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MA 242.003 • Day 14- January 25, 2013 • Chapter 10, sections 10.3 and 10.4
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MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

Jan 08, 2018

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Page 1: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

MA 242.003

• Day 14- January 25, 2013• Chapter 10, sections 10.3 and 10.4

Page 2: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.
Page 3: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.
Page 4: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.
Page 5: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

Section 10.3Arc Length and Curvature

• To describe the acceleration r’’(t) it turns out that the crucial idea is CURVATURE of the curve.

Page 6: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

Section 10.3Arc Length and Curvature

• To describe the acceleration r’’(t) it turns out that the crucial idea is CURVATURE of the curve.

• Compare the unit tangents for– 1. a straight line– 2. a curved line

Page 7: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.
Page 8: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

•Curvature of a straight line is then ZERO

•Curvature of a non-straight line is then NON-ZERO

Page 9: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

•Curvature of a straight line is then ZERO

•Curvature of a non-straight line is then NON-ZERO

•Problem: The number for the curvature depends on choice of parameter.

Page 10: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.
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The Solution

• Everyone must use the same parameter for the curve

Page 15: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

The Solution

• Everyone must use the same parameter for the curve

• The only unique parameter along the curve is the arc length parameter s

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Solution:

Page 20: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

(Continuation of problem)

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Solution:

Page 24: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

(continuation of problem)

Page 25: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

Solution:

Page 26: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

(continuation of problem)

Page 27: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.
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(continuation of problem)

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So now we have a good definition of curvature.

Page 33: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

So now we have a good definition of curvature.

The problem now is that we CANNOT do the calculation for general curves because we cannot compute

Page 34: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

So now we have a good definition of curvature.

The problem now is that we CANNOT do the calculation for general curves because we cannot compute

Page 35: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

So now we have a good definition of curvature.

The problem now is that we CANNOT do the calculation for general curves because we cannot compute

Page 36: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.
Page 37: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.
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New calculation of the curvature of a helix:

Page 55: MA 242.003 Day 14- January 25, 2013 Chapter 10, sections 10.3 and 10.4.

The Unit Normal Vector N(t)

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