Q
`
`nun
`
`Thomas
`
`Samsqu Exhibit 1020
`
`Page 1 of 8
`
`Page 1 of 8
`
`Samsung Exhibit 1020
`
`

`

`GEORGE B. THOMAS, JR.
`
`Dcpumnen! of .‘Ifalhe‘malics
`
`.‘lIassacimseth's Institute of Technology
`
`Menln Park, Culifvrniu ~ L-unr‘ltm - Drm Mills: Ontario
`
`CALCULUS
`
`AND
`
`ANALYTIC
`
`GEOMETRY
`
`i
`
`I
`
`FOURTH EDITION
`
`"
`Mist.“ nun .-
`
`.
`
`'
`
`
`‘ ~ ~» wt»~ .-".“:'.1>07§-‘_"A.-5'-
`.
`
`-.
`
`I
`
`" ‘ "
`
`‘
`
`_
`
`" -r
`
`Page 2 of 8
`
`A
`VV
`
`ADDISON-WESLEY PUBLISHING COMPANY
`Rim-trim; Mussuhzlsulis
`
`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-
`Wealey 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.
`
`
`
`Page 3 of 8
`
`Page 3 of 8
`
`

`

`
`
`11.1 THE POLAR COORDINATE SYSTEM
`
`We know that a point. can be located in 9. plane by
`giving its abscissa. and ordinate relative to a. given
`coordinate system. Such r- and y-coordinates are
`called Cartesian coordinates, in honor of the French
`mathematician-philosopher ché Descartes“ (159(3-
`1650), who is credited with discovering this method
`of fixing the position of s, point in a. plane.
`
`3/38 .- E":
`
`/ xi";
`r).._—L _
`
`
`
`1L1
`
`y = 2.: — 3 with at Z 1 is another ray; its vertex is L1, 5}.
`
`Another useful way to locate a point in a plant:
`by polar coordinales (see Fig. 11.1). First. we. fix
`an origin 0 and an initial ray‘l‘ from 0. The point P
`has polar coordinates r, 6, with
`
`= directed distance from 0 to P,
`
`(19.)
`
`and
`
`9 = directed angle from initial my to 0P.
`
`(lb)
`
`the angle 5 is positive when
`As in trigonometry,
`measured counterclockwise and negative when mea-
`sured clockm’se (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 -= ‘3}, 6 = 30°.
`It also has coordinates r =— 2. = —330'='. or .“ = 2,
`8 .= 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 ertlngm see World of
`.lIat/iemciics, Vol. 1, pp. 235—253.
`TA ray is a. half-line consisting of a vertex and points
`of a limo on one side of tlzv vortex. For vxamgle, the
`origin and positive z-axis is a my. The points on the. line
`
`POLAR
`
`}OORDINATES
`
`CHAPTER 11
`
`vr_.___.‘.,_._____._.'.——————-
`
`Page 4 of 8
`
`Page 4 of 8
`
`

`

`5 Polar and C:
`
`5 The 011016 T =
`
`F t
`
`:
`
`§
`
`'
`
`'
`
`
`
`
`
`0.1: =
`
`=
`
`1‘:
`
`(2) 30° + n:
`‘2, 210° J— n 2
`
`I
`
`‘- “re represent
`'
`_ _ulas
`
`
`362
`
`Four coordinates
`
`11.1
`
`- Pgi‘. 30
`
`—\ [if
`/ 10:31}?
`i
`/
`
`zilllfi. 1‘31:
`9—.
`l
`
`".2 T:.- at? = 30°isthe sameasthe raya = —330°.
`
`_..1 0:31}
`
`
`
`210”:
`
`a: 21-3=
`
`11.3 The rays 9 = 30° and 9 = 210’ make a line.
`
`
`
`-
`
`(1— 1, 3-11“:
`
`— w ‘1—2, 30°)
`
`“.4 The terminal ray 6 = rx'fi and its negative.
`
`in Eq. (1a). The ray 6 = 30° and the ray 6 r 210°
`together make up a. complete line through 0 (see
`Fig. 11.3). The point P(2. 210°) '2 units from 0 on
`the ray 0 = 210° has polar coordinates r = 2,
`9 = 210°.
`It can be reached by a person standing
`at 0 and facing out along the initial ray, if he first
`turns 210° counterclockwise, and then goes forward
`
`‘2 units. He would reach the same point. by tumin;
`
`only 30: counterclockwise from the initial my 311?.
`then going backward 2 units. So we say that. th'
`point also has polar coordinates r = —2. 0 = 30°.
`Whenever the angle between two rays is 180°, tr.—
`rays a :tually make a straight line. We then say lhi—f
`either ray is the negative of the other. Points If'
`the ray 19 = a have polar coordinates (r, a); ti"
`r 2 0. Points on the negative ray. 9 = a - 1813’
`have coordinates it, a) with .v 5 0. The origin
`r —* 0.
`{See Fig. 11.4 for the my a = 30° and
`negative. A word of caution: The "negative"
`the ray 6 _— 30" is the ray 6 >— 30°
`180° -= 21‘"
`and not the m3: 6 -— -30°. “Negative” refers to 1':
`directed distance r.)
`There is a great advantage in being able to us—
`both polar and Cartesian coordinates at once.
`7
`do this. we use a common origin and take the inn-
`my as the positive .r-axis, and take the ray 9 = ET '
`as the positive y-zuis. The coordinates, shown
`Fig. 11.5. are then related by the equations
`
`a: -= r cos 6,
`
`y = r sin 6.
`
`‘.
`
`These are the equations that define sir; 9 and cl.» ‘
`when r is positive. They are. also valid if ." is her--
`tive, because
`
`cos {9 + 180°) f
`
`cos 9,
`
`sin ('3
`
`-— [80°] = —sin 9.
`
`so positive .r’s on the 1‘9 — 180°ju-ray correspond 7
`negative r’s assmiated with the G-ray. When 'r : .
`then x — y = 0, and P is the origin.
`If we impose the condition
`
`:‘ -= a
`
`{a constant),
`
`i
`i
`‘
`
`then the locus of P is a circle with center 0
`radius a. and P describes the circle once as 6 rare
`from 0 to 360° (see Fig. 11.6}. Or. the other he; .
`if we let r vary and hold 6 fixed, say
`
`9 = 30¢.
`
`-.
`
`the locus of P is the straight line shown in Fig. 1'.
`
`-_
`
`Page 5 of 8
`
`
`7‘5e adopt the c
`umber.
`"—30 < -,
`
`
`.
`.
`_
`_
`
`c _.e origin, :r = |
`I
`.
`.—.'le same point
`
`Itsrent ways in
`
`it point
`(2, 30°)
`“resent-axiom:
`(
`
`1. ~7150°). The
`La two formula
`
`
`
`(2.-
`(—2i
`
`§1r+
`it?! P
`
`71
`T.
`
`MIG
`
`'
`
`Page 5 of 8
`
`

`

`
`
`11.1
`
`I
`
`The polar coordinate system
`
`363
`
`
`
`The facr, 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, 7r) is on the curve
`
`r2 = 44:2 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., 11') is on the curve
`
`r = a(1
`
`cos 9).
`
`(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 0 = r2/4a2
`from (6)
`into (7) and then solving the resulting
`quadratic equation
`
`for
`
`(2>2+4(2)—4=°
`
`7'
`
`a
`
`= -—2 it
`
`2~/§,
`
`(8)
`
`we do not obtain the point {2a, 1r) as a point of inter-
`section. Thc 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 0 = 0 and in the other case
`
`It is as though two ships describe
`when 6 = 1r.
`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’
`
`(20! 7r):
`
`(r1: 01):
`
`(r1! _01)y
`
`where
`
`’1 = (—2 — ZV/fim,
`cosol=1—fl=3_2\/§.
`a
`
`(9b)
`
`
`
`11.6 The circle r — a is the locus P.
`
`‘Ve adopt the convention that 1' may be any real
`:umber, ~92 < r < :c. Then r = 0 corresponds
`to :r = O, y = 0 in Eqs. (2), regardless of 0. That is,
`
`r = 0.
`
`6 any value.
`
`(5)
`
`is the origin, x = 0, y = 0.
`The same point may be repraented in several
`different- ways in polar coordinates. For example,
`'he point
`(2, 30°), or (2,2rf6), has the following
`representations:
`(2, 30°),
`(2, —330°),
`{—2, 21 °).
`—2, —150°). These and all others are summarized
`:n the two formulae
`
`(2, 30° —2 n 360°),
`(—2, 210° -'- n 360°),
`
`}
`
`=O,:t1,i2,...;
`
`or, if we represent the angles in radians, in the two
`formulas
`
`by turning
`ll ray and
`I that the
`0 = 30°.
`
`5 180°, the
`n say that
`Points on
`
`r, a) with
`a + 180’
`
`: origin is
`3" and its
`gative” c-t'
`l° = 210:
`’ers to thi-
`
`2le to use
`)nce. T-.
`:he initial
`»' 0 = 90°
`shown ll.
`i
`
`(2:-
`
`md cos 5
`
`' is nega-
`
`spond tr-
`
`n r =
`
`n. = 0, :1, 1-2
`
`....
`
`Only the last two of the points (92.) 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
`
`12.4
`
`Then, applying (3), we have
`
`and
`
`u1=ic056+jsin0
`
`“2
`
`icos(9+§g)+isin (6—337)
`
`isinO—jcosd.
`
`Therefore
`
`c?) a an + (aozvuz
`
`= a(i cos 6 +j sin 0) — adii sin 9 — j cos 6)
`
`= a (cos 6 + 0 sin 9)i + a {sin 0 — 6 cos 6'2Ij.
`
`We equate this with zi—. 31 and. since corresponding
`components must be equal, we obtain the parametric
`equations
`
`a: = a (cos6—. 83in 9},
`y = o(sii~.0—9c050).
`
`6.
`i
`"
`
`EXERCEEES
`
`In Exercises 1 through 10, express each of the vectors in
`the form ai 7— bj.
`Indicate all quantities graphically.
`
`1. P1P2, if P1 is the point (1, 3) and P2 is the poi:t
`(2, -1)
`
`« 2. 0P3, if 0 is the origin and P3 is the midpoint of the
`vector P1P2 joining 13(2, -—1) and P2? «1, 3)
`
`3. The vector from the point .412, 3} to the Origin
`
`4. The sum of the vectors H; and CT), given the four
`points A(1, —1}, 8(2, 0),C(—1,3,’1, and DuI——2, 2)
`
`5. A unit vector making an angle oi 30° with the posi-
`tive z-axis
`
`6. The unit vector ohtained by rotatingj through 120“
`in the clockwise direction
`
`I. A unit vector having the same direction as the
`vector 3i — 4i
`
`8. A unit vector tangent to the curve y = :2 at. the
`point (2, 4)
`the
`9. A unit vector normal to the curve 3; = 2:2 at.
`point £12. 4) and pointing from P toward the cou-
`cave side of the curve (that is, an “inner” normal;I
`
`
`
`10. A uni: vector tangent to the involuze of a circ:
`whose parametric equations are giver. in Eq. (6)
`
`Find the lengths of each of the foilowing vectors and
`angle that each makes with the positive x-axis.
`
`ll. i—ij
`
`12.2i—3j
`
`l3. V? i +1
`
`l4. —2i+ 3i
`
`15. 5i— 1'2j
`
`16. —5x — 12j
`
`:
`
`17. Use vcctor methods to determine parametric eq‘gs .
`tions for the trochoid of Fig. 12.6, by taking
`
`'
`
`:
`
`.
`
`R=07’=OY+JY7—C_P.
`
`Lot A, B. C. D be the vcrtices, in order, of a quadrr
`lateral. Let
`:1’. B’, C', D' be the midpoian of ti.-
`Provo tha'
`sides AB, BC. CD, and 0.1, in order.
`.4'B’C’D’ is a parallelogram.
`Hint. First show that W7 = 175' = $30.
`
`19. Using vecwrs. show that :he diagonals 0:” a parallel.—
`gram bisect each other.
`.lfez'hod. Let
`:1 be one vertex and let .1! and N to
`the mideir‘:5 0f the diagonals. Then show thi'
`= A.\'.
`
`12.4 SPACE COORDINATES
`
`Cartesian coordinates
`
`In Fig. 12.17, a system of mutually orthogonal cv.— -
`ordinate axes. Oz. 01,: and 02,
`is indicated. T he
`‘
`system is Called right-handed if a right-threade:
`screw pointing along 02 will advance when the blade ,
`of the screw driver is twisted from 0: to 0y trhroug:
`an angle, say, of 90".
`In the right—handed systez-
`shown, the y- and 2-axes lie in the plane of the pap-e:
`and the .z-axis points out from the paper. Tb
`Cartesian coordinates of a point Pifz, y, z) in spar-f
`may be read from the scales alongr the coordinate
`axes by passing planes through P perpendicular 1
`each axis. All points on the .r-axis have their ;.-
`and z-coordinarcs both zero; that is, they have the
`fomi (1,0, 0). Points in a plane perpendicular r
`the z-axis. say, all have the same value for their
`z-coordinace.
`Thus.
`for example,
`2 = 5 is
`:1;
`equation satisfied by every point
`(z. y, 5) lying i:
`
`IJ‘. (i. z
`
`.‘Zilll:
`
`a ' r Cartesis
`
`‘. lane perpi
`more the xy-I
`
`:1'crsect in ii
`:Lracterized l
`
`' = 0: y = 0:
`:L'ed octants.
`
`z) have:
`,:
`,'_;- f rst octant
`:15 2-1“ the rem
`
`Srindricalcoor
`
`‘_= frequent
`.i'
`.C'zzzates (r,£
`;:uicular, cy
`men there is
`:ézélem. Es
`
`the pola:
`'5‘.
`in the
`j.)
`'tdinate (sen
`2
`-:_ coordinate
`
`I ‘.
`
`y:
`
`
`
`gauze?
`
`Page 7 of 8
`
`Page 7 of 8
`
`

`

`12.4
`
`|
`
`Space coordinates
`
`391
`
`12.1; Cylindrical coordinates.
`
`
`12.19
`
`2 = constant
`
`f '— cnnsmn:
`
`If we hold r = constant and let 9 and z vary, the
`. locus of P(r, 0, z) is then a right circular cylinder of
`radius r and axis along Oz. The locus r = O is just
`the z-axis itself. The locus 0 =. constant is a plane
`containing the e-axis and making an angle 0 with the
`are-plane (Fig. 12.19).
`
`Z
`
`l
`I
`I
`I
`
`::
`
`\ 2- constant
`(0. U. z;
`i
`l
`|
`I
`
`t
`,
`IJI. y, 2;
`
`
`
`I
`9~~u PG” V’ Z]
`~~~
`“~ II). y. 0)
`
`y - mnstam.
`
`‘9'
`
`4
`
`ll, :3
`
`-
`
`:nstuu:
`
`7-—
`
`/
`
`I
`7’
`
`‘
`,
`3.x: “I 02' I
`
`r’
`
`I
`
`fix, 1:. 02!
`
`2.17 Cartesian coordinates.
`
`1 plane perpendicular to the z-axis and 5 units
`cove the :cy-plane. The three planes
`
`
`
`3:2) y=31 3:5
`
`:ersect in the point P(2, 3, 5). The yz-plane is
`macterized by a: = 0. The three coordinate planes
`: = 0, y = 0, z = 0 divide the space into eight cells,
`sled octants. That octant
`in which the points
`:. y, 2) have all three coordinates positive is called
`:efirst octant, but there is no conventional number-
`:g of the remaining seven octants.
`
`cylindrical coordinates
`
`If is frequently convenient. to use cylindrical co-
`rilnates (r, 0,2) to locate a point in space.
`In
`auricular, cylindrical coordinates are convenient
`then there is an axis of symmetry in a physical
`rabiem. Essentially, cylindrical coordinates are
`“.21 the polar coordinates (r, 9}, used instead of
`:. y)
`in the plane z = 0, c0upled with the z-
`::-:Irdinate (see Fig. 12.18). Cylindrical and Cartes-
`;: coordinates are related by the familiar equations
`
`x: rcoso, r2: z2+y2,
`
`y= rsin0,
`tan 0 = WI,
`2 = z.
`
`(1)
`
`rte of a cru-
`in Eq. {6
`
`rectors and ’Ir
`z-axis.
`
`\/§i~l-i
`
`—5i— 123‘
`
`.rametric e; a»
`' taking
`
`C75.
`
`ter, of a quart»
`.idpoints of 1::
`cr. Prove 2':_-.
`
`= gfi.
`
`is of a paralla.»
`
`:t M and N :v
`'hen show IL;
`
`.y, 5) lying ir-
`
`orthogonal c-'-
`.dicated. The
`
`'
`
`right-threads:
`when the blad:
`
`to 0y throug':
`landed syster;
`1e of the paper
`: paper.
`The
`, y, z) in space ‘
`the coordinate
`
`:rpendicular ts.
`have their 9-
`they have the
`rpendicular tr.
`ralue for their
`z = 5 is at.
`
`?
`
`‘
`
`7
`
`.
`
`Page 8 of 8
`
`Page 8 of 8
`
`