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# Rod and Spring Approximation - PowerPoint PPT Presentation

Rod and Spring Approximation. Steve Gutstein Eric Freudenthal Ali Jalal-Kamali Vladik Kreinovich David Morgenthaler * University of Texas at El Paso and *Lockheed-Martin.

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### Rod and Spring Approximation

Steve Gutstein

Eric Freudenthal

Ali Jalal-Kamali

David Morgenthaler*

University of Texas at El Paso and *Lockheed-Martin

This work is supported by the NSF, DHS, Texas Instruments, Microsoft, Calculex, and the Computing Alliance of Hispanic Serving Institutions.

Our opinions are not necessarily shared by our funders.

• Procedural Fluency – The ability to execute (mathematical) procedures with competence

• Conceptual Understanding – A sufficiently reflective understanding of which (mathematical) procedure is suitable for a given problem and why to facilitate discovery and understanding of additional problems.

• Productive Disposition – The confidence to apply and develop mathematical solutions for novel problems, when suitable

Linear Best Fit w/ Least Square Error(linear regression)

• High school curriculum requires student use

• But algebraic proof is painful and doesn’t enlighten

• Exposes dilemma. Either

• Expose students to proof. Lessons:

• Math is hard, proofs don’t explain.

• Only geniuses understand them

• No advantage to following proof, just memorize.

• Tell students to consult magician within the calculator

• Worse: memorize an opaque & intricate algorithm

• Lesson: math is magic spells. Only wizards understand magic.

• Both messages disempower

X1

X2

X3

10

p = Raster((20,15))

samples = [( 5.0, 4.0), (10.0,10.0),

(15.0,10.0)]

n = Sx = Sy = Sxy = Sxx = 0

for x,y in samples:

x *= 2; y *= 2

Sx += x; Sy += y

Sxy += x*y; Sxx += x*x

n += 1

p.set((x,y), blue)

m = (n*Sxy – Sx*Sy) / (n*Sxx – Sx**2)

b = (Sy – m*Sx) / n

for x in range(0, 40):

p.set((x, m*x+b), red)

Y2 = Y3

4

Y1

b = 2

5

10

15

m=0.6

mxi+b - yi

-1

+2

-1

Fi =

Ti = xiFi

-5

+20

-15

Balancing forces

ΣFi = 0

Σ(mxi+ b – yi) =0

mΣxi+Σb - Σyi=0

Let Sx=Σxi ;Sy= Σyi

mSx+bn – Sy=0

Balancing torques

ΣTi = 0

Σxi(mxi + b – yi) =0

mΣxi2+ bΣxi - Σxiyi=0

Let Sxx=Σxi2;Sxy=Σxiyi

mSxx+bSx – Sxy= 0

Why does this matter? Least Squares

• Data from our intervention

• Media-Propelled Computational-Thinking

http://iMPaCT-STEM.org

• Engages students in quantitative problem solving

• No sexy graphics. Just simulation & plotting of kinematics

• All math & physics transparent & accessible

• From surveys of college freshmen: