Student Solutions Manual

DAN CLEGG • BARBARA FRANK
.Student Solutions Manual for .
MULTIVARIABLE . CALCULUS SEVENTH EDITION
DAN CLEGG Palomar College
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I ·ll
D PREFACE
This Student Solutions Manual contains detailed solutions to selected exercises in the text Multivariable Calculus, Seventh Edition (Chapters 10-17 of Calculus, Seventh. Edition, and Calculus: Early Transcendentals, Sevent,h Edition) by James Stewart. Specifically, it includes solu tions to the odd-numbered exercises in each chapter section, review section, True-False Quiz, and Problems Plus section. Also included are all solutions to the Concept Check questions.
Because of differences between the regular version and the Early Transcendentals version of the text, some references are given in a dual format. In these cases, readers of the Early Transcendentals text should use the references denoted by "ET."
Each solution is presented in the context of the corresponding section of the text. In general, solutions to the initial exercises involving a new concept illustrate that concept in more detail; this knowledge is then utilized in subsequent solutions. Thus, while the intermediate steps of a solution are given, you may need to refer back to earlier exercises in the section or prior sections for addition al explanation of the concepts involved. Note that, in many cases, different routes to an answer may exist which are equally valid; also, answers can be expressed in different but equivalent forms. Thus, the goaJ of this manual is not to give the definitive solution to each exercise, but rather to assist you as a student in understanc#ng the concepts of the text and learning how to apply them to the. chal lenge of solving a problem.
We would like to thank James Stewart for entrusting us with the writing of this manual and offer ing suggestions and Kathi Townes of TECH-arts for typesetting and producing this manual as well as creating the illustrations. We also thank Richard Stratton, Liz Covello, and Elizabeth Neustaetter of Brooks/Cole, Cengage Learning, for their trust, assistance, and patience.
DAN C LEGG
Cape Fear Community College
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D ABBREVIATIONS AND SYMBOLS
indicates the use of ljHospital 's Rule.
' indicates the use ofFormu'Ja j in the Table of Integrals in the back endpapers.
indicates the use of the substitution { u = sin x, if.u = cos x dx}.
indicates the use of the substitution { u = cos x, du = - sin x dx}.
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D · CONTENTS
10.1 Curves Defined by Parametric Equations
10.2 Calculus with Parametric Curves 7
1 0.3 Polar Coordinates 13
10.4 Areas and Lengths in Polar Coordinates 20
10.5 Conic Sections 26
Review 35
11.1 Sequences 45
11.2 Series 51
11 .4 The Comparison Tests 62
11.5 Alternating Series 65
11 .6 Absolute Convergence and the Ratio and Root Tests 68
11 .7 Strategy for Testing Series 72
11.8 Power Series 74
11.10 Taylor and Maclaurin Series 83
11.1 1 Applications of Taylor Polynomials 90
Review 97
12.1 Three-Dimensional Coordinate Systems 111
12.2 Vectors 114
12.3 The Dot Product 119
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viii o CONTENTS
12.5 Equations of Lines and Planes 128
. 12.6 C,ylinders and Quadric Surfaces 135
Review 140
13.1 Vector Functions and Space Curves 151
13.2 Derivatives and Integrals of Vector Functions 157
13.3 Arc Length and Curvature 161
13.4 Motion in Space: Velocity and Acceleration 168
Review 173
14.1 Functions of Several Variables 183
14.2
14.3
14.5 The Chain Rule 207
14.6 Directional Derivatives and the Gradient Vector 213
14.7 Maximum and Minimum Values · 220
14.8 Lagrange Multipliers 229
15.1 Double Integrals over Rectangles 247
15.2 Iterated Integrals 249
15.3 Double Integrals over General Regions 251
15.4 Double Integrals in Polar Coordinates 258
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15.5
15.6
15.7
15.8
15.9
15.10
Surface Area 267
Triple Integrals 269
Review 289
16.1 Vector Fields 303
16.2 Line Integrals 305
16.4 Green's Theorem 313
16.6 Parametric Surfaces and Their Areas 321
16.7 Surface Integrals 328
16.8 Stokes' Theorem 333
Review 337
17.1 Second-Order Linear Equations 345 ·
17.2 Nonhomogeneous Linear Equations 347
17.3 Applications Of Second-Order Differential Equations 350
17.4 Series Solutions 352
CONTENTS D ix
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10 D PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.1 Curves Defined by Parametric Equations
1. X= t2 + t, y = t2 - t, -2 S t S 2
t - 2 - 1 0 l 2
X 2 0 0 2 6
y 6 2 0 0 2
3. x = cos2 t, y = 1 - sin t, 0 S t S 7r /2
t 0 7r/6 7r/3 7r/ 2
X 1 3/ 4 1/ 4 0
y 1 1/ 2 1 - 1 ~ 0.13 0
5. X = 3 - 4t, y = 2 - 3t
(a) t - 1 0 1 2
X 7 3 -1 - 5
y 5 2 -1 - 4
(b) X= 3 - 4t =} 4t = -X + 3 =} t = -~X +~ ' SO
y = 2 - 3t = 2 - 3(-~x + ~ ) = 2 +~X - £ =} y =~X - ~
1. x = 1 - e, v = t - 2, - 2 s t s 2
(a) t - 2 - 1 0 1 2
X -3 0 1 0 -3
y - 4 -3 - 2 - 1 0
(b) y = t - 2 =} t = y + 2, SO X= 1 - t 2 = 1 - (y + 2? =}
x = - (y + 2) 2 + 1, or x =:= - y 2 - 4y -e 3, with - 4 S y S 0
y
y
y
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X
X
X
9. X = Vt, y = 1 - t (a)
t 0 1 2 3 4
X 0 1 1.414 1.732 2
y 1 0 - 1 - 2 .:....3
(b) x = Vt =} t = x 2 =} y = 1- t = 1 - x 2
. Since t ~ 0, x ~ 0.
So the curve is the right half of the parabola y = 1 - x 2 •
11. (a) x = sin ~0, y =cos ~(:1, -1r ~ (:1 ~ 1r.
x2 + y2 = sin2 ~8 + cos2 ~8 = 1. For -1r ~ 0 ~ 0, we have
- 1 ~ x ~ 0 and 0 ~ y ~ 1. For 0 <. 8 ~ 1r, we have 0 < x ~ 1
and 1 > y ~ 0. The graph is a semicircle.
13. (a) x = sint, y = csct, 0 < t < ~- y = csct = ~ = !. . . smt x
For 0 < t < ~.we haveO < x < 1 andy > 1. Thus, the curve is the
portion of the hyperbola y = 1/x withy> 1.
15. (a) x = e2 t =}. 2t = ln x =} t. = ~ ln x.
y = t+1 = ~lnx+1.
17. (a) x =sinh t, y = cosh t '=} y2 - x2 = cosh2 t - sinh2 t = 1. Since
y = cosh t ~ 1, we have the upper branch of the hyperbola y2 - x 2 = 1.
y (0, I) ,f = O
X
'\,,,,, 0 X
(b) y
(b) y
0 X
. 19. x = 3 + 2 cost, y = 1 + 2 sin t, 1r / 2 ~ t ~ 31r / 2. By Example 4 with r = 2, h = 3, and k = 1, the motion of the particle
takes place on a circle centered at (3, 1) with a radius of2. As t goes from -¥- to 3;, the particle starts at the point (3, 3) and
moves counterclockwise along the circle (x - 3)2 + (y - 1)2 = 4 to (3, - 1) [one-half of a circle].
21. x = 5sint, y=2 cost =} sint=~, cost=~· sin2 t+cos2 t =1 =} (~r + (~r = l.The motionofthe particle takes place on an ellipse centered at (0, 0). As t goes from - 1r to 511', the particle starts at the point (0, -2) and moves
clockwise around the ellipse 3 times.
23. We must have 1 ~ x ~ 4 and 2 ~ y ~ 3. So the graph of the curve must be contained in the rectangle (1, 4] by [2, 3].
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SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS D 3
25. When t = - 1,-(x,y) = (0, - 1). As t increases to 0, x decreases to - 1 andy
(-1,0) X
increases to 0. As t increases from 0 to 1, x increases to 0 and y increases. to 1.
As { increases beyo~d 1, both x and y increase. Fort < - 1, x is positive and
decreasing and y is negative and increasing. We could achieve greater accuracy
by estimating x- and y-values for selected values oft from the given graphs and
plotting the correspo~ding points.
t=O t=-1
27. When t = 0 we see that x = 0 and y = 0, so the curve starts at the origin. As t
increases from 0 to~. the graphs show that y increases from 0 to 1 while x
increases from 0 to 1, decreases to 0 and to - 1, then increas.es back to 0, so we
arrive at the point (0, 1). Sinlilarly, as t increases from ~ to 1, y decreases from 1
y
X
to 0 while x repeats its pattern, and we arrive back at the origin. We could achieve greater accuracy by estimating x- and
y-values for selected values of t from the given graphs and plotting the corresponding points.
29. Use y = t and x = t - 2sin 7ft with at-interval of [-7r, 7r].
31. (a) x = X1 + (x2 - x1)t, y = YI + (y2- Y1)t, 0 5 t 5 1. Clearly the curve passes through P1(X1, Yl) when t = 0 and
through P2(x2 , y2) when t = 1. For 0 <.t < 1, x is strictly between x1 and x2 and y is strictly between y1 and y2. For.
every value oft, x andy satisfy the relation y- Y1 = y2 - YI (x - x1), which is the equation of the Line through
X2 - Xl
H (x1, Yl) and P2(;,;2, y2).
Finally, any point (x, y) on that line; satisfies y- Yl = x - x 1 ; if we call that common value t , then the given . Y2 - Y1 X2 - X I ,
parametric equations yield the point (x, y); ancf any (x, y) on .th~ line between P1(x1, y1) and P2(x2, y2) yields a value of
t in [0, 1]. So the given parametric equations exactly specify the line segment from P1 (x1, y1) to P2(x2, y2).
(b) x = -2 + (3- ( -2)]t = -2 + 5t andy = 7 + ( - 1 - 7)t = 7 - Bt for 0 5 t 5 1.
33. The circle x 2 + (y - 1 )2 = 4 has center .(0, 1) and radius 2, so by Example_ 4 it can be represented by x = 2 cost,
y = 1 + 2 sin t, 0 5 t 5 27r. This representation gives us the circle with a counterclocifwise orientation starting at (2, 1).
(a) To get a clockwise orientation, we could change the equations to x = 2 cost, y =. 1- 2 sin t, 0 5 t 5 27r.
(b) To get three times around in the counterclockwise direction, we use the original equations ~ = 2 cost, y = 1 + 2 sin t with
the domain expanded to 0 5 _t 5 67r.
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4 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(c) To start at (0, 3) us ing the original equations, we must have X1 = 0; that is, 2 cos t = 0. Hence, t = ~. So we use
x = 2cost, y = 1 + 2sint, ~ ~ t ~ 3 ;.
Alternatively, if we want t to start at 0, we could change the equations of the curve. For example, we could use
x = - 2 sin t , y = 1 + 2 cos t , 0 ~ t ~ 1r.
35. Big circle: It's centered at (2, 2) 'with a radius of 2, so by Example 4, parametric equations are
x = 2 + 2cost, y = 2 + 2 sin t,
Small circles: They are centered at (1, 3) and (3, 3) with a radius ofO.l. By Example 4, parametric equations are
and
(left)
(right)
0 ~ t ~ 27T
0 ~ t ~ 27T
Semicircle: It's the lower half of a circle centered at (2, 2) with radius 1. By Example 4, parametric equations are
x = 2 + 1 cost, . y = 2 + 1 sin t, . . .
To get all four graphs on the same screen with a typical graphing calculator, we need to change the last t-interval to[O, 21r] in
order to match the others." We can do this by changing t to 0.5t. This change gives us the upper half. There are several ways to
get the lower half-one is to change the"+" to a "-" in the y-assigr1ment, giving us
x = 2 + 1 cos(0.5t), y = 2 - 1 sin (0.5t ), · 0 ~ t ~ 27T
37. (a)x = t 3 :::? t = x 113 ,so y=t2 = x213 • (b) x = t6 :::?·. t = x 116 , so y= t4 = x416 = x213 •
We get the entire curve y = x213 traversed in a left to
right direction.
Since x = t 6 2': 0, we only get the right half of the ' .
curve y = x213 •
(c) x =e-st = (e- t)3 [so e- t = xl/3],
y = e-2t = (e-t )2 = .(xl /3)2 = x2/3.
If t < 0, then x and y are both larger than 1. !f t > 0, then x and y
are between 0 and 1. Since x > 0 and y > 0, the curve never quite
reaches the origin.
39. The case ¥ < (} < 7T is illustrated. 0 has coordinates ( rO, r) as in Example 7,
and Q has coordinates (rB, r + r cos(1r- B)) = (rB, 7"(1- cos B))
[sincecos(1r - a) = cos7rc;osa + sin7T sin a· = -cos a ], so.P has
coordinates (rB- rsin(1r - B),r(1- cos·B)) = (r(B :- sin B),r(l - cos B))
[sincesin(1r - a) = sin1rcosa - cos1r.sin a = sin a ]. Again we have the
parametric equations x = r( (} - sin 8), y = r(l - cos B).
y
y
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X
SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS D 5
41. It is apparent that x = JOQJ andy= IQPI = JSTJ. From the diagram,
X = I OQI = a cos B and y = I STI = b sine. Thus, the parametric equations are
x = a cos B and y = b sin B. To eliminate B we rearrange: sin B = y / b =>
sin2 B = (y/bl and cosO= x/a => cos2 B = (x/a)2 • Adding the two
equations: sin2 B + cos2 B = 1 = x 2 /a2 + y2 /b2 . ,Thus, we have an ellipse.
y
X
43. C = (2a cot B, 2a), so the x-coordinate of Pis x = 2a cot B. Let B = (0, 2a).
Then L.OAB is a rightangle and L.OBA = B, so JOAI '= 2asinB and
A = ( (2a sin B) cos B, (2a sin B) sin B). Thus, they-coordinate of P
is y = 2asin2 B.
( -3, 0) and approximately ( - 2.1, 1.4).
-4
(b) A collision point occurs when x1 =.xz and Yl = Y2 for the same t. So solve the equations:
3sint= - 3+cost (1)
2 cost = 1 + sin t (2)
From (2), sin t = 2 cost- 1. Substituting into (1), we get 3(2 cost - 1) = - 3 +cost => 5 cost = 0 (*) =>
cost = 0 => t = t or 3 ;. We check that t = 3
2..,. satisfies (1) imd (2) butt = t does not. So the only collision point
occurs when t = 3 2", and this gives the point ( - 3, 0). [We could check our work by graphing Xt and x 2 together as
functions oft and, on another plot, y1 and y2 as functions oft. If we do so, we see that the only value oft for which both
pairs of graphs intersect is t ~ 3 ;.]
(c) The circle is centered at (3, 1) instead of ( - 3, 1). Th~re are sti ll 2 intersection points: (3, 0) and (2.1, 1.4), but there are
no collision points, since(*) in part (b) becomes 5 cost = 6 => cost = ~ > 1.
47. x = t 2 , y = t 3 - ct. We use a graphing device to produce the graphs for various values of c with -7r ::; t ::; 7r. Note that all
the members of the family are symmetric about the x -axis. For c < 0, the graph does not cross itself, but for c = 0 it has a
cusp at (0, 0) and for c > 0 the graph crosses itself at x = c, so the loop grows larger as c increases.
- I
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49. x = t + a cost, y = t + a sin t, a > 0. From the first figure, we see that
curves roughly follow the li ne y = x, and they start having loops when a
is between 1.4 and 1.6. The loops increase in s ize as a increases.
While not required, the following is· a solution to determine the exact values for which the curve has a loop,
that is, we seek the values of a for which there exist parameter values t and u such that t < u and
(t+acost,t+asint) = (u + acosu,u+asinu).
y In the diagram at the left, T denotes the point (t, t), U the point (u, u),
and P the point (t +a cost, t +a sin t) = (u + acosu, u + asinu).
Since PT = PU = a, the triangle PTU is isosceles. Therefore its base
angles, a = LPTU and {3 = LPUT are equal. Since a .= t - i and
'!I 4
X
(3 = 27r - 3.;' - u = 5 .;' - u, the relation a = /3. implies that
u + t = s; (1).
Since TU = distance{{t, t), (u, u)) = J2(u- t)2 = y'2 (u- t), we see that
cos a= ~TU = (u - t) / v'2, sou - t = v'2acosa, that is, PT a
p
X
u - t = v'2acos(t- i) (2). Nowcos(t- ~) = sin[f - (t- .;f)] = sinC4,.. - t),
so we can rewrite (2) as u - t = v'2 asine4,. - t) (2'). Subtracting (2') from (1) and T u - t u-t U
J2 J2 dividing by 2, we obtain t = 3
4,.. - ~a sine:- t), or 3 4,..- t = 72 sine4,.. - t) (3). 1-Ji(u -r)--!
Since a> 0 and t < u, it follows from (2') that sin(34" - t) > 0. Thus from (3) we see that t < 3 4,... [We have
' implicitly assumed that 0 < t < 1r by the way we drew our diagram, but we lost no generality by doing so since replacing t
by t + 27r merely increases x andy by 21r. The curve's basic shape repeats every time we change t by 21r.) Solving for a in
v'2(3'~~'-t) · v'2z (3), we get a = . ( 3! ) . Write z = ~,.. - t . Then a = -.- , where z > 0. Now sin z < z for z > 0, so a > v'2.
sm 4 - t sm z
[As z --+ o+, that is, as t --+ e4,..)- , a --+ y'2] .
51. Note that all the Lissajous figures are symmetric about the x -axis. The parameters a and b s imply stretch the graph in the
x- and y-directions respectively. For a = b = n = 1 the graph is simply a circle with radius 1. For n = 2 the graph crosses
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SECTION 10.2 CALCULUS WITH PARAMETRIC-CURVES 0 7
itself at the origin and there are loops above and b~low the x-axis. In general, the figures have n - 1 points of intersection,
all of which are on the y-axis, and a total of n closed loops.
1.1
-1.1
- 1.1
- 2.1
10.2 Calculus with Parametric Curves
1. x = t sin t, y = t 2 + t ::} dy . dx . dy dy I dt 2t + 1 -d = 2t + 1, -d = t cost+ sm t , and -d = d l d = . . t t x x t t cos t + sm t
z . 3 dy = - 3t2 dx = 4- 2t d dy = dyldt = -3tz -Wh -- 1 3. x = 1 + 4t - t , y = 2 - t ; t = 1. dt ' dt , an dx dx I dt 4 - 2t . en t '
(a, b) = (2, 3)
(a, b) = (3, 2)
( x, y) = ( 4, 1) and dy I dx = - ~, so an equa.tion of the tangent to the curve at the point corresponding to t = 1 is
y- 1 = -~(x- 4), oq/ = -~x + 7.
5. x = t cos t, y = t sin t; t = 1r. dy . dx . ' dy dy I dt t cos t + sin t -d = tcost+smt, -d = t(-smt)+cost,and-d = d ld = . . t t x x t -tsm t+cost
When t = 1r, (x, y) = ( -1r, 0) and dy I dx. = - 1r I( -1) = 1r, so an equation of the tangent to U1e curve at the point
•
7. (a) x = 1 + ln t, y = t2 + 2; (1, 3). dy dx 1 dy dyldt 2t 2 dt = 2t, dt = t' and dx = dxldt = 1l t = 2t · At(1•3),
x=l+ ln t = 1 ::} ln t = O =? t =1 an~ : = 2,so an equationofthetangentis y-3=2(x - l),
ory = 2x + 1.
(b)x = 1 + ln t =? ln t=x -1 =? t=e"'-l,soy=t2.+2=(e"'-1 )
2 + 2 = e2 " -
2 +2,andy'=e2x-z . 2.
At (1, 3), y' = e2(l)- z · 2 = 2, so an equation of the tangent is y - 3 = 2(x - 1), or y = 2x + 1.
9. x = 6sin t, y = t2+t; (0,0).
dy dy 1 dt 2t + 1 . - = d ld = --. The potnt (0, 0) corresponds tot = 0, so the dx x t 6cost
slope of the tangent at that point is· lJ. An equation of the tangent is therefore
y - 0 = i(x- 0), or y = ix.
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8 D CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
2 2 dy _ dy I dt _ 2t + 1 _ 1
.!_ 11. x = t + 1• Y = t + t '* dx - dx fdt - 2t - + 2t
!!:._ (dy) dt dx
dx/ dt
The curve is CU when ~~ > 0, that is, when t < 0.
dy dyfdt - te- t + e- t = e-t (1- t) = e- 2t( 1
_ t) 13. x = et, y = te-t ::::} dx = dxjdt = et et '*
d(~) . d
2 t(- 1) + (1 - t)( - 2e-
2 t) = e -
2 t(-1 - 2 + 2t) = e-3t(2t- 3). The curve is CU when
dx2 dx/dt et et
d 2 y 0 th . h t 3 dx2 > , at ts, w en > 2 .
15. x = 2sint, y = 3cost, 0 < t < 21r.
. i_OO(dy - - 3 2 dy _ dy/ dt _ -3sint _ -~ d2 y _ dt dx _ -2sec t _ -~ 3 dx- dxjdt - 2cost - 2 tant, so dx2 - dxjdt - 2cos t - 4 sec t.
The curve is CU when sec3 t < 0 ::::} sect < 0 ::::} cost < 0 ::::} i < t < 3; .
. d d 17. X = t3
- 3t, y = t 2 - 3. dy = 2t, SO dy = 0 ¢> t = 0 ¢>
t ' t
(x, y) = (0, - 3). dxd = 3t2 - 3 = 3(t + 1)(t - 1), so d:dx = 0 <=>
t . t
t = - 1 or 1 <=> (x, y) = (2, -2) or ( -2, -2). The curve has a horizontal
tangent at (0, - 3) and vertical tangents at (2, - 2) and ( -2, -2).
19. x = cos 9, y = cos 39. The whole curve is traced out for 0 ~ 9 ~ 1r.
~~ = - 3'sin 39, so ~~ = 0 <=> sin 39 = 0 <=> 38 = 0, 1r, 21r, or 37r <=>
'e =0, ~ . 2 3.,.,or1r <=> (x,y) = (1,1), (~ , -1), (-~ , 1) , or(-1,-1 ).
dx . dx d9 = - sLD 8, so dO = 0 <=> sin 8 = 0 <=> 8 = 0 or 1r <=>
dy d.x (x,y_) = (1, 1) or (-1,-1). Both dO and dO equal 0 when 8 = 0 and 1r.
2
(1,1) 9 = 0
.., fi d th I I 9 0 fi d li dy lim - 3 sin 38 11 lim _g cos 38 9 hi h . th I h 0 10 n e s ope w ten = , we n m dx = . n = ,.. = , w c 1s e same s ope w en = 1r. o~o o~o - sm" O-•O -cos"
Thus, the curve has horizontal tangents at ( ~ , -1) and (- t, 1), and there are no vertical tangents.
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SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES 0 9
21. From the graph, it appears that the rightmost point on the curve :J: = t - t 6 , y = et
is about (0.6, 2). To find the exact coordinates, we find the value oft for which the
graph has a vertical tangent, that is, 0 = dxl dt = 1 - 6t5 {::} t = 11 W.
Hence, the rightmost point is
23. We graph the curve x = t 4 - 2t3
- 2e, y = t 3 - tin ~he viewing rectangle [-2, 1.1] by [- 0.5, 0.5]. This rectangle
corresponds approximately tot E [ -1, 0.8] .
o.s
- 0.5
We estimate that the curve has horizontal tangents at about ( - 1, - 0.4) and ( - 0:11, 0.39) and vertica·L tangents at
dy dyl dt· 3t2- 1 . about (0, 0) and ( -0.19, 0.37). We calculate -d = dxl d =
4 3 6 2 4 . The honzontal tangents occur when
X t t - t - t ·
dy I dt = 3t2 - 1 = 0 1=} t = ± ~, so both horizontal tangents are shown in our graph. The vertical tange~ts occur when
dxldt = 2t(2t2 - 3t- 2) = 0 {::} 2t(2t + 1)(t - . 2) = 0 {::} t = 0, -~or 2. It seems that we have missed one vertical
tangent, and indeed if we plot the curve on the t-interval [-1.2, 2.2] we see that there is another vertical tangent at ( -8, 6).
25. x = cos t , y = sin tcos t . dxl dt = - sint, dyl dt = - sin2 t + cos2 t = cos 2t.
(x,y )= (O, O) {::} cost = .O. {::} tisanoddmultipleoflf.Whent=~,
dxl dt = - 1 and dyl dt = - 1, so dyl dx = 1. When t = 3; , dxl dt = 1 and
dy I dt = - 1. So dy I dx = - 1. Thus, y = x andy = - x are both tangent to the
curve at (0, 0).
27. x = rB - d sinB, y = r - dcosB.
dx dy , . dy d sinB (a) dB = r- dcosll, dB = dsme, so dx = r _ dcosB '
(b) IfO < d < r, then jdcosllj:::; d <. r , so r- dcosfi ;::: r- d > 0. This shows that dxldll never vanishes,
·so the trochoid can have no vertical tangent if d < r .
29. X = 2t3' y = 1 + 4t - t2 ---'- dy - dy I dt = 4 - 2t N 1 . dy = 1 ~ 4 - 2t = 1. ......,._ _,.. ·dx - dxl dt 6t 2 • ow so ve dx .,_,. 6t2 .....,.
6t 2 + 2t - 4 = 0 {::} 2(3t - 2)(t + 1) = 0 {::} t = ~ or t := - 1. 1ft = ~ . the point is(~~ ' 2 :) , and ift = -1,
the point is ( -2, -4).
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31. By symmetry of the ellipse about the x- andy-axes,
A= 4J 0 a ydx = 4 j~12 bsinB (-a sin B) dB= 4ab I0"
12 sin2 B dB= 4ab Io"12 ~(1- cos 28) d8
, /2 I
= 2ab[8- ~ sin2B]~ = 2ab(~) = 1rab
33. The curve x = 1 + et, y = t- t 2 = t(1 - t) intersects the x-axis when y = 0,
that is, when t = 0 and t = 1. The corresponding values of x are 2 and 1 + e.
The shaded area is given by
1::1+e (YT- YB) dx = 1::1
[y(t)- OJ x'(t) dt = I;(t - t2)et dt
y
= J; tet dt- I 0 1
t 2et dt = f0 1 tet dt- [fet] ~ + 2 j; tet dt [Formula 97 or parts]
1 [ ' 1 = 3 Io tet dt- (e - 0) = 3 (t- 1)et] 0
- e [Formula 96 or parts]
= 3[0 - (-1)] - e = 3 ~ e
35. x = r8- dsin8, y = r- dcosB.
A= I;~r y dx = J;rr (r- dcosB)(r - dcos 8) dB= I;rr (r2 - · 2drcos B + d~ cos2 B) dB
= [r2B- 2dr sine+ ~d2 (8 + ~ .sin 28) ]~7r = 21rr2 + 1rd2
37. x=t+e~t, y=t-e-t, O~t~· 2. dxjdt= 1- e- tanddyjdt=1+e-t,so
(dx/dt) 2 + (dyjdt? = (1- e-t)2 + (1 + e- t? = 1- 2e-t + e-2 t + 1 + 2e-t + e- 2
t = 2 + 2e-2t .
Thus, L = I: './(dx/dt)2 + (dyjdt) 2 dt =I: v2 + 2e- 2 t dt ~ 3.1416.
39 . . x = t- 2sint, y = 1 - 2 cos t, 0 ? t ~ 47r. &cjdt = 1 -2 cost and dyjdt = 2sint, so
(dx/dt) 2 + (dyjdt) 2 = (1- 2 cos t)2 + (2 sin t)2 = 1- 4cos t + 4 cos2 t + 4sin2 t:::;; 5- 4cost.
Thus, L = I: './(dx/dt)2 + (dyjdt) 2 dt = Io4 " v5- 4cost dt ~ 26.7298.
41. x = 1 + 3e, y = 4 + 2t~, o ~ t ~ 1. ~x/dt ~ 6t and dyjdt = 6t2, so (dxjdt? + (dy /dt) 2 = 36t2 + 36t4
1 . 1 ~
Thus, L = Ia './36t2 + 36t4 dt = Ia 6t .Jl+i2·dt = 61 JU Gdu) [u = 1 + t 2, du = 2tdt]
= 3[~u312J: = 2(23 ;
2 - 1) = 2(2J2 - 1)
43. x ~ t sin t, y =. t cost, . 0 ~ t ~ 1. ~~ = t cost + sin t and ~~ = - t sin t + cos t, so
( ~~ y + ( ~; y = e cos2 t + 2t sin t cost + sin2 t + t2 sin2 t ~ 2t sin t cost + cos2 t
= t 2 (cos2 t + sin2 t) + sin2 t + cos2 t = t2 + l.
Thus, L .·= f0 1 vt2 + 1 dt ~ [~tvt~ + 1 + ~ ln(t + vt2 + 1 ) ] ~ = 4J2 + ~ ln(1 + J2).
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45.
47.
( ~~ ) 2 + ( !fJd = [e'(cos t - sin t)f + [et(sin t +cos t)f
= (e') 2 (cos2 t - 2 cos t sin t + sin2 t) '
+ (et)2 (sin2 t+2sint cost+cos2 t
= e2t(2 cos2 t + 2 sin2 t ) = 2e2t
Thus, L = J~" .J2e2t dt = f0" -12 et dt =· -12 [ e' ]~ = -/2 (e,.- 1).
1.4 The figure shows the curve x = sin t + sin 1. 5t, y = cost for 0 .:::; t 5 47r.
. dx/dt = cost+ 1.5 cos 1.5t and dy/ dt = -sin t, so
-2.1 f->o.o--------,.£-f->'<-----~-12.1 (dx/dt)2 + (dy / dt? = cos2 t + 3cost cosl.5t + 2.25cos2 1.5t + sin2 t .
Thus, L = J0 4
" v'1 + 3 cost cos 1.5t + 2 .. 25 cos2 l .5t dt ~ 16.7102.
49. x = t - et, y = t + et, -6 5 t 5 6.
(~~)2 + (!!Jd = (1- et? + (1 + et? = (1 - 2et + e2t) + (1.+ 2et + e2t) = 2 + 2e2t, soL= f~6 v'2 + 2e2t dt.
Set f(t ) = v'2 + 2e2 t. Then by Simpson's Rule with n = .6 and D.t = ·e-~-e) = 2, we get
L ~ j[J( - 6) + 4/( -4) + 2/( -2) + 4/(0) + 2/(2) + 4/(4) + /(6)] ~ 612.3053.
51 . x = sin2 t , y '= cos2 t, 0 5 t 5 37r.
(dx / dt? + (dy/dt? = (2sin t cos t) 2 + ( - 2 cos tsin t? = 8 sin2 t cos2 t = 2 sin2 2t ==>
Distance = J0 3
" V'i lsin 2tl dt = 6 -/2 J; 12 ~in 2t dt [by symmetry) = -3 -/2 [cos 2t]: 12
= -3 -/2 (- ~ - 1) ~ 6 -/2.
The fu ll curve is traversed as t goes from 0 to ¥, because the curve is. the segment of x + y = 1 that lies in the fi rst quadrant
(since x , y 2:: 0), and this segment is completely traversed as t goes from 0 to¥ · Thus, ·£ ~ f0" 12 sin 2t dt = -/2, as above.
53. x = asin8, y = bcos O, ·a$ 8 5 21r.
(~7)2 + (!!Jd· = (a cosO?+ ( - bsin8? .,;, a 2 cos2 8 + b2 sin2 8 = a 2 (1- sin2 B)+ b2 sin2 8 . 2
= a 2 - (a2
- b2 ) sin2 8 = a 2
- c2 sin2 8 = a 2 (1-~ sin2 8) = a 2 (1 - e2 sin2 8)
SoL = 4 j 0 "12 ..jaz (1- e2 sin2 0) dO [hy '?·mmctty) = 4a J;'12 Vl - e2 sin2 8 dO.
55. (a) x = 11 cos t - 4 cos(11t/ 2), y = 11 sin t- 4sin(llt/2).
Notice that 0 5 t 5 21r does not give the complete curve because
x(O) f:. x(21r). In fact, we must take t E [0, 47r] in order to obtain the
complete curve, since the first term in each of the parametric equations has
period 27r and the second has period 1ii2 = ~~, and the least common
integer multiple of these two numbers is 47r.
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(b) We use the CAS to find the derivatives dx/dt and dy/dt, and then use Theorem 6 to find the arc lengtll. Recent versions
of Maple express the integratJ;" J(dxjdt) 2 + (dyjdt) 2 dt as 88E(2 V2 i), where E (x) is the elliptic i~tegral
{1 v/1- x2t2 Jo v'f=t2 dt and i is the imaginary number yCI. I
Some earlier versions of Maple (a5 well as Mathematica) cannot do the integral exactly, so we use the command
evalf (Int (sqrt (diff (x, t) - 2+diff (y, t) -2), t=O .. 4*Pi)); to estimate the length, and find that the arc
length is approximately 294.03. Derive's Para_ar c_length function in the utility fi le Int_apps simplifies the
integral to 11 }~4,. J -4 cost cos(l~t ) - 4 sin t sin(l~t) + 5 dt.
57. x=tsint, y=tcost, O~t~Tr/2. dxjdt=tcost +sintanddyjdt= -tsint +cost,so
(dx/dt? + (dyjdt)2 = t 2 cos2 t + 2tsin t cost+ sin2 t + e sin2 t- 2tsin t cos t+ cos2 t
= t 2 (cos2 t + sin2 t) +.sin2 t + cos2 t = t2 + 1
59. x = 1 + tet, y = (t2 + l) et, 0 ~ t ~ 1.
(~~ )2 + (!fltf = (te1 + et)2 + [(e + I )et + et(2tW = [et(t + 1}f + [et (e + 2t + 1W
= e2t(t + 1? + e2t(t + 1)4 = e21 (t + 1)2[1 + (t + 1)2 ), so
( 3 (u 4) = 2Tr J 4 T Ju ( fg du) [
u = Dt 2 + 4, t 2 = (u- 4)/9, J
du = 18t dt, so t dt = · f.J tlu
- 2!:. [~u5/2 - §.u3/2] 13 - 2!:. • .L [3u5/2 - 20u3/2] 13 - 81 5 3 4 - 81 15 4
= 1;~5 [(3. 132 v'13- 20. 13 VI3) - (3 ·.32 - 20. 8)] = 1;~5 (247 VI3 + 64)
63. x = acos3 8, y =a sin3 8, 0 ~ B ~ I· (~~)2 + (;¥tt)2 = ( - 3acos2 8 sinG?+ (3a sin 2 8 cos 0)2 = 9a2 sin2
(I cos2 0.
S = .{0" 12 2Tr · a sin3
() · 3asin 0 cos() d() = 61ra2 J0rr/ 2 sin~ 0 cos 0 dO = ~1ra2 [sin5 0] ~12 = ~1ra2
65. X= 3t2, y = 2t3, 0 ~ t ~ 5 . ::} C~l + (~/} = (6t? + (6t2)2 = 36t2(1 + t2) ::}
s = .r; 21rx J(dx/dt)2 + (dyJdt)2 dt = J; 27r(3e)6t vt + t 2 d.t = l81r J; t2J1 + t2 2tdt
[ u = 1 +t2,] = 18Trj·2G(u3/2- u1/2) du = 18Tr ['du5/2- ~u3/2] 26
du = 2t dt 1 5 3 1
= 187r [ ( ~ . 676 J26 - ~ . 26 J26) - ( ~ - ~)] = ¥ 7r ( 949 J26 + 1)
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SECTION 10.3 POLAR COORDINATES 0 13
67. If f' is continuous and j' (t) i= 0 for a$ t $ b,. then either!' (t) > 0 for all tin (a, b) or j' (t) < 0 for all t in [a, bj. Thus, f . . '
is monotonic (in fact, strictly increasing or stri<:tly decreasing) on (a, b]. It follows that f has an inver;se. Set F =g o f- 1 ,
that is, define F by F(x) = g(r1 (x)). Th~n x = f(t) ~ r 1 (x) = t , soy= g(t) = g(f-1 (x)) = F(x).
- 1 ( dy) d<P _ d -1 ( dy ) _ 1 [ d ( dy ) ] dy dy 1 dt i1 69· (a)¢> = tan dx ~ dt - -dt tan dx - 1 + (dyjdx)2 dt dx · But dx = dx/dt = ± ~
d ( dy ) d ( if ) fix - xif d¢> 1 ( fl± - xfl ) xy - xif ' . . · dt di = dt ;; = xz ~ dt = 1 + (iJ/x) 2 x2 = xz +iF. Usmg the Cham Rule, and tne
"a·cttllat s-- rot (d:r;)2+(!!11..)2dt d .• - /(d:r;)2+(!!11..)2 ('2+ •2)1/2 b h ! i J
0 dt dt ~ 'Jt - y dt dt = x y , we ave t at
d¢> _ d¢>/dt _ (xii- xfl ) 1 _ xii - xiJ . _- I d¢> 1-1 xy- xy 1- l±ii - xiJI ds - ds/dt - :i;2 + y2 (±2 + y2)1/2 - (~2 + 1;2)3/2 · So"'- - ds - (±2 + y2)3/2 - (±2 + y2)3/2 ·
d !( ) . 1 .. 0 d . dy .. d2y (b) x = x_an y = x ~ x = 'x = an y = dx' y = dx2'
- II· (d~y/dx2 ) - 0 . (dyjcb:) i - ld2 y/ dx2 1
So"' - (1 + (dyjdx)2 )312 - (1 + (dyjdx)2] 312 •
71. X = 8 - sin8 ~ :i; = 1----' cosB -~ X = sin8, and y = 1 - cos8 ~ iJ =sine ~ y = cos8. Therefore,
icosB - cos2 e- sin2 el icosB - (cos2 8 + sin2 8)1 JcosB - II . "' = = . 2 = . The top of the arch 1s
[(1 - cos B)2 + sin2 B)S/2 (1 - 2 cos e + cos2 e + sm 8)312 (2 - 2 cos 8)3/2
characterized by a horizontal tangent, and from Example 2(b) in Section 10.2, the tangent is horizontal when B = (2n - l )1r,
d b · B · ti . fi Jcos 1r -11 1-1- 11 1 sotaken=1 an su stttute =1rmto 1e ex~resston or ~~;: ~t= (2 _ 2 cos1r)3/ 2 =
12 _
2 (-
1 ))3/ 2 =4·
73. The coordinates of T are .(r cosB, r sin B) . Since TP was unwound from
arc T A, T P has length rB. Also LPTQ = LPT R - LQT R = t 1r - 8,
soP has coordinates x = r cos B + r8 cos(~1r ~B) = r(cos B + B sin 8),
y = rsin B - rBsin(~1r - 8) = r(sin8- 8cos9).
10.3 Polar Coordinates
X
1. (a) (2, i) By adding 21r to i, we obtain the point ( 2, 7;). The direction
opposite i is 4 3.,., so (- 2, ~) is a point that satisfies the r < 0
· requirement.
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(b) (1, - 3 ;)
;)
r > 0: (- (- 1) , 1J +1r) = (1, 3 ;)
r < 0: (-1, ~ + 21r) = (- 1, 5 2,.)
(-t,f)
y = 1 sin 1r = 1(0) = 0 give us
the Cartesian coordin~tes ( -1, 0).
x = 2cos(- 2 3"' ) = 2(-~) = - 1 and
y = 2sin( - 2 ;) = 2( -~) ~ ~VS
give us ( -1, -V3).
y = - 2sin 34" = -~( 4) = - J2
gives us ( J2, -J2).
5. (a) x = 2 andy = - 2 => r = J22 + (-2)2 = 2 J2 and 8 = tan- 1 ( :n := -~ . Since (2, -2) is in the fourth
quadrant, the polar coordinates are (i) (2 J2, 7 ;) and (ii) (-2 J2, 3;).
(b) x = - 1 andy = V3 => r = J< -1)2 + ( v'3)2 = 2 and 8 = tan~1 ( :1) = 2
:;. Since_( -1, VS) is in the second
quadrant, the polar coordinates are (i) (2, ~)and (ii) ( - 2, 6;).
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7. r ~ 1. The curve r = 1 represents a circle with center
0 and radius 1. So r ~ 1 represents the region on or
outside the circle. Note that 8 can take on any value.
11 2 < < 3 s.,.. < B < r1r · r •a--a
r=2
9. r ~ 0, 7r/4 S, B S, 37r/ 4 . .
8 = IC represents a line through 0.
0= 37T 4
0='!!.. 4
13. ·Converting the p~lar coordinates (2, 1r / 3) and ( 4, 21r / 3) to Cartesian coordinates gives us (2 cos f, 2 sin f) = (1, J3) and
( 4 cos 2; , 4 sin 2;) = (- 2, 2 J3). Now use the distance !ormula.
15. r 2 = 5 <* x2 + y2 = 5, a circle of radius .j5 centered at the origin.
17.r =2 cos8 ~ r 2 = 2rcos8 <* x2 + y2 =2x <* x 2 -2x+ 1 +y2 = 1 <* (x-1)2 +y2 = 1,acircleof
radius 1 centered at (1, 0). The first two equations are actually equivalent since r 2 = 2r cos() ~ r(r - 2 cos 8) = 0 ~
r = 0 or r = 2 cos 8. But r = 2 cos 8 gives the point r = 0 (the pole) when 8 = 0. Thus, the equation r = 2 cos 8 is
equivalent to the compound condition ( r = 0 or r = 2 cos 8).
the origin with foci on the x -axis.
21. y = 2 <* r s in B = 2 ¢? 2
r=-- sin8
¢? r = '2csc8
23.y=1 +3x ¢? rsin8 = 1 +3rcos8 ¢? rsin8-3rcosB = 1 ¢? r(sin8-3cos8)=1 <*
1 r = -sin::--8:::--- -3-=-co_s_e-=
25. x 2 +y2 =2cx ¢? r 2 =2crcos8 ¢? r 2 -2crcos8=0 <* r(r-2ccos8) = 0 ¢? r = Oorr=2ccos8.
r = 0 is included in r = 2c cos 8 when 8 = ~ + n1r, so the curve is represented by the single equation r ~ 2c cors ().
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27. (a) The description leads jmmediately to the polar equation 9 = ~.and the Cartesian equation y,; tan (~) x = 7:i xis
slightly more difficult to derive.
(b) The. easier description here is the Cartesian equation x = 3.
29. r = - 2 sin9
2'11' 0
(J
4
./ o=f ,..:/ 2 (4.f)
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39. r = 1--2sin8
45. r = 1 + 2cos28
IJ
0=!!.
__________ ..• ----- 6
(3, ..-j6)
47. For(}=:= 0, 7T, and 27T, r has its minimum ;value of about 0.5. For 0 =~and 3 2", r attains its maximum va lue of2.
We see that the graph has a similar shape for 0 ~ (} ~ 7T and 7T ~ 8 ~ 27T.
. r 2
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49. x = r cos 8 = ( 4 + 2 sec 8) cos 8 = 4 cos 8 + 2. Now, r -+ oo =>
(4 + 2sec8)-+ 00 => 8-+ (~) - or 8 -+ e2"' )+ [since we need only
consider 0 $ 8 < 271' ], so lim x = lim ( 4 cos 8 + 2) = 2. Also, . r -oo 8 -+-rr/2-
T -+ -oo => ( 4 + 2 sec B) -+ -oo => 8 -+ ( ~) + or 8 -+ ( 3;)-, so
lim x = lim ( 4 cos 8 + 2) = 2. Therefore, lim x = 2 => x = 2 is a vertical asymptote. r--oo 8--+rr/2+ T"--+±oo
(6,0)
51 . To show that x = 1 is an asymptote we must prove lim x = 1. r - ±oo x= l
x = (r) cosO= (sin 8 tan 8) cos 8 = sin2 ().Now; r -+ oo => sinO tan8-+ oo =>
8 -+ (f) - ,so lim x= lim sin2 8 = l.Aiso,r-+-oo => sinBtanB-->-oo => r --+oo 0-+'Tr/2- .
e-+ (~) +,so lim X = lim sin2 {;I,;, 1. Therefore, lim X= 1 . => X= 1 is . r ...... - oo 0--+1fj 2+ ,·-±oo
a vertical asymptote. Also notice that X = sin2 e ~ 0 for a ll e, and X = sin2 {;I $ 1 for all e. And Xi= 1, since the curve is not
defined at odd multiples of ~. Therefore, the curve lies entirely within the vertical strip 0 $ x < 1.
53. (a) We see that the curve r = 1 + csin 8 crosses itself at the origin, where r = 0 (in fact the inner loop corresponds to
negative r-values,) so we solve the equation of the lima~on for r = 0 <=> c sin 8 = - 1 <=> sin 8 = - 1/ c. Now if
lei < 1, then this equation has no solution and hence there is no inner loop. But if c < - 1, then on the interval (0, 27T)
the equation has the two solutions 8 = sin-1 ( -1/c) and 8 = 1r- sin- 1
( - 1/c), and if c > 1, the solutions are
(} = 1r + sin - 1 (1/c) and ()= 21r- sin- 1 (1(c). Tn each case, r < 0 for (} between the two solutions, indicating a loop.
(b) For 0 < c < 1, the dimple (if it exists) is characterized by the fact that y has a local maximum at 8 = 3 ; . So we
determ.ine for what c-valu~s d 2
~ is negative at(} = 32~, since by the Second Derivative Test tllis indicates a maximum: ~ . .
dy d2 y y = rsin8 =sinO + c sin2
(} => dB = cos8 + 2csinfJ cos8 = cosfJ + csin 28 =>. - ., = - sin8 + 2ccos 28. dB~
At 8 = 3 2,.., this is equal to - ( - 1) + 2c( -1) = 1 - 2c, which is negative only for c > %. A similar argument shows that
for - 1 < c < 0, y only has a local minimum at e = ~ (indicating a dimple) for c < -~ . .
55. r = 2 sin (} => x = r cos(} = 2 sin fJ cos 8 = sin 28, y = 1· sin 8 = 2 sin2 B =>
dy = dyjd(J = 2 · 2 sin fJ cosO= sin 28 = tan 2B dx dxjdB cos2fJ·2 cos28
7T dy ( 7r) 7T rr; · When 8 = 6, d.'t =tan 2 · 6 =tan 3 = v 3. [Another method: Use Equation 3.]
57. r= l / 8 => x=rcos8 = (cosfJ) j B, y=rsin8=(sinfJ) / 8 =>
dy dyjd8 sin.O( - 1/ 82 ) + (1/ 8) cos 8 82 - sin (}+(} cosO dx = dxj dO = cos8(- 1/ fJ2 ) - (1/ 8) sinB · (}2 = - cosO - Osin B
.dy - 0 + 7r(- 1) - 'Tr
When 8 = 1r, dx = -(-1) -1r(O) = l = -1r.
'
SECTION 10.~ POLAR COORDINATES 0 19
59. r = cos 2() => x = r cos() = cos 2() cos(}, y = r sin() = cos 2(} !lin(} =>
1f dy When() = 4, dx
dy dy / df) cos 2(} cos(}+ sin(} ( -2 sin 2()) dx = dx/ d() = cos 2() (- sin(})+ cos() ( -2 sin 2())
-,0,!-( J2---:2 /:---27) +_('-:-v-2-:::2 /,.--!2 )~( -_2;,_) - _- J2_2 - 1 O(-J2/ 2) + (J2/ 2}(-2) - -J2 - .
61 . r = 3cos0 => x = rcosO = 3cos0 cosO,_ y = rsinO = 3cos0 sinO => ,
~ = -3sin2 9 +3cos2 ,() = 3cos20 = 0 => 2() = ~ or 3; <=> 0 = %or 3 ;.
So the tangent is horizontal at ( ~, %) and ( - -32, 3 ;) [same as ( -32 , -:a:)] .
~~ = - 6 sin() cos~= -3 sin 20 = 0 => 29 = 0 or 1r <=> 0 = 0 or~· So the tangent is vertical at (3, 0) and (o, ~).
63. r = 1 + cos 0 · => x = r cos 9 = cos() (1 +cos 9), y = r sin 0 = sin 0 (1 + cos 9) =>
·.~ = (1 +cosO) cosO- sin2 0 = 2cos2 0 + cos9 - 1 = (2cos9 - 1) (cos(J + 1) = 0 => cos()=~ or - 1 =>
; ) .
~~ = - ( 1 + cos 8) sin (} - cos ()sin() = - sin (} ( 1 + 2 cos 9) = 0 => sin 8 = 0 or cos 9 = - ~ =>
() = 0, 1r, 2 ;, or 4
; => vertical tangent ~t (2, 0), (!, 2;), and (!, -t;).
Note that the ~gent is horizontal, not vertical when 8 = 7r, since e : j :: = 0.
65. r = a sin() + b cos 9 => r 2 = ar sin() + br cos 8 '=> x2 + y2 = ay + bx =>
x 2
- bx + (~b? + y2 - ay + ( ~a)2 = (~b)2 + (ta? => (x - ~b)2 + (y - ~a? = Ha2 + b2
) , and this is a circle
with center (!b, ~a) and radius. ~.Ja2 + b2 •
67. r = 1 + 2 sin(8 / 2): The parameter intei-val is [0, 47r]. 69. r = e•inB - 2cos(40).
The parameter interval is [0, 21l'j.
71. r = 1 + cos999 8. The parameter interval is [0, 2tr).
1.1 /
- 1.1
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73. It appears that the graph of r = 1 + sin(~ - "if) is the same shape as
the graph of r = 1 + sine, but ro,tated counterclockwise about the
origin by "if. Similarly, the graph of r = 1 + sin( e ....,. ~) is rotated by·
i . In general, the graph of r = f ( e - a) is the same shape as that of
r = f (e), but rotated counterclockwise through a about the origin.
That is, for any point (ro, eo) on the curve r = j(e), the point
(ro, Bo +a) is on the curve r = j(e- a), since ro = f(Oa) = f((eo +a)- a).
-0.9
r = l +s~(o - ~_) r = 1 + sin(ll - ~)
75. Consider curves with polar equation r = 1 + c cos e, where c is a real number. If c = 0, we get a circle of radius 1 centered at
the pole. For 0 ~ c s 0.5, the curve gets slightly larger, moves right, and flattens out a bit on the left side. For 0.5 < c < 1,
the left side has a dimple shape. For c = 1, ,the dimple becomes a cusp. For c > 1, there is an internal loop. For c 2: 0, the
rightmost point on the cu;ve is (1 + c, 0). For c < 0, the curves are reflections through the vertical axis of the curves
with c > 0.
tan ¢ - tane 77. tan'lj! = tan(¢- e)= r/> e
1 + tan tan
dy dyjde 1 + dx ta11e 1 + dxj de tane
_ ~ - ~ tan e ~ (~ ~in e + r cos e) - tan e ( ~ cos B - r sine)
- : + :~ tane - (:;.cos e - r sine) + tan B ( :; sine + r cos e) _ rcos2 B +rsin2 e r - dr 2 iJ dr . 2 e = dr I dO
de cos . + dB sm
c = 2
case dr dr sin2 e de case+ de. case
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SECTION 10.4 A~~S AND LENGTHS IN POLAR COORDINATES 0 21
3. r 2 = 9sin28, r ~ 0, ·a~ 8::; 7rj2.
1 ~/2 1~/2 / 2
A - lr2 d8 - 1(9sin2B)d8 -l!. [- lcos2B)~ --l!.(- 1 - 1)=l!. - 2 - 2 -2 2 0 - 4 2 0 . 0
7. 1' = 4 + 3sin8, -%::; () ::; %·
= ~ (16 + 9sin2 8) dO -rr/2
[by Theorem 4.5.6(b) [ET 5.5.7(b)]]
r /2 = ~·2}0 [16 + 9 ·~(1 -cos28))d~ [by Theorem 4.5.6(a) [ET 5.5.7(a)]] .
rr / 2 '
= ( 4~" - O)- (0 _ 0) = 4~"
9. The area is bounded by r = 2 sin 8 for 8 = 0 to 8 = 7r. (2, 'TT/ 2)
A= 1"' ~r2 d8 := ~ 1 -rr (2 sinfJ)
2 dB= ~ 1"' 4sin2 Bd8
=21" ~(1 - cos2fJ)dB ~ [e - ~sin2eJ: =1r
Also, note that th is is a circle with radius 1, so its area is 1r(1)2 = 1r.
11. A = 12
rr(9+12.cos8+4cos2 8)d8
= ~ 12
.,.. [9 + 12cos8 + 4 · t(1 .+ cos 28)] dB _(_l ,_'TT--:) H-=-- - --'-:..._:..1----+
{2-rr = ~ lo (11 + 12 cos 8 + 2 cos 28) d8 = Hne + 12 sin O+ sin 28) ~71'
= H227r) = 111r
13. A = r 71' ~r2 d8= r rr t{2 + sin4fJ)2 dfJ = ~ r"(4+4si~4fJ+sin2 4fJ)d8 Jo lo · Jo · 3
= ~ fo2
1 2"
= ~[(97r- 1) - (- 1)) = ~7('
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1 ( 3 9 1 . 9] 21r 1 ( ) 3 = 2 2 + 2o sm 10 o = 2 3'1T = 27T
-1.4
17. The curve passes through the pole when r = 0 ==?- 4 cos 30 = 0 ==?- cos 39 = 0 ==?- 39 = ~ + 1rn =?-
9 = % +·in. The part of the shaded loop above the polar axis is traced out for
9 = 0 to 9 = 7T /6,' so we'll use -7T / 6 and 7T /6 as our limits of integration. r = 4 cos 30
19. r = 0 ==?- sin49 = 0 ==?- 48 = 1rn ==?- 9 = ~n. r=sin 40
r'4 . r'4 r '4 A = Jo Hsin48)
2 dfJ = t Jo sin
2 48d8 = t Jo t(1-cos89)d8
0 . . 0 0
1 (9 1 · 89] "' I 4 1 (,.) . 1 = 4 - B SlD 0 = 4 4 = Tii7T
21. This is a lima~on, with inner loop traced r = 1 + 2 sin 8 (rect.)
3 out between 8 = 7 ; and 1 ~" [found by
solving r = 0].
17T ------· 0=6 .
[
31r/2 . [37f/2 [ 37r/2 A = 2 !(1 + 2sin8? d8 = (1 + 4sin8 + 4sin2 8) d8 = [1 +4sin8 + 4· t{1- cos29)] d9
71r /6 77r /6 71r /6 . .
= [9 - 4cos0 + 20- ~n28J~:~~ = (9 ;)- C; + 2 v'3 - 4) = 7T - ¥
23. 2 cos B = 1 => cos 8 = ~ => 8 = :g. or 5 ; .
A = 2 f0,. 13 ~[(2 cos8)2
- 12 ) dB= J0"
13 (4cos2 8 - l) d8
= J; ; a { 4 [ ~ (1 + cos 28)] - 1} d8 = J; /S ( 1 + 2 cos 28) d8
= [8 + sin 28)~13 = i + 4 r = 2cos0
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SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES 0 23
25. To find the area inside the l~miniscate r2 = 8 cos 29 and outside the circle r = 2,
we first note that the two curves intersect when r2 = 8 cos 29 and ~ = 2, . I
that is, when cos2B = ~-For -1r < B ::; 1r, cos2B = ~ {::} 2B = ±7r/ 3
or ±511' /3 {::} B = ±1r / 6 or ±57r / 6. The figure shows that the desired area is
4 times the area between the curves from 0 to 1r /6. Thus,
A = 4 J0" /. 6 [~(8 cos2B)- ~(2)2] dB = 8 J0"
16 (2 cos 2B - 1) dB
[ ] rr~ .
A= 2 f0" 13 ~ [(3 cos 8?- (1 +cos 0)2
) dB
= J0"' 13(8 cos2 8-'- 2 cos B- 1) dB = f0"
13[4(1 +cos 2B) - 2 cos B - 1) dB
= J0.,. 13 (3 + 4cos28- 2cosB) dB= [3B + 2sin2B- 2sinf!] ~13
= 1T +v'3-.J3=1T
29. v'3cosB = sinB => J3 = sinBB => tanB = J3 => B = i· . cos
A - f "' 13 1 (sin B)2 dB + 1'-"' 12 1 (J3 cos B)2 d() -Jo 2 hr/ 3 . 2
- f"' / 3 1. 1(1 - cos28) dB+ f "/2 1 · 3 · 1(1 + cos2B) dB - Jo 2 2 rr/3 2 2
- 1 [B - 1 · 28]"13 +!! [B +.! · 28]1f12 - 4 2 sm o 4 2 sm 7f/3
= t [ ( i - 4) - o] + ~ [ (~ + o)- ( f + 4)] _ .l!.. _ _il + .!!: _ 1fl _ 5rr _ fl - 12 16 . 8 16 - 24 4
31. sin 2B = cos 2B => sin 2
B = 1 => tan 28 = 1 => 28 = ~ => cos2B
B=. f =>
A= 8 · 2 f0"/8'~ sin 2 2B dB= 8 f0"' 18 H1- cos 48) dB
= 4[9- t sin48]~18 = 4(f - i ·1) = ~ - 1
33. sin2B = cos2B => tan2B = 1 => 2B = ~ => (} = f
A= 4 f0"' 18 ~sin 2B dB [since r 2 = sin 2B)
= f0"' 18 2 sin2B dB= [-;- cos 28] ; 18
= -~ v'2- ( - 1) = 1- ~ Y2.
r = J3cos8
.... ···B=i
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24 • 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
35. The darker shaded region (from 0 = 0 to 0 = 27T / 3) represents ~ of the desired area plus ~ of the area of the inner loop.
From this area, we' ll subtract~ of the area of the inner loop (the lighter shaded region from 0 = 27r /3 to 0 = 7r), and then
double that difference to obtain the desired ·area.
A = 2 [J;.,. 13 H~ +~os0) 2
dO- g.,.13 H~ +cos 0) 2
dB]
= £0 2
.,. 13 (~+cosO+ cos2 0) dB- f2: 13 (~ + cos 0"+ cos2 0) dB
. . = J0211'/S [~+cos 0 + ~(1 + cos28)) dO
- I2:/3 a + cosO+ ~(1 + cos28)) d(J
= [~ + sin 8 + ~ + sin 20] 2
" 13
_ [~ + sin 8 + ~ + sin 20 1.,. 4 2 4 0 4 2 4 211'/ 3
- (1!: + .il + 1!: - .:.::2) - (!!: + !!:) + ( 1!: + .il + 1!: - .:.::2) -6 2 3 8 4 2 6 2 3 8
= ~ + ~ J3 = H 7T+3v'3)
37. The pole is a point of intersection.
1 + sin 8 = 3 sin 0 => 1 = 2 sin (J => sin 0 = ~ =>
(J =~ or 5 ;.
The other two ~oints of intersec~ion ~e ( ~, i) and ( f, 56").
39. 2 sin 20 = 1 =? sin 20 = ~ =? 2o = 1!: &.,. 13.,. or 11" 6' 6 ' 6 ' 6 °
By symmetry, the eight points of intersection are given by
(1 8) h (} _ 11' 5,. 13rr d 1771" d , , w ere - 12, 12, 12, an 12, an
[;.
[There are many ways to describe these points.]
41. The pole is a point of intersection. sin (J = sin 20 = 2 sin(} cos 0 <=>
sinO (1 - 2 cos 9) = 0 <=> sinO= 0 or coslJ = t =>
8 = 0, 7T, t. or - t => the other intersectio~ points" are ( ~. t )
and ( ~, 2; ) [by symmen:'].
r = 2sin20
r =sin20
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43.
SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES 0 25
. 3.4 3 y = 2x
-1.4 1.4
- 0.3 -3
From th~ first graph, we see that the pole .is one point of intersection. By zooming in or usin~ the cursor, we find the 0-values
of the intersection points to be a:~ 0.88786 ~ 0.89 and :r - a ~ 2.25. (The first of these values may be more easily
estimated by plotting y = 1 +sin x andy = 2x in rectangular coordinates; see the second graph.) By symmetry, the total
area contained is twice the area contained in the first quadrant; that is,
A= 21o. ~(20)2 dO+ 2[ 1r/Z ~(1 +sin 0.)2 d8 = 1"' 402.d8 + l1r/~ [1 + 2sinU + H1- cos 2B)] d8
= (~03] ~ + [8 - 2cosB + (~0 - t sin2B)J: 12
= ~a3 + [(~+~)-(a- 2cosa + ~a: - t sin2o:)) ~ 3.4645
45. L = 1b Jr2+(d7·jd0)2dB= fo7r }(2cos0)2 +(-2sinB.)2d8
= 17r V 4(cos2 B + sin2 B) dB = 1"' ..f4d8 = [2B]; = 21r
As a check, note that the curve is a circle of radius 1, so its circumference is 27r(1) = 21r.
= 12
1f BV02 +4d0
Now let u = 02 + 4, so that du = 20d8 [Ode= t du] and
49. The curve r = cos4 (0 / 4) is completely traced with 0 :::; () :::; 47r.
r 2 + (drfd8)2 = [cos4 (B/4)]2 + [4cos3 (B/ 4) · (-sin(B/4)) · t ]2
= cos8.(B /4) + cos6 (8 /4) sin2 (B / 4)
= cos6 (8/4)[cos2 (B/4) +sin2 (0/4)] = cos6 (B/4)
L = f0 41
"' jcos3 .(0/4) j rj,B .
= 2 f~" cos8 (B/4)dB [sincecos3 (B/4) ~ 0 for O :S B $ 27r] = 8J;12 cos3 udu [u = tB]
= 8 J;12(1 - sin2 u) cosudu = 8 f0 1 (1- x2
) dx
x = sinu, ·] dx = cosudu
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51. One loop of the curve r = cos 28 is traced with - 1r I 4 ~ 8 ~ 1r I 4.
r 2 + (~~) 2
= cos2 28 + ( - 2 sin 28)2 = cos2 28 + 4sin2 28 = 1 + 3 sin2 28 ::::} L /"' 14
Vl + 3 sin2 28 d8:::::: 2.4221. -?r/4 -
53. The curver = sin(6sin9) is completely traced with 0 ~ 8 ~ 1r. r = sin{6sin9) ::::} ~ = cos{6sin9)_-6cos8, so
r2 : (~;y =s~n2(6sin8)+36cos2 8cos2 (6sin0) ::::} L L" Jsin2 (6sin8)+36cos2 8pos2(6sinO)d8:::::8.0091.
55. (a) From (10.2.6),
s =I: 21ryJ(dxld0)2 + (dyld9) 2 dO
[from the derivation of Equation 10.4.5]
(b) The curve r 2 = cos 28 goes through the pole when cos 28 = 0 ::::}
20 = ~ ::::} 8 = ~. We'll rotate the curve from () = 0 to 8 = ~ and double
this value to obtain the total surface area generated.
· dr ( dr ) 2
10.5 Conic Sections
(o. f) ? = cos28
1. x2 = 6y and x2 = 4py ::::} 4p = 6 ::::} p = ~
The vertex is (0, 0), the focus is (0, ~),and the directrix
isy = -~.
'\
The vertex is (0, O),the focus is ( -~, 0), and the
directrix is x = ~.
y= -~
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5. (x + 2)2 = 8 (y- 3). 4p = 8, sop= 2. The vertex is
( - 2, 3), the focus is ( -2, 5), and the directrix is y = 1.
y
-------------------------- __ ,y_~_!_
X
7. y2 + 2y + 12x + 25 = 0 =l>
y2 + 2y + 1 = - 12x- 24 =;..
(y + 1)2 = - 12(x + 2). 4p = - 12, sop= -3.
The vertex is ( - 2, -1), the focus is ( -5, - 1), and the
directrix is x = 1.
I jx=l
9. The equation has the form y2 = 4px, where p < 0. Since the parabola passes through (- 1, 1), we have 12 = 4p( - 1), so
4p = - 1 and an equation is y 2 = - x or x = - y 2 • 4p = - 1, sop = -i and the focus is ( -;i , 0) while the directrix
. 1 ISX = 4·
y
2
ellipse is centered at {0, 0), with vertices at (0, ±2). The foci are (0, ±v'2).
-..fi
-2
x2 Y2 13. x 2 + 9y2 = 9 <=> 9 + l = 1 =l> a = v'9 = 3, 15. 9x2
- 18x + 4y2 = 27 <=>
b = Vi.= 1, c = ~ = v'9=I = J8 = 2v'2.
The ellipse is centered at {0, 0), with vertices (± 3, 0).
The foci are (±2v'2, 0).
- 3
9(x- 1)2 + 4y2 = 36 <=> (x- 1)2 y2
4 + g- = 1 =}
vertices (1 , ± 3}, foci (1, ±J5)
y (1, 3)
3 X
(1,- 3)
2 2
17. The center is (0, 0), a= 3, and b = 2, so an equation is ~ + Y 9
= 1. c = ~ = J5, so the foci are (0, =;I=J5).
© 2012 Ceng3ge LC3JT1ing. All Rights Reser\'ed. Muy not be scuMed. copied, or duplicated, or pos1cd to a publicly access ible wc:bsite, in whole or in pan.
28 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLARCOORDINATES
yz xz 19.
25 - 9 = 1 ::::? a = 5, b = 3, c = J25 + 9 = J34 ::::?
center (0, 0), vertices (0, ±5), foci (0, ±J34), asymptotes y = ±~x.
Note: It is helpful to draw a 2a-by-2b rectangle whose center is the center of
the hyperbola. The asymptotes are the extended diagonals of the rectangle.
x2 y2 21 . x2
c = JlOO + 100 = 10J2 => center (0, 0), vertices (±10, 0),
foci (±10 J2, 0), asymptotes y = .±f§x = ±x
23. 4x2 - y2 - 24x - 4y + 28 = 0 <=>
4(x2 - 6x + 9) - (y2 + 4y + 4) = -28 + 36- 4 <=>
4(x '_ 3)2 - (y + 2)2 = 4 <=> (x - 3)2 - (y + 2)2 = 1 => 1 4
a = v'f = 1, b = v'4 = 2, c = Vf+4 = J5 ::::?
center (3, - 2), vertices (4, - 2) and (2, -2), foci (3 ± J5, -2),
asymptotes y + 2 = ±2(x- 3).
y
X
25. x2 = y + 1 <=> x2 = 1(y + 1 ). This is an equation of a parabola with 4p = 1, so p = i. The vertex is (0, -1) and the
focus is (0, - ~).
27. x 2 = 4y- 2y2 <=> x 2 + 2y2 - 4y = 0 <=> x 2 + 2(y2
- 2y + 1) = 2 <=> x 2 + 2(y - 1)2 = 2 <=>
x; + (y ~ 1) 2
= l..This is an equation of an ellipse with vertices a~ (±J2, 1) . The foci ~eat (±.,J2=1, 1) = (±1, 1).
29. y2 + iy = 4x2 + 3 <=> y2 + 2y + 1 = 4x2 + 4 <=> (y + 1? - 4x 2 = 4 <=> (y: 1
) 2
- x2 = 1. This is an equation
of a hyperbola with vertices (0, - 1 ± 2) = {0, 1) and (0, -3). The foci are at (0, - 1 ± J4+T) = (0, - 1 ± J5).
31. The parabola wi~ vertex {0, 0) and focus (1 , 0) ?pens to the right and hasp= 1, so its equation is y 2 = 4px, or y 2 = 4x. '
33. The distance from. U1e focus ( -4, 0) to the directrix x = 2,is 2 - ( - 4) = 6, so the distance from the focus to the vertex is
~(6) = 3 and the vertex is ( - 1, 0). Since the focus is to the left of the vertex, p = - 3. An equation is y 2 .= 4p(x + 1) =>
y2 = -12(x + 1).
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SECTION 10.5 CONIC SECTIONS 0 29
35. A parabola with vertical axis and vertex (2, 3) has equation y- 3 = a(x- 2?. Since it passes through (I; 5), we have
5-3=a(I-2)2 '* a =2,soanequationisy -3 =2(x-2?.
37. The ellipse with foci (±2, 0) and vertices (± 5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 2,
x2 1 2 so b2 = a2
- c2 = 25- 4 = 21. An equation is 25
+ ; 1
= 1.
39. Since the vertices are (0, 0) and (0, 8), the ellipse has center (0, 4) with a vertical axis and a = 4. The foci at (0, 2) and (0, 6)
are 2 units from the center, soc= 2 and b = .,Ja2 - c2 = .,j42 - 22 = v'f2. An equation is (x - 2 0)
2 + (y- 4)
2 = 1 '*
b a2
_41. An equation of an ellipse with center (-1,4) and vertex (- 1,0) is (x !2
1 )
2
= 1. The focus (-1,6) is 2 units
from the center, soc= 2. Thus, b2 + 22 = 42 '* b2 = 12 and the equation is (x + 1? + (y - 4) 2
= 1 . , 12 16 .
2 2
43. An equation of a hyperbola with vertices (±3, 0) is ~2 ....:. ; 2 = 1. Foci (±5, 0) '* c = 5 and 32 + b2 =52 '* . 2 2
b2 = 25 - 9 = 16, so the equation is ~ - r6 = 1.
45. The center of a hyperbola with vertices ( -3, - 4) and ( -3, 6) is ( -3, 1), so a = 5 and a.n equation is
(x + 3)2 . · ·
b2 = l.Focl(- 3, - 7)and(- 3,9)-* c =8,so52 +b2 =82 '* b2 =64-25=39andthe
( - 1}2 {x + 3)2
. . 2 • 2
47. The center of a hyperbola with vertices (±3, 0) is (0, 0) , so a = 3 and an equation is ~2 - t2 = 1.
b . 2 2 Asymptotes y = ±2x '* - = 2 '* b = 2(3) = 6 and the equation is ~ - JL = 1.
a 9 36
49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance
a - c from it) while the farthest point is the other vertex (at a distance of a+ c). So for this lunar orbit,
(a- c)+ (a+ c) = 2a = (1728 + 110) + (1728 + 314), or a = 1940; and (a+ c)- .(a - c) = 2c = 314 - 110,
x2 y2 or c = 102. Thus, b2 = a2
- c2 = 3,753,196, and the equation is 3
, 763
, 600
+ 3
, 753
, 196
=I.
51. (a) Set up the coordinate system so that A is ( -200, 0) and B is (200, 0) .
IPAI-\PBI = (1200)(980) = 1,176,000 ft = 21~ mi = 2a -* a= 1~5 , and c ~ 200 so
b2 - 2 - 2 - 3,339,375 - c a - 121
121x 2
1,500,625 121y
2 - 1
3,339,375 - .
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(b) Due north of B => x = 200 => (121}(200) 2
1,500,625 121y
3,339,375 = 1 => 539
53. The function whose graph is the upper branch of this hyperbola is concave upward. The function is
2 2 55. (a) If k > 16, then k - 16 > 0, and xk + -k y = 1 is an ellipse since it is the sum of two squares on the left side.
- - - 16
2 2 (b) IfO < k < 16, then k - 16 < 0, and xk + k ~
16 = 1 is a hypel"bola since it is the difference of two squares on the
left side.
(c) If k < 0, then k - 16 < 0, and there is no cw11e since the left side is the sum of two negative tenns, which cannot equal I .
(d) In case (a), a 2 = k, b2 = k- 16, and c2 = a2 - b2 = 16, so the foci are at (±4, 0}. Jn case (b), k - 16 < 0, so a 2 = k,
b2 = 16 - k , and c2 = a2 + b2 = 16, and so again the foci are at (±4, 0}.
57. x2 = 4ptJ => 2x = 4py' => y' = ; , so the tangent line at (x0 , y0 ) is . p
2
y- :; = ~; (x - xo). This line passes through the point (a, - p) on the
d. . x6 xo ( ) 4 2 2 2 2 2 1rectnx, so - p - 4 p =
2 p a - xo => - p - x0 = axo - x0 <=>
y
x5 - 2axo - 4p2 = 0 <=> x5 - 2axo + a 2 = a2 + 4p2 <=> (a,-p)
(xo- a)2 = a2 + 4p2 <=> x0 =a± Ja2 + 4p2 • The slopes of the tangent lines at x = a± Ja2 + 4p2
a± ..ja2 + 4p2
p , so the product of the two s lopes ts
a+ Ja2 + 4p2 a - Ja2 + 4p2 a2 _ . (a2 + 4p2) -4p2. -----'----'--- . = = -- = -1 2p 2p 4p2 4p2 '
showing that the tangent lines are perpendicular.
y= - p
2 2
59. 9x2 + 4y2 = 36 <=> ~ + Y 9
= 1. We use the parametrization x = 2 cost, y = 3 s in t, 0 ~ 't ~ 21r. ·The circumference
is given by
L =I:" J(dx/dt)2 + (dyjdt)2 dt = ]~2" J( - 2 s in t)2 + (3 cost)2 dt =I:" )4sin2 t + 9cos2 t dt
= I:" v 4 + 5 cos2 t dt
Now use Simpson's Rule with n = 8, 11t = 2 1T-
0 = ~.and f(t) = v'4 + 5cos2 t to get 8 . 4
L ~ ss = ~ [J(o) + 4f(*) + 2f(~) + 4Jen + 2J(1r) + 4/( 5 ;) + 21en + 4tCn + J(21T)J ~ 15.9.
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x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1 =? - = --- =? y = ±- Jx2- a2.
a b2 b2 a2 a
A= 2 r !!_ J x 2 ·_ a2 dx ~ 2b [~ J x2 - a2 - 42
In Jx + J x2 - a2J ] c
la a a 2 2 a· b
= -[cvc2 - a2 -a2 ln jc + vc2 - a2 j + a2 ln!al) a
Since a2 + b2 = c2 , c2 - a2 = b2
, and J c2 - a2 = b.
=!!. [cb- a2 1n(c + b)+ a2 !na) = !!_ [ch + a2 (!na - !n(b +c))) a a
= b2 c/a + abln[a/(b +c)), where c2 = a2 + b2 .
SECTION 10.5 CONIC SECTIONS 0 31
x2 Y2 63. 9x2 + 4y2 = 36 <::> '4 + g = 1 =? a= 3, b = 2. By symmetry, x = 0. By Example 2 in Section 7.3, the area o_fthe
top half of the ellipse is H 1rab) = 37T. Solve 9x2 + 4y2 = 36 for y to get an equation for the top half of the ell ipse:
9x2 +4y2 =36 <::> 4y2 =36-9x2 ~ y2 = ~(4-x2 ) =? y = ~v'4-x2 .Now
1 1b 1 1 12 1 ( 3 )
2 3 12 y = - - [f(x))2 dx =- - - V 4- x2 dx= - (4 -x2 )dx
A a 2 37T - 2 2 2 87T - 2
y
3 ' 12 2 3 [ 1 3] 2
3 ( 16) 4 = - · 2 ( 4 - x ) dx = - 4x - - x = - - = - 87T 0 47T 3 0 47T 3 7T
so the centroid is (0, 4/ rr).
' ffi . . . I' . I •x 2
y 2
1 65. Dt erenttatmg tmp tctt y, a2 + b2 = =? [y =f. 0]. Thus, the slope ofthe tangent
line at Pis - b:x1 . The slope of F1P is ___)f];_ and of F2P is ____lt!__ By the formula from Problems Plus, we have a Y1 X1 + c X1 - c
Yl b2Xt -Xt_+_c + -a2_y_1 _ a2 yt + b2x 1(x1 +c) _ a2b2 + b2cx1
tan a = _.:........;.---:-:.-----'-'-=-- 1 _ b2x1Y1 - a2y1 (x1 + c) - b2XtY1 - C2XtY1 + a 2CIJ1
a2yt (Xt +c)
- a 2y'f. - b2
x1 ( x1 -c) - a 2 b2 + b2 cx1 b2 (ext - a2) b2
= a2y1 (x1- c) - b2X 1Y1 = c2x1y1 - a2C1Jt = CIJ1 (cx1 - a2) = CIJ1
Thus, a = {3.
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10.6 Conic Sections in Polar Coordinates
1. The. directrix x = 4 is to the right of the focus at the origin, so we use the form with "+ e cos(}" in the denominator.
ed l. 4 4 (See Theorem 6 and Figure 2.) An equation is r = (} = 2
1 (} = 2
(}. ' 1+ ecos 1+ 2 cos +cos
3. The directrix y = 2 is above the focus at the origin, so Wf! use the form w ith "+ e sin(}" in the denominator. An equation is
_ ed _ 1.5(2) _ 6 r - 1 + e sin () - 1 + 1.5 sin(} - 2 + 3 sin(}·
5. The vertex ( 4, 3Jr /2) is 4 units below the focus at the origin, so the directrix is 8 units below the focus ( d = 8), and we
use the form with "- e sin(}" in the denominator. e = 1 for a parabola, so an equation is
ed 1(8) 8 r = 1 - e sin () = 1 - 1 sin (} = 1 - sin (} ·
7. The directrix r = 4 sec (} (equ ivalent to r cos(} = 4 or x· = 4) is to the right of the focus at the origin, so we w ill use the form
with"+ e cos()" in the denominator. The distance from the focus to the directrix is d = 4, so an equation is
ed ~{4) 2 4 r = 1 + ecosB = 1 + ~cos(}· 2 = 2 +cos((
9. r = 4
= 4
/ 5
where e = .1 and ed = .1 => d = 1. 5 - 4sinB 1/5 . 1 - ~sinB' 5 &
(a) Eccentricity = e = ~
(b) Since e = ~ < 1, the conic is an ellipse.
(c) Since"- e sin (}" appears in the denominator, the directrix is below the focus
at the origin, d = IFl l = 1, so an equat'ion of the directrix is y = - 1.
(d) The vertices are (4, ¥) and(~ , 3 ;).
2 1/3 2/ 3 ' 2 11. r = 3 + 3 sinB · . 1/~ = l + 1 sinB ' wheree=1and ed= 3 => d= ~.
(a) Eccentricity= e = 1
(b) Since e = .1, Lhe conic is a parabola.
(c) Since"+ e sin(}" appears in the denom inator, the directrix is above the focus
at the origin. d = IFll = j , so an equation oflhe directrix is y = i· (d) T~e vertex is at ( ~, ¥), midway between .the focus and directrix.
y
y y = 2/ 3
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SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES 0 33
13. r = 6
+ cos 1+ 3 cos0
(a) Eccentricity= e = t (b) Since e = t < 1, the conic is an ellipse.
(c) Since "+ e cos e" appears in the denominator, the directrix is to the right of
the focus at the origin. d = !Fll = ~.so an equation of the directrix is
x = ~-
(d) The vertices are ( ~, 0) and ( t, 1r), so the center is midway between them~
that is, ( fn, 1r) .
3 1/ 4 3/ 4 3 15. r = 4 - 8 cos 9 . 1 I 4 = 1 - 2 cos 9' where e = 2 and ed = 4
(a) Eccentricity = e = 2
(b) Since e = 2 > 1, the conic is a hyperbola.
d - 1 - s ·
(c) Since "- e cos 0" appears in the denominator, the directrix is to the left of
the focus at the origin. d'= IFll = ~.so an equation of the directrix is
X - 3 . - - s· (d) The vertices are ( -~. 0) and (i, 1r), so the center is midway between them,
that is, (~ , 1r).
. The eccentricity 1 - sm
e = 2 > 1, so the conic is a hyperbola. Since" -e sin B" appears in the
denominator, the directrix is below the focus at the origin. d = IFl l = ~.
so an equation of the directrix is y = - ~ . The vertices are ( - 1, ~) and
( 1 3'~~" ) h . "d b th h . ( 2 3" ) 3 , 2 , sot e center IS m1 way etween l'!m, t at 1s, 3 , 2 .
(b) By the discussion that precedes Example 4, the equation
. 1 IS T = ( 3 ) • 1 - 2sin e- ;
I
(f,1T)
9 :
-3
-2
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19. Fore < 1 the curve is an ellipse. It is nearly circular when e is close to 0. As e
increases, the graph is stretched out to the right, and grows larger (that is, its
right-hand focus moves to the right while its left-hand focus remains at the
origin.) At e = 1, the curve becomes a parabola with focus at the origin.
21.IPFI= e iPLI => r =e[d-rcos(7r-9)J = e(d+rcos8) =*
r(1 - ecos~) = ed ed => r=---....,.
1- ecosO
23. IPFI = eiP ll => r = e[d- rsin(9 -1r)] = e(d+r sinO) =>
r(1 - esin 9) = ed => ed r = ---""7
.. 1- esin9
x=-d
l • y= - d
25. We are given e = 0.093 and a = 2.28 x 108 . By (7), we have
a(1 - e2 ) 2.28 x 108 (1- (0.093)2] . 2.26 x 108
r = = ~ -::-----::-=::------:: 1 + ecos 8 1 + 0.093 cosO 1 + 0.093cos8
X
X
27. Here 2a = length of major axis = 36.18 AU => a = 18.09 AU and e = 0.97. By (7), the equation of the orbit is
18.09(1 - (0.97) 2
] 1.07 B (&) th · d" fr th h · r = ~ 8
. y , e maxtmum 1stance om e comet to t e sun 1s 1 + 0.97 cos 9 1 + 0.97 cos
18.09(1 + 0.97) ~ 35.64 AU or about 3.314 billion miles.
29. The minimum distance is at perihelion, where 4.6 x 107 = r = a(1 - e) = a(1 - 0.206) = a(0.794} =>
a = 4.6 x 107/ 0.794. So the maximum distance, which is at aphelion, is
r = a(1 + e) = ( 4.6 x 107/ 0.794) (1.206} ~ 7.0 x 107 km.
31. From Exercise 29, we have e = 0.206 and a(l- e)= 4.6 x 107 km. Thus, a= 4.6 x 107/ 0.794. From (7), we can write the
. . 1 - e2 •
. So smce + ecos
dr a(1- e2 )esin8 -= => d9 (1 + ecos8)2
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CHAPTER 10 REVIEW 0 35
the length of the orbit is
L = 12
,. .Jl(t e 2 + 2~)~sO dO:::::: 3.6 x 108 km
0 0 +ecos
This seems reasonable, since ,Mercury's orbit is nearly circular, and the circumference of a circle of radius a
is 21ra :::::: 3.6 x 108 km.·
10 Review CONCEPT CHECK
1. (a) A parametric curve is a set of points of the form (x, y) = (J(t), g(t)), where f and g are continuous funct ions of a
variable t.
(b) Sketching a parametric curve, like sketching the.graph of a function, is difficult to do in general. We can plot points on the
curve by finding f(t) and g(t) for various values oft, either by hand or with a calculator or computer. Sometimes, when
f and g are given by formulas, we can eliminate t from the equations x = j(t) andy = g(t) to get a Cartesian equation
relating x andy. It may be easier to graph that equation than to work with the original formulas for x andy in terms oft.
2. (a) You can find : as a function oft by calculating~~ = ~~J~~ [if dxjdt f: 0].
(b) Calculate the area as I: y dx = I: g(t) j'(t)dt [or I; g(t) !' (t)dt ifthe leftmost point is (!({3), g({J)) ~ther
than (f(a),g(a))].
3. (a) L = I: J(dxjdt)2 + (dy jdt)2 dt =I: vlf'(t))2 + [.g'(t))2 dt
(b) s =:: I: 27ryJ(dxfdt)2 + (dyfdt) 2 dt =I: 21rg(t) v lf'(t)F + [g'(t)J2 dt .
4. (a) See Figure 5 in Section 10.3.
(b) x = rcosO, y = r sinO
(c) To find a polar representation (r, 0) with r ;?. 0 and 0 s 0 < 27r, firs t calculate r = ·Jx2 + y2. TI1en 0 is specified by
cosO= xjr and sinO= yjr.
d dy !:.._ (y) !:.._ ( r ::;in 8) 5. (a) Calculate Jx = ~ = ~ = ..::::llJ:;:-- -
dO dO (x) d() (rcosB)
(~)sin(} + r cos(}
(b) calculate A= I: !r2 dO= I: tlfCOW d8
(c) L =I: J(dx/d8.)2 + (dy/d0)2 dO= I: Jr2 + (dr/d0)2 dO= .r: J[f(0))2 + [!'(0)]2 dO
6. (a) A parabola is a set of points in a plane whose distances from a fixed point F (t11e focus) and a fixed line l (the 'directrix)
are equal.
(b) x2 = 4py; y2 = 4px
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7. (a) An ellipse is a set of points in a plane the sum of whose distances from two fixed points (the foci) is a constant.
2 2 (b) X + _Y_ = 1.
a2 a2 - c2
8. (a) A hyp~rbola is a set of points in a plane the difference of whose distances from two fixed points (the foci) is a constant.
This difference should be interpreted as the larger distance minus the smaller distance.
2 2 (b) ::__ - _ Y_ = 1
a2 c2- a2
a
9. (a) Jfa conic section has focus F and corresponding directrix l, then the eccentricity e is the fixed ratio /PFI / /Pl / for points
P of the conic section.
(b) e < 1 for an ellipse; e > 1 for a hyperbola; e = 1 for a parabola.
ed . ed · ed ed (c) x = d: r = . x = - d:.r = . y = d: 1' = . . y = - d: r = -:----:--::
1 + e cos (} 1 - e cos (} 1 + e sm 8 1 - e sin 8 ·
1. False.
3. False.
. 5. True.
TRUE-FALSE QUIZ
Consider the curve defined by x = f(t) = (t - 1)3 and y= g(t) = (t- 1)2 : Then g'(t) = 2(t - 1), so g'(1) = 0,
but its graph bas a vertical tangent when t = 1. Note: The statement is true iff' (1) =I= 0 when l (1) = 0.
For example, iff ( t) = cost and g( t) = sin t for 0 ::; t :S 4rr, then the curve is a circle of radius I, hence its length
is 2rr, but f0 4
" J [!'(t)]2 + [g'(t))2 dt = .{0 4
" J(- sin t)2 +(cos t)2 dt = f0 4
" 1 dt = 4rr, since as t increases
from 0 to 4rr, the circle is traversed twice.
The curve r = 1 - sin 28 is unchanged if we rotate it through 180° about 0 because
1 - sin 2(8 + rr) = 1 - sin(28 + 2rr) = 1 - sin 28. So it's unchanged if we replacer by -r. (See the discussion
after Example 8 in Section 10.3.) In other words, it's the same curve as r = -(1 -sin 20) = sin 28 - 1.
7. False. The first pair of equations gives the portion of the parabola y = x2 with x ;::: 0, whereas the second pair of equations
traces out the whole parabola y = x2 •
9. True. By rotating and translating ~he parabola, we can assume it has an equation of the form y = cx2 , where c > 0.
Tile tangent at the point (a, ca2 ) is the line y - ca2 = 2ca(x - a); i.e., y = 2cax - ca2
• This tangent meets
the parabola at the points (x, cx2 ) where cx2 = 2cax - ca2
• This equation is equivalent to x2 = 2a~ - a2
[since c > 0]. But x2 = 2ax - a2 <=> x2 - 2ax + a 2 = 0 <=> (x- a)2 = 0 <=> x =a <=>
( x, cx2 ) = (a, ca2
). This shows that eacb tangent meets the parabola at exactly one point.
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EXERCISES
1. X = t 2 + 4t, y = 2 - t, - 4 ~ t ~ 1. ~ = 2 - y, SO
X = {2 - y)2 + 4{2- y) = 4- 4y + y2 + B- 4y = y2 - By+ 12 {::}
x + 4 = y 2 - By+ 16 = (y - 4)2 . This is part of a parabola with vertex
( -4, 4), opening to the right.
3. y = sec() = ~() = .! . Since 0 ~ () ~ 1r /2, 0 < x ~ 1 and y ;::: 1. COS X
This is part of the hyperbola y = 1/x.
5. Three different sets of parametric equations for the curve y = -IX are
(i) X = t, y = Vt (ii) X = t 4
, y = t 2
(iii) x = tan2 t, y = tan t, 0 ~ t < 1r /2
There are many other sets of equations that also give this curve.
y
y
\ ,1,1),0-0 0
7. (a) The Cartesian coordinates are x = 4 cos 2; = 4 (- ~) = -2 and
y = 4sin ;,. = 4( 4) = 2 J3, that is, the point ( - 2,2 J3).
X
(b) Give~ x = - 3 and y= 3, we haver = J( -3)2 + 32 = JIB = 3 v'2. Also, tan() = ~ '* tan() = ~3 , and since
( -3, 3) is in the second quadrant,(}= ' 34". Thus, one set of polar coordinates for ( -3, 3) is (3 V2, 3,;'"), and two others are
(3 '2 1111") and (- 3 '2 7,.) y ~, 4 v. .L.i, 4 .
9. r = 1 - cosO. This cardioid is
symmetric about the polar axis.
r
2
7T 27T 0
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11. r = cos 30. This is a
three-leaved rose. The curve is
traced twice.
sym·metric about the pole and
both the horizontal and vertical
axes.
r
1T
2
3 15. r = =} e = 2 > 1, so the conic is a hyperbola. ·de = 3 =>
t + 2sin£1
d = ~ and the form "+2 s in(}" imply that the directrix is above the focus at
the origin and bas equation y = ~-The vertices are (1, ~) and ( - 3, 3 ;) .
(2, 1r)
2 . 17. x+y=2 ¢:> rcosll+rsinll = 2 ¢:> r(cos0+sin 0)=2 ¢:> r= - - - --
cos(J + s in£1
As (} __, 0, r --.. 1. In the first figure,
there are an infinite number of
x -intercepts at x = 1rn, n a nonzero
integer. These correspond to pole
points in the second figure. -0.25
2 dy dx 1 dy dy / dt 2t 2 21. x = lnt, y = 1 + t ; t = 1. dt = 2t and dt = t' so dx = dxjdt = 1/t = 2t ·
When t = 1, (x , y) = (0, 2) and dyjdx = 2.
23. r = e-0 => y = rsin£1 = e-0 s in(} and x = T"cos O = e- 0 cosO =>
dy dyjd£1 *sin(} +rcosO _ -e- 0 sin(}+ e-0 cos(} - e0 sine- cosO dx = dx j d(} = ~;cos(} - rsin(} - -e-0 cosB - e-0 sinO · - e0 = cos B+sine ·
dy 0 - (- 1) 1 When (} = 1r, - = = - = - 1.
dx -1 + 0 -1
(2,0)
-1
- 0.75
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25. x = t + sin t, y = t - cost ~ dy dyjdt 1 + sint dx = d.xjdt = 1 +cos t
{1+cost) cost- {1 +sint)(- sint)
d (dy) dt dx
(1 + cost)3
3 2 . 27. We graph the curve x = t - 3t, y = t + t + 1 for -2.2 :=::; t :=::; 1.2.
By zooming in or using a cursor, we find that the lowest point is about
(1.4, 0. 75). To find the exact values, we find the t-value at which
dyjdt=2t+1 = 0 ¢=> t=~~ <=> , (x,y),;,(lj,~).
29. x·= 2acost- acos2t ~ dx = -2asin t + 2asin2t = 2asint(2cost - 1) = 0 ¢:> dt
sint = 0 or cost = ~ ~ t = O,·t, 1r, orr;.
y = 2asin t - a sin 2t ~ dy = 2a cost - 2a cos 2t = 2a(l +cost- 2 cos2 t) = 2a(1 -cos t)(1 + 2 cost) = 0 =} dt
t 0 271" 4" = '3,or3.
Thus the graph has vertical tangents where t = t. 1r and ~">3" , and horizontal tangents where t = 2 ; and 4
;. To determine
what the slope is where t = 0, we use !'Hospital's Ru;e to evaluate lim ddy/jddt = 0, so there is a. horizontal tangent there. t->0 X t
t X y y
(-3a,O) (a,O)
)C
7r - 3a 0 4 7r -~a -~a 3 2 51r ~a _ :.::]:a 3 2
31. The curve r 2 = 9 cos 50 has 10 "petals." For instance, for - -fu :=::; 8 :=::; {'0 , there are two petals, one with r > 0 and one
with r < 0.
33. The curves intersect when 4 cos 8 = 2 ~ cos 8 = ~ ~ 8 = ± i
for -1r :=::; 8 :=::; 1r. The points of intersection are (2 , i) and (2,- ~).
r = 4cos 0
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35. The curves intersect where 2 sin () = sin() + cos () =>
sin 0 = cos 0 => 0 = "i, and also at the origin (at which 0 = 3 4rr
on the second curve).
4 ~(sinO+ cos 0)2 dO
= Io"14 (1- cos 20) dO+~ I:;t (1 + sin2fJ) dO
8=f
' = [0 ~ ~ sin20]~14 + [~0-:- t cos2BJ!~~4 = ~(1r -1) r = 2 sin 8 r = sin 0 + cos 8
L =I~ .J(dxjdt)2 + (dyjdt) 2 dt = 1~ .J(6t)2 + {6t2)2 dt =I: V36t_2 + 36t4 dt = I~ y'36t2 v'1 + t2 dt
=I~ 6\t! v1 + t 2 dt = 6 I~ t v'f+t2 dt = 6 I1 5
u 1 ;
2 (~du)
= 6. ~ · ~ [u312J: = 2(5312 - 1) = 2(5VS -1)
39. L =I:" Jr2 + (drjdB)2 dB = I:" }(1/0)2 + (-1/02 )2 dB= L2
"' 7 dB
~ [-#+! +In (B + J B2 + 1 )] 2,. = #+I - v' 471"2 + 1 + In ( 271" + ~ ) () 7r: 271" 71" + v' 71"2 + 1
11"
s = I1 4
27ry.J(dxfdt)2 -t. (dyfdt)2 dt = I1 4 21r{~t3 + ~c2) J (2/0)2 + (t2- t-3)2 dt
= 21r J4 (.!t3 + .!c 2) . l(t2 + t - 3)2 dt = 21r f 4 (.!t5 + !! + l t-5) dt - 21r [...!..t6 + l!t - 1c4) 4
- 471·295 1r 1 3 2 v 1 3 6 2 - 18 6 8 1 - 1024
43. For all c except - 1, the curve ·is asymptotic to the line x = 1. For
c < -1, the curve bulges to the right near y = 0. As c increases, the
bulge becomes smaller, until at c = - 1 the curve is the straight line x = 1.
As c continues to increase, the curve bulges to the left, until at c = 0 there
is a cusp at the origin. For c > 0, there is a loop to the left of the origin,
whose size and roundness increase as c increases. Note that the x-intercept
of the curve is always -c.
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2 2 . 45. ~ + 'lL = 1 is an ellipse with center (0, 0).
9 8 47. 6y2 + X - 36y + 55 = 0 ~
6(y2 - 6y + 9) = - (x + 1) ~
a = 3, b = 2 J2, c = 1 =?
foci (±1, 0), vertices (±3, 0).
y J-2 2
directrix x = -~. y
- 3 3 X
0 X
49. The ellipse with foci (±4, 0) and vertices (±5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 4,
2 2
- 42 = 9. An equation is ~5 + Y 9
= 1.
2 ?
51 . The center of a hyperbola with foci {0, ±4) is (0, 0), so c= 4 and an equation is y 2
- x: = 1. . a b
a 3 · The asymptote y = 3x has slope 3, sob = l - ::::} a= 3b and a2 + b2 = c2
=? {3b? + b2 = 42 =?
b2 = * and so a2 = 16 - ~ = L5 2
• Thus, an equation is - 1 - - -
1 = 1, or -
" " 72 5 8 5
53. x2 = -(y - 100) has its vertex at (0,100), so one of the vertices ofthe ellipse is (0, 100). Another form of the equation of a
parabola is x 2 = 4p(y - 100) so 4p(y - 100) = - (y- 100) =? 4p = -1 ::::} p = -~.Therefore the shared focus is
found at .(0, 3~9 ) so 2c = 3~9 -0 ::::} c = 3~9 and the center of the ellipse is (0, 3~9 ). So a= 100- 3~9 = 4~1 and
2 2 . 2 ( 39!))2
b2 __ a2 _ c2 __ 401 8 - 2
399 x y - 8 ----,=--- = 25. So the equation of the e ll ipse is /1 + a 2 = 1 =?
x 2 (8y - 399? = 1 or 25 + 160,801 ·
55. Directrix x = 4 =? d = 4, so e = ~ ed 4
=? r - - ..,..--- -., - 1 +ecosfJ- 3+cosfJ'
57. (a) Jf (a, b) lies on the curve, then there is some parameter value t1 such that~ = a and 3 tr ~ =b. l ft1 = 0,
1 + tl 1 + t l
the point is (0, 0), which lies on the line y = x. lft1 =I 0, then the point corresponding tot= ..!. is given by tl
3(1/tl ) 3ti 3(1/td 3tl x = 1 + (1/t
1 ) 3 = t~ + 1 = b, Y = 1 + (1/h)3 = tt + 1 = a. So (b, a) also lies on the curve. [Another way to see
this is to do part (e) first; the result is immediate.] The curve intersects the line y = x when ~3 = 3t 2
3 1 + t 1 +t ::::}
t = t 2 ::::} t = 0 or. 1, so the points are (0, 0) and ( ~, ~).
<D 2012 Cengogc Leamins. ;\II RightS Rescr.,.cd. May noc be scanned, COJ')ict.l. or duplicated, or postt.'tlto u publicly llCCessiblc website, in whole or in (mrt.
42. D . CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(b) dy = (1 +t 3
)(Gt) - 3t2(3t 2
) = Gt - 3t 4
=Dwhen6t - 3t4 =3t(.2-t3 )=D ·=> t=Dort= 3'2,so thereare dt (1+t3)2 (1+t3)2 y.r.
horizontal tangents at (0, D) and ( V2', ?-'4). Using the symmetry from part (a), we se

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