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I feel like I’m I feel like I’m diagonally parked in a diagonally parked in a parallel universe” parallel universe”
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“I feel like I’m diagonally parked in a parallel universe”

Jan 01, 2016

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Hilary Reed
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Page 1: “I feel like I’m diagonally parked in a parallel universe”

““I feel like I’m diagonally I feel like I’m diagonally parked in a parallel parked in a parallel

universe”universe”

Page 2: “I feel like I’m diagonally parked in a parallel universe”

A)A) IntroductionIntroductiona.a. SymbolsSymbolsb.b. OperOperatioationsnsc.c. Central TendenciesCentral Tendencies

B)B) Linear AlgebraLinear AlgebraC)C)Correlation/Regression AnalysisCorrelation/Regression Analysis D)D) Applied CalculusApplied Calculus

Math ReviewMath ReviewMonday June 7 2003Monday June 7 2003

Page 3: “I feel like I’m diagonally parked in a parallel universe”

B)B)System of equationsSystem of equationsBasic Math ReviewBasic Math Review

a)a) 33xx - - yy = -7 = -755yy + 5 = -5 + 5 = -5xx

b)b) 3x + 4y = 23x + 4y = 22y = 4 - 3/2x 2y = 4 - 3/2x

Page 4: “I feel like I’m diagonally parked in a parallel universe”

D)D) Applied CalculusApplied CalculusBasic Math ReviewBasic Math Review

Rate of change (slope): Rate of change (slope): y/y/x or (yx or (y22-y-y11)/(x)/(x22-x-x11))HereHere

y/y/x is constant regardless of the “limit”x is constant regardless of the “limit”

0

20

40

60

80

100

120

140

160

0 1000 2000 3000 4000

Broadway Address

Street Number

Page 5: “I feel like I’m diagonally parked in a parallel universe”

D)D) Applied CalculusApplied CalculusBasic Math ReviewBasic Math Review

How long does it take to fill one beaker (1L)?How long does it take to fill one beaker (1L)?

V(t)

t

⎥⎦

⎤⎢⎣

⎡=dt

dVmlt /1000

dV

dt

⎡ ⎣ ⎢

⎤ ⎦ ⎥=1000ml / t

Page 6: “I feel like I’m diagonally parked in a parallel universe”

D)D) Differential equationsDifferential equationsBasic Math ReviewBasic Math Review

A differential equation is an equation in which A differential equation is an equation in which one or more unknowns depend on its/their one or more unknowns depend on its/their rate rate of changeof change (or that of other variables included in (or that of other variables included in the equations)the equations)

2

2 )()(

dt

txd

dt

tdva

where

maF

==

=

Newton’s principiaNewton’s principia

Page 7: “I feel like I’m diagonally parked in a parallel universe”

D)D) Definition of a derivativeDefinition of a derivativeBasic Math ReviewBasic Math Review

x

xfxxfxf

x −+

=→

)()()( lim

00

The derivative of a function The derivative of a function ((xx)) at a point at a point “a” is the slope of the straight line tangent “a” is the slope of the straight line tangent to to ((xx)) at “a” at “a” instantaneous rate of instantaneous rate of change!change!One is pushing to limit to “0”: the slope is One is pushing to limit to “0”: the slope is close to real as close to real as ((xx)) approaches 0 approaches 0

Page 8: “I feel like I’m diagonally parked in a parallel universe”

D)D) Definition of a derivativeDefinition of a derivativeBasic Math ReviewBasic Math Review

x

xfxxfxf

x −+

=→

)()()( lim

00

The derivative of a function The derivative of a function ((xx)) = = f’f’((xx))Important derivatives:Important derivatives:ff((xx) = C ) = C f’f’((xx) = 0 ) = 0 ff((xx) = ) = xxn n f’f’((xx) = n) = nxxn-1n-1 ff((xx) = ) = eexx f’f’((xx) = ) = eexx ff((xx) = ln) = lnxx f’f’((xx) = 1/) = 1/xx

Page 9: “I feel like I’m diagonally parked in a parallel universe”

D)D) Maxima, MinimaMaxima, MinimaBasic Math ReviewBasic Math Review

One of the great applications of calculus One of the great applications of calculus (particularly in economics) is to determine the (particularly in economics) is to determine the “maxima” and “minima” of functions.“maxima” and “minima” of functions.The derivatives of the maxima and minima = 0The derivatives of the maxima and minima = 0

The function neither increase nor decreases!The function neither increase nor decreases!

ff((xx) = ) = xx33 – 3 – 3xx22 – 24 – 24x + x + 55

f’f’((xx) = 3) = 3xx22 – 6 – 6xx – 24 – 243 (3 (xx22 – 2 – 2xx – 8) – 8)3 (3 (xx + 2)( + 2)(xx –4) –4)

f’f’((xx)) vanishes (reaches a critical point) only when vanishes (reaches a critical point) only when f’f’((xx) = 0) = 0

Page 10: “I feel like I’m diagonally parked in a parallel universe”

D)D) Maxima, MinimaMaxima, Minima

-200

-150

-100

-50

0

50

100

150

200

250

-8 -6 -4 -2 0 2 4 6 8 10

f”f”((xx) < 0 ) < 0 maximum maximumf”f”((xx) > 0 ) > 0 minimum minimum

Page 11: “I feel like I’m diagonally parked in a parallel universe”

D)D) Application: DerivativesApplication: DerivativesJi = - DS ([Ci ]/z)

DS = D0 2

J Cz=0 = -3 D0 (8 o

C) ([Ci ]/z)z=0

0

5

10

15

20

25

30

35

0 500 1000 1500 2000

CO2

Depth (cm)

0 20 40 60 80 100 120

CH4

CO2 (uM)CH4 (uM)

Page 12: “I feel like I’m diagonally parked in a parallel universe”

D)D) Application: DerivativesApplication: DerivativesJi = - DS ([Ci ]/z)

DS = D0 2

J Cz=0 = -3 D0 (8 o

C) ([Ci ]/z)z=0

y = -0.0002x6 + 0.017x5 - 0.682x4 + 13.528x3 - 137.88x2 + 666.46x + 531.18

R2 = 0.9798

1000

1100

1200

1300

1400

1500

1600

1700

1800

0 5 10 15 20 25 30 35

Depth (cm)

CO2 (uM)

Poly.(CO2(uM))

Page 13: “I feel like I’m diagonally parked in a parallel universe”

““A River (used to) Run Through It”: Part IA River (used to) Run Through It”: Part I

Page 14: “I feel like I’m diagonally parked in a parallel universe”

Reservoirs and soil erosionReservoirs and soil erosion

0

5

10

15

20

25

30

35

40

45

50

0% 2% 4% 6% 8% 10% 12%Fo (%)

Depth (cm)

12 14 16 18 20

(C/N)a

Fo

C/N

1200

Page 15: “I feel like I’m diagonally parked in a parallel universe”

D)D) Application: IntegralApplication: Integral

0

5

10

15

20

25

30

35

40

45

50

0.0 0.2 0.4 0.6 0.8 1.0J-Co (gC/m2.yr)

Depth (cm)

1200

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 10 20 30 40 50 60 70 80 90

Age (years)

Jo (gC/m2.yr)

Page 16: “I feel like I’m diagonally parked in a parallel universe”

D)D) Application: IntegralApplication: Integral(x) = Polynomial (6th degree)

∫ (xn) = [n(n+1)/(n+1)] + C

y = -2E-10x 6 + 4E-08x 5 - 3E-06x 4 + 0.0001x3 - 0.0013x 2 + 0.0136x + 0.3082

R2 = 0.9795

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 10 20 30 40 50 60 70 80 90

Age (years)

Jo (gC/m2.yr)

Integral between two limits (0 and 85 yrs)∫ (xn)100 - ∫ (xn)0

C - C