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Maths for Petroleum Engineering
2012-2013
Karl Stephen
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Aims
Session 1.
Review of Trig and Calculus
Session 2
Deriving important equations
Session 3
Solving important equations
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Session 1 Overview
Trigonometry
Calculusdifferentiation
Calculusintegration
Log and Exponential functions
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Overview
Trigonometry
Calculusdifferentiation
Calculusintegration
Log and Exponential functions
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Sine, cosine and tangent
definitions
(x,y)
0
x
yr
r
xcos
r
ysin
x
ytan
1sincos
22
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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
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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
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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(
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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
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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(
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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
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Summation of angles
ab
1. bababa sincoscossin)sin(
2. bababa sinsincoscos)cos(
3. aaa cossin2)2sin(
4.
1cos2sin21sincos)2cos( 2222 aaaaa
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Overview
Trigonometry
Calculusdifferentiation
Calculusintegration
Log and Exponential functions
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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
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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
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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
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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
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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
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Simple rules: linearity
)x(g)x(fy
)x('g)x('fdx
dy
If
And if)x(f.cy
)x('f.cdx
dy
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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
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How is it calculated?
The derivative - Example 1.
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
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How is it calculated?
The derivative - Example 1.
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)
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The derivative - Example 2.
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
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The derivative - Example 2.
)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(
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The derivative - Example 2.
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!
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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
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Results
Can we show that:
)xsin()x('f 00 )xcos()x(f then
)xcos()x('f 00 )xsin()x(f then
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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
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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
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The Chain Rule
Combined functions: y=f(u); u=g(x)
dx
du
du
df)x('y
))x(g(fy
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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
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The Product Rule
Products
)x(g)x(f)x(y
)x('g)x(f)x(g)x('f)x('y
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The Product Rule: example
Products
e.g.
)x(g)x(f)x(y
)x('g)x(f)x(g)x('f)x('y
xcosxsiny
dx
)x(cosdxsinxcos
dx
)x(sind)x('y
xsinxsinxcosxcos)x('y
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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(
)x('g)x(f)x(g)x('f)x('y
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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
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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
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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
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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
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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
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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:
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Mass balance
Fluid flow in 1-D:
flux, vin vout
Taking limits as Dt and Dx vanish:
x
)(
t
C
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Example
x
)x(f)xx(flim
x
f 000x D
D
D
222 yxy2x)yx()y,x(f
Think y is a constant
y2x2xf
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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
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Overview
Trigonometry
Calculusdifferentiation
Calculusintegration
Log and Exponential functions
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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
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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
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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
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Integrals - approximation
Definition:
Area=Sh((f(x0 +nh)+f(x0+(n-1)h))/2y
x
y=f(x)
a ta+h
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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
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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
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Example solutions:
inverse of differentiation E.g.
Cx1n
adxax 1nn
Cxcosxdxsin
Cxtandx)xtan1( 2
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Difficult integrals
xdxcosxsinn
dxxsinx
How do we solve:
or:
Substitution
Solve by parts
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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
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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
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Integrals - solution by parts
From the derivative of a product,
y(x)=f(x)g(x):
Integrating both sides by x gives:
)x('g)x(f)x(g)x('f)x('y
dxdx
dgfdxgdx
dfg.f
dxgdx
dfg.fdx
dx
dgf
Product rule
Rearranged
Easier to solve?
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Integrals - solution by parts
Example: dxxsinx
xsindx
dg
;xf
xcosg;1dx
df
xdxcosxcosxdxxsinxdxg
dxdfg.fdx
dxdgf
xsinxcosxdxxsinx
Reduce order
of x to zero
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Overview
Trigonometry
Calculusdifferentiation
Calculusintegration
Log and Exponential functions
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Log function - definition
Definition
natural logarithm
xlndx
x
1x
1
1.0 x
y=1/x
y
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Log function - relationships
Known results and limits
01ln
xasxln
0xasxln
x
y=1/x
y
x
yln x
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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!
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Log Function addition
So
If x=1 then
Therefore
Also
Cxlnaxln
CC1lnaln
xlnalnaxln
xlnaxln a
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Exponential function - definition
Definition:
e=2.718
derivative:
xe)xexp(y
xe)x('y
ylnx
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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
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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
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Unconverted data
x
-4 -2 0 2 4 6
y
0
1000
2000
3000
4000
5000
y=exp(2x)
y=10x
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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
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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
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Log:log plots
ln x
ln y
cxlnmyln
cxlnmey
m
kxy
12
22
xlnxln
ylnylnm
22 xlnmylnc
cxlnm eey
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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
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Log:linear (semilog) plots
x
ln y
cmxyln
cmxey
22
mxylnc
12
22
xx
ylnylnm
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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
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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
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See also
http://www.mathcentre.ac.uk/
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