1. Maths Tools

7/27/2019 1. Maths Tools

1/73

Maths for Petroleum Engineering

2012-2013

Karl Stephen

[email protected]


2/73

Aims

Session 1.

Review of Trig and Calculus

Session 2

Deriving important equations

Session 3

Solving important equations


3/73

Session 1 Overview

Trigonometry

Calculusdifferentiation

Calculusintegration

Log and Exponential functions


4/73

Overview

Trigonometry


Calculusintegration



5/73

Sine, cosine and tangent

definitions

(x,y)

0

x

yr

r

xcos

r

ysin

x

ytan

1sincos

22


6/73

Sine function

(x,y)

0

x

ytan

x

yr

sin x

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8

sin x

r

ysin


7/73

Sine and cosine functions

(x,y)

r

xcos

r

ysin

0

x

ytan

x

yr

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8

sin x

cos x


8/73

Sine and cosine,

some properties

x

ytan

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8

sin x

cos x

)sin()sin(

)cos()cos(

)2

sin()cos(

)2

cos()sin(

)sin()2sin( )cos()2cos(


9/73

Tangent function

(x,y)

0

x

ytan

x

yr

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 1 2 3 4 5 6 7


10/73

Tangent properties

x

ytan

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 1 2 3 4 5 6 7

)tan()tan(

)tan()tan(


11/73

Applications: co-ordinate change

a

b

r

(x,y) or (r,)

x

y

cosrx

sinry

0

Co-ordinates:

Vector, r (r, ):

22 yxr (Pythagoras)

x

ytan 1


12/73

Summation of angles

ab

1. bababa sincoscossin)sin(

2. bababa sinsincoscos)cos(

3. aaa cossin2)2sin(

4.

1cos2sin21sincos)2cos( 2222 aaaaa


13/73

Overview

Trigonometry


Calculusintegration



14/73

How fast were you going?

t

s

t

sv

t2

s2

1

11

t

sv

s1

t1

12

122

tt

ssv

t9

s9s8

t8

1ii

1iii

tt

ssv

1ii

1iii

tt

ssv

Even smaller increments?

What is the instantaneous

Rate of change?

Dis

tance,s

time, t


15/73

The derivative - definition

Derivative,

rate of change of function f(x) at point x0

Graphical interpretation

slope of tangent

Notation

f'(x) = f(x) =dxd

dxdf

y

xx 0

= f(x)

x0


16/73

How is it calculated?

The derivative - calculation

y

x

y = f(x)

x0 +h

f(x0+h)

x0

f(x0)

00

00

x)hx(

)x(f)hx(fm,gradient

Newton Quotient

00

000h0

x)hx(

)x(f)hx(flim)x('f


17/73

Darcys Law

p1p0

aq

x x + Dx

Dx

xppkaq 01

f

f D

A porous substance has a permeability of1 Darcy if, in 1 second, 1

cubic centimetre of a gas or liquid with a viscosity of1 centipoise

will flow through a section 1 centimetre thick with a cross section of

1 square centimetre, when the difference between the pressures onthe two sides of the section is 1 atmosphere.

x

ppkv

a

q 01

f

ff

D


18/73

p1p0

av

x x + Dx

Darcys Law- as a gradient

p1p0

a

x x + Dx

xd

dpkv

f

f

Let the length of the sample tend to zero.

x

pplim

kv

010x

f

f D

D


19/73

Simple rules: linearity

)x(g)x(fy

)x('g)x('fdx

dy

If

And if)x(f.cy

)x('f.cdx

dy


20/73

The derivative - Example 1.

Example, f(x) = ax2

00

00

0h0 x)hx(

)x(f)hx(flim)x('f

00

2

0

2

00h0

x)hx(

)x(a)hx(alim)x('f

hax)hhx2x(alim)x('f

2

0

2

0

2

00h0

)ahax2(lim)x('f 00h0


21/73



y

x

y = ax2

)hax2(limx)hx(

)x(f)hx(flim)x('f 00h

00

000h0

x0 +h

f(x0+h)

x0

f(x)

gradient


22/73



y

x

y = ax2

000h

00

000h0 ax2)hax2(lim

x)hx(

)x(f)hx(flim)x('f

x0 +h

f(x0+h)

x0

f(x)


23/73


Example, f(x) = axn

00

000h0

x)hx()x(f)hx(flim)x('f

00

n

0

n

0

0h0 x)hx(

)x(a)hx(alim)x('f


24/73


)hx)...(hx).(hx).(hx()hx( 0000n

0

)h,x(ghnhxx)hx( 21n

0

n

0

n

0

00

n

0

n

00h0

x)hx(

)x(a)hx(alim)x('f

iin

0

n

0i

i

n

0 hxa)hx(

n

n

1n

01n

22n

02

1n

01

n

00

n

0 hahxa...hxahxaxa)hx(


25/73


Example, f(x) = axn

h

))h,x(ghhnx(alim)x('f

21n

00h0

))h,x(hgnx(alim)x('f1n

00h0

1n

00 anx)x('f

00

n

0

21n

0

n

00h0x)hx(

)x(a))h,x(ghhnxx(alim)x('f

Works for n real also!


26/73

More examples

1n

00 anx)x('f

2/1x3)x(f

x

1x1x3x2)x(f

2

37

x

1)x(f

2

11

x

1x1)x('f

2

11

2

1

x5.1x2

13)x('f

2/33

26

x2

1

x

2x9x14)x('f

;ax)x(f n


27/73

Results

Can we show that:

)xsin()x('f 00 )xcos()x(f then

)xcos()x('f 00 )xsin()x(f then


28/73

Gradients

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8

sin x

cos x

)xcos()x('f 00 )xsin()x(f then


29/73

Calculating more

complicated derivatives

Chain rule; y=f(g(x))

Products; y=f(x)g(x)

Quotients; y=f(x)/g(x)

2)x(siny

xcosxsiny

x

xcosy

Examples


30/73

The Chain Rule

Combined functions: y=f(u); u=g(x)

dx

du

du

df)x('y

))x(g(fy


31/73

The Chain Rule - example

e.g.

dx

du

du

df)x('y ))x(g(fy

2)x(siny

xcosxsin2xcosu2dx

)x(sinddu

)u(d)x('y2

xsinu;uy 2


32/73

The Product Rule

Products

)x(g)x(f)x(y

)x('g)x(f)x(g)x('f)x('y


33/73

The Product Rule: example

Products

e.g.

)x(g)x(f)x(y


xcosxsiny

dx

)x(cosdxsinxcos

dx

)x(sind)x('y

xsinxsinxcosxcos)x('y


34/73

The Quotient Rule

Quotients

)x(g

1).x(f

)x(g

)x(f)x(y

)')x(g

1)(x(f

)x(g

1)x('f)x('y Product rule

2)x(g

)x('g)x(f)x(g

1)x('f)x('y Chain rule

2

))x(g(



35/73

The Quotient Rule: example

Example:

dx

)x(dxcos

x

1

dx

)x(cosd)x('y

1

Product rule

x

xcosy

Chain rule 2x1

xcosx

1xsin)x('y

)xcosxsinx(

x

1)x('y

2


36/73

Higher Order derivatives

Second order derivative

the gradient of the gradient of f(x).

n orders, notation:-

00

000h2

2

x)hx(

)x('f)hx('flim)x(''f

dx

fd)x(f

dx

d

dx

d

)x(fdx

fd nn

n


37/73

Taylors Series

A function f(x+h) can be expanded to:

This infinite series can be truncated:

......dx

fd!3

hdx

fd!2

hdxdfh)x(f)hx(f

3

33

2

22

2

22

dx

fd

!2

h

dx

dfh)x(f)hx(f


38/73

y=sin(x) sin(0)=0

2 2 3 3

2 3(0 ) (0) ......

2! 3!

dy h d y h d yy h y h

dx dx dx

cos( )

dy

xdx 2

2sin( )

yx

x

3

3cos( )y x

x

0

1x

dy

dx

2

2

0

0x

y

x

3

3

0

1x

yx

2 3 3 5 7

sin( ) 0 .1 .0 ( 1) ......2! 3! 3! 5! 7!

h h h h h

h h h

5913 11173 order


39/73

Partial derivative

Let z=f(x,y)

Partial deriviative:

gradient of f wrt x or ywhile y or x are fixed

respectively.

y*

x

*)y,x(f*)y,xx(flim

x

f0x

D

D

D


40/73

Mass balance

Fluid flow in 1-D:

flux, vin vout

x

)()(lim

t

)C()C(lim

inout

0x

ttt

0t

D

D

D

D

D

DxDy

Dz

Make instantaneous and local:


41/73

Mass balance

Fluid flow in 1-D:

flux, vin vout

Taking limits as Dt and Dx vanish:

x

)(

t

C


42/73

Example

x

)x(f)xx(flim

x

f 000x D

D

D

222 yxy2x)yx()y,x(f

Think y is a constant

y2x2xf


43/73

Partial and ordinary derivatives

Ordinary derivatives used for y(t)

functions of one variable

Partial derivatives are used for f(x,y,z,)

functions of several variables

However, if u=f(x,y) and x=x(t) and y=y(t),

du f dx f dy

dt x dt y dt

Chain rule


44/73

Overview

Trigonometry


Calculusintegration



45/73

How far did you go?

t

v

vts

Average velocity

t2

v2

111 tvs

v1

t1

)tt(vss 12212 )tt(vss 1iii1ii

Average velocity

111 tvs

velocity

time

i

1iii

i

1iin )tt(vsss)tt(vss 1iii1ii

111 tvs

Actual velocity?

velocity

time

i

1ii1ii

i

1iin 2/)tt)(vv(sss

Start with the velocity curve then approximate


46/73

Integrals - definition

Integral = area under the curve

e.g. total distance travelled at a certain speed

Inverse of the derivative

y

x

y=f(x)

x0 x0 +h


47/73

Integrals - approximation

Approximate

Area=(hf(x0)+0.5h(f(x0 +h)-f(x0))

Area=h((f(x0 +h)+f(x0))/2

y

x

y=f(x)

x0 x0 +h

f(x0 +h)

f(x0)

Explain areas


48/73

Integrals - approximation

Definition:

Area=Sh((f(x0 +nh)+f(x0+(n-1)h))/2y

x

y=f(x)

a ta+h


49/73

Integrals - definition

Approximation improved:

y

x

y=f(x)

x1 x2x1+h

2

0 0 0

11

( ) lim 0.5 ( ( ) ( ( 1) )x N

h

nx

area f x dx h f x nh f x n h


50/73

Indefinite and definite Integrals

The Indefinite Integral is

The Definite Integral is

C)x(gdx)x(f )x(f)x('g where

b

a

)a(g)b(gdx)x(f

x

g(x)

Same

derivative

ynamicwhats the difference here? Relate actual area to abstract indefin


51/73

Example solutions:

inverse of differentiation E.g.

Cx1n

adxax 1nn

Cxcosxdxsin

Cxtandx)xtan1( 2


52/73

Difficult integrals

xdxcosxsinn

dxxsinx

How do we solve:

or:

Substitution

Solve by parts


53/73

Integration by substution

2 2 2

1 1 1

( ) ( ( )) ( ), ( ( ))

u x x

u x x

dF u dF u x dF u duarea F u x du dx dx

du dx du dx

2 2

1 1

, ( ( )) ( ) ( )

u x

u x

duarea F u x f u du f u dxdx

Inverse

chain rule


54/73

E.g.

Solution by substitution

2

1

x

x

nxdxcosxsinArea

u)u(fand;xsinuset

1n

1

1n

2

u

u

n

x

x

n

uu1n

1

duudxdx

du

uArea

2

1

2

1

11n21n xsinxsin

1n

1Area


55/73

Integrals - solution by parts

From the derivative of a product,

y(x)=f(x)g(x):

Integrating both sides by x gives:


dxdx

dgfdxgdx

dfg.f

dxgdx

dfg.fdx

dx

dgf

Product rule

Rearranged

Easier to solve?


56/73

Integrals - solution by parts

Example: dxxsinx

xsindx

dg

;xf

xcosg;1dx

df

xdxcosxcosxdxxsinxdxg

dxdfg.fdx

dxdgf

xsinxcosxdxxsinx

Reduce order

of x to zero


57/73

Overview

Trigonometry


Calculusintegration



58/73

Log function - definition

Definition

natural logarithm

xlndx

x

1x

1

1.0 x

y=1/x

y


59/73

Log function - relationships

Known results and limits

01ln

xasxln

0xasxln

x

y=1/x

y

x

yln x


60/73

Log function - relationships

addition and multiplication

xlnalnaxln

ulnaxlnf(x)let

dx

du

du

)u(lnd

(x)f'hent

where u=ax

Cxlndxx

1dx)x('f)x(f

au

1

dx

du

du

)u(lnd

(x)f'hent aax1

au

1

dx

du

du

)u(lnd

(x)f'hent x1

aax

1

au

1

dx

du

du

)u(lnd

(x)f'hent Chain

Rule!


61/73

Log Function addition

So

If x=1 then

Therefore

Also

Cxlnaxln

CC1lnaln

xlnalnaxln

xlnaxln a


62/73

Exponential function - definition

Definition:

e=2.718

derivative:

xe)xexp(y

xe)x('y

ylnx


63/73

dx

dy)x('y

dy/dx

1

dx

dy)x('y

dy/)y(lnd

1

dy/dx

1

dx

dy)x('y

y/1

1

dy/)y(lnd

1

dy/dx

1

dx

dy)x('y xe

y/1

1

dy/)y(lnd

1

dy/dx

1

dx

dy)x('y

Derivative of exponential

xe)xexp(y


64/73

Log versus ln

x10y

x10ln

ylnylogylog 10

x10ln10lnxyln

ln always means natural log

log often means log10 but can mean ln

When writing use ln or log10


65/73

Unconverted data

x

-4 -2 0 2 4 6

y

0

1000

2000

3000

4000

5000

y=exp(2x)

y=10x


66/73

Converting datanatural log

x

-4 -2 0 2 4 6

y

e-9e-8e-7e-6e-5e-4e-3e-2e-1e0e1e2e3e4e5e

6

e7e8e9e10e11

y=exp(2x)

y=10x


67/73

Converting datalog 10

x

-4 -2 0 2 4 6

y

0.0001

0.001

0.01

0.1

1

10

100

1000

10000

100000

y=exp(2x)

y=10x


68/73

Log:log plots

ln x

ln y

cxlnmyln

cxlnmey

m

kxy

12

22

xlnxln

ylnylnm

22 xlnmylnc

cxlnm eey


69/73

Log:log plots

ln x

cxlnmdx

dyln

12

1xx2xx

xlnxln

dx

dyln

dx

dyln

m

22

xlnmylnc

dx

dyln mkx

dx

dy

constx1m

ky 1m


70/73

Log:linear (semilog) plots

x

ln y

cmxyln

cmxey

22

mxylnc

12

22

xx

ylnylnm


71/73

Log:linear (semilog) plots

x

22 mxylnc

dx

dyln

cmxdx

dyln

12

1xx2xx

xx

dx

dyln

dx

dyln

m

cmxedx

dy

constem

1

y

cmx


72/73

Next time?

Applications 1. Deriving flow equations

Mass balance

Diffusivity Equation

Vector Calculus

Application 2. Solution of flow equations

Steady State flow

Pressure solution and averaging

Numerical Simulation

Buckley-Leverret Theory

Exponential Integral


73/73

See also

http://www.mathcentre.ac.uk/

Documents

1. Maths Tools