`
`fisleseett
`
`Samsung Exhibit 1020
`
`Page | of 8
`
`Page 1 of 8
`
`Samsung Exhibit 1020
`
`

`

`
`
`GEORGE B. THOMAS, JR.
`
`Depariment of Mathematics
`
`Massachusetis Institute of Technology
`
`A
`vv
`
`ADDISON-WESLEY PUBLISHING COMPANY
`
`Reading, Massachusetts
`
`Menlo Park, California - London - Don Mills, Ontario
`
`CALCULUS
`
`AND
`
`ANALYTIC
`
`GEOMETRY
`
`FOURTH EDITION
`
`Sredeeege}
`
`RB
`
`eT Oo NISL eae oeee Bane
`
`Page 2 of 8
`
`Page 2 of 8
`
`

`

`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 #
`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.
`
`
`
`Page 3 of 8
`
`Page 3 of 8
`
`

`

`
`
`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 aninitial rayt from O. The point P
`has polar coordinatesr, 6, with
`= directed distance from O to P,
`
`(1a)
`
`and
`
`9 = directed angle frominitial ray to OP.
`(1b)
`As in trigonometry,
`the angle 4 is positive when
`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 point 2 units from the origin, 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 ray. The points on the line
`
`POLAR
`SOORDINATES
`
`CHAPTER 11
`
`Page 4 of 8
`
`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 2- and y-coordinates are
`called Cartesian coordinates, in honor of the French
`mathematician-philosopher René Descartes* (1596-
`1650), whois credited with discovering this method
`of fixing the position of a point in a plane.
`
`P| r, B)
`
`Me \eLS
`
`0
`
`
`>
`———
`initial ray
`
`11.1
`
`y = 2c+3withe = lis another ray; its vertex is (1, 5).
`
`Page 4 of 8
`
`

`

`32
`
`Poter coordinates
`
`11.1
`
`» P(2, 30
`
`The vay 9 = 30° is the same as the ray @ = —330°.
`
`w b= 30°
`
`
`
`7Pi2, 210°)
`¢
`#=210°
`
`
`
`5 Polar and C;
`
`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
`oN al
`point also has polar coordinates r = —2, @ = 30°.
`=ae \e=30°
`Whenever the angle between two rays is 180°, th:
`0—_ peewee
`j
`initial ray
`rays actually make a straightline. We then say ths:
`}
`ée=0°
`either ray is the negative of the other. Points cz
`the ray @= a have polar coordinates (7, a) Wit
`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 6 = 30°+180° = 210
`and not the ray § = —30°. “Negative” refers to tt
`directed distance r.)
`There is a great advantage in being able to us
`both polar and Cartesian coordinates at once.
`T
`do this, we use a commonorigin and take the init:s
`ray as the positive z-axis, and take the ray = or
`as the positive y-axis. The coordinates, shown
`Fig. 11.5, are then related by the equations
`11.3. The rays @ = 30° and @ = 210° makealine.
`6 The circle r -
`z=reosé, y=rsiné.
`3
`
`
`
`
`11.4 The terminal ray @ =
`
`7/6 and its negative.
`
`These are the equations that define sin § and co: *
`whenr is positive. They are alsovalid if r is nez=
`tive, because
`
`cos (9+ 180°) =
`
`—cos 4,
`
`sin (6
`
`- 180°) = —sin 4,
`
`so positive r’s on the (6 + 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
`
`y=a_ (aconstant),
`
`
`We adopt the ¢
`Bccber, —0 <}
`
`p= = 0,y = Oir
`T=
`
`
`“ne origin, x = |
`©
`. “he same point
`
`tf-rent ways in
`
`He point
`(2, 30°)
`hexresentations:
`(
`2, —150°). The
`
`-"2e two formula
`
`
`
`4
`
`(2, 30° + ne
`—2, 210° + n;
`
`> 2 we represent
`
`in Eq. (1a). The ray 6 = 30° and the ray 6 = 210°
`together make up a complete line through O (see
`Fig. 11.3). The point P(2, 210°) 2 units from O 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
`
`|
`|
`|
`
`then the locus of P is a circle with center O ace
`radius a, and P describes the circle once as 9 vam
`from 0 to 360° (see Fig. 11.6). On the other haze
`if we let r vary and hold 6 fixed, say
`@ = 30°,
`
`”
`
`the locus of P is the straight line shownin Fig. 11
`
`Page 5 of 8
`
`
`
`
`Page 5 of 8
`
`

`

`oy turning
`al ray and
`, that the
`6 = 30°.
`; 180°, the
`n say tha:
`Points on
`r,a) with
`a + 180°
`: origin is
`)° andits
`mative” oF
`)° = 210°
`ers to the
`
`le to use
`mee. Te
`che initia!
`"#8 = 90°
`shown in
`=
`
`(2)
`
`ind cos 4
`is nega-
`
`spond te
`
`nr= 0,
`
`
`
`
`
`—p——m
`
`i
`———___
`\
`
`o
`
`Suey
`
`11.5 Polar and Cartesian coordinates.
`
`
`
`11.6 The circle r = a is the locus P.
`
`We adopt the convention that r may be anyreal
`number, —o <7 < x. Then r = 0 corresponds
`coz = 0. y = Oin Eas. (2), regardless of 6. That is,
`
`r=0,
`
`#@ any value,
`
`(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,...
`
`11.1
`
`|
`
`The polar coordinate system
`
`363
`
`
`
`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(1 — 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 by first substituting cos @ = r?/4a?
`from (6)
`into (7) and then solving the resulting
`quadratic equation
`
`for
`
`= —2
`
`He
`
`2V2,
`
`(8)
`
`@)+4G)-e=
`Rin
`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 = 0 and in the other case
`when @= 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),
`
`(4, 61);
`
`(ri, —41),
`
`(9a)
`
`where
`
`n= (—2 T 2V/2)a,
`(9b)
`cos 6; = 1 — “1 = 3 — av.
`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.
`
`Page 6 of 8
`
`Page 6 of 8
`
`

`

`390
`
`Vectors and parametric equations
`
`|
`
`Then, applying (3), we have
`
`sad
`
`uy = icos@+jsiné
`7
`
`ua =
`
`icos (9+ 22) +isin (+)
`
`= isin #@ —jcos@.
`
`Therefore
`
`OP = au + (a0)uz2
`afi cos 6+ j sin @} — ad{isin @ — j cos 4)
`
`a (cos §+- @sin $i++ a (sin § — G cos Aj.
`
`10. A unit vector tangent to the involute of a circ= 4
`whose parametric equations are given in Eq. (6)
`
`Find the lengths of each of the following vectors and th
`angle that each makes with the positive z-axis.
`
`lL. i+j
`
`14. —2i+ 3j
`
`12.
`
`15.
`
`2i — 3j
`
`si + 12j
`
`13, V3i+j
`
`16. —5i — 12
`
`17. Use vector methods to determine parametric equs |
`tions for the trochoid of Fig. 12.6, by taking
`
`R = OP = OM + MC ~ CP.
`
`|
`
`j
`
`In the right-handed system |
`
`18. Let A, 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
`Prove ths:
`sides AB, BC, CD, and DA,
`in order.
`equations
`A’B'C' D' is s parallelogram.
`x=a(cos@— 6 sin @),
`Hint. First show that A’B’ = DC’ = 4AC.
`{6)
`y=a (sin 6 — 8 cos 9).
`19. Using vectors, showthat the diagonals of a parallel
`Tiane perp
`gram bisect each other.
`meve the ry
`Method. Let A be one vertex and let M and Nte
`the midpoints of the diagonals. Then show ths
`AM = AN.
`
`EXERCISES
`
`In Exercises 1 through 10, express each of the vectors in
`the form ai + dj.
`Indicate all quantities graphically.
`
`1. Pi Pe, 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 Pi P2 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 AB 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 = x? 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
`
`3
`
`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 will advance when the blade |
`of the screwdriver is twisted from Ox to Oy throug:
`an angle, say, of 90°.
`shown, the y- and 2-axes lie in the plane of the paper
`and the z-axis points out from the paper.
`The
`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 z-coordinates both zero; that is, they have the
`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, z= 5 is ac
`equation satisfied by every point
`(z, y, 5) lying ir
`
`stant
`
`iz, 0,0
`//
`
`x”
`
`Cartesis
`
`wtersect in t!
`taracterized|
`aaiae 0, y= 0,
`ei-d octants.
`= ¥, 2) have |
`te trst octant.
`ne of the rem
`
`Oeimdrical coor
`
`- is frequent
`ic—nates (r,
`macticular, cy
`wien there is
`cczblem. Es
`= the pola:
`> ¥)
`in the
`>ordinate (se
`<= coordinate
`
`wt
`
`y :
`
`
`
`BEST
`
`Page 7 of 8
`
`Page 7 of 8
`
`

`

`_¥, 5) lying in
`
`OP.
`
`‘er, of a quas>-
`idpoints of ts
`er. Prove the
`
`= 3AC.
`is of a parallcs=
`
`t+ Mand V >»
`hen show tis:
`
`2
`
`(r, y, 0
`
`2.17 Cartesian coordinates.
`
`« plane perpendicular to the z-axis and 5 units
`sSove the zy-plane. The three planes
`
`z= 2, y = 3, a= 5
`scersect in the point P(2,3,5). The y2-plane is
`snaracterized by sx = 0. The three coordinate planes
`== 0,y = 0,2 = 0 divide the spaceinto eight cells,
`sled octants. That octant
`in which the points
`=. y, 2) have all three coordinates positive is called
`s2¢ Jirst octant, but there is no conventional number-
`ag of the remaining seven octants.
`
`Page 8 of 8
`
`12.4
`
`|
`
`Space coordinates
`
`391
`
`
`12.19
`
`12.18 Cylindrical coordinates.
`
`2=constant
`
`reconstant
`
`2= constant
`|
`|
`|
`i
`
`.
`:
`(0, ¥, 2)
`
`
`
`24
`
`|
`I
`|
`!
`
`,
`(0, 0, 2)
`
`(2, 0, 2)
`
`ym constant
`
`ite of a cht»
`in Eq. (6
`
`vectors anc tty
`axis.
`
`V3i-+j
`
`—i5i — 135
`
`I
`1
`sonstant — HW.Pry)
`=
`.
`/
`ie
`7 ~~._(0, y, 0)
`
`if
`:
`Tametric eo2s-
`’ taking
`7
`* es
`al
`(x, 0, 0},7
`
`orthogonal c-- _
`dicated. Th=
`right-threade=
`Cylindrical coordinates
`vhen the blad-
`= is frequently convenient to use cylindrical co-
`to Oy throug:
`dinates (r, 6,2) to locate a point in space.
`In
`ianded syster:
`varticular, cylindrical coordinates are convenient
`1e of the paper
`raen there is an axis of symmetry in a physical
`; paper.
`The
`czoblem. Essentially, cylindrical coordinates are
`, Y¥, z) in space
`st the polar coordinates (r, 9), used instead of
`the coordinates
`=.y)
`in the plane z= 0, coupled with the z
`rpendicular to |
`~ordinate (see Fig. 12.18). Cylindrical and Cartes-
`have their y-
`—
`If we hold r = constant and let 9 and z vary, the
`== coordinates are related by the familiar equations
`they have the
`_locus of P(r, @, z) is then a right circular cylinder of
`rpendicular tc
`z=rcosd,
`r?= zg? + y?
`radius r and axis along Oz. The locus r =0is just
`ralue for their
`|
`the z-axis itself. The locus ¢ =constant is a plane
`y = rsin 6,
`tan § = y/z,
`{1)
`z= 5 is an
`containing the ¢-axis and making an angle @ with the
`2=> 2,
`xe-plane (Fig. 12.19).
`
`Page 8 of 8
`
`