Geol 351 Geomath - West Virginia Universitypages.geo.wvu.edu/~wilson/geomath/Calculus/calclec8.pdf · Geol 351 Geomath Tom Wilson, Department of Geology and Geography tom.h.wilson

Geol 351 Geomath

Tom Wilson, Department of Geology and Geography

[email protected]

Dept. Geology and GeographyWest Virginia University

Integral calculus - continued

Due dates


• Hand in indefinite integrals problems (16).

• Finished reading chapter 9 ?

•Next week we’ll go after questions 9.9 and 9.10 look them over. Due dates forthcoming.

• The computer version of Question 9.7 will be assigned reviewed next time so look it over.

• Final exam will consist of in-class activities during the last week of class on April 25th and April 27th

Objectives for the day


• Return homework/discuss

• Calculating lithostatic pressure (Sv) an integration

problem – problem introduction.

• Evaluating the area covered by a function.

• Use Waltham excel files to illustrate integral

relationships

• Distinguish between the indefinite and definite

integrals

• Continue to sharpen integration and problem

solving skills with in-class/take home and

assessment activities.

Questions 8.13-14


Could you

actually present

the details of the

derivative??

A good self test


1 2

1 2

max max

1 2

( ) ( )

t t

dS dS

dt dt

S Se e

How do you solve for t?

Can you show that …

Solve for t

1

2

1 2

ln

1 1t

Give it a try!

Alternative views of the fault problem


sin cos

tew Lm

sin sin

t zw Lm

Two ways to minimize the “work” w=Lm

cos

em

Approach 1 Approach 2

t

We locate the zeros for the first derivative and know

they locate either a maximum or minimum


At /4

Approach 1

Second derivatives


2 4 4

2 3 3

sin cos

sin cos

d w

d

At 45o or /4, value

is positive so work

is a minimum

Approach 1

Where does the derivative =0


At /2

2sin sin sin

t z t zw Lm

Approach 2

2nd derivative at /2 = 2 (positive) so a minimum


The analysis gave us

two possible

solutions: one at 45o

and the other at 90o.

+ at /2=2

Approach 2

The vertical compressive stress (Sv) can be estimated using the

following integral which yields a result in units of F/A


Sv=

t

e

m

L

( )* *b

a

forcedensity z g dz

area

oforce m g Vg =Mass x g (FORCE) of overburden/unit area

Equivalent to gh =force/area

Subsurface density varies continuously but is

sampled at discrete intervals in the density log


P(z)

In continuous form, this is an integral of (z).

( )b

va

S z gdz

1

n

v i

i

S g z

In discrete form, this is a sum of I’s x …

Water column, ~1.03gm/cm3

Estimated density increase based on nearby logs

Log data 0.5ft sample rate

For now – let’s look at the in-class/take home

activity and assessment question


Group discussion for a few minutes. Turn in next time.

For the definite integral


2

2 2

2

2 2

bb

a a

xxdx

b a

y=x

x=a x=b

The area of the larger triangle

minus the smaller one.

Area of small

triangle =1/2 base

x height= a2/2.

Use class page link to Waltham Excel file

Integ.xlsx to experiment with these ideas


Consider the integral of the function y=x. Compare areas

estimated by summing a set of rectangles and that obtained

by the actual integral.

Another comparison

These are all definite integrals


2 2 & y x x dx Approximation versus explicit integration

In these discrete approximations we are just

adding the areas of little rectangles


x

( )f x

( )A f x x

Total Area under this Curve ≈ 1

( )n

i

f x x

Discrete and analytic estimates of

the integral/ the area


In Waltham’s integ.xls file set n = 2

These examples provide illustrations of the ways computers can compute integrals. There’s usually some error, but we

can make that as small as we want by decreasing x.

Discrete integral computations allow you to solve

problems that don’t have simple integrals


A simple integral with easily derived exact solution

11

1

n nx dx xn

the power rule for integrals


3

3

d xc

dx

You can see that the derivative

3

3

d x

dx

This is an easy one. We just use the power law (in reverse: add 1 and divide by the new exponent) to find that

2x=

In general then the integral 11

1

n nx dx xn

The special case


1 ln( )x dx x k

1

1

nxy

n

1

1

nn x

x dx kn

ndyx

dxIn general if then

Thus

When n = -1, you get 1/0!

See earlier lecture slides illustrating relationship between 1/x and ln(x).

1x dx

0

0

xc

As an indefinite integral


1 ln( )x dx x k We add that constant to allow us to

accommodate arbitrary starting conditions.

2 2

11

1 ln( )x x

xxx dx x

As definite integral over defined limits

Definite versus indefinite integrals


103 3 3

102

44

10 4312

3 3 3

xx dx

2y x

32

3

xx dx c

The indefinite

integral

Indefinite integrals provide general solutions

without specified range of integration


32

3

xx dx C

2y x

Basic integration rules


• power rule

• sum rule (distributive like the derivative)

• multiplication by a constant simply carries through as

with the derivative

• special case for

• integration of exponential functions

• integrals involving roots (use power rule and chain

rule)

• indefinite and definite evaluations of integrals

1x dx

3/22( )

3x adx x a C

Trig integrals in definite form


32

2

cos( )a da

3sin( ) sin( )2 2

3

2

2

sin( )a

Before you evaluate this, draw a picture of the cosine and ask yourself what the area will be over this range

What is the area under the cosine from /2 to 3/2

Rules overview


( )af x dx

. . 3sin(x)dx=3 sin(x)dxe g

Given

where a is a constant; ( )a f x dx

a cannot be a function of x.

The constant factor rule for integration. If it’s not a function of x – pull it out.


( ) ( ) ( )f x g x h x dx

( ) ( ) ( )f x dx g x dx h x dx

11

1

n nx dx x Cn

Just as with derivatives, distribute the integration through

the sum or difference of terms

The power rule has to be applied in reverse: add

one to the exponent and divide the function by

the exponent plus 1.

An integration problem to look over for next time


Detachment

horizon

Detached rock forced

along the fault into a fold

We approximate the

shape of the deeper

fold as 24

425

s

xy

2

125

d

xy

We approximate the

shape of the

shallower detached

fold as

-x +x

Here we use two quadratics to represent deviations in relief of the

2nd order (shallow) fold relative to the 1st order (deep) regional fold


Note that the limits used here coincide with the area for which relief on the upper blue

detached fold is greater than the orange fold in the lower unit (-5kft to 5kft)

244

25

x

2

125

x

The fold in the

overlying unit is a

fault propagation

fold. The Upper sheet

has additional

shortening associated

with this fold.

A structural geology problem cast in terms of

calculus concepts


Detachment

horizon

Given the analytic shapes of the deeper

fold and shallower detached fold, how do

we calculate the excess area in this cross

sectional view?

244

25s

xy

2

125

d

xy

Calculate the area between these two curves


Evaluate ( )

x

s dx

y y dx

2 25

5

44 1

25 25

x xdx

This is a definite integral. The area (or difference of areas

in this case) is computed only over a certain limited range

corresponding to the extent of the shallow detached fold.

Take a few minutes for group discussion.

Due end of next class.

Another geological application (see Section 9.6)


Estimate the volume of material ejected during repeated eruptions of a volcano – in this case Mt. Fuji?

1

N

i

i

V V

2

1

N

i

i

r z

max

min

2Z

iZ

V r dz

Sum of flat circular disks

A sum of volume elements

Volume element z


max

min

2Z

iZ

V r dz

2

ir dz

ridz

ri

is the volume of a disk having radius r and thickness dz.

=total volume

The sum of all disks with thickness dz

Area

Radius

2

1

N

i

i

V r z

go to let z dz


2 2400 800400

3 3

z zr km

32

0

400 800400

3 3

z zV km

3 3 3

0 0 0

400 800400

3 3

z zV dz dz dz

Waltham notes that for Mt. Fuji, r2 can be approximated by the following polynomial

To find the volume we evaluate the definite integral


32 1.5

0

400 800400

6 1.5 3

z zz

600 1600 1200

3200 628km

32

0iV r dz

The “definite” solution


We know Mount Fuji is 3,776m (3.78km). So, does the integral underestimate the volume of Mt. Fuji? This is what

happens when you carry the calculations on up …Radius of Mt. Fuji

0

0.5

1

1.5

2

2.5

3

3.5

4

0 5 10 15 20 25

Radius (km)

Ele

vati

on

(km

)

It works out pretty good though since the elevation at the foot of Mt.

Fuji is about 600-700 meters.

Lastly – take a look at these assessment

problems and hand in before leaving


Looking ahead


•Hand in indefinite integrals problems before leaving.

• Hand in the short assessment activity before leaving•We’ll take some time in the next class to go over the

lithostatic/hydrostatic pressure and fold area problems. They will be collected at the end of next class.

• Review questions 9.6 and 9.7 for next time•Have a look at problems 9.9 and 9.10

• 9.9 and 9.10 are tentatively due __TBA__.• The computer version of Question 9.7 will also be due on the

______to be announced______.

Documents

Geol 351 Geomath - West Virginia Universitypages.geo.wvu.edu/~wilson/geomath/Calculus/calclec8.pdf · Geol 351 Geomath Tom Wilson, Department of Geology and Geography tom.h.wilson