#### Meeting Three

U. Memphis, CE 2131
Excerpt: ... CoplanarVectorsand CartesianRepresenta1on Op#mistTheglassishalffull. PessimistTheglassifhalfempty. EngineerTheglassisoverdesigned. Review Twoforcescanbecombinedusingthe parallelogramlawtoformaresultant Aresultantcanbebrokenupintoits componentsusingthegeometryofthesystem andsometrig 2 Cartesian Coordinate s 23 January 2009 1 CartesianCoordinates Thereisaspecialcasewhentwocomponents thatareperpendiculararecombined 3 Cartesian Coordinate s 23 January 2009 CartesianCoordinates Forexample,considertwoforcesF1andF2 4 Cartesian Coordinate s 23 January 2009 2 CartesianCoordinates Again,wecanextendtheirlinesofac1onto thepointwheretheyintersect 5 Cartesian Coordinate s ...

#### relativity_2005_lecture_2

East Los Angeles College, PHY 314
Excerpt: ... Lecture 2: Coordinate Transforms and Dot Products PHY 314 - Relativity Edward Daw edaw@shef.ac.uk Office: room D28 Office hours: Fridays, 10:00 - noon extension: 2-4353 Recap on Vectors .we discussed vectors and their behaviour under coordinate transformations. Under a transformation of coordinates, the vector itself is unaffected. The coordinates of the vector do change when you transform between coordinate systems Recap on Components and Basis Vectors .or in some other coordinate system Suppose you know the transformation between the coordinates (x,y) in one coordinate system and the coordinates (x,y) in another. We studied two examples last time: Transform from 2d cartesian coordinate s to plane polar coordinates .and another example: transformation from one set of cartesian coordinate s to another which is rotated through an angle with respect to the former one. Lets use the Einstein Summation Convention Transformation of Basis Vectors between Coordinate Systems If you know the tra ...

#### LEC1NOTES

Washington, FM 360
Excerpt: ... LECTURE NOTES 1 Plot the following functions using a Cartesian coordinate system: 3X1 - 5X2 = 11 X1 + 2X2 = 11 CARTESIAN COORDINATE SYSTEM 25 20 15 X2 10 5 0 0 5 10 X1 15 20 25 Determine the simultaneous solution to this system of linear equations: graphically, analytically. What are the slopes and vertical intercepts of these two straight lines? Indicate the region on the graph corresponding to the inequalities: 3X1 - 5X2 11 X1 + 2X2 11 Plot the quadratic equation: X2 = -2 X12 + 8 X1 - 3.5 Analytically determine the two points at which this equation intercepts the horizontal axis. What is the equation for the instantaneous slope of the curve, dX2/dX1? What is the numerical value of the slope at the point on the curve where x1 = 1? Find the maximum point (X1, X2) of this curve analytically. ...

#### guide

Emory, CHEM 731
Excerpt: ... NOTES ON NORMAL MODE ANALYSIS CODE The source codes are dpes_shell.f, normal.f plus a user-supplied potential code which is called by dpes_shell.f. The input to the user code is the Cartesian coordinate s of all atoms. (See deps_shell.f for details.) To run normal.f the Cartesian coordinate s of the stationary point must be known. These are other information are given in equil.dat. Samples files are given for H2O. ...

#### m2334.Manif.a3sols

Allan Hancock College, ICS 334
Excerpt: ... MACQUARIE UNIVERSITY MATH233/334 (2006D1) / Tanya Schmah Solutions to Manifolds Assignment 3. 1. (Exercise 5.5 from the lecture notes) The vector field X = + , defined on all of S 2 except the north and south poles, appeared in Assignment 2. Prove that X has no continuous extension to S 2 . What this means is: though, by defining X at the poles, it is possible to extend X to be a vector field on all of S 2 , this can never be done in such a way that the extended vector field is continuous. Solution. X has no continuous extension to S 2 because the limit of X(x, y, z) as (x, y, z) tends to (1, 0, 0) does not exist, and the limit of X(x, y, z) as (x, y, z) tends to (-1, 0, 0) does not exist (we only need to prove this for one of the points). There are two general methods for proving this: use cartesian coordinate s in R3 , or use local coordinate systems that cover the poles. Since the most likely candidates for local coordinate systems (including stereographic projection) are defined in terms of cartesian c ...

#### cm08ln_polar

Delaware, PHYSICS 419
Excerpt: ... PHYS 419: Classical Mechanics Lecture Notes POLAR COORDINATES A vector in two dimensions can be written in Cartesian coordinate s as ^ ^ r = xx + y y (1) ^ ^ where x and y are unit vectors in the direction of Cartesian axes and x and y are the components of the vector, see also the figure. It is often convenient to use coordinate systems other than the Cartesian system, in particular we will often use polar coordinates. ^ These coordinates are specified by r = |r| and the angle between r and x, see the figure. The relations between the polar and Cartesian coordinate s are very simple: x = r cos and r= x2 + y 2 y = arctan . x y = r sin ^ ^ The unit vectors of polar coordinate system are denoted by r and . The former one is defined accordingly as ^ r= Since ^ ^ r = r cos x + r sin y, ^ ^ ^ r = cos x + sin y. ^ ^ ^ ^ The simplest way to define is to require it to be orthogonal to r , i.e., to have r = 0. This gives the condition cos x + sin y = 0. 1 r r (2) The simplest solution is x = - sin and ...

#### ps04v2

Caltech, PH 106
Excerpt: ... Physics 106a Problem Set 4 Due Oct 28, 2004 Version 2 October 26, 2004 These problems cover the material on analytical mechanics in Hand and Finch Chapters 1 and 2 and Section 2.1 and 2.2 of the lecture notes Lagrangian mechanics with generalized coordinates, variational calculus and variation dynamics with constraints applied via Lagrange multipliers. Please again write down the rough amount of time you are spending on each problem. When we say "solve using unconstrained generalized coordinates", that means you should find the set of generalized coordinates such that the constraints no longer appear (e.g., for a point on a sphere, this would be spherical coordinates with motion allowed only in and ). In other cases, you will be asked to use cartesian coordinate s and incorporate constraints via Lagrange multipliers. Clarifications since v. 1: Explicitly indicate that calculus of variations should be used to do problem 1. Clarify what is meant by "uncoupling" in Problem 3. 1. Use the calculus of va ...

#### lecture10-04

Wisc La Crosse, M 182
Excerpt: ... M182 Lecture Worksheet, Sec. 10.4 Name: Polar Coordinates: Polar coordinates are an alternative to our usual Cartesian coordinate system. The polar coordinates of a point are given in terms of a distance (r) and a direction (). The angle is measured in radians counter-clockwise from the positive x-axis. Points are expressed P (r, ). The polar and Cartesian coordinate systems are linked by the following pairs of equations: (1) (2) x = r cos x2 + y 2 = r 2 y = r sin tan = y . x 1. Example: Plot the points whose polar coordinates are given. (a) (1, 3/4) (b) (2, 3) (c) (2, -/4) (d) (-1, /3) (e) (-2, 0) 2. Example: Convert the following points from polar to Cartesian coordinate s. (a) (4, /6) (b) (-2, /3) (c) (1, 3/2) 3. Example: Convert the following points from Cartesian to polar coordinates. (a) (1, 1) (b) (1, -1) 4. Example (Graphing in Polar): Sketch and Describe the following polar curves: (a) r = 4 (b) r = -4 (c) = 1 5. Example: Describe the region in the plane consisting of points which satisfy the ...

#### lecPart

University of Florida , CAP 4730
Excerpt: ... Jorg's Graphics Lecture Notes 4a. Space Partition 1 4a. Space Partition Barycentric coordinates, Convex hull If n + 1 points P0 , . . . , Pn in Rn are in general position (span the space) then they define a unique coordinate system with barycentric coordinates (0 , 1 , . . . , n ), i = 1. Special barycentric coordinates, the Cartesian coordinate s arise when one point is made special and called origin, and the difference vectors to the other points are orthogonal. We then drop 0 since it multiplies the point 0. In 2D, the barycentric coordinates of Q with respect to P1 , P2 , P3 relate to Cartesian coordinate s via -+ x P1 y P1 1 x P2 y P2 1 x x P3 0 Q y P3 1 = Qy 1 2 1 +-+ + -+ +- +- -+- Point Q is strictly inside the triangle if and only if all i > 0. Generalization: Q convex hull (P1 , . . . , Pn ) : Q= 1= n i=1 i Pi n i=1 i , i 0. Jorg's Graphics Lecture Notes 4a. Space Partition 2 Collision Detection Eulerian (fixed coordinate system) approach. Quad-tree, k-D tree, BSP ...

#### l-23

Furman, MATH 21
Excerpt: ... Lecture 23: Cylindrical and Spherical Coordinates 23.1 Cylindrical coordinates If P is a point in 3-space with Cartesian coordinate s (x, y, z) and (r, ) are the polar coordinates of (x, y), then (r, , z) are the cylindrical coordinates of P . If f : R3 R is continuous on a region in space described by D in Cartesian coordinate s and by T in cylindrical coordinates, then f (x, y, z)dxdydz = D T f (r cos(), r sin(), z)rdrddz. Example If V is the volume of the region D bounded by the two surfaces z = x2 + y 2 and z = 2 x2 y 2 , then, changing to cylindrical coordinates, 1 2 0 2r 2 1 2 V = D dxdydz = 0 r2 rdzddr = 0 0 2(r r3 )ddr = , where the evaluation of the nal double integral follows as in a previous example. 23.2 Spherical coordinates We may describe a point P in 3-space using coordinates (, , ) where is the distance from P to the origin, is the polar coordinate angle for the projection of P onto the horizontal plane, and is the angle between t ...

#### APPM1360Spring07Final

Excerpt: ... was able to study this curve in 225 BC then surely you can: (a) Write its equation in cartesian coordinate s. You may leave the equation in implicit form. (b) Find the cartesian coordinate s of the points P1 , P2 and P3 , where the spiral passes through the positive xaxis. (c) Find the area enclosed by the spiral between P1 and P2 . (d) Set up, but DO NOT evaluate, the integral describing the total (P1 x P3 ) length of the spiral. d cos(x). dx P1 P2 P3 Good Luck! ...

#### W8.2

Michigan State University, PHY 183
Excerpt: ... hen write the position of the Cartesian components of center of gravity of an object as 1 X= M ! " x! (r )dV ; 1 Y= M ! " y ! (r )dV ; 1 Z= M ! " z ! (r )dV V V V If the density is constant for the whole object we can write ! ! ! 1 ! R= " rdV = V V rdV (for constant ! ) " MV Or X= 1 ! xdV; VV Y= 1 ! ydV; VV Z= 1 ! zdV VV 6 February 26, 2008 Physics for Scientists&Engineers 1 Position of the Center of Gravity-1d In a one-dimensional case, the integration just extends over 1 coordinate axis. The mass-density is then expressed as mass-per-unit-length ! (x) Center of mass coordinate in 1d-case: 1 X= M x2 x1 " x! (x)dx and M = " ! (x)dx x1 x2 You can show that for constant density in 1d-case: X = 1 (x2 ! x1 ) 2 February 26, 2008 Physics for Scientists&Engineers 1 7 Three-Dimensional Non- Cartesian Coordinate Systems In chapter 1, we introduced a three-dimensional orthogonal coordinate system For some applications it is useful to represent the positio ...

#### Math2210Midterm3Review

Utah, MATH 2210
Excerpt: ... Math2210 Midterm 3 Review Problems Summer, 2007 Kelly MacArthur Chapter 12 Review (pg 673) Sample Test Problems #27-29 all Chapter 13 Review (pg 728-729) Concepts Test Problems #1-5 all, 11, 13, 14, 16, 17 Sample Test Problems #1-12 all, 15, 16, 20 The exam covers sections 12.9, 13.1-13.4 and 13.6-13.8. You will NOT need a calculator for this exam. You can bring one 8.5 x 11 inch piece of paper of notes (on both sides, if you want) to use as a reference. Topics Lagrange Method Double Integrals Cartesian coordinate s Polar coordinates Area Triple Integrals Cartesian coordinate s Cylindrical coordinates Spherical coordinates Volume Surface Area ...

#### exam1_2006

University of Florida , EML 6282
Excerpt: ... EML 6282 - Robot Geometry 2 1 February 2006 Exam 1 55 minutes ; open book and notes The coordinates of two planes are given as [10 ; 2, 4, 0] and [0 ; 0, -2, 4] where the first term has units of meters and the last three terms are dimensionless. Determine the coordinates of the point on the line formed by the intersection of the two planes that is closest to the point (1 ; 0, 0,10). The Cartesian coordinate s of the point have units of meters. ...

#### ccc1989

Villanova, CCC 1989
Excerpt: ... Curvilinear Coordinates and Curvature Curvilinear Coordinates and Curvature by bob jantzen [January 1989] University of Rome "La Sapienza", Dipartimento di Fisica, Corso di Fisica Teorico (Ruffini) A brief motivation for how connection and curvature arise by considering transforming from Cartesian coordinate s and ordinary derivatives to spherical coordinates and the corresponding covariant derivatives. Written as a preference to existing notes to be reused. ccc1989.pdf: 5 pages, 125K Page 1 ...

#### final-study

Clemson, M 206
Excerpt: ... (d) Setup Setup Setup Setup a a a a double double double double integral integral integral integral for for for for the the the the area in Cartesian coordinate s. area in polar coordinates. mass in Cartesian coordinate s. mass in polar coordinates. 8 2. The region is the inside of the triangular plate with vertices (-2, 0), (4, 0) and (4, 6) and the mass density is (x, y) = 2x + y. (a) (b) (c) (d) Setup Setup Setup Setup a a a a double double double double integral integral integral integral for for for for the the the the area in Cartesian coordinate s. area in polar coordinates. mass in Cartesian coordinate s. mass in polar coordinates. 3. The region is the inside of the triangular plate with vertices (-2, -2), (4, -2) and (4, 6) and the mass density is (x, y) = 2x + y. (a) (b) (c) (d) Setup Setup Setup Setup a a a a double double double double integral integral integral integral for for for for the the the the area in Cartesian coordinate s. area in polar coordinates. mass in Cartesian coordinate s. mass in ...

#### reln_dot_prod_Cartsn

S.F. State, METR 402
Excerpt: ... A B = Ax Bx + A y By + AzBz Dot product expressed in Cartesian coordinate s ...

#### lec_18_ece_220_f_2007_with_figs

Wisconsin, ECE 220
Excerpt: ... e integrated magnetic field from equation (11) by rewriting it as F12 = I2 d 2 B12 (14) C2 and B12 is B12 = 0 4 I1 d C1 aR 0 I1 = 2 R 4 1 d C1 1 aR R2 (15) Again, as can be seen in equations (13) and (15), the magnetic field behaves a lot like the electric field in that its magnitude decreases as the square of the distance between the current-carrying elements. Vector Magnetic Potential Since the divergence of B is zero, then we know that B can be expressed as the curl of a vector, since the divergence of the curl is always zero. Thus we can write B= Thus B= A (17) where A is the magnetic vector potential. We can then insert equation (17) into a Ampere's Law, equation (4), and obtain ( A) = 0 J (18) ( A) = 0 (16) Now the curl of the curl operation in equation (18) can be changed to ( A) - 2 A = 0 J (19) This identity is especially helpful in cartesian coordinate s. Note that in equation (19) the second term on the left-hand side of the equation is del squared of a vector, not a ...

#### m2334.Manif.assign3

Allan Hancock College, ICS 334
Excerpt: ... MACQUARIE UNIVERSITY MATH233/334 (2006D1) / Tanya Schmah Manifolds Assignment 3. Due in class Friday 19 May (Week 10) 1. (Exercise 5.5 from the lecture notes) The vector field X = + , defined on all of S 2 except the north and south poles, appeared in Assignment 2. Prove that X has no continuous extension to S 2 . What this means is: though, by defining X at the poles, it is possible to extend X to be a vector field on all of S 2 , this can never be done in such a way that the extended vector field is continuous. 2. Let X(x, y, z) = (-y, x, 1) and Y (x, y, z) = (x, y, z). Compute LX Y in two ways: (i) by finding the flow of X and computing d dt Y ; t t=0 (ii) by computing the Jacobi-Lie bracket [X, Y ]. 3. Let x0 be the point in S 2 given in cartesian coordinate s as x0 = ( 1 , 0, 2 ates, dy(x0 ) = (0, 1, 0). What is dy(x0 ) in spherical coordinates? 3 2 ). In cartesian coordin- 4. Let (x, y) = (yex , y 2 ) (in cartesian coordinate s). Compute dx and x ( dx). 5. Choose a topic about which you wi ...

#### Supplement_to_Worksheet_4

Washington, MATH 126
Excerpt: ... Supplement to Worksheet 4 Worksheet 4 is slightly ahead of the material in lecture. However, it still is a useful tool to give you an introduction to the concepts pertaining to polar coordinates. Please read through this supplement before beginning Worksheet 4. Introduction There is more than one way to describe a location in a coordinate system. You are all familiar with the Cartesian coordinate method. Cartesian method: (x, y) 1. Stand on the origin. 2. First, walk x units on the x-axis. 3. Then, walk y units parallel to the y-axis. This is kind of like describing how to get somewhere by driving along streets. However, in some scenarios it is more convenient to give the location in terms of an angle and a radius. For example, imagine you are firing a cannon (you need to know where to aim it and how powerful to shoot it). We call this the Polar coordinate method. Polar coordinates: (r, ) 1. Stand on the origin facing the positive x-axis. 2. Rotate counterclockwise by the angle . 3. Walk (or fire your cannon) ...

#### lect4b_2

Rutgers, PHYSICS 504
Excerpt: ... Last Latexed: February 5, 2009 at 17:08 1 504: Lecture 4b Last Latexed: February 5, 2009 at 17:08 2 Physics 504, Lecture 4, Part 2, Lecture 5 Feb. 2, 5, 2009 Copyright c 2009 by Joel A. Shapiro 1 Curvilinear Coordinates Many of our notions in vector calculus are developed in the context of cartesian coordinate s describing a Euclidean space. We will use r i , i = 1, 2, . . . D as cartesian coordinate s describing a D dimensional Euclidean space. (Note the index on the coordinates has been written as a superscript rather than as a subscript this is in preparation for discussing Minkowski space soon and curved spaces, possibly, later in your career.) Usually we will take D = 3. Being a Euclidean space, the distance s between the points labelled with {r i} and {r i + r i } is given by Pythagoras: (s)2 = i But there are many situations in which it is useful to use coordinates other than cartesian, not to mention spaces which are other than Euclidean. So we need to develop the expressions ap ...

#### lect4b

Rutgers, PHYSICS 504
Excerpt: ... Last Latexed: February 5, 2009 at 17:08 1 Physics 504, Lecture 4, Part 2, Lecture 5 Feb. 2, 5, 2009 Copyright c 2009 by Joel A. Shapiro 1 Curvilinear Coordinates Many of our notions in vector calculus are developed in the context of cartesian coordinate s describing a Euclidean space. We will use r i , i = 1, 2, . . . D as cartesian coordinate s describing a D dimensional Euclidean space. (Note the index on the coordinates has been written as a superscript rather than as a subscript this is in preparation for discussing Minkowski space soon and curved spaces, possibly, later in your career.) Usually we will take D = 3. Being a Euclidean space, the distance s between the points labelled with {r i} and {r i + r i } is given by Pythagoras: (s)2 = i (r i )2 . Vectors are discribed in terms of unit vectors ei , and a displacement is a vector r = i r i ei . Scalar and vector elds can be considered functions of r or functions of the D coordinates {r i }, and the gradiant and lap ...

#### lec_6_relativity_comps_summary

East Los Angeles College, PHY 314
Excerpt: ... Last Time Relativity & Cosmology PHY 314 - Relativity Edward Daw edaw@shef.ac.uk Lecture 6 - Christoffel Symbols from the Metric Components .we derived an expression for the covariant derivative of the components of the metric tensor. The covariant derivative transforms like a tensor, so we can use the usual rules to write down its transformation law: The reason why covariant derivatives are useful is they provide a powerful mechanism for proving things in arbitrary coordinate systems. The first part of this mechanism is, in cartesian coordinate systems, they become things we recognize. For example: FLAT SPACE ONLY: Proof Mechanism The second part of this system for proving things is to note that once equations are written in terms of tensors, those equations are valid in ANY coordinate system in a flat space, even a curvilinear coordinate system. So the mechanism is: 1) Derive a result in a cartesian coordinate system where things are familiar. 2) Write that result in terms of objects with tensor tran ...

#### relativity_lecture_6

East Los Angeles College, PHY 314
Excerpt: ... Relativity & Cosmology PHY 314 - Relativity Edward Daw edaw@shef.ac.uk Lecture 6 - Christoffel Symbols from the Metric Components Last Time .we derived an expression for the covariant derivative of the components of the metric tensor. The covariant derivative transforms like a tensor, so we can use the usual rules to write down its transformation law: The reason why covariant derivatives are useful is they provide a powerful mechanism for proving things in arbitrary coordinate systems. The first part of this mechanism is, in cartesian coordinate systems, they become things we recognize. For example: FLAT SPACE ONLY: Proof Mechanism The second part of this system for proving things is to note that once equations are written in terms of tensors, those equations are valid in ANY coordinate system in a flat space, even a curvilinear coordinate system. So the mechanism is: 1) Derive a result in a cartesian coordinate system where things are familiar. 2) Write that result in terms of objects with tensor tra ...