth

You’ll find the derivative z

ode

dz

Evaluation of the derivative at each of these points yields

0.51.5

1.51.5

1.51.5

0.40.19

1.5

0.4 0.098

1.5

0.4 0.036

1.5

zod

e edz

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.

Slopes of the cosine are a little more obvious


0, -0.707 (at 0.785), -1 (at 1.571), -0.707 (at 2.356)


(radians)0.

00.

51.

01.

52.

02.

53.

03.

54.

04.

55.

05.

56.

0

cos

()

-1.0

-0.5

0.0

0.5

1.0Cosine function

-sin()

Calculation of these

slopes, point-by-

point, outline the

negative sine function

For y=x2


x0 1 2 3 4 5 6

y

0

5

10

15

20

25

30

35

40

y=x2

Derivative of a constant: i.e. y-a


y=a

y

x

What’s the slope?

So derivative of a constant is 0

The y is zero so y/x and dy/dx are always 0.


The book works through the differentiation of y = x2, so let’s try y =x4.

4)( dxxdyy

multiplying that out -- you get ...432234 )()(4)(64 dxdxxdxxdxxxdyy


432234 )()(4)(64 dxdxxdxxdxxxdyy



Remember the idea of the dy and dx is that they represent differential changes that are infinitesimal - very small.

So if dx is 0.0001 (that’s 1x10-4) then (dx)2 = 0.00000001 (or 1x10-8) (dx)3 = 1x10-12 and (dx)4 = 1x10-16.

So even though dx is very small, (dx)2 is orders of magnitude smaller



so that we can just ignore all those terms with (dx)n where n is greater than 1.

dxxxdyy 34 4

Our equation gets simple fast

Also, since y =x4, we have

dxxydyy 34

dxxdy 34

and then -


34xdx

dy

Divide both sides of this equation by dx to get

dxxdy 34

This is just another illustration of what you already know as the power rule,


1 nnaxdx

dyis

Just as a footnote, remember that the constant factors in an expression carry through the differentiation.

This is obvious when we consider the derivative -

baxy 2

which - in general for naxy

+ =

The shift does not change the slope (derivative).


bdxxadyy 2)(

bdxxdxxadyy )2( 22

axdxbaxdyy 2)( 2

axdxydyy 2

)2( xadx

dy

Examining the effects of differential increments in y and x we get the following

Where y=ax2+b

Alternatively


2

2

2

( )

=

=

=2

dy d ax b

dx dx

dax db

dx dx

dax

dx

ax

=0 since b is a constant.

Line with slope 2a

and 0 intercept.

Distribute the d/dx operator


Don’t let negative exponents fool you. If n is -1, for example, we still have

1 nnaxdx

dy

2 axdx

dy

or just


)()()( xgxfxy Given the function -

what is dx

dy?

dx

dg

dx

df

dx

dy

We just differentiate f and g individually and take their sum, so that

Distribute the d/dx operator


Take the simple example

)()( 42 baxcxy

- what is dx

dy?

baxcxy 42We can rewrite

Then just think of the derivative operator as being a distributive operator that acts on each term in the sum.


Where

then -

On the first term apply the power rule

What happens to

baxcxy 42

2 4( )dy d

x c ax bdx dx

2 4dy dx dc dax db

dx dx dx dx dx and

2dx

dxdc

dx?

What is the derivative of a flat line?


Successive differentiations yield

Thus -

342 axxdx

dy

2 4

32 0 4 0

dy dx dc dax db

dx dx dx dx dx

dyx ax

dx


)()( 42 baxcxdx

d

dx

dy

Differences are treated just like sums

so that

is just

342 axxdx

dy


Differentiating functions of functions -

Given a function 22 )1( xy we consider

)()1( 2 xhx write 2hy compute hhdh

d

dh

dy22

Then compute xxdx

d

dx

dh212 and

take the product of the two, yielding dx

dh

dh

dy

dx

dy.


xxdx

dh

dh

dy

dx

dy2).1(2. 2

)1(4 2 xx

22 )1( xy

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


22 )1( xyWhere

xxdx

dy2.)1(2 12

Derivative of the quantity squared viewed from the outside.

Again use power rule to differentiate the inside term(s)


Using a trig function such as

)2sin( axy

let axh 2

then dx

dh

dh

dy

dx

dy.

Which reduces to aaxdx

dy2).2cos( or just

)2cos(2 axadx

dy

(the angle is another function 2ax)

A brief look at derivatives of trig functions.

Consider dsin()/d.

Start with the following -


sin( ) sin( )

sin( ) sin cos cos sin

sin( ) sin cos cos sin

cos( ) cos cos sin sin

cos( ) cos cos sin sin

identities

Take notes as we go through this and the derivative of the cosine in class.

We end up with


0

sincos lim

3 5 7

sin ...3! 5! 7!

When is small (such as in ), sin~

We can also see this graphically using arc length relationships


In general if

))...))))((...(((( xqihgfy

then

dx

dq

di

dh

dh

dg

dg

df

df

dy

dx

dy........


How do you handle derivatives of functions like

)()()( xgxfxy

?

or

)(

)()(

xg

xfxy

The products and quotients of other functions


fgy

Removing explicit reference to the independent variable x, we have

))(( dggdffdyy

Going back to first principles, we have

Evaluating this yields

dfdgfdggdffgdyy

Since dfdg is very small we let it equal zero; and since y=fg, the

above becomes -

Product rule applied to straight line formula

with its constants


y=ax+b

y

x

b is the intercept

The slope a=y/x is a constant

dy/dx=xda/dx+adx/dx +db/dx = a

b, the intercept is a constant that just gets added to the ax and shifts it up or down. The slope does not change.

Product rule


&

dy gdf fdg

dy df dgg f

dx dx dx

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

g

fy


2g

dxdg

fdx

dfg

g

f

dx

d

The quotient rule states that

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

1 fgg

fy


fhy

We could let h=g-1 and then rewrite y as

Its derivative using the product rule is just

dx

dhf

dx

dfh

dx

dy

dh = -g-2dg and substitution yields

2g

dxdg

f

g

dxdf

dx

dy


2g

dxdg

f

g

dxdf

g

g

dx

dy

Multiply the first term in the sum by g/g (i.e. 1) to get >

Which reduces to

2g

dxdg

fdx

dfg

dx

dy

the quotient rule

Next time we’ll use Excel to demonstrate that the rules

noted below accurately characterize slope variations.


xxde

edx

( )cxcx cxdAe d cx

Ae cAedx dx

This is an application of the rule for differentiating exponents and the chain rule

Basically indestructible in this form

( )( )

h xh xde dh

edx dx

Rewrite the function

Take derivative of the exponent

For a function like , this is not the case. Calculating the derivative becomes a little more complex.

cxAe


xxde

edx

( )cxcx cxdAe d cx

Ae cAedx dx

Basic rule for differentiating exponential functions

Sketch and discuss

Rewrite the exponential function and multiply it by the derivative of the exponent – a two-step process.

Second derivative?


2

2

cxd Ae

dx

cxcxdAe

cAedx

2

2

cx cxd Ae dcAe

dx dx and

Follow up on carrying the constants through


Use the product rule to differentiate a simple function like

2y ax

dy dg dff g

dx dx dx

….


If we finish these today

hand in otherwise bring in

for discussion next time

Next time we’ll put the rule to the test using Excel


0

czdc e

dz

In the lab exercise c = 1.

derivative

Look over problems 8.13 and 8.14


•Bring questions to class next time•No due date set at present for these two problems.


Next time we’ll continue with exponentials and logs, but also have a look at question 8.8 in

Waltham (see page 148).

xexi . )( 2

)sin(.3 )( 2 yii

)tan(.xx.cos(x) )( 2 xziii

24 17)ln(.3 )( Biv

Find the derivatives of

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.

Looking ahead


continue reading Chapter 8 – Differential Calculus

Look over problems 8.13 and 8.14

geology 351 - geomathematics an introduction to...

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