example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. repeat for k...

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Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3 m Age = 1500 years Age = 3000 years Age = 7950 years For k = 3000years/m Age = 3000 years Age = 6000 years Age = 15900 years kz a notation symbolic

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Page 1: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m

1m

2m

5.3m

Age = 1500 years

Age = 3000 years

Age = 7950 years

For k = 3000years/m

Age = 3000 years

Age = 6000 years

Age = 15900 years

kzanotationsymbolic

Page 2: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

The geologist’s use of math often turns out to be a periodic and necessary endeavor. As time goes by you may find yourself scratching your head pondering once-mastered concepts that you suddenly find a need for.

This is often the fate of basic power rules.

Evaluate the following

xaxb =

xa / xb =

(xa)b =

xa+b

xa-b

xab

Page 3: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

Question 1.2a Simplify and where possible evaluate the following expressions -

i) 32 x 34

ii) (42)2+2

iii) gi . gk

iv) D1.5. D2

Page 4: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

Exponential notation is a useful way to represent really big numbers in a small space and also for making rapid computations involving large numbers - for example,

the mass of the earth is

5970000000000000000000000 kg

the mass of the moon is 73500000000000000000000 kg

Express the mass of the earth in terms of the lunar mass.

Page 5: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

Can you write the following numbers in exponential notation (powers of 10)?

Density of the earth’s core is 11000 kg/m3

The volume of the earth’s crust is 10000000000000000000 m3

The mass of the earth’s crust is 28000000000000000000000kg

Differences in the acceleration of gravity on the earth’s surface (and elsewhere) are often reported in milligals. 1 milligal =10-5 meters/second2.

Express the acceleration due to gravity of the earth (9.8m/s2) in milligals.

Page 6: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

The North Atlantic Ocean is getting wider at the average rate vs of 4 x 10-2 m/y and has width w of approximately 5 x 106 meters.

1. Write an expression giving the age, A, of the North Atlantic in terms of vs and w.

2. Evaluate your expression. When did the North Atlantic begin to form?

Page 7: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

41.394054.300113.01099.110289.1 237411 xxxxdxT

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

0

1000

2000

3000

4000

5000

T

4th

Polynomial

Here is another polynomial which again attempts to fit the

near-surface 100km. Notice that this 4th order equation (redline plotted in graph) has three bends or turns.

Page 8: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

In sections 2.5 and 2.6 Waltham reviews negative and fractional powers. The graph below illustrates the set of curves that result as the exponent p in

0aaxy p

Is varied from 2 to -2 in -0.25 steps, and a0 equals 0. Note that the negative powers rise quickly up along the y axis for values of x less than 1 and that y rises quickly with increasing x for p greater than 1.

X2

X1.75

X-2

X-1.75

x

0 1 2 3 4 50

200

400

600

800

1000

1200

Y

Power Laws

?01.0 isWhat

?01.0 isWhat

42

2

2

2

Page 9: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

Power Laws - A power law relationship relevant to geology describes the variations of ocean floor depth as a function of distance from a spreading ridge (x).

02/1 daxd

Spreading Ridge

0 200 400 600 800 1000

X (km)

0

1

2

3

4

5

D (km)

Ocean Floor Depth

What physical process do you think might be responsible for this pattern of seafloor subsidence away from the spreading ridges?

Page 10: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

zx 2 6.0

This equation assumes that the initial porosity (0.6) decreases by 1/2 from one kilometer of depth to the next. Thus the porosity () at 1 kilometer is 2-1 or 1/2 that at the surface (i.e. 0.3), (2)=1/2 of (1)=0.15 (I.e. =0.6 x 2-2 or 1/4th of the initial porosity of 0.6.

Equations of the typecxaby

Are referred to as allometric growth laws or exponential functions.

Page 11: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

zx 2 6.0In the equation

a = ?

b = ?

c = ?

cxaby

The constant b is referred to as a base.

Recall relationships like y=10a. a is the power to which the base 10 is raised in order to get y.

Page 12: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

xy 10

By definition, we also say that x is the log of y, and can write

xy x 10loglog

So the powers of the base are logs. “log” can be thought of as an operator like x and which yields a certain result. Unless otherwise noted, the operator “log” is assumed to represent log base 10. So when asked what is

45y where,log y

We assume that we are asking for x such that

4510 x

Page 13: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

Sometimes you will see specific reference to the base and the question is written as

45y where,log10 y

y10log leaves no room for doubt that we are

specifically interested in the log for a base of 10.

ypow 10

One of the confusing things about logarithms is the word itself. What does it mean? You might read log10 y to say -”What is the power that 10 must be raised to to get y?”

How about this operator? -

The power of base 10 that yields () y

653.1log10 y

Page 14: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

5 6 7 8 9 10

Richter Magnitude

0

100

200

300

400

500

600

Num

ber

of e

arth

quak

es p

er y

ear

Observational data for earthquake magnitude (m) and frequency (N, number of earthquakes per year)

What would this plot look like if we plotted the log of N versus m?

Page 15: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

5 6 7 8 9 10

Richter Magnitude

0.01

0.1

1

10

100

1000

Num

ber

of e

arth

quak

es p

er y

ear

cbmN log

The Gutenberg-Richter Relation

-b is the slope and c is the intercept.

Page 16: Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m 1m 2m 5.3m Age = 1500 years Age = 3000 years

In general, base

numberbase

10

10

log

)(log number) some(log

or

b

ab

10

10

log

)(log alog

Try the following on your own

?3log

)7(log 7log

10

103

8log8

21log7

7log4