Samsung Exhibit 1021
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`Samsung Exhibit 1021
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`
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`- aeeenreare-paren — “a
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`GEORGE B. THOMAS, JR.
`Department of Mathematics
`Massachusetis Institute of Technology
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`ADDISON-WESLEY PUBLISHING COMPANY
`Reading, Massachusetts
`Menlo Park, California - London - Don Mills, Ontario
`
`CALCULUS
`
`AND
`
`ANALYTIC
`GEOMETRY
`
`FOURTH EDITION
`
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`Page 2 of 8
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`This book is in the
`Addison-Wesley Series in Mathematics
`
`Second printing, December 1969
`
`Copyright © 1968, Philippines copyright 1968 by Addison-
`Wesley Publishing Company, Inc. All rights reserved. No
`part of
`this publication may be reproduced, stored in a
`retrieval system, or transmitted,
`in any form or by any
`means, electronic, mechanical, photo-copying, recording, or
`otherwise, without the prior written permission of the pub-
`lisher. Printed in the United States of America. Published
`simultaneously in Canada.
`Library of Congress Catalog
`Card No. 68-17568.
`
`
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`Page 3 of 8
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`
`
`11.1 THE POLAR COORDINATE SYSTEM
`We knowthat a point can be located in a plane by
`giving its abscissa and ordinate relative to a given
`coordinate system. Such z- and y-coordinates are
`called Cartesian coordinates, in honor of the French
`mathematician-philosopher René Descartes* (1596-
`1650), whois eredited with discovering this method
`of fixing the position of a point in a plane.
`
`Pir, 8
`
`
`
`
`y = 2c+3withe > lis another ray; its vertex is (1, 5).
`
`Another useful way to locate a point in a plane
`is by polar coordinates (see Fig. 11.1). First, we fix
`an origin O and an initial rayt from O. The point P
`has polar coordinates r, 6, with
`r = directed distance from O to P,
`
`(1a)
`
`and
`
`@ = directed angle frominitial ray to OP.
`
`(1b)
`
`the angle 4 is positive when
`As in trigonometry,
`measured counterclockwise and negative when mea-
`sured clockwise (Fig. 11.1). But the angle associated
`with a given point is not unique (Fig. 11.2). For
`instance, the point2 units fromtheorigin, along the
`ray 6 = 30°, has polar coordinates r = 2, 6 = 30°.
`It also has coordinates r = 2, # = —330°, orr = 2,
`6 = 390°.
`There are occasions when we wish to allow r to be
`negative. That’s why we say “directed distance”
`
`* For an interesting biographical account together with
`an excerpt from Descartes’ own writings, see World of
`Mathematics, Vol. 1, pp. 235-253.
`+ A ray is a half-line consisting of a vertex and points
`of a line on one side of the vertex. For example, the
`origin and positive z-axis isa tay. The points on theline
`
`POLAR
`JOORDINATES
`
`CHAPTER 11
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`2 ynits. He would reach the same point by turning
`only 30° counterclockwise from the initial ray anc
`then going backward 2 units. So we say that the
`point also has polar coordinates r = —2, @= 30°.
`Whenever the angle between two rays is 180°, th:
`rays actually make a straightline. We then say the
`either ray is the negative of the other. Points cz
`the ray = a have polar coordinates (r, a) wit
`é=
`r > 0. Points on the negative ray,
`a — 180°
`have coordinates (r,a) with r < 0. The origin
`r= 0.
`(See Fig. 11.4 for the ray @ = 30° and i:
`negative. A word of caution: The “negative”
`the ray @ = 30° is the ray @ = 30°+180° = 210°
`and not the ray 6 = —30°. “Negative”refers totr
`directed distance r.)
`There is a great advantage in being able to u=
`both polar and Cartesian coordinates at once.
`T
`do this, we use a commonorigin and take the initis
`ray as the positive x-axis, and take the ray 9 = ae
`as the positive y-axis. The coordinates, shown
`Fig. 11.5, are then related by the equations
`
`
`
`
`
`s Polar and C;
`
`6
`
`Thecircle r :
`
`®
`
`
`We adopt the ¢
`smmber, —0 <3
`- = y= Oir
`
`r=
`
`
`<2 origin, z = (
`-he same point
`_
`
`fif=rent ways in
`
`He point
`(2, 30°)
`mcresentations:
`(
`2, —150°). The
`“ae two formula
`
`reosé, y=rsin é.
`
`3
`
`These are the equations that define sin § and cos *
`whenr is positive. They are also valid if r is nez=
`tive, because
`
`cos (¢ + 180°)
`
`sin (6 — 180°)
`
`cos 6,
`
`=
`= —sin 4,
`
`so positive r’s on the (# + 180°)-ray correspond 3
`negative r’s associated with the é@-ray. When r = +
`then z = y = 0, and P is theorigin.
`If we impose the condition
`
`r=a (a constant),
`
`|
`
`then the locus of P is a circle with center O sce
`radius a, and P describes the circle once as @ vars
`from 0 to 360° (see Fig. 11.6), On the other haze
`if we let r vary and hold 6 fixed, say
`
`Poter coordinates
`
`V1.1
`
`
`
`11.2 Thersy9 = 30°is the same as the ray @ = —330°.
`
`aw b= 30°
`
`
`
`
`
`— ~°<(=2, 30°)
`
`11.4 The terminal ray @ = 7/6 andits negative.
`
`in Eq. (la). The ray @ = 30° and the ray § = 210°
`together make up a complete line through O (see
`Fig. 11.3). The point P(2, 210°) 2 units from 0 on
`the ray @= 210° has polar coordinates r = 2,
`§ = 210°.
`It can be reached by a person standing
`at O and facing out along the initial ray, if he first
`turns 210° counterclockwise, and then goes forward
`
`Page 5 of 8
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`
`
`(2, 30° + ni
`—2,210°+ ni
`
`> 2 we represent
`
`(2, $9 + 2
`(—2, $7 + 2n
`
`@ = 30°,
`
`24
`
`the locus of P is the straight line shown in Fig. 113.
`
`
`
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`
`
`by turning
`al ray and
`ry that the
`@= 30°.
`3 180°, the
`n say tha:
`Points on
`r,a@) with
`a + 180°
`> origin is
`)° andits
`tative” of
`Y= 210
`ers to the
`
`rle to use
`mee. Te
`she initia!
`"8 = 90°
`shown in
`
`
`
`WW |
`
`The polar coordinate system
`
`The fact that the same point may be represented
`in several different ways in polar coordinates makes
`added care necessary in certain situations.
`For
`example, the point (2a, 7) is on the curve
`r? = 4a? cos 6
`
`(6)
`
`even though its coordinates as given do not satisfy
`the equation, because the same point is represented
`by (—2a,0) and these coordinates do satisfy the
`equation. The same point (2a, 7) is on the curve
`
`r = a(l — cos 8),
`
`(7)
`
`and hence this point should be included among the
`points of intersection of the two curves represented
`by Eqs. (6) and (7). But if we solve the equations
`simultaneously byfirst substituting cos @ = r?/4a?
`from (6)
`into (7) and then solving the resulting
`quadratic equation
`
`()+4G)-«=
`
`a .
`
`for
`11.6 The circle r = ais the locus P.
`H+
`
`
`= —242vV2, (8)
`
`We adopt the convention that r may be anyreal
`number, —o <r < x. Then r= 0 corresponds
`coz = 0,y = Oin Eas. (2), regardless of 6. Thatis,
`
`r=0, @anyvalue,
`
`(5)
`
`is the origin, z = 0, y = 0.
`The same point may be represented in several
`different ways in polar coordinates. For example,
`the point
`(2, 30°), or (2,2/6), has the following
`representations:
`(2, 30°),
`(2, —330°),
`(—2, 210°),
`—2, —150°). These andall others are summarized
`in the two formulas
`
`(2, 30° + n 360°),
`(—2, 210° + n 360°),
`
`n = 0, +1, +2,...;
`
`or, if we represent the angles in radians, in the two
`formulas
`
`n= 0, +1, +2,...
`
`we do not obtain the point (2a, 7) as a point of inter-
`section. The reason is simple enough: The point is
`not on the curves “simultaneously” in the sense of
`being reached at the “same time,”since it is reached
`in the one case when 6 = O andin the other case
`when 6= 7.
`It is as though two ships describe
`paths that intersect at a point, but the ships do not
`collide because they reach the point of intersection
`at different
`times!
`The curves represented by
`Eqs. (6) and (7) are shown in Fig. 11.9(c). They
`are seen to intersect at the four points
`
`(0, 0),
`
`(2a, 7),
`
`(v1, 63),
`
`(ri, —41),
`
`(9a)
`
`where
`
`n= (—2 T 2V/2)a,
`(9b)
`cos 6, = 1 — = 3 — 2v72.
`Onlythe last two of the points (9a) are found from
`the simultaneous solution; the first two are disclosed
`only by the graphs of the curves.
`
`spond to
`nr= 0,
`
`(3
`
`: O and
`0 varies
`x hand.
`
`(4
`
`ig. 11,4.
`
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`Find the lengths of each of the following vectors and th:
`angle that each makes with the positive z-axis.
`3r
`_ or)...
`.
`
`u2 = 1cos| @— >J]7 Jsin OS:
`(x, 0, z
`13. W3it+j
`Wee
`12, 2— 3
`
`390
`
`Vectors and parametric equations
`
`12.4
`
`Then, applying (3), we have
`
`sad
`
`uy = icos§+jsin#@
`
`isin @ —jceosé.
`
`Therefore
`
`OP = au + (a8)uz
`afi cos 6+ j sin @) — a@f{isin & — j cos @)
`a(cos§+ @sin 9)i-+ a (sin § — G cos Aj.
`
`to the involute of a circ= 4
`10. A unit vector tangent
`whose parametric equations are given in Eq. (6)
`
`14. —21+3j
`
`15. 51+ 13
`
`16. —5i — 12j
`
`17. Use vector methods to determine parametric equs §
`tions for the trochoid of Fig. 12.6, by taking
`
`R = OP = OM + NC —CP.
`
`(z, y, 5) lying ir
`
`18. Let 4, B,C, D be the vertices, in order, of a quadr-
`We equate this with zi+ yj and, since corresponding
`lateral. Let A’, B’, C’, D’ be the midpoints of th |
`components must be equal, we obtain the parametric
`sides AB, BC, CD, and DA, in order. Prove ths
`equations
`A’B'C' D’ is 8 parallelogram.
`x=a(cos@—+> 6 sin 8),
`Hint. First show that A’B’ = DC’ = 3AC.
`{8)
`y=@ (sin 6 — @ cos 9).
`19. Using vectors, showthat the diagonals of a parallel
`Tlane perp
`gram bisect each other.
`weve the ry}
`Method. Let A be one vertex and let MW and N te
`the midpoints of the diagonals. Then show th=
`AM = AN,
`
`EXERCISES
`
`In Exercises 1 through 10, express each of the vectors in
`the form ai + j. Indicate all quantities graphically.
`
`1. Pi Po, if Pi is the point (1,3) and Pe is the point
`(2, —1)
`-2. OP3, if O is the origin and P3 is the midpointof the
`vector PiP2 joining P:(2, —1) and P2{—4, 3)
`3. The vector from the point A(2, 3) to the origin
`4. The sum of the vectors 1B and CD, given the four
`points A({1, —1), B(2, 0), C(—1, 3}, and D(—2, 2)
`5. A unit vector making an angle of 30° with the posi-
`tive z-axis
`6. The unit vector obtained byrotating j through 120°
`in the clockwise direction
`
`~I
`
`. A unit vector having the same direction as the
`vector 3i — 4j
`8. A unit vector tangent to the curve y = x7 at the
`point (2, 4)
`9. A unit vector normal to the curve y = xr? at the
`point (2, 4) and pointing from P toward the con-
`cave side of the curve (that is, an “inner” normal)
`
`12.4 SPACE COORDINATES
`
`Cartesian coordinates
`
`In Fig. 12.17, a system of mutually orthogonal cc- |
`ordinate axes, Oz, Oy, and Oz,
`is indicated. Th:
`system is called right-handed if a right-threadec |
`screw pointing along Oz wil! advance when the blade |
`of the screwdriver is twisted from Ox to Oy throug:
`an angle, say, of 90°.
`In the right-handed system |
`shown, the y- and 2-axeslie in the plane of the paper
`and the z-axis points out from the paper. Th:
`Cartesian coordinates of a point P(x, y, z) in space
`may be read from the seales along the coordinat=
`axes by passing planes through P perpendicular +
`each axis. All points on the z-axis have their i-
`and 2-coordinates both zero; that is, they have th-
`form (z,0,0). Points in a plane perpendicular t
`the z-axis, say, all have the same value for their
`z-coordinate.
`Thus,
`for example, 2= 5 is ac
`equation satisfied by every point
`
`stant
`
`(x, 4, 0
`Jf
`
`a”
`
`a7 Cartesis
`
`wrersect in t])
`tiaracterized |
`ae 0, y= 0,
`ea-d octants.
`=. 4,2) have;
`= crst octant.
`nz of the rem
`
`Deimdrical coor
`
`= is frequent
`w=nates (r, é
`
`sarticular, cy
`wien there is
`cczblem. Es
`“s- the pola:
`=.¥)
`in the
`>ordinate (sei
`<= coordinate
`
`z:
`
`y :
`
`SES.
`
`
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`24
`
`Space coordinates
`
`391
`
`12.18 Cylindrical coordinates.
`
`Pte
`
`@= constant
`
`
`12.19
`
`2=constant
`
`re constant
`
`If we hold + = constant and let 9 and z vary, the
`locus of P(r, @, z) is then a right circular cylinder of
`radius r and axis along Oz. The locus r = 0 is just
`the z-axis itself. The locus @ =constant is a plane
`containing the ¢-axis and making an angle @ with the
`xe-plane (Fig. 12.19).
`
`_
`(0, 0, z)
`
`2= constant
`|
`|
`|
`i
`
`.
`(0, y, 2)
`
`
`
`y= constant
`
`!
`!
`!
`
`\|
`
`I'
`
`=
`
`2, 0, 2}
`
`4
`
`(x, 0, 0,7
`
`1
`sonstant — OH.Pyz
`Z
`4
`~~.
`4
`Boni
`~~~ _ (0, y, 0)
`>
`
`ir, y, OF
`
`2.17 Cartesian coordinates.
`
`« plane perpendicular to the z-axis and 5 units
`ssove the zy-plane. The three planes
`
`t2=2, y=3, 2=5
`ssersect in the point P(2,3,5). The yz-planeis
`snaracterized by s = 0. The three coordinate planes
`== 0,y = 0,2 = 0 divide the space into eight cells,
`sled octanis. That octant
`in which the points
`=. y, 2) haveall three coordinates positive is called
`the Jirst octant, but there is no conventional number-
`ag of the remaining seven octants.
`
`Cylindrical coordinates
`= is frequently convenient to use cylindrical co-
`scdinates (r, 8,2) to locate a point in space.
`In
`gatticular, cylindrical coordinates are convenient
`raen there is an axis of symmetry in a physical
`csoblem. Essentially, cylindrical coordinates are
`st the polar coordinates (r, #}, used instead of
`z.y) in the plane z= 0, coupled with the <z
`ordinate (see Fig. 12.18). Cylindrical and Cartes-
`42 coordinates are related by the familiar equations
`
`r? = z?+ y?,
`z=reosd,
`tang= y/z,
`y=rsing,
`22> 2,
`
`(1)
`
`
`
`ite of a cotm
`in Eq. (6
`
`vectors anc “ty
`axis.
`
`VBi-+j
`—5i — 13j
`
`Tametric eous-
`’ taking
`
`CP.
`
`ler, of a quas>-
`idpoints of =>:
`er. Prove the
`
`= 3AC.
`ls of a parallcs-
`
`tt M and V >»
`‘hen show ths
`
`b
`
`—
`
`orthogonal c-
`dicated.
`Th=-
`right-threade:
`vhen the blad-
`to Oy throug
`1anded syster=
`ae of the paper
`: paper.
`The
`,¥, Z) in space -
`the coordinate
`rpendicular tc —
`have their y-
`|
`they have the
`rpendicular tc
`value for their
`z= 5 is an
`1 ¥, 5) lying in
`
`|
`
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