A Tutorial on Active Contours
|
|
- Lester Kennedy
- 5 years ago
- Views:
Transcription
1 Elham Sakhaee March 4, 14 1 Introduction I prepared this technical report as part of my preparation for Computer Vision PhD qualifying exam. Here we discuss derivation of curve evolution function for some well-known energy functionals. Parametric Curves A parametric curve is represented by C(s) (x(s),y(s)) with s being the arc length parametrization. Beitaspacecurveoraplanarcurve, onecanchangearc length parametrization to any arbitrary parametrization p using the following: ds dx +dy ds 1+( dy dx ) dx 1+y dx or ds ( dx dp ) +( dy dp ) dp x p +ypdp (1) WeknowthatcurvelengthisdefinedtobeL dsinarclengthparametrization. Replacing from the given infinitesimal length element ds we get L x p +y pdp C p dp Let s first show that minimizer of such length functional is κ. n L C p dp x p +y pdp () 1
2 Euler-Lagrange Formula L C(p) { L x d L dp x p L y d L dp y p d x p dp x p +yp d y p dp x p +yp x pp x p +yp x p xpp+xpypypp x p +y p x p +y p y pp x p +yp xpypxpp+y p ypp x p +y p x p +y p x ppy p+x py pp (x p +y p )3 y ppx p y px pp (x p +y p )3.( y p ).(x p ) x ppx p +xppy p x p xpp xpypypp (x p +y p )3 y ppx p +yppy p xpypxpp y p ypp (x p +y p )3 y ppx p y p x pp. n κ. n (x p +yp) 3 The above formula is the length minimization problem for a 3D curve. 3 Geodesic Active Contours x ppy p xpypypp (x p +y p )3 y ppx p xpypxpp (x p +y p )3 As we discussed in Differential Geometry tutorial, geodesics are minimal length space curves lying on a surface and connecting two points on that given surface. (3) Figure 1: a geodesic curve on a cone (blue) The formula derived in(3) is for a length minimization problem, however in geodesic active contours we are interested in finding a minimizer that takes the image characteristics (e.g. edges) into account. Let s define a general edge detector function g(c(s)) which can be any function that decreases at edges and increases elsewhere. 1 For example g(c(s)) 1+ I or more traditionally g(c(s)) I The energy functional that we are looking for would be: E α C p dp+λ g( I(C(p)) ) dp (4)
3 Let s define L α C p + λg(c) C p. It can be shown through Maupertuis Principle and Fermat s Principle from dynamicall systems, that minimizing the above energy functional is equivalent to minimizing intrinsic problem g(c(p)) C p dp. The details of proof of this equivalence is out of scope if this report, we refer the reader to ([1]) section.. and appendix A. Asdiscussedabove, solutionof(4)isgivenbyageodesiccurveinriemannian space. In order to minimize L R g(c(p)) C p dp we search for its gradient descent direction which is a way of minimizing the weighted length via steepest descent method. To do so we need to compute Euler-Lagrange formula. g(c(s))ds g(c(p)) C p dp we note that g p g x x p + g y y p g xx p +g y y p g(x(p),y(p))( x p +yp) dp L R C(p)... y ppx p y px pp (x p +y p )3 L R x(p) g x. x p +yp d dp ( x p x p +yp).g(x(p),y(p)) L R y(p) g y. x p +yp d dp ( y p x p +yp).g(x(p),y(p)) g x ( y p )+(g x ( y p )+g y x p )( y p ) y ppx p y px pp g x (x p )+(g x ( y p )+g y x p )(x p ) (x p +y p )3 (κ.g(c(p)) < g(c(p)), n > ) ( y p ) (κ.g(c(p)) < g(c(p)), n >)(x p ) (5) C t (κ.g < g, n >) n Where the n is the inward normal vector at any point of the curve. The geometric intuition of this gradient descent equation is that if we want to reach at a curve with minimal length the best way to evolve the curve C is to follow C(p) t (κ.g(c(p)) < g(c(p)), n >) n since it is the steepest descent equation and will lead to local minimum length curve. The detected object is the steady state solution of (5) when C(p) t. 4 Level-set framework Suppose a planar curve C(t) is represented by cross section of a 3D surface. 3
4 Figure : A curve can be implicitly defined as a level set of a surface Such representation of curve C is advantageous since it is independent of parametrization and hence is intrinsic. On the zero level-set plane, assume φ(x,y,t) is negative inside the zero level-set and positive outside, and by definition zero on the front(curve) itself. One can think of signed distance function as a potential φ. 4.1 Leve-set properties By definition (above) φ(x,y,t i ) is zero along the curve C for any fixed t i. Taking derivative with respect to t, we will get φ x x t +φ y y t hence [x t,y t ] T φ. But we know that the vector [x t,y t ] T is the tangent to the curve, therefore from [x t,y t ] T. φ we conclude that φ is normal to the curve. Accordingly, one can replace the inward normal n by normalized gradient of zero level-set as: φ φ n (6) Now, let s turn our attention to curvature of a level-set. From geometry of curves and surfaces we know that C s is the tangent to curve and C ss is the curvature of the curve. Given above definition for level-set, curvature of a level-set is also φ s in arc length parametrization. φ s s (φ xx s +φ y y s ) (φ xx x s +φ xy y s )x s +φ x x ss +(φ yx x s +φ yy y s )y s +y ss φ y φ xx x s +φ xy y s x s +φ yy y s+ < φ,c ss > (7) In order to simplify this equation we note that C ss κ n and φ φ n which allows for replacing x s with φy φ and y s with φx φ. Therefore we get the curvature of level-set to be κ φ xxφ y φ xy φ x φ y +φ yy φ x φ 3 (8) 4
5 One can simply work this out and get the well-known equation for curvature of the level-set, κ, to be κ. φ φ div( φ φ ) (9) 4. Level-set evolution Now we need to derive evolution equation of the embedding function φ such that evolving curve (C(t)) is represented by evolution of zero level-set of φ. by definition of zero level-set, we have φ(x,y,t) taking total derivative with respect to t we get φ x x t + φ y y t + dφ dt φ t (φ x x t +φ y y t ) [x t,y t ] T φ C t φ (1) In section (3) we derived C t (κg < g, n >) n. Replacing this in (1) we get: φ t (κg < g, n >) n. φ (κg < g, n >) φ φ. φ (κg < g, n >) φ κ φ g+ < g, φ > Replacing κ from (9) we finalize level-set evolution equation φ t as: (11) φ t div( φ ) φ g+ < g, φ > (1) φ 5 Closed Active Contours Beyond Gradient Vector Flow [] and traditional snakes, there are closed active contours that align themselves with the contour of the smoothed gradient vector field [3]. In order to intuitively understand why this alignment improves object detection, we can draw the following image: Figure 3: contour normal being aligned with image gradient 5
6 which shows that boundaries are more likely to lie where gradient of the object is aligned with the contour normal. Maximizing the alignment of the normal vector and gradient of the image is equivalent to minimizing the negative of the inner product of the two. 5.1 a simple case of closed active contours In case of closed active contours the functional to be minimized can be written as[3]: E ρ(< v, n >)ds (13) here n is the inward normal vector and v (u,v) is a vector flow. For now let s assume ρ(α) α. In the next section we deal with a more robust case where ρ(α) α and general case ρ(α). To derive the minimizer of the following energy functional: < v, n > ds [u(x,v),v(x,y)] T.[ y s,x s ]ds [u(x,y),v(x,y)] T 1. C p [ y p,x p ] C p dp [u(x,y),v(x,y)] T.[ y p,x p ] dp v(x,y)x p u(x,y)y p dp (14) d using chain rule for dp v(x,y) and d dpu(x,y), we find the Euler-Lagrange formula to be: { E E C(p) x d dp v(x,y) E y d dp u(x,y) { vx x p u x y p (v x x p +v y y p ) v y x p u y y p +(u x x p )+u y y p ) { (ux +v y )y p (u (u x +v y )x x +v y ). n p if one chooses v I E C(p) I. n { ux y p v y y p v y x p +u x x p (15) As described in section(3), the object edges are steady state solution of I. n E when C(p) since n we conclude that the object edges are I which are zero crossings of Laplacian of the image. 6
7 5. Generalized Closed Active Contours Now assume that ρ(α) α and note the fact u sign(u).u < v, n > ds v(x,y)x p u(x,y)y p dp { E sign(< v, n C >).(vx x p u x y p ) d dp (sign(< v, n >).v(x,y)) sign(< v, n >).(v y x p u y y p ) d dp (sign(< v, n >).u(x,y)) We also note that d dp sign(g(p)) g.δ(g(p)) and is assuming g(p) { sign(v(x,y)xp u(x,y)y p ).( u x y p v y y p ) sign(v(x,y)x p u(x,y)y p ).(v y x p +u x x p ) sign(v(x,y)x p u(x,y)y p )div(v). n Now Let s derive the evolution equation for a generic function ρ(α). ψ ρ (C) ρ(α)ds ρ(α) C p dp ψ C dρ dα. α C C p d dp ( C p C p.ρ(α)+ dρ dα. α C p C p ) We notice the following unities:... C p 1 C p C p.c p t unit tangent α 1 C p C p < v, t >. n (16) (17) (18) (19) 6 Region Based Active Contours Figure 4: Region-Based Image Segmentation 7
8 We assume that there are two regions in the image. we represent their average intensities with µ 1 and µ. The goal is to find the evolution equation of the contours in terms of known values µ 1 and µ and I(x,y). First Let s pose Green s theorem which we find useful in relating line integral along curve C to double integral on the region Ω enclosed by curve C L dx+m dy M x L y dx dy () C Just keep in mind what follows is Chan-Vese segmentation model WITHOUT regularization terms. Although not the exact Chan-Vese model [4] but gives an idea how region-based models can be worked out. Exact Chan-Vese segmentation model is given by: F(µ 1,µ,C) γ.length(c)+ν.area+λ 1 Ω (I µ 1 ) dxdy +λ Ω/ (I µ ) dxdy (1) And is minimzed through level set formulation which is beyond scope of this tutorial. So without regularization terms and setting λ 1 λ 1, energy functional of segmentation model would be: F(C) 1 (I µ 1 ) 1 dxdy + (I µ ) dxdy () Ω/ Opening the integrals, we get: F(C) 1 (I µ 1 ) 1 dxdy + (I µ ) dxdy (I µ ) dxdy (3) Ω and finally: 1 F(C) (µ µ 1) I(µ 1 µ ) dxdy µ +µ 1 (µ µ 1 )(I ) dxdy (4) We now apply Green s theorem to show that evolution equation is F(C) (µ µ 1 )(I µ+µ1 ). n Let M x f(x,y) and L y f(x,y) by Green s theorem we get: C 8
9 f(x,y) dxdy M x L y dxdy C L dx+m dy (5) Now use the fact that dx x s ds and dy y s ds replacing this into (5) we get L.x s +M.y s ds < [ M,L] T, n > ds (6) C From closed active contours section, we know that minimizer of C < v, n > ds is div(v). n,therefore minimizer of above functional is: C F(C) C div([ M,L]). n ( M x +L y ). n f(x,y). n (7) which is the desired result for evolving the regions. References [1] Caselles, V., Kimmel, R., Sapiro, G. (1997). Geodesic active contours. International journal of computer vision, (1), [] Xu, C., Prince, J. L. (1998). Snakes, shapes, and gradient vector flow. Image Processing, IEEE Transactions on, 7(3), [3] Kimmel, R., Bruckstein, A. M. (3). Regularized Laplacian zero crossings as optimal edge integrators. International Journal of Computer Vision, 53(3), [4] Chan, T. F., Vese, L. A. (1). Active contours without edges. Image processing, IEEE transactions on, 1(),
Segmentation using deformable models
Segmentation using deformable models Isabelle Bloch http://www.tsi.enst.fr/ bloch Téécom ParisTech - CNRS UMR 5141 LTCI Paris - France Modèles déformables p.1/43 Introduction Deformable models = evolution
More informationOutline Introduction Edge Detection A t c i ti ve C Contours
Edge Detection and Active Contours Elsa Angelini Department TSI, Telecom ParisTech elsa.angelini@telecom-paristech.fr 2008 Outline Introduction Edge Detection Active Contours Introduction The segmentation
More informationMATH H53 : Final exam
MATH H53 : Final exam 11 May, 18 Name: You have 18 minutes to answer the questions. Use of calculators or any electronic items is not permitted. Answer the questions in the space provided. If you run out
More informationMath 10C - Fall Final Exam
Math 1C - Fall 217 - Final Exam Problem 1. Consider the function f(x, y) = 1 x 2 (y 1) 2. (i) Draw the level curve through the point P (1, 2). Find the gradient of f at the point P and draw the gradient
More informationChapter 2. First-Order Partial Differential Equations. Prof. D. C. Sanyal
BY Prof. D. C. Sanyal Retired Professor Of Mathematics University Of Kalyani West Bengal, India E-mail : dcs klyuniv@yahoo.com 1 Module-2: Quasi-Linear Equations of First Order 1. Introduction In this
More informationNonlinear Diffusion. Journal Club Presentation. Xiaowei Zhou
1 / 41 Journal Club Presentation Xiaowei Zhou Department of Electronic and Computer Engineering The Hong Kong University of Science and Technology 2009-12-11 2 / 41 Outline 1 Motivation Diffusion process
More informationLecture No 1 Introduction to Diffusion equations The heat equat
Lecture No 1 Introduction to Diffusion equations The heat equation Columbia University IAS summer program June, 2009 Outline of the lectures We will discuss some basic models of diffusion equations and
More informationExact Solutions of the Einstein Equations
Notes from phz 6607, Special and General Relativity University of Florida, Fall 2004, Detweiler Exact Solutions of the Einstein Equations These notes are not a substitute in any manner for class lectures.
More informationPrint Your Name: Your Section:
Print Your Name: Your Section: Mathematics 1c. Practice Final Solutions This exam has ten questions. J. Marsden You may take four hours; there is no credit for overtime work No aids (including notes, books,
More informationMath 115 ( ) Yum-Tong Siu 1. Euler-Lagrange Equations for Many Functions and Variables and High-Order Derivatives
Math 115 006-007 Yum-Tong Siu 1 Euler-Lagrange Equations for Many Functions and Variables and High-Order Derivatives The Case of High-Order Derivatives. The context is to find the extremal for the functional
More informationarxiv: v1 [cs.cv] 17 Jun 2012
1 The Stability of Convergence of Curve Evolutions in Vector Fields Junyan Wang* Member, IEEE and Kap Luk Chan Member, IEEE Abstract arxiv:1206.4042v1 [cs.cv] 17 Jun 2012 Curve evolution is often used
More information(a) The points (3, 1, 2) and ( 1, 3, 4) are the endpoints of a diameter of a sphere.
MATH 4 FINAL EXAM REVIEW QUESTIONS Problem. a) The points,, ) and,, 4) are the endpoints of a diameter of a sphere. i) Determine the center and radius of the sphere. ii) Find an equation for the sphere.
More informatione x3 dx dy. 0 y x 2, 0 x 1.
Problem 1. Evaluate by changing the order of integration y e x3 dx dy. Solution:We change the order of integration over the region y x 1. We find and x e x3 dy dx = y x, x 1. x e x3 dx = 1 x=1 3 ex3 x=
More informationMATH 332: Vector Analysis Summer 2005 Homework
MATH 332, (Vector Analysis), Summer 2005: Homework 1 Instructor: Ivan Avramidi MATH 332: Vector Analysis Summer 2005 Homework Set 1. (Scalar Product, Equation of a Plane, Vector Product) Sections: 1.9,
More informationENGI Partial Differentiation Page y f x
ENGI 3424 4 Partial Differentiation Page 4-01 4. Partial Differentiation For functions of one variable, be found unambiguously by differentiation: y f x, the rate of change of the dependent variable can
More informationModulation-Feature based Textured Image Segmentation Using Curve Evolution
Modulation-Feature based Textured Image Segmentation Using Curve Evolution Iasonas Kokkinos, Giorgos Evangelopoulos and Petros Maragos Computer Vision, Speech Communication and Signal Processing Group
More informationCN780 Final Lecture. Low-Level Vision, Scale-Space, and Polyakov Action. Neil I. Weisenfeld
CN780 Final Lecture Low-Level Vision, Scale-Space, and Polyakov Action Neil I. Weisenfeld Department of Cognitive and Neural Systems Boston University chapter 14.2-14.3 May 9, 2005 p.1/25 Introduction
More information1. (a) (5 points) Find the unit tangent and unit normal vectors T and N to the curve. r (t) = 3 cos t, 0, 3 sin t, r ( 3π
1. a) 5 points) Find the unit tangent and unit normal vectors T and N to the curve at the point P 3, 3π, r t) 3 cos t, 4t, 3 sin t 3 ). b) 5 points) Find curvature of the curve at the point P. olution:
More informationReview Sheet for the Final
Review Sheet for the Final Math 6-4 4 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the absence
More informationMath 210, Final Exam, Spring 2012 Problem 1 Solution. (a) Find an equation of the plane passing through the tips of u, v, and w.
Math, Final Exam, Spring Problem Solution. Consider three position vectors (tails are the origin): u,, v 4,, w,, (a) Find an equation of the plane passing through the tips of u, v, and w. (b) Find an equation
More informationDisclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.
Disclaimer: This is meant to help you start studying. It is not necessarily a complete list of everything you need to know. The MTH 234 final exam mainly consists of standard response questions where students
More informationMAT 211 Final Exam. Fall Jennings.
MAT 211 Final Exam. Fall 218. Jennings. Useful formulas polar coordinates spherical coordinates: SHOW YOUR WORK! x = rcos(θ) y = rsin(θ) da = r dr dθ x = ρcos(θ)cos(φ) y = ρsin(θ)cos(φ) z = ρsin(φ) dv
More informationAllen Cahn Equation in Two Spatial Dimension
Allen Cahn Equation in Two Spatial Dimension Yoichiro Mori April 25, 216 Consider the Allen Cahn equation in two spatial dimension: ɛ u = ɛ2 u + fu) 1) where ɛ > is a small parameter and fu) is of cubic
More information12/19/2009, FINAL PRACTICE VII Math 21a, Fall Name:
12/19/29, FINAL PRATIE VII Math 21a, Fall 29 Name: MWF 9 Jameel Al-Aidroos MWF 1 Andrew otton-lay MWF 1 Oliver Knill MWF 1 HT Yau MWF 11 Ana araiani MWF 11 hris Phillips MWF 11 Ethan Street MWF 12 Toby
More informationMath 417 Midterm Exam Solutions Friday, July 9, 2010
Math 417 Midterm Exam Solutions Friday, July 9, 010 Solve any 4 of Problems 1 6 and 1 of Problems 7 8. Write your solutions in the booklet provided. If you attempt more than 5 problems, you must clearly
More informationWORKSHEET #13 MATH 1260 FALL 2014
WORKSHEET #3 MATH 26 FALL 24 NOT DUE. Short answer: (a) Find the equation of the tangent plane to z = x 2 + y 2 at the point,, 2. z x (, ) = 2x = 2, z y (, ) = 2y = 2. So then the tangent plane equation
More informationImage Analysis. Feature extraction: corners and blobs
Image Analysis Feature extraction: corners and blobs Christophoros Nikou cnikou@cs.uoi.gr Images taken from: Computer Vision course by Svetlana Lazebnik, University of North Carolina at Chapel Hill (http://www.cs.unc.edu/~lazebnik/spring10/).
More information+ 2gx + 2fy + c = 0 if S
CIRCLE DEFINITIONS A circle is the locus of a point which moves in such a way that its distance from a fixed point, called the centre, is always a constant. The distance r from the centre is called the
More informationChapter 3. Riemannian Manifolds - I. The subject of this thesis is to extend the combinatorial curve reconstruction approach to curves
Chapter 3 Riemannian Manifolds - I The subject of this thesis is to extend the combinatorial curve reconstruction approach to curves embedded in Riemannian manifolds. A Riemannian manifold is an abstraction
More informationExercise 1 (Formula for connection 1-forms) Using the first structure equation, show that
1 Stokes s Theorem Let D R 2 be a connected compact smooth domain, so that D is a smooth embedded circle. Given a smooth function f : D R, define fdx dy fdxdy, D where the left-hand side is the integral
More information7a3 2. (c) πa 3 (d) πa 3 (e) πa3
1.(6pts) Find the integral x, y, z d S where H is the part of the upper hemisphere of H x 2 + y 2 + z 2 = a 2 above the plane z = a and the normal points up. ( 2 π ) Useful Facts: cos = 1 and ds = ±a sin
More informationMATH 2203 Exam 3 Version 2 Solutions Instructions mathematical correctness clarity of presentation complete sentences
MATH 2203 Exam 3 (Version 2) Solutions March 6, 2015 S. F. Ellermeyer Name Instructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation.
More informationMultivariable Calculus Notes. Faraad Armwood. Fall: Chapter 1: Vectors, Dot Product, Cross Product, Planes, Cylindrical & Spherical Coordinates
Multivariable Calculus Notes Faraad Armwood Fall: 2017 Chapter 1: Vectors, Dot Product, Cross Product, Planes, Cylindrical & Spherical Coordinates Chapter 2: Vector-Valued Functions, Tangent Vectors, Arc
More informationContents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9
MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Contents Multiple Integrals 3 2 Vector Fields 9 3 Line and Surface Integrals 5 4 The Classical Integral Theorems 9 MATH 32B-2 (8W)
More information1. For each function, find all of its critical points and then classify each point as a local extremum or saddle point.
Solutions Review for Exam # Math 6. For each function, find all of its critical points and then classify each point as a local extremum or saddle point. a fx, y x + 6xy + y Solution.The gradient of f is
More informationDerivatives and Integrals
Derivatives and Integrals Definition 1: Derivative Formulas d dx (c) = 0 d dx (f ± g) = f ± g d dx (kx) = k d dx (xn ) = nx n 1 (f g) = f g + fg ( ) f = f g fg g g 2 (f(g(x))) = f (g(x)) g (x) d dx (ax
More informationThe University of British Columbia Final Examination - December 17, 2015 Mathematics 200 All Sections
The University of British Columbia Final Examination - December 17, 2015 Mathematics 200 All Sections Closed book examination Time: 2.5 hours Last Name First Signature Student Number Special Instructions:
More informationCalculus of Variations and Computer Vision
Calculus of Variations and Computer Vision Sharat Chandran Page 1 of 23 Computer Science & Engineering Department, Indian Institute of Technology, Bombay. http://www.cse.iitb.ernet.in/ sharat January 8,
More informationMathematical Tripos Part IA Lent Term Example Sheet 1. Calculate its tangent vector dr/du at each point and hence find its total length.
Mathematical Tripos Part IA Lent Term 205 ector Calculus Prof B C Allanach Example Sheet Sketch the curve in the plane given parametrically by r(u) = ( x(u), y(u) ) = ( a cos 3 u, a sin 3 u ) with 0 u
More informationMTH4101 CALCULUS II REVISION NOTES. 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) ax 2 + bx + c = 0. x = b ± b 2 4ac 2a. i = 1.
MTH4101 CALCULUS II REVISION NOTES 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) 1.1 Introduction Types of numbers (natural, integers, rationals, reals) The need to solve quadratic equations:
More informationSOME PROBLEMS YOU SHOULD BE ABLE TO DO
OME PROBLEM YOU HOULD BE ABLE TO DO I ve attempted to make a list of the main calculations you should be ready for on the exam, and included a handful of the more important formulas. There are no examples
More informationMAT 211 Final Exam. Spring Jennings. Show your work!
MAT 211 Final Exam. pring 215. Jennings. how your work! Hessian D = f xx f yy (f xy ) 2 (for optimization). Polar coordinates x = r cos(θ), y = r sin(θ), da = r dr dθ. ylindrical coordinates x = r cos(θ),
More information(b) Find the range of h(x, y) (5) Use the definition of continuity to explain whether or not the function f(x, y) is continuous at (0, 0)
eview Exam Math 43 Name Id ead each question carefully. Avoid simple mistakes. Put a box around the final answer to a question (use the back of the page if necessary). For full credit you must show your
More informationAPPM 2350 Final Exam points Monday December 17, 7:30am 10am, 2018
APPM 2 Final Exam 28 points Monday December 7, 7:am am, 28 ON THE FONT OF YOU BLUEBOOK write: () your name, (2) your student ID number, () lecture section/time (4) your instructor s name, and () a grading
More informationA Pseudo-distance for Shape Priors in Level Set Segmentation
O. Faugeras, N. Paragios (Eds.), 2nd IEEE Workshop on Variational, Geometric and Level Set Methods in Computer Vision, Nice, 2003. A Pseudo-distance for Shape Priors in Level Set Segmentation Daniel Cremers
More informationExercises for Multivariable Differential Calculus XM521
This document lists all the exercises for XM521. The Type I (True/False) exercises will be given, and should be answered, online immediately following each lecture. The Type III exercises are to be done
More information1 First and second variational formulas for area
1 First and second variational formulas for area In this chapter, we will derive the first and second variational formulas for the area of a submanifold. This will be useful in our later discussion on
More informationMath 265H: Calculus III Practice Midterm II: Fall 2014
Name: Section #: Math 65H: alculus III Practice Midterm II: Fall 14 Instructions: This exam has 7 problems. The number of points awarded for each question is indicated in the problem. Answer each question
More informationx + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the
1.(8pts) Find F ds where F = x + ye z + ze y, y + xe z + ze x, z and where T is the T surface in the pictures. (The two pictures are two views of the same surface.) The boundary of T is the unit circle
More informationCalculus of Variations Summer Term 2014
Calculus of Variations Summer Term 2014 Lecture 5 7. Mai 2014 c Daria Apushkinskaya 2014 () Calculus of variations lecture 5 7. Mai 2014 1 / 25 Purpose of Lesson Purpose of Lesson: To discuss catenary
More informationMA3D9. Geometry of curves and surfaces. T (s) = κ(s)n(s),
MA3D9. Geometry of 2. Planar curves. Let : I R 2 be a curve parameterised by arc-length. Given s I, let T(s) = (s) be the unit tangent. Let N(s) be the unit normal obtained by rotating T(s) through π/2
More informationMLC Practice Final Exam
Name: Section: Recitation/Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages 1 through 13. Show all your work on the standard
More informationNumerical Methods of Applied Mathematics -- II Spring 2009
MIT OpenCourseWare http://ocw.mit.edu 18.336 Numerical Methods of Applied Mathematics -- II Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
More informationMATH 353 LECTURE NOTES: WEEK 1 FIRST ORDER ODES
MATH 353 LECTURE NOTES: WEEK 1 FIRST ORDER ODES J. WONG (FALL 2017) What did we cover this week? Basic definitions: DEs, linear operators, homogeneous (linear) ODEs. Solution techniques for some classes
More informationPage Points Score Total: 210. No more than 200 points may be earned on the exam.
Name: PID: Section: Recitation Instructor: DO NOT WRITE BELOW THIS LINE. GO ON TO THE NEXT PAGE. Page Points Score 3 18 4 18 5 18 6 18 7 18 8 18 9 18 10 21 11 21 12 21 13 21 Total: 210 No more than 200
More informationFinal Exam. Monday March 19, 3:30-5:30pm MAT 21D, Temple, Winter 2018
Name: Student ID#: Section: Final Exam Monday March 19, 3:30-5:30pm MAT 21D, Temple, Winter 2018 Show your work on every problem. orrect answers with no supporting work will not receive full credit. Be
More informationSOLUTIONS TO THE FINAL EXAM. December 14, 2010, 9:00am-12:00 (3 hours)
SOLUTIONS TO THE 18.02 FINAL EXAM BJORN POONEN December 14, 2010, 9:00am-12:00 (3 hours) 1) For each of (a)-(e) below: If the statement is true, write TRUE. If the statement is false, write FALSE. (Please
More informationPRACTICE PROBLEMS. Please let me know if you find any mistakes in the text so that i can fix them. 1. Mixed partial derivatives.
PRACTICE PROBLEMS Please let me know if you find any mistakes in the text so that i can fix them. 1.1. Let Show that f is C 1 and yet How is that possible? 1. Mixed partial derivatives f(x, y) = {xy x
More informationArchive of Calculus IV Questions Noel Brady Department of Mathematics University of Oklahoma
Archive of Calculus IV Questions Noel Brady Department of Mathematics University of Oklahoma This is an archive of past Calculus IV exam questions. You should first attempt the questions without looking
More informationSec. 1.1: Basics of Vectors
Sec. 1.1: Basics of Vectors Notation for Euclidean space R n : all points (x 1, x 2,..., x n ) in n-dimensional space. Examples: 1. R 1 : all points on the real number line. 2. R 2 : all points (x 1, x
More informationFaculty of Engineering, Mathematics and Science School of Mathematics
Faculty of Engineering, Mathematics and Science School of Mathematics GROUPS Trinity Term 06 MA3: Advanced Calculus SAMPLE EXAM, Solutions DAY PLACE TIME Prof. Larry Rolen Instructions to Candidates: Attempt
More informationMath 53 Homework 5 Solutions
14. #: dw dt = w = 14. #7: s = t = Math Homework Solutions dx dt + w (t t 1+t t (1 t) ) (1+t) e (1 t)/(1+t). dy dt + w dz dt = tey/z x z ey/z xy z ey/z s + s = (x y)4 (st) (x y) 4 t = (x y) 4 (st t ).
More informationCS 468. Differential Geometry for Computer Science. Lecture 17 Surface Deformation
CS 468 Differential Geometry for Computer Science Lecture 17 Surface Deformation Outline Fundamental theorem of surface geometry. Some terminology: embeddings, isometries, deformations. Curvature flows
More information1MA6 Partial Differentiation and Multiple Integrals: I
1MA6/1 1MA6 Partial Differentiation and Multiple Integrals: I Dr D W Murray Michaelmas Term 1994 1. Total differential. (a) State the conditions for the expression P (x, y)dx+q(x, y)dy to be the perfect
More informationEdge Detection. CS 650: Computer Vision
CS 650: Computer Vision Edges and Gradients Edge: local indication of an object transition Edge detection: local operators that find edges (usually involves convolution) Local intensity transitions are
More informationCURRENT MATERIAL: Vector Calculus.
Math 275, section 002 (Ultman) Fall 2011 FINAL EXAM REVIEW The final exam will be held on Wednesday 14 December from 10:30am 12:30pm in our regular classroom. You will be allowed both sides of an 8.5 11
More informationReview for the First Midterm Exam
Review for the First Midterm Exam Thomas Morrell 5 pm, Sunday, 4 April 9 B9 Van Vleck Hall For the purpose of creating questions for this review session, I did not make an effort to make any of the numbers
More informationPractice Final Solutions
Practice Final Solutions Math 1, Fall 17 Problem 1. Find a parameterization for the given curve, including bounds on the parameter t. Part a) The ellipse in R whose major axis has endpoints, ) and 6, )
More informationReview problems for the final exam Calculus III Fall 2003
Review problems for the final exam alculus III Fall 2003 1. Perform the operations indicated with F (t) = 2t ı 5 j + t 2 k, G(t) = (1 t) ı + 1 t k, H(t) = sin(t) ı + e t j a) F (t) G(t) b) F (t) [ H(t)
More informationMultiple Integrals and Vector Calculus (Oxford Physics) Synopsis and Problem Sets; Hilary 2015
Multiple Integrals and Vector Calculus (Oxford Physics) Ramin Golestanian Synopsis and Problem Sets; Hilary 215 The outline of the material, which will be covered in 14 lectures, is as follows: 1. Introduction
More informationMathematical Notation Math Calculus & Analytic Geometry III
Name : Mathematical Notation Math 221 - alculus & Analytic Geometry III Use Word or WordPerect to recreate the ollowing documents. Each article is worth 10 points and can e printed and given to the instructor
More informationAn improved active contour model based on level set method
1 015 1 ( ) Journal of East China Normal University (Natural Science) No. 1 Jan. 015 : 1000-5641(015)01-0161-11, (, 0006) :,., (SPF),.,,.,,,,. : ; ; ; : O948 : A DOI: 10.3969/j.issn.1000-5641.015.01.00
More informationMath 210, Final Exam, Practice Fall 2009 Problem 1 Solution AB AC AB. cosθ = AB BC AB (0)(1)+( 4)( 2)+(3)(2)
Math 2, Final Exam, Practice Fall 29 Problem Solution. A triangle has vertices at the points A (,,), B (, 3,4), and C (2,,3) (a) Find the cosine of the angle between the vectors AB and AC. (b) Find an
More informationFormulas for probability theory and linear models SF2941
Formulas for probability theory and linear models SF2941 These pages + Appendix 2 of Gut) are permitted as assistance at the exam. 11 maj 2008 Selected formulae of probability Bivariate probability Transforms
More informationCANONICAL EQUATIONS. Application to the study of the equilibrium of flexible filaments and brachistochrone curves. By A.
Équations canoniques. Application a la recherche de l équilibre des fils flexibles et des courbes brachystochrones, Mem. Acad. Sci de Toulouse (8) 7 (885), 545-570. CANONICAL EQUATIONS Application to the
More informationMajor Ideas in Calc 3 / Exam Review Topics
Major Ideas in Calc 3 / Exam Review Topics Here are some highlights of the things you should know to succeed in this class. I can not guarantee that this list is exhaustive!!!! Please be sure you are able
More informationIdeas from Vector Calculus Kurt Bryan
Ideas from Vector Calculus Kurt Bryan Most of the facts I state below are for functions of two or three variables, but with noted exceptions all are true for functions of n variables..1 Tangent Line Approximation
More informationDirection of maximum decrease = P
APPM 35 FINAL EXAM PING 15 INTUTION: Electronic devices, books, and crib sheets are not permitted. Write your name and your instructor s name on the front of your bluebook. Work all problems. how your
More informationMath 4263 Homework Set 1
Homework Set 1 1. Solve the following PDE/BVP 2. Solve the following PDE/BVP 2u t + 3u x = 0 u (x, 0) = sin (x) u x + e x u y = 0 u (0, y) = y 2 3. (a) Find the curves γ : t (x (t), y (t)) such that that
More informationFinal Review Worksheet
Score: Name: Final Review Worksheet Math 2110Q Fall 2014 Professor Hohn Answers (in no particular order): f(x, y) = e y + xe xy + C; 2; 3; e y cos z, e z cos x, e x cos y, e x sin y e y sin z e z sin x;
More informationName: Instructor: Lecture time: TA: Section time:
Math 222 Final May 11, 29 Name: Instructor: Lecture time: TA: Section time: INSTRUCTIONS READ THIS NOW This test has 1 problems on 16 pages worth a total of 2 points. Look over your test package right
More informationActive Contours Snakes and Level Set Method
Active Contours Snakes and Level Set Method Niels Chr. Overgaard 2010-09-21 N. Chr. Overgaard Lecture 7 2010-09-21 1 / 26 Outline What is an mage? What is an Active Contour? Deforming Force: The Edge Map
More informationModule 2: First-Order Partial Differential Equations
Module 2: First-Order Partial Differential Equations The mathematical formulations of many problems in science and engineering reduce to study of first-order PDEs. For instance, the study of first-order
More informationLine, surface and volume integrals
www.thestudycampus.com Line, surface and volume integrals In the previous chapter we encountered continuously varying scalar and vector fields and discussed the action of various differential operators
More informationSolutions to Sample Questions for Final Exam
olutions to ample Questions for Final Exam Find the points on the surface xy z 3 that are closest to the origin. We use the method of Lagrange Multipliers, with f(x, y, z) x + y + z for the square of the
More informationMathematical Notation Math Calculus & Analytic Geometry III
Mathematical Notation Math 221 - alculus & Analytic Geometry III Use Word or WordPerect to recreate the ollowing documents. Each article is worth 10 points and should be emailed to the instructor at james@richland.edu.
More informationFinal Exam May 4, 2016
1 Math 425 / AMCS 525 Dr. DeTurck Final Exam May 4, 2016 You may use your book and notes on this exam. Show your work in the exam book. Work only the problems that correspond to the section that you prepared.
More informationKillingFusion: Non-rigid 3D Reconstruction without Correspondences. Supplementary Material
KillingFusion: Non-rigid 3D Reconstruction without Correspondences Supplementary Material Miroslava Slavcheva 1, Maximilian Baust 1 Daniel Cremers 1 Slobodan Ilic 1, {mira.slavcheva,maximilian.baust,cremers}@tum.de,
More informationPractice Problems for the Final Exam
Math 114 Spring 2017 Practice Problems for the Final Exam 1. The planes 3x + 2y + z = 6 and x + y = 2 intersect in a line l. Find the distance from the origin to l. (Answer: 24 3 ) 2. Find the area of
More informationLecture 11: Differential Geometry
Lecture 11: Differential Geometry c Bryan S. Morse, Brigham Young University, 1998 2000 Last modified on February 28, 2000 at 8:45 PM Contents 11.1 Introduction..............................................
More informationMATH 32A: MIDTERM 2 REVIEW. sin 2 u du z(t) = sin 2 t + cos 2 2
MATH 3A: MIDTERM REVIEW JOE HUGHES 1. Curvature 1. Consider the curve r(t) = x(t), y(t), z(t), where x(t) = t Find the curvature κ(t). 0 cos(u) sin(u) du y(t) = Solution: The formula for curvature is t
More informationTopics. CS Advanced Computer Graphics. Differential Geometry Basics. James F. O Brien. Vector and Tensor Fields.
CS 94-3 Advanced Computer Graphics Differential Geometry Basics James F. O Brien Associate Professor U.C. Berkeley Topics Vector and Tensor Fields Divergence, curl, etc. Parametric Curves Tangents, curvature,
More informationTotal Variation Theory and Its Applications
Total Variation Theory and Its Applications 2nd UCC Annual Research Conference, Kingston, Jamaica Peter Ndajah University of the Commonwealth Caribbean, Kingston, Jamaica September 27, 2018 Peter Ndajah
More informationMotion Estimation (I) Ce Liu Microsoft Research New England
Motion Estimation (I) Ce Liu celiu@microsoft.com Microsoft Research New England We live in a moving world Perceiving, understanding and predicting motion is an important part of our daily lives Motion
More informationLEVEL SET BASED MULTISPECTRAL SEGMENTATION WITH CORNERS
LEVEL SET BASED MULTISPECTRAL SEGMENTATION WITH CORNERS WENHUA GAO AND ANDREA BERTOZZI Abstract. In this paper we propose an active contour model for segmentation based on the Chan-Vese model. The new
More informationMATH 452. SAMPLE 3 SOLUTIONS May 3, (10 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic.
MATH 45 SAMPLE 3 SOLUTIONS May 3, 06. (0 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic. Because f is holomorphic, u and v satisfy the Cauchy-Riemann equations:
More informationMath 115 ( ) Yum-Tong Siu 1. Canonical Transformation. Recall that the canonical equations (or Hamiltonian equations) = H
Math 115 2006-2007) Yum-Tong Siu 1 ) Canonical Transformation Recall that the canonical equations or Hamiltonian equations) dy = H dx p dp = dx H, where p = F y, and H = F + y p come from the reduction
More informationWeighted Minimal Surfaces and Discrete Weighted Minimal Surfaces
Weighted Minimal Surfaces and Discrete Weighted Minimal Surfaces Qin Zhang a,b, Guoliang Xu a, a LSEC, Institute of Computational Mathematics, Academy of Mathematics and System Science, Chinese Academy
More informationMaxima and Minima. (a, b) of R if
Maxima and Minima Definition Let R be any region on the xy-plane, a function f (x, y) attains its absolute or global, maximum value M on R at the point (a, b) of R if (i) f (x, y) M for all points (x,
More information1. If the line l has symmetric equations. = y 3 = z+2 find a vector equation for the line l that contains the point (2, 1, 3) and is parallel to l.
. If the line l has symmetric equations MA 6 PRACTICE PROBLEMS x = y = z+ 7, find a vector equation for the line l that contains the point (,, ) and is parallel to l. r = ( + t) i t j + ( + 7t) k B. r
More information