Transcript No Slide Title

Geomath: final review week
tom.h.wilson
[email protected]
Dept. Geology and Geography
West Virginia University
Tom Wilson, Department of Geology and Geography
I’m giving you a bunch of problems to look over and
review today and Thursday
Some of the additional slides you see in today’s slide
set are simple reminders of some of the processes we
often go through to set up an integral. These are
approaches and concepts you are already familiar
with, but keep these approaches in mind. We’ll spend
more time on integration this Thursday.
The exam will be drawn from previous homework (& inclass) exercises along with example problems from the
text discussed in detail by Waltham
Current grade is based on homework
and the mid term.
Additional grades include problems 9.9 and 9.10 due today, an extra credit
exercise, term report if you notified me, and the final exam.
Current grade - Geology 351
6
5
Number
4
3
2
1
0
30
Tom Wilson, Department of Geology and Geography
40
50
70
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Grade
80
90
100
Transforming the discrete sum into an integral
Represent a complex shape by the sum of a lot of
more simply shaped objects.
The volume of each
of the little disks
(see below)
represents a Vi
N
V   Vi
i 1
Vi   ri2zi
N
   ri 2 z
i 1
V 
Zmax
Zmin
 ri2dz
ri
 ri2dz
dz
is the volume of a single
disk having radius r and
thickness dz.
ri
Disk
Area
Radius
V 
Zmax
Zmin
 ri2dz =total volume
The sum of all disks with thickness dz
The key to a successful integration is often
obtained by reducing the number of variables
Zmax
In the volume integral V  Z  ri2dz we have two variable r and z.
min
r2 
400 z 800 z

 400km2
3
3
Expression of r2 interns of z makes it possible to easily solve this integral
 400 z 800 z

2
V  

 400km 
0
3
 3

3
V 
3
0
3 800 z
3
400 z
dz   
dz    400dz
0
0
3
3
The integral in this case is a definite one and we are not done with
the problem until we evaluate the integral at the limits
3
V    ri 2 dz
0
3
 400 z 800 z



 400 z 
1.5 3
 6
0
2
1.5
  600 1600 1200
 200  628km
3
Remember that the most likely errors in our work occur in the math
and algebra: so take the time to double check your results!
Li
There are different ways of quantifying strains:
The elongation e 
Lf
L
Li
Lf
Li
is the stretch
If we consider two deformations L1 and L2, the final strain

L1  L2
Li
L
L2
For the total natural strain, the
 1
Li Li  L1
reference point continually changes
Notation is always a problem, so read problems carefully. Don’t assume symbols
used in one problem mean the same thing as those used in another.
The true strain or total natural strain, can be totaled. It
is the sum of an infinite number of infinitely small
extensions and is also known as the logarithmic strain
The sum is
expressed as a
definite integral
 
Lf
Li
dL
L
 ln L L
Lf
i
 ln( L f )  ln( Li )
 Lf 
 ln  
 Li 
  ln(S )
Where S is known as the stretch or relative
elongation and is also sometimes denoted as 
  ln( )
note that e  
Computers can really help you turn abstract problems
into more concrete, easy-to-visualize ones.
Calculate and plot to get a picture of the functions you are analyzing.
Tom Wilson, Department of Geology and Geography
We also noted that integral evaluation can be simplified by
dividing the problem into parts, as in the example of Earth mass.
We can simplify the problem and still obtain a useful result.
Approximate the average densities
11,000
kg/m3
4,500
kg/m3
N
M   4 ri 2 i r
i 1
R
M   4 r 2  dr
0
It is difficult to carry out the integration from 0 to R so we divide the Earth
into two regions and simplify the variable of integration in each region.
We can simplify the problem and still obtain a useful result.
Approximate the average densities
11,000
kg/m3

0
4 r 2 .11000dr  
6371
3480
3480
4
  r 3 1
3
0
3480
44000 3

r
3
0
M   4 r 2  dr
0
4,500
kg/m3
3480
R
4
  r3  C
3
4 r 2 .4500dr
6371
4
  r 3 2
3
3480
6371
18000 3

 r 2
3
3480
The result – 6.02 x
1024kg is close to the
generally accepted value
of 5.97 x 1024kg.
z kW
Q
20 km3
Radiogenic heat flow
i. Determine the heat generation rate at 0, 10, 20, and 30
km from the base of the crust
0 kW
kW
0 3
20 km
km
10 kW
kW
 0.5 3
3
20 km
km
Q is the heat generation rate so this is a simple substitution for specific
values of z (remember origin is important) and calculation.
ii. What is the heat generated in a in a small boxshaped volume z thick and 1km x 1km surface?
Heat generation rate has units of power (kW) per unit
volume, so total heat generated is QV. Use units to guide
your computations.
V  Az
V  zkm3
V
The heat generated in the volume element z
QV  QzkW
This is a differential quantity so there is no need to integrate
iii & iv. Calculate total heat generated in the vertical
column
In this case the sum extends over a large
n
range of z, so
q   Qi z
i 1
However, integration is the way to
go.
q
z
0

z
dz
20
1 z
zdz
20 0
Computation of the flow rate at the surface requires integrating
the input to heat flow from the base of the crust to the surface
v. Heat flow at surface
z
q   Qdz
0
q
2 30
z
 22.5kW
40 0
vi. To generate 100MW of power
100, 000kW
 4444km 2
22.5 kW 2
km
Remember how this get’s turned into a calculus problem?
At the base of a cliff you are standing on top of geological Unit A. The cliff
face is formed along a normal fault (nearly vertical). The top of Unit A is
also exposed at the top of the cliff face. You walk a distance x = 200 feet
away from the fault scarp. Looking back toward the cliff, you use your
Brunton and measure and note that the top of the cliff is 23o above the
horizon. What is the offset along this fault?
Top of Unit A
h
Cliff
Face

Top of Unit A
x
h=85feet
You computed gv at arbitrary x along the surface, but what
are the derivatives of gv wrt x and z and what do they mean?
gv 
Tom Wilson, Department of Geology and Geography
GMz
x
2
z
2

3
2
What have you computed in taking those
derivatives?
Tom Wilson, Department of Geology and Geography
Problems due today: 9.9 and 9.10
Volume of the earth – an oblate spheroid
  2 
r  r 1   z 2  
  rp  
2
2
e
In this equation r varies from re, at the equator, to r=0 at
the poles. z represents distance along the earth’s rotation
axis and varies from –rp to rp.
The equatorial radius is given as 6378km and the polar
radius, as 6457km.
Tom Wilson, Department of Geology and Geography
Problem 9.10
t  t0 exp( x / X )
In this problem, we return to the
thickness/distance relationship for the
bottomset bed.
Problems 9.9 and 9.10 are due today
Tom Wilson, Department of Geology and Geography
Start reviewing materials for the final!
… current to-do list
1. Problems 9.9 and 9.10 are due today
2. Keep reviewing! The final is next week.
3. Bring questions for Thursday
Tom Wilson, Department of Geology and Geography