Rod and Spring Approximation

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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.

Goals of STEM courses
• 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?
• 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: