Cones, Pyramids and Spheres
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Arc Length. Surface Area
Calculus 2 Lia Vas Arc Length. Surface Area. Arc Length. Suppose that y = f(x) is a continuous function with a continuous derivative on [a; b]: The arc length L of f(x) for a ≤ x ≤ b can be obtained by integrating the length element ds from a to b: The length element ds on a sufficiently small interval can be approximated by the hypotenuse of a triangle with sides dx and dy: p Thus ds2 = dx2 + dy2 ) ds = dx2 + dy2 and so s s Z b Z b q Z b dy2 Z b dy2 L = ds = dx2 + dy2 = (1 + )dx2 = (1 + )dx: a a a dx2 a dx2 2 dy2 dy 0 2 Note that dx2 = dx = (y ) : So the formula for the arc length becomes Z b q L = 1 + (y0)2 dx: a Area of a surface of revolution. Suppose that y = f(x) is a continuous function with a continuous derivative on [a; b]: To compute the surface area Sx of the surface obtained by rotating f(x) about x-axis on [a; b]; we can integrate the surface area element dS which can be approximated as the product of the circumference 2πy of the circle with radius y and the height that is given by q the arc length element ds: Since ds is 1 + (y0)2dx; the formula that computes the surface area is Z b q 0 2 Sx = 2πy 1 + (y ) dx: a If y = f(x) is rotated about y-axis on [a; b]; then dS is the product of the circumference 2πx of the circle with radius x and the height that is given by the arc length element ds: Thus, the formula that computes the surface area is Z b q 0 2 Sy = 2πx 1 + (y ) dx: a Practice Problems. -
A Technology to Synthesize 360-Degree Video Based on Regular Dodecahedron in Virtual Environment Systems*
A Technology to Synthesize 360-Degree Video Based on Regular Dodecahedron in Virtual Environment Systems* Petr Timokhin 1[0000-0002-0718-1436], Mikhail Mikhaylyuk2[0000-0002-7793-080X], and Klim Panteley3[0000-0001-9281-2396] Federal State Institution «Scientific Research Institute for System Analysis of the Russian Academy of Sciences», Moscow, Russia 1 [email protected], 2 [email protected], 3 [email protected] Abstract. The paper proposes a new technology of creating panoramic video with a 360-degree view based on virtual environment projection on regular do- decahedron. The key idea consists in constructing of inner dodecahedron surface (observed by the viewer) composed of virtual environment snapshots obtained by twelve identical virtual cameras. A method to calculate such cameras’ projec- tion and orientation parameters based on “golden rectangles” geometry as well as a method to calculate snapshots position around the observer ensuring synthe- sis of continuous 360-panorama are developed. The technology and the methods were implemented in software complex and tested on the task of virtual observing the Earth from space. The research findings can be applied in virtual environment systems, video simulators, scientific visualization, virtual laboratories, etc. Keywords: Visualization, Virtual Environment, Regular Dodecahedron, Video 360, Panoramic Mapping, GPU. 1 Introduction In modern human activities the importance of the visualization of virtual prototypes of real objects and phenomena is increasingly grown. In particular, it’s especially required for training specialists in space and aviation industries, medicine, nuclear industry and other areas, where the mistake cost is extremely high [1-3]. The effectiveness of work- ing with virtual prototypes is significantly increased if an observer experiences a feeling of being present in virtual environment. -
Lateral and Surface Area of Right Prisms 1 Jorge Is Trying to Wrap a Present That Is in a Box Shaped As a Right Prism
CHAPTER 11 You will need Lateral and Surface • a ruler A • a calculator Area of Right Prisms c GOAL Calculate lateral area and surface area of right prisms. Learn about the Math A prism is a polyhedron (solid whose faces are polygons) whose bases are congruent and parallel. When trying to identify a right prism, ask yourself if this solid could have right prism been created by placing many congruent sheets of paper on prism that has bases aligned one above the top of each other. If so, this is a right prism. Some examples other and has lateral of right prisms are shown below. faces that are rectangles Triangular prism Rectangular prism The surface area of a right prism can be calculated using the following formula: SA 5 2B 1 hP, where B is the area of the base, h is the height of the prism, and P is the perimeter of the base. The lateral area of a figure is the area of the non-base faces lateral area only. When a prism has its bases facing up and down, the area of the non-base faces of a figure lateral area is the area of the vertical faces. (For a rectangular prism, any pair of opposite faces can be bases.) The lateral area of a right prism can be calculated by multiplying the perimeter of the base by the height of the prism. This is summarized by the formula: LA 5 hP. Copyright © 2009 by Nelson Education Ltd. Reproduction permitted for classrooms 11A Lateral and Surface Area of Right Prisms 1 Jorge is trying to wrap a present that is in a box shaped as a right prism. -
Using History of Mathematics to Teach Volume Formula of Frustum Pyramids: Dissection Method
Universal Journal of Educational Research 3(12): 1034-1048, 2015 http://www.hrpub.org DOI: 10.13189/ujer.2015.031213 Using History of Mathematics to Teach Volume Formula of Frustum Pyramids: Dissection Method Suphi Onder Butuner Department of Elementary Mathematics Education, Bozok University, Turkey Copyright © 2015 by authors, all rights reserved. Authors agree that this article remains permanently open access under the terms of the Creative Commons Attribution License 4.0 International License Abstract Within recent years, history of mathematics dynamic structure of mathematics [1, 2, 3, 4, 5, 6, 7]. It is (HoM) has become an increasingly popular topic. Studies stressed that history of mathematics will enrich the repertoire have shown that student reactions to it depend on the ways of teachers and develop their pedagogical content knowledge they use history of mathematics. The present action research [7, 8]. Jankvist [5] (2009) explains the use of history of study aimed to make students deduce volume rules of mathematics both as a tool and as a goal. History-as-a-tool frustum pyramids using the dissection method. Participants arguments are mainly concerned with learning of the inner were 24 grade eight students from Trabzon. Observations, issues of mathematics (mathematical concepts, theories, informal interviews and feedback forms were used as data methods, formulas), whereas history-as-a-goal arguments collection tools. Worksheets were distributed to students and are concerned with the use of history as a self-contained goal. the research was conducted in 3 class hours. Student views From the history-as-a-goal point of view, knowing about the on the activities were obtained through a written feedback history of mathematics is not a primary tool for learning form consisting of 4 questions. -
Calculus Terminology
AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential -
Uniform Panoploid Tetracombs
Uniform Panoploid Tetracombs George Olshevsky TETRACOMB is a four-dimensional tessellation. In any tessellation, the honeycells, which are the n-dimensional polytopes that tessellate the space, Amust by definition adjoin precisely along their facets, that is, their ( n!1)- dimensional elements, so that each facet belongs to exactly two honeycells. In the case of tetracombs, the honeycells are four-dimensional polytopes, or polychora, and their facets are polyhedra. For a tessellation to be uniform, the honeycells must all be uniform polytopes, and the vertices must be transitive on the symmetry group of the tessellation. Loosely speaking, therefore, the vertices must be “surrounded all alike” by the honeycells that meet there. If a tessellation is such that every point of its space not on a boundary between honeycells lies in the interior of exactly one honeycell, then it is panoploid. If one or more points of the space not on a boundary between honeycells lie inside more than one honeycell, the tessellation is polyploid. Tessellations may also be constructed that have “holes,” that is, regions that lie inside none of the honeycells; such tessellations are called holeycombs. It is possible for a polyploid tessellation to also be a holeycomb, but not for a panoploid tessellation, which must fill the entire space exactly once. Polyploid tessellations are also called starcombs or star-tessellations. Holeycombs usually arise when (n!1)-dimensional tessellations are themselves permitted to be honeycells; these take up the otherwise free facets that bound the “holes,” so that all the facets continue to belong to two honeycells. In this essay, as per its title, we are concerned with just the uniform panoploid tetracombs. -
About This Booklet on Surface Areas and Volumes
About This Booklet on Surface1 Areas and Volumes About This Booklet on Surface Areas and Volumes This Particular Booklet has been designed for the students of Math Class 10 (CBSE Board). However, it will also help those who have these chapters in their curriculum or want to gain the knowledge and explore the concepts. This Booklet explains: 1. Cube, Cuboid and Cylinder 2. Cone and Frustum 3. Sphere and Hemisphere 4. Combination of solids This Booklet also covers: 1. MCQs 2. Questions with Solutions 3. Questions for practice and 4. QR Codes to scan and watch videos on Surface Areas and Volumes QR codes when scanned with mobile scanner take you to our YouTube channel Let’sTute (www.youtube.com/letstute) where you can watch our free videos (need internet connection) on the topic. For Surface Areas and Volumes, in total, we have 8 videos which are accessible on several other platforms as described on the back cover of this Booklet. However, if there is any query, feel free to connect with us on the details given on the last page. Some other documents are also available: About Let’s Tute Let’sTute (Universal Learning Aid) is an E-learning company based in Mumbai, India. (www.letstute.com) We create content for Mathematics, Biology, Physics, Environmental Science, Book-Keeping & Accountancy and also a series on value education known as V2lead 2 13 .1 Cube, Cuboid and Cylinder Scan to watch the video on INTRODUCTION Surface area and Volume Surface Area: It is the sum of total exposed area of a three dimensional solid object. -
Comparing Surface Area & Volume
Name ___________________________________________________________________ Period ______ Comparing Surface Area & Volume Use surface area and volume calculations to solve the following situations. 1. Find the surface area and volume of each prism. Which two have the same surface area? Which two have the same volume? 4 in. 22 in. 10 in. C A 11 in. B 18 in. 10 in. 6 in. 10 in. 6 in. 2. Sketch and label two different rectangular prisms that have the same volume. Find the surface area of each. 3. Do the following cylinders have the same surface area, the same volume, or the same surface area and volume? r = 4 cm r = 2 cm h = 9 cm h = 24 cm 4. A cylinder has a surface area of 125.6 mm2 and a volume of 100.48 mm3. If another cylinder has a radius of 5 mm and the same volume, what would be the height? 5. A cylinder has a radius of 9 yd and a height of 3 yd. A second cylinder has a radius of 6 yd and a height of 12 yd. Which statement is true? a. The cylinders have the same surface area and the same volume. b. The cylinders have the same surface area and different volumes. c. The cylinders have different surface areas and the same volume. d. The cylinders have different surface areas and different volumes. 6. The surface area of a rectangular prism is 2,116 mm2. If you double the length, width, and height of the prism, what will be the surface area of the new prism? How much larger is this amount? 7. -
Why Cells Are Small: Surface Area to Volume Ratios
Name__________________________________ Why Cells are Small: Surface Area to Volume Ratios Today’s lab is about the size of cells. To understand this, we first have to understand surface area to volume ratios. Let’s start with surface area. Get a box of sugar cubes. The surface area of the cube is calculated by finding the area (length x width) of one side. But we want to know the whole surface. On a cube, there are 6 sides, so we multiply that area x 6. So Surface area of a cube = length x width x 6 Volume is another simple calculation – we just multiply length x width x height. Build the following four structures from your sugar cubes, and fill out the table below. The one we’re most interested in, though is the surface area to volume ratio, which is just Surface area ÷ volume A B C D 1 UNIT 2 UNITS 3 4 UNITS OR CM UNITS Figure Total number Surface area of Volume of Surface area to Total surface of Cubes figure figure Volume Ratio of individual (=6 x length x (= length x ( = SA / V) cubes* width) width x height) (= 6 * # cubes) A 1 6 1 6 1 B 8 24 8 3 48 C 27 54 27 2 162 64 96 64 1.5 384 D *because surface area of one cube is 6 Surface Area to Volume Lab Questions 1. What happened to the surface area as the size increased? It increased. 2. What happened to the volume as the size increased? It also increased, but not as quickly. -
Surface Area of 3-D Objects: Prisms: a Prism Is a Polyhedron with Two Opposite Faces That Are Identical Polygonal Regions Called the Bases
Surface Area of 3-d Objects: Prisms: A prism is a polyhedron with two opposite faces that are identical polygonal regions called the bases. The vertices of the bases are joined with segments to form the lateral faces which are parallelograms. If the lateral faces are rectangles, the prism is called a right prism, otherwise, it’s called an oblique prism. oblique right The surface area is the sum of the areas of the two bases and the areas of the lateral faces. Right rectangular prism: Right triangular prism: 5 ft 7” 4 ft 5 ft 3 ft 3” 4” 1 2 SA..243 =( )( ) + 247( )( ) + 237( )( ) SA. .=( 3)( 5) + 2 ( 3)( 4) +( 4)( 5) + ( 5) 2 bottom rectangle square top and bottom two sides two sides two triangles 2 =24 + 56 + 42 = 122in =15 + 12 + 20 + 25 = 72 ft 2 A cube has a total surface area of 384 square units. What is the length of one of its edges? The surface area of a cube is 6s2 , so 6s22= 384 s = 64 s = 8 units . If the length of the side of a cube is multiplied by 4, how will the surface area of the larger cube compare to the surface area of the original cube? 2 2 2 6s is the original surface area. 6( 4ss) = 16 6 is the new surface area, so the larger cube has 16 times the original surface area. Pyramids: A pyramid is a polyhedron formed by connecting the vertices of a polygonal region called the base to another point not in the plane of the base called the apex using segments. -
Surface Areas and Volumes
CHAPTER 13 SURFACE AREAS AND VOLUMES (A) Main Concepts and Results • Cuboid whose length = l, breadth = b and height = h (a) Volume of cuboid = lbh (b) Total surface area of cuboid = 2 ( lb + bh + hl ) (c) Lateral surface area of cuboid = 2 h (l + b) (d) Diagonal of cuboid = lbh222++ • Cube whose edge = a (a) Volume of cube = a3 (b) Lateral Surface area = 4a2 (c) Total surface area of cube = 6a2 (d) Diagonal of cube = a 3 • Cylinder whose radius = r, height = h (a) Volume of cylinder = πr2h (b) Curved surface area of cylinder = 2πrh (c) Total surface area of cylinder = 2πr (r + h) • Cone having height = h, radius = r and slant height = l 1 (a) Volume of cone = πrh2 3 (b) Curved surface area of cone = πrl 16/04/18 122 EXEMPLAR PROBLEMS (c) Total surface area of cone = πr (l + r) (d) Slant height of cone (l) = hr22+ • Sphere whose radius = r 4 (a) Volume of sphere = πr3 3 (b) Surface area of sphere = 4πr2 • Hemisphere whose radius = r 2 (a) Volume of hemisphere = πr3 3 (b) Curved surface area of hemisphere = 2πr2 (c) Total surface area of hemisphere = 3πr2 (B) Multiple Choice Questions Write the correct answer Sample Question 1 : In a cylinder, if radius is halved and height is doubled, the volume will be (A) same (B) doubled (C) halved (D) four times Solution: Answer (C) EXERCISE 13.1 Write the correct answer in each of the following : 1. The radius of a sphere is 2r, then its volume will be 4 8πr3 32 (A) πr3 (B) 4πr3 (C) (D) πr3 3 3 3 2. -
History of Mathematics #4 1. Before 11 Pm, Sunday, February 5. Go to The
History of Mathematics #4 1. Before 11 pm, Sunday, February 5. Go to the Forum “Archimedes” and make at least one substantive contribution by 11 pm, Sunday, February 5, and at least one substantive response to others’ postings before class on Tuesday, February 7. Write about the following: (a) Where in the K–16 curriculum do the various results of Archimedes discussed in Dunham make their appearance? (b) Where in the K–16 curriculum do the proofs of these results make their appear- ance? (c) Do you think we should present Archimedes’ proof of the area of a circle in middle school? in high school? 2. Before Tuesday, February 7. (a) Read Dunham Chapter 4, MTA Sketch 7, and Boyer Chapter 8. If you aren’t able to to read all of Boyer, at least read the sections in Chapter 8 on The Method, Volume of a sphere, and Recovery of The Method. (b) Think about the following questions for discussion at the Centra session: i. Study Euclid’s proof by exhaustion of the volume of a circular cone (XII.10). Refer also to XII.7 and XII.5. Pretty amazing! ii. On pages 93–94 of Dunham it is asserted that the perimeter of a regular poly- gon inscribed in a circle is less than the circumference of the circle, and also that the perimeter of a regular polygon circumscribed about a circle exceeds the circumference of the circle. How might these statements be justified? iii. Try making and measuring some perimeters of inscribed and circumscribed polygons using Wingeom or Geometer’s Sketchpad.