DOCSLIB.ORG

01. Coordinate Geometry of the Straight Line

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
01. Coordinate Geometry of the Straight Line 1. COORDINATE GEOMETRY OF A STRAIGHT LINE Lines on the Cartesian Plane A straight line, drawn on a Cartesian Plane can be described by an equation. Such equations have a 3 general form and vary depending on where the line x=-3 x=2 x=4 cuts the axes and its degree of slope. 2 Horizontal Lines 1 Horizontal lines are all parallel to the x-axis. Hence, their equations resemble the equation of the x-axis. -4 -3 -2 -1 0 1 2 3 4 5 We recall that the equation of the x-axis is y = 0. This -1 is because all points on this axis have a y-coordinate -2 of zero, regardless of their different x-coordinates. -3 So, the horizontal line which cuts the y-axis at 2 has equation y = 2. Also, the horizontal line that cuts the y-axis at −1 has equation y = −1, and so on. In general, the equation of a vertical line is xb= (where b is a constant). This line cuts the 4 horizontal axis at b and all points on the line have an x-coordinate of b. Different points, though, y=3 3 have different y-coordinates. y=2 2 The gradient of a straight line 1 To determine the equation of a line other than a horizontal or a vertical line, we must know how to -4 -3 -2 -1 0 1 2 3 4 5 measure its gradient. The gradient of a line is a y= -1 measure of its slope or steepness. It is defined as a -1 ratio of its vertical displacement or Rise, to its -2 horizontal displacement or Run. To determine the magnitude of these displacements, we can use: In general, the equation of a horizontal line is • direct measurement OR (where a is a constant). This line would ya= • count the number of units on the vertical and cut the vertical axis at a and all points on the line horizontal lines when the line is drawn on the have a y-coordinate of a. Different points, though, Cartesian plane or a grid. have different x-coordinates. In the diagram shown, we are given measurements. Vertical Lines A Vertical lines are all parallel to the y-axis. Hence, their equations resemble the equation of the y-axis. We recall that the equation of the y-axis is x = 0. This 3 cm is because all points on this axis have an x-coordinate of zero, regardless of their different y-coordinates. So, the vertical line which cuts the x-axis at 2 has B 5 cm equation x = 2. Also, the vertical line that cuts the x- Hence, we can determine the gradient as follows: axis at −3 has equation x = −3 and so on. Copyright ©2019.Some Rights Reserved. faspassmaths.com �������� ������������ 3 Calculating gradient �������� = = ���������� ������������ 5 The gradient of a straight line can be found when the following information is given: Positive and negative gradient The gradient of a line can be positive or negative, (i) By reading off, the vertical and horizontal depending on the direction of its vertical and displacements from a given line, drawn on a horizontal displacements. These differences are best grid, we can obtain this ratio directly. shown on the Cartesian Plane. A positive gradient is Rise Gradient = obtained when both displacements, are in the same Run direction. A negative gradient is obtained when the 8 = signs of the displacements are in opposite directions. 4 = 2 (ii) Given the coordinates of any two points on the line and using, yy- gradient = 21. xx21- Example, A = (1, 2 ) and B = (3, 6) 6 − 2 4 Gradient = = = 2 3−1 2 Positive and negative gradients can also be (iii) Calculating the tangent of the angle, θ, recognised by observing the angle, measured in an which the line makes with any horizontal anticlockwise direction, between the line and any and where θ is measured in an anticlockwise horizontal that intersects the line. It is often direction. convenient to use the x-axis as the horizontal line. Lines that are inclined at an acute angle to the horizontal, have positive gradients. Gradient = tanq Lines that are inclined at an obtuse angle to the horizontal, have negative gradients. Gradient=°= tan 45 1 Copyright ©2019.Some Rights Reserved. faspassmaths.com (iv) Obtaining the gradient, � from the general equation of a straight line, � = �� + �, For example, y + 2x +1 = 0. We re-write the equation in this form y = −2x – 1 where m = −2 Hence, the gradient is − 2 Gradient of horizontal and vertical lines Horizontal lines are said to have no steepness. For any ‘Run’, the ‘Rise’ is always zero. So, the gradient 1 2 ; If m and m are the gradients of two perpendicular of a horizontal line is = zero. <=> lines, then Vertical lines are said to have maximum steepness. m × m = −1 For any ‘Rise’, the ‘Run’ is always zero. The gradient 1 2 <?@A of a vertical line is → ∞, which is undefined. ; General form of the equation of a straight line We can verify these results using the definition, All straight lines have a general equation of the form, gradient = tan θ. y=+ mx c , where m is the gradient and c is the For horizontal lines, θ = 00 and tan 00= 0. intercept on the vertical axis. For vertical lines, θ = 900 and tan 900 is undefined. Parallel Lines yx=+32 Parallel lines have the same slope and so must have the same gradient. The converse is also true. That is, This line has gradient lines that have the same gradients are parallel. We 3 and cuts the y-axis at can use this fact to prove that any two straight lines 2. are parallel or not. Sometimes we may have to rearrange the equation to obtain this general form. If m1 and m2 are 210yx+-= the gradients of 21yx two lines that are =-+ 11 parallel, then yx=-+ m1 = m2. 22 This line has gradient C − and cuts the y-axis D 1 at . Perpendicular Lines 2 The product of the gradients of perpendicular lines is equal to −1. The converse is also true. That is, if the product of the gradients of any two lines is −1, then To determine the equation of a straight line, we the lines are perpendicular to each other. We can use require (i) The coordinates of one point, (� , � ) on the line. this fact to prove that and two straight lines are C C (ii) The gradient, m of the line. perpendicular or not. yy- (iii) Then we use the equation 1 = m xx- 1 Copyright ©2019.Some Rights Reserved. faspassmaths.com Example 1 Solution Determine the equation of the straight line that The line yx=+23 has a gradient of 2. passes through the point (– 1, 4) and has gradient C The line 210yx+-= has a gradient of − . of – 2. D Since, 1 Solution 2 × − = −1 2 yy- Using 1 = m , we have The lines are perpendicular to each other. xx- 1 y - 4 Example 4 =-2 x --( 1) Find the equation of the line passing through O yx-42=-( + 1) and perpendicular to the line 342yx=+. yx=-22+ Solution By expressing the equation 342yx=+in the form Alternative Method 4 2 Using � = �� + �, we can substitute y = x + � = −1, � = 4 ��� � = −2 to solve for �. 3 3 4 = −2(−1) + � L The gradient of the given line is deduced to be . 4 = 2 + � M � = 2 ∴ the gradient of the line perpendicular to the given 3 � = −2� + 2 line is =- (since the product of the gradients of 4 Example 2 perpendicular lines is -1) Find the equation of the line joining (2,1) to (3,4). yy- We can find the equation using 1 = m and take xx- Solution 1 The gradient is first calculated using: (x1, y1) = (0,0) yy21- y − 0 3 gradient = , where = − xx21- x − 0 4 4y 3x (x1, y1) = (2,−1) and (x2 , y2 ) = (3,4) = − 4 − (−1) 5 3 m = = = 5 y = − x 3− 2 1 4 Then, we proceed to find the equation using Example 5 yy- 1 = m and either of the two given points Find the equation of the line passing through xx- 1 (1,- 1) and parallel to yx=31-. y - ( - 1) = 5 x - 2 y + 1 = 5 ( x - 2 ) Solution By comparison with the general form of an equation, y + 1 = 5 x - 10 y=+ mx c the gradient of yx=31- is deduced as y = 5 x - 11 3. Hence the gradient of the required line is also 3 Had we used the other point, the equation would have since parallel lines have equal gradients. been the same. yy- We can find the equation using 1 = m and xx- 1 Example 3 (x , y ) = (1,−1) Determine whether the lines y = 2x + 3 and 1 1 2y + x −1 = 0 are parallel or perpendicular. Copyright ©2019.Some Rights Reserved. faspassmaths.com y − (−1) The midpoint of a straight line = 3 x −1 Once we know the coordinates of two points on a y +1 = 3(x −1) straight line we can find the mid-point of the line. y +1 = 3x − 3 If A is and B is , then the y = 3x − 4 (xy11, ) (xy22, ) midpoint, M, of the straight line joining A to B, is Example 6 æöxxyy1212++ ç÷, . Find the equation of the straight line that passes èø22 through the point (2,- 2) and inclined at 45° to Example 8 the horizontal axis. State the coordinates of the midpoint of the line joining A(−1, 4) and B(3, 6).
Load more

Recommended publications
  • Descriptive Geometry Section 10.1 Basic Descriptive Geometry and Board Drafting Section 10.2 Solving Descriptive Geometry Problems with CAD
    10 Descriptive Geometry Section 10.1 Basic Descriptive Geometry and Board Drafting Section 10.2 Solving Descriptive Geometry Problems with CAD Chapter Objectives • Locate points in three-dimensional (3D) space. • Identify and describe the three basic types of lines. • Identify and describe the three basic types of planes. • Solve descriptive geometry problems using board-drafting techniques. • Create points, lines, planes, and solids in 3D space using CAD. • Solve descriptive geometry problems using CAD. Plane Spoken Rutan’s unconventional 202 Boomerang aircraft has an asymmetrical design, with one engine on the fuselage and another mounted on a pod. What special allowances would need to be made for such a design? 328 Drafting Career Burt Rutan, Aeronautical Engineer Effi cient travel through space has become an ambi- tion of aeronautical engineer, Burt Rutan. “I want to go high,” he says, “because that’s where the view is.” His unconventional designs have included every- thing from crafts that can enter space twice within a two week period, to planes than can circle the Earth without stopping to refuel. Designed by Rutan and built at his company, Scaled Composites LLC, the 202 Boomerang aircraft is named for its forward-swept asymmetrical wing. The design allows the Boomerang to fl y faster and farther than conventional twin-engine aircraft, hav- ing corrected aerodynamic mistakes made previously in twin-engine design. It is hailed as one of the most beautiful aircraft ever built. Academic Skills and Abilities • Algebra, geometry, calculus • Biology, chemistry, physics • English • Social studies • Humanities • Computer use Career Pathways Engineers should be creative, inquisitive, ana- lytical, detail oriented, and able to work as part of a team and to communicate well.
    [Show full text]
  • 1-1 Understanding Points, Lines, and Planes Lines, and Planes
    Understanding Points, 1-11-1 Understanding Points, Lines, and Planes Lines, and Planes Holt Geometry 1-1 Understanding Points, Lines, and Planes Objectives Identify, name, and draw points, lines, segments, rays, and planes. Apply basic facts about points, lines, and planes. Holt Geometry 1-1 Understanding Points, Lines, and Planes Vocabulary undefined term point line plane collinear coplanar segment endpoint ray opposite rays postulate Holt Geometry 1-1 Understanding Points, Lines, and Planes The most basic figures in geometry are undefined terms, which cannot be defined by using other figures. The undefined terms point, line, and plane are the building blocks of geometry. Holt Geometry 1-1 Understanding Points, Lines, and Planes Holt Geometry 1-1 Understanding Points, Lines, and Planes Points that lie on the same line are collinear. K, L, and M are collinear. K, L, and N are noncollinear. Points that lie on the same plane are coplanar. Otherwise they are noncoplanar. K L M N Holt Geometry 1-1 Understanding Points, Lines, and Planes Example 1: Naming Points, Lines, and Planes A. Name four coplanar points. A, B, C, D B. Name three lines. Possible answer: AE, BE, CE Holt Geometry 1-1 Understanding Points, Lines, and Planes Holt Geometry 1-1 Understanding Points, Lines, and Planes Example 2: Drawing Segments and Rays Draw and label each of the following. A. a segment with endpoints M and N. N M B. opposite rays with a common endpoint T. T Holt Geometry 1-1 Understanding Points, Lines, and Planes Check It Out! Example 2 Draw and label a ray with endpoint M that contains N.
    [Show full text]
  • Machine Drawing
    2.4 LINES Lines of different types and thicknesses are used for graphical representation of objects. The types of lines and their applications are shown in Table 2.4. Typical applications of different types of lines are shown in Figs. 2.5 and 2.6. Table 2.4 Types of lines and their applications Line Description General Applications A Continuous thick A1 Visible outlines B Continuous thin B1 Imaginary lines of intersection (straight or curved) B2 Dimension lines B3 Projection lines B4 Leader lines B5 Hatching lines B6 Outlines of revolved sections in place B7 Short centre lines C Continuous thin, free-hand C1 Limits of partial or interrupted views and sections, if the limit is not a chain thin D Continuous thin (straight) D1 Line (see Fig. 2.5) with zigzags E Dashed thick E1 Hidden outlines G Chain thin G1 Centre lines G2 Lines of symmetry G3 Trajectories H Chain thin, thick at ends H1 Cutting planes and changes of direction J Chain thick J1 Indication of lines or surfaces to which a special requirement applies K Chain thin, double-dashed K1 Outlines of adjacent parts K2 Alternative and extreme positions of movable parts K3 Centroidal lines 2.4.2 Order of Priority of Coinciding Lines When two or more lines of different types coincide, the following order of priority should be observed: (i) Visible outlines and edges (Continuous thick lines, type A), (ii) Hidden outlines and edges (Dashed line, type E or F), (iii) Cutting planes (Chain thin, thick at ends and changes of cutting planes, type H), (iv) Centre lines and lines of symmetry (Chain thin line, type G), (v) Centroidal lines (Chain thin double dashed line, type K), (vi) Projection lines (Continuous thin line, type B).
    [Show full text]
  • A Historical Introduction to Elementary Geometry
    i MATH 119 – Fall 2012: A HISTORICAL INTRODUCTION TO ELEMENTARY GEOMETRY Geometry is an word derived from ancient Greek meaning “earth measure” ( ge = earth or land ) + ( metria = measure ) . Euclid wrote the Elements of geometry between 330 and 320 B.C. It was a compilation of the major theorems on plane and solid geometry presented in an axiomatic style. Near the beginning of the first of the thirteen books of the Elements, Euclid enumerated five fundamental assumptions called postulates or axioms which he used to prove many related propositions or theorems on the geometry of two and three dimensions. POSTULATE 1. Any two points can be joined by a straight line. POSTULATE 2. Any straight line segment can be extended indefinitely in a straight line. POSTULATE 3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center. POSTULATE 4. All right angles are congruent. POSTULATE 5. (Parallel postulate) If two lines intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. The circle described in postulate 3 is tacitly unique. Postulates 3 and 5 hold only for plane geometry; in three dimensions, postulate 3 defines a sphere. Postulate 5 leads to the same geometry as the following statement, known as Playfair's axiom, which also holds only in the plane: Through a point not on a given straight line, one and only one line can be drawn that never meets the given line.
    [Show full text]
  • Geometry Course Outline
    GEOMETRY COURSE OUTLINE Content Area Formative Assessment # of Lessons Days G0 INTRO AND CONSTRUCTION 12 G-CO Congruence 12, 13 G1 BASIC DEFINITIONS AND RIGID MOTION Representing and 20 G-CO Congruence 1, 2, 3, 4, 5, 6, 7, 8 Combining Transformations Analyzing Congruency Proofs G2 GEOMETRIC RELATIONSHIPS AND PROPERTIES Evaluating Statements 15 G-CO Congruence 9, 10, 11 About Length and Area G-C Circles 3 Inscribing and Circumscribing Right Triangles G3 SIMILARITY Geometry Problems: 20 G-SRT Similarity, Right Triangles, and Trigonometry 1, 2, 3, Circles and Triangles 4, 5 Proofs of the Pythagorean Theorem M1 GEOMETRIC MODELING 1 Solving Geometry 7 G-MG Modeling with Geometry 1, 2, 3 Problems: Floodlights G4 COORDINATE GEOMETRY Finding Equations of 15 G-GPE Expressing Geometric Properties with Equations 4, 5, Parallel and 6, 7 Perpendicular Lines G5 CIRCLES AND CONICS Equations of Circles 1 15 G-C Circles 1, 2, 5 Equations of Circles 2 G-GPE Expressing Geometric Properties with Equations 1, 2 Sectors of Circles G6 GEOMETRIC MEASUREMENTS AND DIMENSIONS Evaluating Statements 15 G-GMD 1, 3, 4 About Enlargements (2D & 3D) 2D Representations of 3D Objects G7 TRIONOMETRIC RATIOS Calculating Volumes of 15 G-SRT Similarity, Right Triangles, and Trigonometry 6, 7, 8 Compound Objects M2 GEOMETRIC MODELING 2 Modeling: Rolling Cups 10 G-MG Modeling with Geometry 1, 2, 3 TOTAL: 144 HIGH SCHOOL OVERVIEW Algebra 1 Geometry Algebra 2 A0 Introduction G0 Introduction and A0 Introduction Construction A1 Modeling With Functions G1 Basic Definitions and Rigid
    [Show full text]
  • 13 Graphs 13.2D Lengths of Line Segments
    MEP Pupil Text 13-19, Additional Material 13 Graphs 13.2D Lengths of Line Segments In a right-angled triangle the length of the hypotenuse may be calculated using Pythagoras' Theorem. c b cab222=+ a Worked Example 1 Determine the length of the line segment joining the points A (4, 1) and B (10, 9). Solution y (a) The diagram shows the two points and the line segment that joins them. 10 B A right-angled triangle has been 9 drawn under the line segment. The 8 length of the line segment AB (the 7 hypotenuse) may be found by using 6 Pythagoras' Theorem. 5 8 4 AB2 =+62 82 3 2 =+ 2 AB 36 64 A 6 1 AB2 = 100 0 12345678910 x AB = 100 AB = 10 Worked Example 2 y C Determine the length of the line 8 7 joining the points C (−48, ) 6 (−) and D 86, . 5 4 Solution 3 2 14 The diagram shows the two points 1 and a right-angled triangle that can –4 –3 –2 –1 0 12345678 x be used to determine the length of –1 the line segment CD. –2 –3 –4 –5 D –6 12 1 13.2D MEP Pupil Text 13-19, Additional Material Using Pythagoras' Theorem, CD2 =+142 122 CD2 =+196 144 CD2 = 340 CD = 340 CD = 18. 43908891 CD = 18. 4 (to 3 significant figures) Exercises 1. The diagram shows the three points y C A, B and C which are the vertices 11 of a triangle. 10 9 (a) State the length of the line 8 segment AB.
    [Show full text]
  • Line Geometry for 3D Shape Understanding and Reconstruction
    Line Geometry for 3D Shape Understanding and Reconstruction Helmut Pottmann, Michael Hofer, Boris Odehnal, and Johannes Wallner Technische UniversitÄat Wien, A 1040 Wien, Austria. fpottmann,hofer,odehnal,[email protected] Abstract. We understand and reconstruct special surfaces from 3D data with line geometry methods. Based on estimated surface normals we use approximation techniques in line space to recognize and reconstruct rotational, helical, developable and other surfaces, which are character- ized by the con¯guration of locally intersecting surface normals. For the computational solution we use a modi¯ed version of the Klein model of line space. Obvious applications of these methods lie in Reverse Engi- neering. We have tested our algorithms on real world data obtained from objects as antique pottery, gear wheels, and a surface of the ankle joint. Introduction The geometric viewpoint turned out to be highly successful in dealing with a variety of problems in Computer Vision (see, e.g., [3, 6, 9, 15]). So far mainly methods of analytic geometry (projective, a±ne and Euclidean) and di®erential geometry have been used. The present paper suggests to employ line geometry as a tool which is both interesting and applicable to a number of problems in Computer Vision. Relations between vision and line geometry are not entirely new. Recent research on generalized cameras involves sets of projection rays which are more general than just bundles [1, 7, 18, 22]. A beautiful exposition of the close connections of this research area with line geometry has recently been given by T. Pajdla [17]. The present paper deals with the problem of understanding and reconstruct- ing 3D shapes from 3D data.
    [Show full text]
  • Mel's 2019 Fishing Line Diameter Page
    Welcome to Mel's 2019 Fishing Line Diameter page The line diameter tables below offer a comparison of more than 115 popular monofilament, copolymer, fluorocarbon fishing lines and braided superlines in tests from 6-pounds to 600-pounds If you like what you see, download a copy You can also visit our Fishing Line Page for more information and links to line manufacturers. The line diameters shown are compiled from manufacturer's web sites, product catalogs and labels on line spools. Background Information When selecting a fishing line, one must consider a number of factors. While knot strength, abrasion resistance, suppleness, shock resistance, castability, stretch and low spool memory are all important characteristics, the diameter of a line is probably the most important. As long as these other characteristics meet your satisfaction, then the smaller the diameter of the line the better. With smaller diameter lines: more line can be spooled onto the reel, they are usually less visible to the fish, will generally cast better, and provide better lure action. Line diameter measurements provided by manufacturers are expressed in thousandths of an inch (0.001 inch) and its metric system equivalent, hundredths of a millimeter (0.01 mm). However, not all manufacturers provide line diameter information, so if you don't see it in the tables, that's the likely reason why. And some manufacturers now provide line diameter measurements in ten-thousandths of an inch (0.0001 inch) and thousandths of a millimeter (0.001 mm). To give you an idea of just how small this is, one ten-thousandth of an inch is less than 3% of the diameter of an average human hair.
    [Show full text]
  • LECTURE 3: LINES and PLANES in THREE-DIMENSIONAL SPACE 1. Lines in R We Can Now Use Our New Concepts of Dot Products and Cross P
    LECTURE 3: LINES AND PLANES IN THREE-DIMENSIONAL SPACE MA1111: LINEAR ALGEBRA I, MICHAELMAS 2016 1. Lines in R3 We can now use our new concepts of dot products and cross products to describe some examples of writing down geometric objects efficiently. We begin with the question of determining equations for lines in R3, namely, real three-dimensional space (like the world we live in). Aside. Since it is inconvenient to constantly say that v is an n-dimensional vector, etc., we will start using the notation Rn for n-dimensional vectors. For example, R1 is the real-number line, and R2 is the xy-plane. Our methods will of course work in the plane, R2, as well. Instead of beginning with a formal statement, we motivate the general idea by beginning with an example. Every line is determined by a point on the line and a direction. For example, suppose we want to write an equation for the line passing through the point A = (1; 3; 5) and which is parallel to the vector (2; 6; 7). Note that if I took any non-zero multiple of v, the following argument would work equally well. Now suppose that B = (x; y; z) is an arbitrary point. Then −! −! B is on L () AB points in the direction of v () AB = tv; t 2 R 8 x = 1 + 2t <> () (x − 1; y − 3; z − 5) = (2t; 6t; 7t) () y = 3 + 6t for t 2 R: :>z = 5 + 7t These are the parametric equations for L. The general formula isn't much harder to derive.
    [Show full text]
  • Perpendicular and Parallel Line Segments
    Name: Date: r te p a h Perpendicular and C Parallel Line Segments Practice 1 Drawing Perpendicular Line Segments Use a protractor to draw perpendicular line segments. Example Draw a line segment perpendicular to RS through point T. S 0 180 10 1 70 160 20 150 30 140 40 130 50 0 120 180 6 0 10 11 0 170 70 20 100 30 160 80 40 90 80 150 50 70 60 140 100 130 110 120 R T © Marshall Cavendish International (Singapore) Private Limited. Private (Singapore) International © Marshall Cavendish 1. Draw a line segment perpendicular to PQ. P Q 2. Draw a line segment perpendicular to TU through point X. X U T 67 Lesson 10.1 Drawing Perpendicular Line Segments 08(M)MIF2015CC_WBG4B_Ch10.indd 67 4/30/13 11:18 AM Use a drawing triangle to draw perpendicular line segments. Example M L N K 3. Draw a line segment 4. Draw a line segment perpendicular to EF . perpendicular to CD. C E F D © Marshall Cavendish International (Singapore) Private Limited. Private (Singapore) International © Marshall Cavendish 5. Draw a line segment perpendicular to VW at point P. Then, draw another line segment perpendicular to VW through point Q. Q V P W Line Segments 68 and Parallel Chapter 10 Perpendicular 08(M)MIF2015CC_WBG4B_Ch10.indd 68 4/30/13 11:18 AM Name: Date: Practice 2 Drawing Parallel Line Segments Use a drawing triangle and a straightedge to draw parallel line segments. Example Draw a line segment parallel to AB . A B 1. Draw a pair of parallel line segments.
    [Show full text]
  • A Geometric Introduction to Spacetime and Special Relativity
    A GEOMETRIC INTRODUCTION TO SPACETIME AND SPECIAL RELATIVITY. WILLIAM K. ZIEMER Abstract. A narrative of special relativity meant for graduate students in mathematics or physics. The presentation builds upon the geometry of space- time; not the explicit axioms of Einstein, which are consequences of the geom- etry. 1. Introduction Einstein was deeply intuitive, and used many thought experiments to derive the behavior of relativity. Most introductions to special relativity follow this path; taking the reader down the same road Einstein travelled, using his axioms and modifying Newtonian physics. The problem with this approach is that the reader falls into the same pits that Einstein fell into. There is a large difference in the way Einstein approached relativity in 1905 versus 1912. I will use the 1912 version, a geometric spacetime approach, where the differences between Newtonian physics and relativity are encoded into the geometry of how space and time are modeled. I believe that understanding the differences in the underlying geometries gives a more direct path to understanding relativity. Comparing Newtonian physics with relativity (the physics of Einstein), there is essentially one difference in their basic axioms, but they have far-reaching im- plications in how the theories describe the rules by which the world works. The difference is the treatment of time. The question, \Which is farther away from you: a ball 1 foot away from your hand right now, or a ball that is in your hand 1 minute from now?" has no answer in Newtonian physics, since there is no mechanism for contrasting spatial distance with temporal distance.
    [Show full text]
  • Beginner S Guide to Moths of the Midwest Geometers
    0LGZHVW5HJLRQ86$ %HJLQQHU V*XLGHWR0RWKVRIWKH0LGZHVW*HRPHWHUV $QJHOOD0RRUHKRXVH ,OOLQRLV1DWXUH3UHVHUYH&RPPLVVLRQ Photos: Angella Moorehouse ([email protected]). Produced by: Angella Moorehouse with the assistance of Alicia Diaz, Field Museum. Identification assistance provided by: multiple sources (inaturalist.org; bugguide.net) )LHOG0XVHXP &&%<1&/LFHQVHGZRUNVDUHIUHHWRXVHVKDUHUHPL[ZLWKDWWULEXWLRQEXWFRPPHUFLDOXVHRIWKHRULJLQDOZRUN LVQRWSHUPLWWHG >ILHOGJXLGHVILHOGPXVHXPRUJ@>@YHUVLRQ $ERXWWKH%(*,11(5¶6027+62)7+(0,':(67*8,'(6 Most photos were taken in west-central and central Illinois; a fewDUH from eastern Iowa and north-central Wisconsin. Nearly all were posted to identification websites: BugGuide.netDQG iNaturalist.org. Identification help was provided by Aaron Hunt, Steve Nanz, John and Jane Balaban, Chris Grinter, Frank Hitchell, Jason Dombroskie, William H. Taft, Jim Wiker,DQGTerry Harrison as well as others contributing to the websites. Attempts were made to obtain expert verifications for all photos to the field identification level, however, there will be errors. Please contact the author with all corrections Additional assistance was provided by longtime Lepidoptera survey partner, Susan Hargrove. The intention of these guides is to provide the means to compare photographs of living specimens of related moths from the Midwest to aid the citizen scientists with identification in the field for Bio Blitz, Moth-ers Day, and other night lighting events. A taxonomic list to all the species featured is provided at the end along with some field identification tips. :(%6,7(63529,',1*,'(17,),&$7,21,1)250$7,21 BugGuide.net LNaturalist.org Mothphotographersgroup.msstate.edu Insectsofiowa.org centralillinoisinsects.org/weblog/resources/ :+,&+027+*8,'(7286( The moths were split into 6 groups for the purposes of creating smaller guides focusing on similar features of 1 or more superfamilies.
    [Show full text]