{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecnotes2

lecnotes2 - 13.472J/1.128J/2.158J/16.940J COMPUTATIONAL...

This preview shows pages 1–5. Sign up to view the full content.

13.472J/1.128J/2.158J/16.940J COMPUTATIONAL GEOMETRY Lecture 2 Kwanghee Ko T. Maekawa N. M. Patrikalakis Massachusetts Institute of Technology Cambridge, MA 02139-4307, USA Copyright c 2003 Massachusetts Institute of Technology Contents 2 Differential geometry of curves 2 2.1 Definition of curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2.1.1 Plane curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2.1.2 Space curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2 Arc length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 Tangent vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.4 Normal vector and curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.5 Binormal vector and torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.6 Serret-Frenet Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Bibliography 16 Reading in the Textbook Chapter 1, pp.1 - pp.3 Chapter 2, pp.36 - pp.48 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Lecture 2 Differential geometry of curves 2.1 Definition of curves 2.1.1 Plane curves Implicit curves f ( x, y ) = 0 Example: x 2 + y 2 = a 2 It is difficult to trace implicit curves. It is easy to check if a point lies on the curve. Multi-valued and closed curves can be represented. It is easy to evaluate tangent line to the curve when the curve has a vertical or near vertical tangent. Axis dependent. (Difficult to transform to another coordinate system). Example: x 3 + y 3 = 3 xy : Folium of Descartes (see Figure 2.1a) Let f ( x, y ) = x 3 + y 3 - 3 xy = 0 , f (0 , 0) = 0 ( x, y ) = (0 , 0) lies on the curve Example: If we translate by (1,2) and rotate the axes by θ = atan ( 3 4 ), the hyperbola x 2 4 - y 2 2 = 1, shown in Figure 2.1(b), will become 2 x 2 - 72 xy +23 y 2 +140 x - 20 y +50 = 0. Explicit curves y = f ( x ) One of the variables is expressed in terms of the other. Example: y = x 2 It is easy to trace explicit curves. It is easy to check if a point lies on the curve. Multi-valued and closed curves can not be easily represented. It is difficult to evaluate tangent line to the curve when the curve has a vertical or near vertical tangent. 2
-3 -2 -1 0 1 2 -3 -2 -1 0 1 2 X Y asymptote line x+y+1=0 multi-valued -3 -2 -1 0 1 2 3 -2 -1 0 1 2 3 4 X Y Figure 2.1: (a) Descartes; (b) Hyperbola. Axis dependent. (Difficult to transform to another coordinate system). Example: If the circle is represented by an explicit equation, it must be divided into two segments, with y = + r 2 - x 2 for the upper half and y = - r 2 - x 2 for the lower half, see Figure 2.2. This kind of segmentation creates cases which are inconvenient in computer programming and graphics. y = + r - x 2 2 y = - r - x 2 2 x y o Figure 2.2: Description of a circle with an explicit equation. Note: The derivative of y = x at the origin x = 0 is infinite, see Figure 2.3. Parametric curves x = x ( t ) , y = y ( t ) , t 1 t t 2 2 D coordinates ( x, y ) can be expressed as functions of a parameter t . Example: x = a cos( t ) , y = a sin( t ) , 0 t < 2 π 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
y x y = x ; y 0 = 1 / 2 x ; as x 0, y 0 → ∞ Figure 2.3: Vertical slopes for explicit curves involve non-polynomial functions. It is easy to trace parametric curves. It is relatively difficult to check if a point lies on the curve. Closed and multi-valued curves are easy to represent.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 16

lecnotes2 - 13.472J/1.128J/2.158J/16.940J COMPUTATIONAL...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online