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Unformatted text preview: 5E11(pp 686695) 1/18/06 9:31 AM Page 686 CHAPTER 11 Parametric curves are used to
represent letters and other symbols on laser printers. See the
Laboratory Project on page 705. P arametric Equations and
Polar Coordinates 5E11(pp 686695) 1/18/06 9:31 AM Page 687 So far we have described plane curves by giving y as a
function of x y f x or x as a function of y x t y
or by giving a relation between x and y that deﬁnes y
0 . In this chapter
implicitly as a function of x f x, y
we discuss two new methods for describing curves.
Some curves, such as the cycloid, are best handled when both x and y are given
in terms of a third variable t called a parameter x f t ,y t t . Other curves, such as the cardioid, have their most convenient description when we use a new
coordinate system, called the polar coordinate system.  11.1
y Curves Defined by Parametric Equations
C
(x, y)={ f(t), g(t)} 0 FIGURE 1 x Imagine that a particle moves along the curve C shown in Figure 1. It is impossible to
describe C by an equation of the form y f x because C fails the Vertical Line Test. But
the x and ycoordinates of the particle are functions of time and so we can write x f t
and y t t . Such a pair of equations is often a convenient way of describing a curve and
gives rise to the following deﬁnition.
Suppose that x and y are both given as functions of a third variable t (called a parameter) by the equations
x ft
y tt
(called parametric equations). Each value of t determines a point x, y , which we can
f t , t t varies and traces out a
plot in a coordinate plane. As t varies, the point x, y
curve C, which we call a parametric curve. The parameter t does not necessarily represent time and, in fact, we could use a letter other than t for the parameter. But in many
applications of parametric curves, t does denote time and therefore we can interpret
x, y
f t , t t as the position of a particle at time t.
EXAMPLE 1 Sketch and identify the curve deﬁned by the parametric equations t2 x 2t y t 1 SOLUTION Each value of t gives a point on the curve, as shown in the table. For instance, if t 0, then x 0, y 1 and so the corresponding point is 0, 1 . In Figure 2 we plot the
points x, y determined by several values of the parameter and we join them to produce
a curve.
t
2
1
0
1
2
3
4 x y y
8
3
0
1
0
3
8 1
0
1
2
3
4
5 t=4
t=3 t=2
t=1 (0, 1)
8 t=0
0 x t=_1
t=_2 FIGURE 2 A particle whose position is given by the parametric equations moves along the curve
in the direction of the arrows as t increases. Notice that the consecutive points marked on 687 5E11(pp 686695) 688 ❙❙❙❙ 1/18/06 9:32 AM Page 688 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES the curve appear at equal time intervals but not at equal distances. That is because the
particle slows down and then speeds up as t increases.
It appears from Figure 2 that the curve traced out by the particle may be a parabola.
This can be conﬁrmed by eliminating the parameter t as follows. We obtain t y 1
from the second equation and substitute into the ﬁrst equation. This gives
x t2 2t y 1 2 2y y2 1 4y 3 and so the curve represented by the given parametric equations is the parabola
x y 2 4y 3.
No restriction was placed on the parameter t in Example 1, so we assumed that t could
be any real number. But sometimes we restrict t to lie in a ﬁnite interval. For instance, the
parametric curve
y 2t y t 1 0 t 4 shown in Figure 3 is the part of the parabola in Example 1 that starts at the point 0, 1 and
ends at the point 8, 5 . The arrowhead indicates the direction in which the curve is traced
as t increases from 0 to 4.
In general, the curve with parametric equations (0, 1)
0 t2 x (8, 5) x x ft y tt a t b has initial point f a , t a and terminal point f b , t b . FIGURE 3 EXAMPLE 2 What curve is represented by the parametric equations x 0 t cos t, y sin t, 2? SOLUTION If we plot points, it appears that the curve is a circle. We can conﬁrm this
impression by eliminating t. Observe that x2 y2 cos 2t sin 2t 1 Thus, the point x, y moves on the unit circle x 2 y 2 1. Notice that in this example
the parameter t can be interpreted as the angle (in radians) shown in Figure 4. As t
increases from 0 to 2 , the point x, y
cos t, sin t moves once around the circle in
the counterclockwise direction starting from the point 1, 0 .
π t= 2 y
(cos t, sin t ) t=0 t=π t
0 (1, 0) x t=2π
t= FIGURE 4 3π
2 EXAMPLE 3 What curve is represented by the parametric equations x 0 t 2? sin 2 t, y cos 2 t, 5E11(pp 686695) 1/18/06 9:32 AM Page 689 S ECTION 11.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ❙❙❙❙ 689 SOLUTION Again we have x2 y2 sin 2 2 t cos 2 2 t 1 so the parametric equations again represent the unit circle x 2 y 2 1. But as t
increases from 0 to 2 , the point x, y
sin 2 t, cos 2 t starts at 0, 1 and moves twice
around the circle in the clockwise direction as indicated in Figure 5.
y t=0, π, 2π
(0, 1) 0 x FIGURE 5 Examples 2 and 3 show that different sets of parametric equations can represent the
same curve. Thus, we distinguish between a curve, which is a set of points, and a parametric curve, in which the points are traced in a particular way.
(_1, 1) y
(1, 1) EXAMPLE 4 Sketch the curve with parametric equations x sin t, y sin 2t. sin t 2 x 2 and so the point x, y moves on the parabola
y x . But note also that, since 1 sin t 1, we have 1 x 1, so the parametric equations represent only the part of the parabola for which 1 x 1. Since sin t
is periodic, the point x, y
sin t, sin 2t moves back and forth inﬁnitely often along the
parabola from 1, 1 to 1, 1 . (See Figure 6.) SOLUTION Observe that y
2 0 x FIGURE 6 x x=cos t
t Module 11.1A gives an animation of the
relationship between motion along a
parametric curve x f t , y t t and
motion along the graphs of f and t as functions
of t. Clicking on TRIG gives you the family of
parametric curves
x a cos bt
y c sin dt
If you choose a b c d 1 and click
START, you will see how the graphs of x cos t
and y sin t relate to the circle in Example 2. If
you choose a b c 1, d 2, you will see
graphs as in Figure 7. By clicking on PAUSE and
then repeatedly on STEP, you can see from the
color coding how motion along the graphs of
x cos t and y sin 2 t corresponds to motion
along the parametric curve, which is called a
Lissajous ﬁgure. y y x FIGURE 7 x=cos t y=sin 2t t y=sin 2t 5E11(pp 686695) 690 ❙❙❙❙ 1/18/06 9:32 AM Page 690 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES Graphing Devices
Most graphing calculators and computer graphing programs can be used to graph curves
deﬁned by parametric equations. In fact, it’s instructive to watch a parametric curve being
drawn by a graphing calculator because the points are plotted in order as the corresponding parameter values increase.
3 SOLUTION If we let the parameter be t
_3 y4 EXAMPLE 5 Use a graphing device to graph the curve x 3 _3 y, then we have the equations
t4 x 3y 2. 3t 2 y t Using these parametric equations to graph the curve, we obtain Figure 8. It would be
possible to solve the given equation x y 4 3y 2 for y as four functions of x and
graph them individually, but the parametric equations provide a much easier method. FIGURE 8 In general, if we need to graph an equation of the form x
metric equations
x tt
yt t y , we can use the para Notice also that curves with equations y f x (the ones we are most familiar with—
graphs of functions) can also be regarded as curves with parametric equations
x t y ft Graphing devices are particularly useful when sketching complicated curves. For
instance, the curves shown in Figures 9 and 10 would be virtually impossible to produce
by hand.
8 _6.5 3 6.5 _1.5 _8 1.5 _3 FIGURE 9 FIGURE 10 x=t+2 sin 2t, y=t+2 cos 5t,
_2π t 2π x=cos tcos 80t sin t,
y=2 sin tsin 80t, 0 t 2π One of the most important uses of parametric curves is in computeraided design
(CAD). In the Laboratory Project after Section 10.2 we will investigate special parametric
curves, called Bézier curves, that are used extensively in manufacturing, especially in the
automotive industry. These curves are also employed in specifying the shapes of letters and
other symbols in laser printers. The Cycloid
An animation in Module 11.1B shows
how the cycloid is formed as the circle
moves. EXAMPLE 6 The curve traced out by a point P on the circumference of a circle as the circle rolls along a straight line is called a cycloid (see Figure 11). If the circle has
radius r and rolls along the xaxis and if one position of P is the origin, ﬁnd parametric
equations for the cycloid. 5E11(pp 686695) 1/18/06 9:32 AM Page 691 S ECTION 11.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ❙❙❙❙ 691 P
P
F IGURE 11
y r
P SOLUTION We choose as parameter the angle of rotation of the circle
0 when P is
at the origin). Suppose the circle has rotated through radians. Because the circle has
been in contact with the line, we see from Figure 12 that the distance it has rolled from
the origin is
OT
arc PT r C (r¨, r) ¨ P Therefore, the center of the circle is C r , r . Let the coordinates of P be x, y . Then
from Figure 12 we see that Q
y x x x OT PQ r y T O TC QC r r sin r sin r1 cos r¨
FIGURE 12 r cos Therefore, parametric equations of the cycloid are
1 x r sin y r1 cos One arch of the cycloid comes from one rotation of the circle and so is described by
0
2 . Although Equations 1 were derived from Figure 12, which illustrates the
case where 0
2, it can be seen that these equations are still valid for other
values of (see Exercise 37).
Although it is possible to eliminate the parameter from Equations 1, the resulting
Cartesian equation in x and y is very complicated and not as convenient to work with as
the parametric equations. A cycloid
B
FIGURE 13 P P
P P
P FIGURE 14 One of the ﬁrst people to study the cycloid was Galileo, who proposed that bridges be
built in the shape of cycloids and who tried to ﬁnd the area under one arch of a cycloid.
Later this curve arose in connection with the brachistochrone problem: Find the curve
along which a particle will slide in the shortest time (under the inﬂuence of gravity) from
a point A to a lower point B not directly beneath A. The Swiss mathematician John
Bernoulli, who posed this problem in 1696, showed that among all possible curves that
join A to B, as in Figure 13, the particle will take the least time sliding from A to B if the
curve is part of an inverted arch of a cycloid.
The Dutch physicist Huygens had already shown that the cycloid is also the solution to
the tautochrone problem; that is, no matter where a particle P is placed on an inverted
cycloid, it takes the same time to slide to the bottom (see Figure 14). Huygens proposed
that pendulum clocks (which he invented) swing in cycloidal arcs because then the pendulum takes the same time to make a complete oscillation whether it swings through a
wide or a small arc. Families of Parametric Curves
EXAMPLE 7 Investigate the family of curves with parametric equations x a cos t y a tan t sin t What do these curves have in common? How does the shape change as a increases? 5E11(pp 686695) ❙❙❙❙ 692 1/18/06 9:32 AM Page 692 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES SOLUTION We use a graphing device to produce the graphs for the cases a 2, 1,
0.5, 0.2, 0, 0.5, 1, and 2 shown in Figure 15. Notice that all of these curves (except
the case a 0) have two branches, and both branches approach the vertical asymptote
x a as x approaches a from the left or right. a=_2 a=_1 a=0 a=0.5 FIGURE 15 Members of the family  11.1 a=1 Sketch the curve by using the parametric equations to plot
points. Indicate with an arrow the direction in which the curve is
traced as t increases.
t2 1. x 1 2. x 2 cos t, y t 3. x 5 sin t, y t 2, 4. x e
■ 5–10 a=2 Exercises
8. x y st, t t,
■ et y 0 t 0 t 2 ■ 11–18 t
t, ■ 2 ■ t ■ 2
■ ■ ■ ■  (a) Sketch the curve by using the parametric equations to plot
points. Indicate with an arrow the direction in which the curve
is traced as t increases.
(b) Eliminate the parameter to ﬁnd a Cartesian equation of
the curve.
5. x 3t 5, 6. x 1 t, 7. x t2 2, y
y
y 2t
5
5 2 t, ■ 3t, t,
■ 1 t2 2 y 2 y t y t ■ 3 ■ ■ ■ 2
3 t 3
t 4 ■ ■ ■ ■ ■  (a) Eliminate the parameter to ﬁnd a Cartesian equation of the
curve.
(b) Sketch the curve and indicate with an arrow the direction in
which the curve is traced as the parameter increases.
11. x sin , 12. x 4 cos , 13. x 2 sin , y cos 14. x 1
2 t, st, 10. x 5 cos t, ■ 4 t, 1 9. x  ■ a=_0.2 When a
1, both branches are smooth; but when a reaches 1, the right branch
acquires a sharp point, called a cusp. For a between 1 and 0 the cusp turns into a loop,
which becomes larger as a approaches 0. When a 0, both branches come together and
form a circle (see Example 2). For a between 0 and 1, the left branch has a loop, which
shrinks to become a cusp when a 1. For a 1, the branches become smooth again,
and as a increases further, they become less curved. Notice that the curves with a positive are reﬂections about the yaxis of the corresponding curves with a negative.
These curves are called conchoids of Nicomedes after the ancient Greek scholar
Nicomedes. He called them conchoids because the shape of their outer branches
resembles that of a conch shell or mussel shell. x=a+cos t, y=a tan t+sin t,
all graphed in the viewing rectangle
_4, 4 by _4, 4 1–4 a=_0.5 sec , y tan , 15. x e t, 16. x ln t, y y cos , 0
y e
y 5 sin , 2 t st, 2 2 2 t 1 2 ■ 5E11(pp 686695) 1/18/06 9:33 AM Page 693 ❙❙❙❙ S ECTION 11.1 CURVES DEFINED BY PARAMETRIC EQUATIONS 17. x cosh t, 18. x
■ 1 y sinh t cos , ■ ■ y ■ 2 cos
■ 25–27  Use the graphs of x
f t and y t t to sketch the
parametric curve x f t , y t t . Indicate with arrows the direction in which the curve is traced as t increases. 1 ■ ■ ■ ■ ■ ■ ■ 25.
19–22 x y Describe the motion of a particle with position x, y as
t varies in the given interval.
 19. x cos 20. x 2 21. x 2 sin t, 22. x cos 2t, ■ y t, cos t, ■ y 1 t, 3 t sin t, cos t, 0
■ ■ t
t ■ 1
1 2
0 3 cos t , 0 y
y ■ sin 693 t 1 1 x ■ t y 1 26. 2 ■ t _1 2 1 4
■ t ■ ■ ■ 1 t 23. Suppose a curve is given by the parametric equations x f t,
y t t , where the range of f is 1, 4 and the range of t is
2 , 3 . What can you say about the curve? 27. x 24. Match the graphs of the parametric equations x f t and
y t t in (a)–(d) with the parametric curves labeled I–IV.
Give reasons for your choices. (a) y
1 1
1 1t t I
y x
2 y 2 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 28. Match the parametric equations with the graphs labeled I–VI.
1 1 1 t 2x t (b) II x y y 2 2 2 Give reasons for your choices. (Do not use a graphing device.)
(a) x t 3 2 t, y t 2 t
(b) x t 3 1, y 2 t 2
(c) x sin 3 t, y sin 4 t
(d) x t sin 2 t, y t sin 3 t
(e) x sin t sin t , y cos t cos t
(f) x cos t, y sin t sin 5t
I 1t 1t y 2x
x 0 (c) y II x 0 III
x y y
2 2 2t 1 III 1 2t IV y y 2x 0 x
x 0 (d) x IV y y 0
2t VI y 2 2 2 V y x 2t 0 2x ; 29. Graph the curve x y 3y 3 y 5. x 5E11(pp 686695) 694 ❙❙❙❙ 1/18/06 9:33 AM Page 694 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES x 5 and x y y 1 2 and ﬁnd their
points of intersection correct to one decimal place. ; 30. Graph the curves y 31. (a) Show that the parametric equations x x1 x2 x1 t y ﬁgure, using the angle as the parameter. Then eliminate the
parameter and identify the curve.
40. If a and b are ﬁxed numbers, ﬁnd parametric equations for the y2 y1 y1 t where 0 t 1, describe the line segment that joins the
points P1 x 1, y1 and P2 x 2 , y 2 .
(b) Find parametric equations to represent the line segment
from 2, 7 to 3, 1 . curve that consists of all possible positions of the point P in
the ﬁgure, using the angle as the parameter. The line segment
AB is tangent to the larger circle.
y A ; 32. Use a graphing device and the result of Exercise 31(a) to draw
a the triangle with vertices A 1, 1 , B 4, 2 , and C 1, 5 . P b
¨ 33. Find parametric equations for the path of a particle that moves O along the circle x 2
y 1 2 4 in the manner described.
(a) Once around clockwise, starting at 2, 1
(b) Three times around counterclockwise, starting at 2, 1
(c) Halfway around counterclockwise, starting at 0, 3 x B ; 34. Graph the semicircle traced by the particle in Exercise 33(c).
; 35. (a) Find parametric equations for the ellipse 41. A curve, called a witch of Maria Agnesi, consists of all pos x 2 a 2 y 2 b 2 1. [Hint: Modify the equations of
a circle in Example 2.]
(b) Use these parametric equations to graph the ellipse when
a 3 and b 1, 2, 4, and 8.
(c) How does the shape of the ellipse change as b varies?
36. Find three different sets of parametric equations to represent the curve y 3 x ,x . 37. Derive Equations 1 for the case 2 sible positions of the point P in the ﬁgure. Show that parametric
equations for this curve can be written as
x 2 a cot 2 a sin 2 y Sketch the curve.
y C y=2a . 38. Let P be a point at a distance d from the center of a circle of A radius r. The curve traced out by P as the circle rolls along a
straight line is called a trochoid. (Think of the motion of a
point on a spoke of a bicycle wheel.) The cycloid is the special
case of a trochoid with d r. Using the same parameter as
for the cycloid and assuming the line is the xaxis and
0
when P is at one of its lowest points, show that parametric
equations of the trochoid are
x r d sin y Sketch the trochoid for the cases d r ¨
O x 42. Find parametric equations for the curve that consists of all pos d cos r and d r. 39. If a and b are ﬁxed numbers, ﬁnd parametric equations for the curve that consists of all possible positions of the point P in the
y sible positions of the point P in the ﬁgure, where OP
AB .
Sketch the curve. (This curve is called the cissoid of Diocles
after the Greek scholar Diocles, who introduced the cissoid as
a graphical method for constructing the edge of a cube whose
volume is twice that of a given cube.)
y A
a B
x=2a b
¨
O P a P P
x O a x 5E11(pp 686695) 1/18/06 9:33 AM Page 695 L ABORATORY PROJECT RUNNING CIRCLES AROUND CIRCLES ; 43. Suppose that the position of one particle at time t is given by
x1 3 sin t y1 0 2 cos t t 2 and the position of a second particle is given by
x2 3 cos t y2 1 sin t 0 t 2 (a) Graph the paths of both particles. How many points of
intersection are there?
(b) Are any of these points of intersection collision points ?
In other words, are the particles ever at the same place at
the same time? If so, ﬁnd the collision points.
(c) Describe what happens if the path of the second particle is
given by
x2 3 cos t y2 1 sin t 0 t 2 44. If a projectile is ﬁred with an initial velocity of v 0 meters per second at an angle above the horizontal and air resistance is
assumed to be negligible, then its position after t seconds is
given by the parametric equations
x v 0 cos y t v 0 sin t 1
2 tt 2 ; ❙❙❙❙ 695 (b) Use a graphing device to check your answers to part (a).
Then graph the path of the projectile for several other
values of the angle to see where it hits the ground. Summarize your ﬁndings.
(c) Show that the path is parabolic by eliminating the
parameter. ; 45. Investigate the family of curves deﬁned by the parametric equations x t 2, y t 3 ct. How does the shape change as c
increases? Illustrate by graphing several members of the family. ; 46. The swallowtail catastrophe curves are deﬁned by the parametric equations x 2 ct 4 t 3, y
c t 2 3 t 4. Graph several of these curves. What features do the curves have in
common? How do they change when c increases? ; 47. The curves with equations x a sin n t, y b cos t are called
Lissajous ﬁgures. Investigate how these curves vary when a,
b, and n vary. (Take n to be a positive integer.) ; 48. Investigate the family of curves deﬁned by the parametric
equations 2 where t is the acceleration due to gravity (9.8 m s ).
(a) If a gun is ﬁred with
30 and v 0 500 m s, when
will the bullet hit the ground? How far from the gun will it
hit the ground? What is the maximum height reached by
the bullet? x sin t c sin t y cos t c sin t How does the shape change as c changes? In particular, you
should identify the transitional values of c for which the basic
shape of the curve changes. L ABORATORY PROJECT
; Running Circles around Circles
In this project we investigate families of curves, called hypocycloids and epicycloids, that are
generated by the motion of a point on a circle that rolls inside or outside another circle.
y 1. A hypocycloid is a curve traced out by a ﬁxed point P on a circle C of radius b as C rolls on the inside of a circle with center O and radius a. Show that if the initial position of P is
a, 0 and the parameter is chosen as in the ﬁgure, then parametric equations of the
hypocycloid are C
b
¨ a
O P (a, 0) A x a b cos b cos x a b
b y a b sin b sin a b
b 2. Use a graphing device (or the interactive graphic in TEC Module 11.1B) to draw the graphs of hypocycloids with a a positive integer and b 1. How does the value of a affect the
graph? Show that if we take a 4, then the parametric equations of the hypocycloid reduce
to
x
Look at Module 11.1B to see how
hypocycloids and epicycloids are formed
by the motion of rolling circles. 4 cos 3 y 4 sin 3 This curve is called a hypocycloid of four cusps, or an astroid.
3. Now try b 1 and a n d, a fraction where n and d have no common factor. First let
n 1 and try to determine graphically the effect of the denominator d on the shape of the
graph. Then let n vary while keeping d constant. What happens when n d 1? 5E11(pp 696705) 696 ❙❙❙❙ 1/18/06 9:54 AM Page 696 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 4. What happens if b 1 and a is irrational? Experiment with an irrational number like s2 or
e 2. Take larger and larger values for and speculate on what would happen if we were to
graph the hypocycloid for all real values of . 5. If the circle C rolls on the outside of the ﬁxed circle, the curve traced out by P is called an epicycloid. Find parametric equations for the epicycloid.
6. Investigate the possible shapes for epicycloids. Use methods similar to Problems 2–4.  11.2 Calculus with Parametric Curves
Having seen how to represent curves by parametric equations, we now apply the methods
of calculus to these parametric curves. In particular, we solve problems involving tangents,
area, arc length, and surface area. Tangents
In the preceding section we saw that some curves deﬁned by parametric equations x f t
and y t t can also be expressed, by eliminating the parameter, in the form y F x .
(See Exercise 67 for general conditions under which this is possible.) If we substitute
x f t and y t t in the equation y F x , we get
tt Fft and so, if t, F , and f are differentiable, the Chain Rule gives
tt
If f t F ft f t F xf t 0, we can solve for F x :
Fx 1 tt
ft Since the slope of the tangent to the curve y F x at x, F x is F x , Equation 1
enables us to ﬁnd tangents to parametric curves without having to eliminate the parameter.
Using Leibniz notation, we can rewrite Equation 1 in an easily remembered form:  If we think of a parametric curve as being
traced out by a moving particle, then dy dt and
dx dt are the vertical and horizontal velocities
of the particle and Formula 2 says that the slope
of the tangent is the ratio of these velocities. 2 dy
dx dy
dt
dx
dt if dx
dt 0 It can be seen from Equation 2 that the curve has a horizontal tangent when dy dt 0
(provided that dx dt 0 ) and it has a vertical tangent when dx dt 0 (provided that
dy dt 0). This information is useful for sketching parametric curves. 5E11(pp 696705) 1/18/06 9:55 AM Page 697 S ECTION 11.2 CALCULUS WITH PARAMETRIC CURVES ❙❙❙❙ 697 As we know from Chapter 4, it is also useful to consider d 2 y d x 2. This can be found
by replacing y by dy dx in Equation 2:  2 dy
Note that
dx 2 d 2y
dt 2
d 2x
dt 2 d2y
dx 2 d
dx d
dt dy
dx dy
dx
dx
dt EXAMPLE 1 A curve C is deﬁned by the parametric equations x t 2, y t 3 3t.
(a) Show that C has two tangents at the point (3, 0) and ﬁnd their equations.
(b) Find the points on C where the tangent is horizontal or vertical.
(c) Determine where the curve is concave upward or downward.
(d) Sketch the curve.
SOLUTION (a) Notice that y t 3 3 t t t 2 3
0 when t 0 or t
point 3, 0 on C arises from two values of the parameter, t
This indicates that C crosses itself at 3, 0 . Since
dy
dx dy dt
dx dt 3t 2 3
2t 3
2 t y
y=œ„ (x3)
3
t=_1
(1, 2)
(3, 0)
0 s3 x 2 dy
dx 2 t=1
(1, _2) FIGURE 1 and y s3 x s3 , so the 3 (b) C has a horizontal tangent when dy dx 0, that is, when dy dt 0 and dx dt 0.
Since dy dt 3t 2 3, this happens when t 2 1, that is, t
1. The corresponding points on C are 1, 2 and (1, 2). C has a vertical tangent when dx dt 2 t 0,
that is, t 0. (Note that dy dt
0 there.) The corresponding point on C is (0, 0).
(c) To determine concavity we calculate the second derivative: x y=_ œ3 (x3)
„ 3 1
t
6 (2 s3 ) the slope of the tangent when t
s3 is d y d x
equations of the tangents at 3, 0 are
y s3. Therefore, the
s3 and t
s3. d
dt dy
dx
dx
dt 3
2 1
t2 1 3 t2 1
4t 3 2t Thus, the curve is concave upward when t 0 and concave downward when t
(d) Using the information from parts (b) and (c), we sketch C in Figure 1.
EXAMPLE 2 (a) Find the tangent to the cycloid x r
sin , y r 1 cos
where
3. (See Example 6 in Section 11.1.)
(b) At what points is the tangent horizontal? When is it vertical?
SOLUTION (a) The slope of the tangent line is
dy
dx dy d
dx d r1 r sin
cos 1 sin
cos at the point 0. 5E11(pp 696705) 698 ❙❙❙❙ 1/18/06 9:55 AM Page 698 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES When 3, we have
x r sin 3 r 3 dy
dx and s3
2 3 y sin 3
1 cos 3 r1 s3 2
11
2 cos r
2 3 s3 Therefore, the slope of the tangent is s3 and its equation is y r
2 r
3 s3 x r s3
2 or s3 x y r s3 2 The tangent is sketched in Figure 2.
y (_πr, 2r) (πr, 2r) (3πr, 2r) (5πr, 2r) π ¨= 3
0 FIGURE 2 2πr 4πr x (b) The tangent is horizontal when dy d x 0, which occurs when sin
0 and
1 cos
0, that is,
2 n 1 , n an integer. The corresponding point on the
cycloid is 2 n 1 r, 2r .
When
2 n , both d x d and d y d are 0. It appears from the graph that there are
vertical tangents at those points. We can verify this by using l’Hospital’s Rule as follows:
lim
l 2n dy
dx lim
l2n 1 sin
cos A similar computation shows that d y d x l
cal tangents when
2 n , that is, when x lim
l2n cos
sin as l 2 n
2 n r. , so indeed there are verti Areas
We know that the area under a curve y F x from a to b is A xab F x d x, where
Fx
0. If the curve is given by parametric equations x f t , y t t and is traversed
once as t increases from to , then we can adapt the earlier formula by using the Substitution Rule for Deﬁnite Integrals as follows:
A or y y b a y dx t t f t dt if f y
,t t t f t dt is the leftmost endpoint EXAMPLE 3 Find the area under one arch of the cycloid x y r1 cos . (See Figure 3.) r sin , 5E11(pp 696705) 1/18/06 9:56 AM Page 699 S ECTION 11.2 CALCULUS WITH PARAMETRIC CURVES y SOLUTION One arch of the cycloid is given by 0 with y
0 r1 x 2πr cos y A and d x
2r FIGURE 3 r2 y 2 r2 y 2 r2 [ 3
2 cos r1 1
2
1
4 r 2( 3 2
2 ) 2 0 2 cos
2 sin r1 r2 y d 699 2 . Using the Substitution Rule
d , we have cos 2 cos [1 0 2 0 1 0  The result of Example 3 says that the area
under one arch of the cycloid is three times the
area of the rolling circle that generates the
cycloid (see Example 6 in Section 11.1). Galileo
guessed this result but it was ﬁrst proved by the
French mathematician Roberval and the Italian
mathematician Torricelli. y y dx 0 r1 ❙❙❙❙ 1 cos
1 d 2 cos cos 2 d
d cos 2 2
0 sin 2 3 r2 Arc Length
We already know how to ﬁnd the length L of a curve C given in the form y
a x b. Formula 9.1.3 says that if F is continuous, then y L 3 2 dy
dx b 1 a Fx, dx Suppose that C can also be described by the parametric equations x f t and y t t ,
t
, where dx dt f t
0. This means that C is traversed once, from left to
right, as t increases from to and f
a, f
b. Putting Formula 2 into Formula
3 and using the Substitution Rule, we obtain
L y dy
dx b 1 a 2 y dx dy dt
dx dt 1 2 dx
dt
dt y C Since dx dt 0, we have Pi _ 1 P™ Pi L 4 y P¡
Pn
P¸
0 FIGURE 4 x 2 dx
dt dy
dt 2 dt Even if C can’t be expressed in the form y F x , Formula 4 is still valid but we obtain
it by polygonal approximations. We divide the parameter interval , into n subintervals
of equal width t. If t0 , t1 , t2 , . . . , tn are the endpoints of these subintervals, then xi f ti
and yi t ti are the coordinates of points Pi xi , yi that lie on C and the polygon with vertices P0 , P1 , . . . , Pn approximates C (see Figure 4).
As in Section 9.1, we deﬁne the length L of C to be the limit of the lengths of these
approximating polygons as n l :
n L lim nl Pi 1 Pi
i1 The Mean Value Theorem, when applied to f on the interval ti 1, ti , gives a number ti* in
ti 1, ti such that
f ti
f ti 1
f ti* ti ti 1
If we let xi xi xi 1 and yi yi yi 1 , this equation becomes
xi f ti* t 5E11(pp 696705) 700 ❙❙❙❙ 1/18/06 9:57 AM Page 700 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES Similarly, when applied to t, the Mean Value Theorem gives a number ti** in ti 1, ti such
that
yi t ti** t
Therefore
Pi 1 Pi 2 xi s yi s f ti* t s f ti* 2 t ti** 2 t ti** 2 2 t 2 t and so
n L 5 s f ti* lim nl 2 t ti** 2 t i1 t t 2 but it is not
The sum in (5) resembles a Riemann sum for the function s f t 2
exactly a Riemann sum because ti* ti** in general. Nevertheless, if f and t are continuous, it can be shown that the limit in (5) is the same as if ti* and ti** were equal, namely, y L sf t 2 2 tt dt Thus, using Leibniz notation, we have the following result, which has the same form
as (4).
f t,
6 Theorem If a curve C is described by the parametric equations x
y tt,
t
, where f and t are continuous on , and C is traversed
exactly once as t increases from to , then the length of C is
2 dx
dt y L 2 dy
dt dt Notice that the formula in Theorem 6 is consistent with the general formulas L
dx 2
d y 2 of Section 9.1.
and ds 2 x ds EXAMPLE 4 If we use the representation of the unit circle given in Example 2 in
Section 11.1, x
then dx dt cos t sin t and dy dt
L y dx
dt 2 0 y 2 0 dt y sin t 0 t 2 cos t, so Theorem 6 gives
2 dy
dt 2 dt y 2 0 ssin 2 t cos 2 t dt 2 as expected. If, on the other hand, we use the representation given in Example 3 in Section 11.1,
x sin 2 t y cos 2 t 0 t 2 5E11(pp 696705) 1/18/06 9:58 AM Page 701 S ECTION 11.2 CALCULUS WITH PARAMETRIC CURVES then dx dt y 2 cos 2 t, d y dt
dx
dt 2 0 2 dy
dt ❙❙❙❙ 701 2 sin 2 t, and the integral in Theorem 6 gives
2 y dt 2 s4 cos 2 2t 0 4 sin 2 2t dt y 2 2 dt 0 4  Notice that the integral gives twice the arc length of the circle because as t increases
from 0 to 2 , the point sin 2 t, cos 2 t traverses the circle twice. In general, when ﬁnding
the length of a curve C from a parametric representation, we have to be careful to ensure
that C is traversed only once as t increases from to .
EXAMPLE 5 Find the length of one arch of the cycloid x r ,y sin r1 cos . S OLUTION From Example 3 we see that one arch is described by the parameter interval 2 . Since 0 dx
d r1 cos dy
d and r sin we have
L y 0  The result of Example 5 says that the length
of one arch of a cycloid is eight times the radius
of the generating circle (see Figure 5). This was
ﬁrst proved in 1658 by Sir Christopher Wren, who
later became the architect of St. Paul’s Cathedral
in London.
y L=8r y sr 2 1 s2 1
2πr x dy
d 2 y d
cos 2 2 cos 2 sr 2 1 0 sin 2 2 cos
ry d 2 0 r 2 sin 2 d s2 1 cos To evaluate this integral we use the identity sin 2x 1 1 cos 2 x with
2
gives 1 cos
2 sin 2 2 . Since 0
2 , we have 0
2
sin 2
0. Therefore r
0 2 0 2 dx
d 2 and so s4 sin 2 cos
L 2r y 2 0 2r 2 FIGURE 5 2 sin 2d 2 2 sin
2r 2 2 sin 2 cos 2 d 2 x, which
and so
2 2 8r 0 Surface Area
In the same way as for arc length, we can adapt Formula 9.2.5 to obtain a formula for
surface area. If the curve given by the parametric equations x f t , y t t ,
,
t
is rotated about the xaxis, where f , t are continuous and t t
0, then the area of the
resulting surface is given by
7 S y dx
dt 2y 2 dy
dt The general symbolic formulas S x 2 y ds and S
9.2.8) are still valid, but for parametric curves we use
ds dx
dt 2 dy
dt 2 dt x2
2 dt x ds (Formulas 9.2.7 and 5E11(pp 696705) ❙❙❙❙ 702 1/18/06 9:59 AM Page 702 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES EXAMPLE 6 Show that the surface area of a sphere of radius r is 4 r 2.
SOLUTION The sphere is obtained by rotating the semicircle x r cos t y r sin t 0 t about the xaxis. Therefore, from Formula 7, we get y 2 r sin t s 2 y r sin t sr 2 sin 2 t 2 S y r sin t r dt 0 0 0 2 r2  11.2
1–2 1. x 3 t, t ■ ■ 3–6 ■ y 2 ■ t4 st e, 6. x 1, 2t 2 5. x cos ■ ■ 7–8 2. x 5t
■ ■ ■ te ,
■ y ■ t e ■ ■ 17. x t3 y 1, t; t 13
3 y y 4 r2 0 2 t ln t ; t ■ ■ y t tan , y 1 2; ; 9–10 ■ y
y ■ ■ ■ ■ ■ ■ ■ ■ 10. x
■ sin t, y
■ ■ ■ ■ 23. x
■ ■ ■ ■ y t2 13. x t e t, y t ■ ■ ■ 15. x 2 sin t, y 3 cos t, 16. x cos 2 t , y cos t , ■ ■ ■ ■ t3
e ■ 12. x t3 t 14. x t 0
0
■ ■ ■ ■ ln t. Then use calculus to ﬁnd t t ■ ■ ■ 2t 3 4 2 t 2, 3 t3 y ■ t
■ 4t
■ 2 t 8t ,
■ y
■ 2t 2 ■ t
■ ■ ■ ■ ■ ■ 25. Show that the curve x cos t, y sin t cos t has two tangents
at 0, 0 and ﬁnd their equations. Sketch the curve.
1 2 cos 2t,
y
tan t 1 2 cos 2t cross itself? Find the equations of
both tangents at that point. 12 t,
ln t, t2 y
y t 27. (a) Find the slope of the tangent line to the trochoid xr
d sin , y r d cos in terms of . (See Exercise 38 in Section 11.1.)
(b) Show that if d r, then the trochoid does not have a
vertical tangent. 1
ln t 2
■ a cos 3 ,
y a sin 3 in terms of . (Astroids are explored in the
Laboratory Project on page 695.) 28. (a) Find the slope of the tangent to the astroid x t
■ t4 26. At what point does the curve x 0, 0 2
2
 Find dy d x and d y d x . For which values of t is the
curve concave upward? t 2, ■ ■ 11–16 4 ■ most point on the curve x te t, y te t. Then ﬁnd the exact
coordinates. What are the asymptotes of this curve? (s3, 1)
■ 11. x ■  Graph the curve in a viewing rectangle that displays all
the important aspects of the curve. ■ sin t ; sin t ■ ; 23–24 ■ 2 sin t ; ■ 1 ; 22. Try to estimate the coordinates of the highest point and the left Find an equation of the tangent(s) to the curve at the given
point. Then graph the curve and the tangent(s).
2 sin 2 t, y y the curve x t 4 t 2, y
the exact coordinates. ■  9. x 3t 2 2 sin 0 (1, s2 ) ■ 2t 3 sin 2 cos 3 , 12 t, ; 21. Use a graph to estimate the coordinates of the leftmost point on
cos 2 ; 1, 1 sec ;
■ 12 t 2 cos , 3t 24. x ■ 2 2t ■ 1 ■ 3 t3 y 20. x 3 sin ■ t 2, 10 19. x 1 t; t t sin 2 , e t, y 8. x 0 18. x Find an equation of the tangent to the curve at the given
point by two methods: (a) without eliminating the parameter and
(b) by ﬁrst eliminating the parameter. ■ 2 r 2 y sin t dt ■  7. x cos 2 t dt  Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to
check your work. t Find an equation of the tangent to the curve at the point
corresponding to the given value of the parameter. 4. x r cos t 2 dt 17–20
t  3. x 2 Exercises Find dy dx.  cos t r sin t ■ ■ ■ ■ 5E11(pp 696705) 1/18/06 10:01 AM Page 703 S ECTION 11.2 CALCULUS WITH PARAMETRIC CURVES 48. Find the length of the loop of the curve x (b) At what points is the tangent horizontal or vertical?
(c) At what points does the tangent have slope 1 or 1?
t3
parallel to the line with equations x 3t 2 30. Find equations of the tangents to the curve x y 2t 3 49. Use Simpson’s Rule with n 4t, y 6 t 2 is the tangent
7t, y 12 t 5? 29. At what points on the curve x curve x parametric equations x 2a cot , y 2a sin 2 for the curve
called the witch of Maria Agnesi. Use Simpson’s Rule with
n 4 to estimate the length of the arc of this curve given by
4
2. 1 that pass through the point 4, 3 .
a cos ,
2 , to ﬁnd the area that it encloses. b sin , 0 32. Find the area bounded by the curve x and the line y 1 t, y t t 0  Find the distance traveled by a particle with position
x, y as t varies in the given time interval. Compare with the
length of the curve. e t, cos t, y
1 and x 0. 2, and the lines y t 51–52 1t 2.5. 33. Find the area bounded by the curve x sin 2 t, 51. x 2 52. x a cos 3 ,
y a sin . (Astroids are explored in the Laboratory Project
on page 695.) 34. Find the area of the region enclosed by the astroid x
3 ■ cos t,
■ y b cos , a e t, t cos t, 40. x
■ ln t,
■ b ■ ■ ■ a sin , 0, is
4a y 2 e 2 sin 2 d s1 0 C AS c a, where a cos 3 , y a sin 3 , 0. 55. (a) Graph the epitrochoid with equations 1 t 2 x 11 cos t 4 cos 11t 2 3 t 3 y 11 sin t 4 sin 11t 2 t , y y t st ■ ■ t 2, y 39. x ■ 4 32
3 y 1 ■ 54. Find the total length of the astroid x where a 38. x 4 ■ where e is the eccentricity of the ellipse (e
c sa 2 b 2 ) . Set up, but do not evaluate, an integral that represents the
length of the curve.
t 2, 0 ■ L  t t ■ 3 r. be the region enclosed by the loop of the curve in
Example 1.
(a) Find the area of .
(b) If is rotated about the xaxis, ﬁnd the volume of the
resulting solid.
(c) Find the centroid of . 37. x t cos t, y ■ 36. Let 37–40 cos 2 t, 0 y 53. Show that the total length of the ellipse x 35. Find the area under one arch of the trochoid of Exercise 38 in Section 11.1 for the case d 3 t 2. 6 to estimate the length of the
6 t 6. e t, t t 3, y 703 50. In Exercise 41 in Section 11.1 you were asked to derive the 1, 31. Use the parametric equations of an ellipse, x y e t, y t 3t ❙❙❙❙ 1, 1 ■ 0 sin t, ■ t ■ t 2 What parameter interval gives the complete curve?
(b) Use your CAS to ﬁnd the approximate length of this curve. 5
■ ■ ■ ■ ■ ■ CAS 56. A curve called Cornu’s spiral is deﬁned by the parametric equations
41–44  Find the length of the curve.
2 41. x 1 42. x a cos 3t , y 3 2t , 4
sin , x 0 y t Ct y y St y 1 a sin cos , t
t et e t, ■ ; 45–47 ■  46. x e
■ t, 0 5 2 t, 0 y ■ ■ t
t
■ 3
■ ■ ■ ■ t e t sin t, 0 ln(tan t),
1
2 t,
■ y y
■ y ■ t2 4e ,
■ 8
■ t
■ 4 u 2 2 du  Set up, but do not evaluate, an integral that represents the
area of the surface obtained by rotating the given curve about the
xaxis. t 3 4 57. x ■ ■ ■ ■ ■ t 2, t
2 58. x 3
■ sin u 2 2 du 57–58 t sin t, cos where C and S are the Fresnel functions that were introduced in
Chapter 5.
(a) Graph this curve. What happens as t l and as t l
?
(b) Find the length of Cornu’s spiral from the origin to the
point with parameter value t. 2 Graph the curve and ﬁnd its length. cos t 47. x ln 1 ■ e t cos t, 45. x ■ y 1 44. x
■ , t 0 0
43. x t 0 sin t,
■ ■ y
■ 4 32
3 , 1 t sin 3 t, y 0 t t ■ ■ 2
3
■
■ ■ ■ ■ ■ 5E11(pp 696705) ❙❙❙❙ 704 1/18/06 10:01 AM Page 704 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 59–61  Find the area of the surface obtained by rotating the given
curve about the xaxis. t 3, 59. x t 2, y
3 0 t
2 1 60. x 3t t, y 3t , 61. x a cos 3 , y a sin 3 , 0 ■ ■ ■ ■ 0 ■ t (b) By regarding a curve y f x as the parametric curve
x x, y f x , with parameter x, show that the formula
in part (a) becomes 1 1
2 ■ ■ ■ ■ ■ ■ d 2 y dx 2
dy dx 2 32 y ■ ; 62. Graph the curve
x 2 cos y cos 2 2 sin P sin 2 If this curve is rotated about the xaxis, ﬁnd the area of the
resulting surface. (Use your graph to help ﬁnd the correct
parameter interval.) ˙
0 63. If the curve x t3 t y 70. (a) Use the formula in Exercise 69(b) to ﬁnd the curvature of 1
t2 t 1 t the parabola y x 2 at the point 1, 1 .
(b) At what point does this parabola have maximum curvature? 2 is rotated about the xaxis, use your calculator to estimate the
area of the resulting surface to three decimal places. 71. Use the formula in Exercise 69(a) to ﬁnd the curvature of the cycloid x
arches. 64. If the arc of the curve in Exercise 50 is rotated about the xaxis, estimate the area of the resulting surface using
Simpson’s Rule with n 4. 66. x et ■ ■ t,
■ 2 t 3, y
y
■ 0 t
0 73. A string is wound around a circle and then unwound while 5 4e t 2, t ■ ■ cos at the top of one of its 1 is
0.
(b) Show that the curvature at each point of a circle of radius r
is
1 r.  Find the surface area generated by rotating the given
curve about the yaxis. 3 t 2, sin , y 72. (a) Show that the curvature at each point of a straight line 65–66 65. x x 1
■ ■ 67. If f is continuous and f t 0 for a
parametric curve x f t , y t t , a
the form y F x . [Hint: Show that f ■ ■ ■ ■ t b, show that the
t b, can be put in
1
exists.] 68. Use Formula 2 to derive Formula 7 from Formula 9.2.5 for the being held taut. The curve traced by the point P at the end of
the string is called the involute of the circle. If the circle has
radius r and center O and the initial position of P is r, 0 , and
if the parameter is chosen as in the ﬁgure, show that parametric equations of the involute are
x r cos sin y T
r 69. The curvature at a point P of a curve is deﬁned as where is the angle of inclination of the tangent line at P,
as shown in the ﬁgure. Thus, the curvature is the absolute value
of the rate of change of with respect to arc length. It can be
regarded as a measure of the rate of change of direction of the
curve at P and will be studied in greater detail in Chapter 14.
(a) For a parametric curve x x t , y y t , derive the
formula
xy
xy
x2 y2 3 2
where the dots indicate derivatives with respect to t, so
x d x dt. [Hint: Use
tan 1 d y d x and Equation 2
to ﬁnd d dt. Then use the Chain Rule to ﬁnd d ds.] cos y case in which the curve can be represented in the form
y F x , a x b. d
ds r sin ¨ O P
x 74. A cow is tied to a silo with radius r by a rope just long enough to reach the opposite side of the silo. Find the area available for
grazing by the cow. 5E11(pp 696705) 1/18/06 10:02 AM Page 705 S ECTION 11.3 POLAR COORDINATES ❙❙❙❙ 705 LABORATORY PROJECT
; Bézier Curves
The Bézier curves are used in computeraided design and are named after the French mathematician Pierre Bézier (1910–1999), who worked in the automotive industry. A cubic Bézier curve
is determined by four control points, P0 x 0 , y0 , P1 x 1, y1 , P2 x 2 , y2 , and P3 x 3 , y3 , and is
deﬁned by the parametric equations
x x0 1 t 3 3x1 t 1 t 2 3x 2 t 2 1 t x3t 3 y y0 1 t 3 3y1 t 1 t 2 3y2 t 2 1 t y3 t 3 where 0 t 1. Notice that when t 0 we have x, y
x, y
x 3 , y3 , so the curve starts at P0 and ends at P3 . x 0 , y0 and when t 1 we have 1. Graph the Bézier curve with control points P0 4, 1 , P1 28, 48 , P2 50, 42 , and P3 40, 5 . Then, on the same screen, graph the line segments P0 P1, P1 P2, and P2 P3. (Exercise 31 in
Section 11.1 shows how to do this.) Notice that the middle control points P1 and P2 don’t lie
on the curve; the curve starts at P0, heads toward P1 and P2 without reaching them, and ends
at P3.
2. From the graph in Problem 1 it appears that the tangent at P0 passes through P1 and the tangent at P3 passes through P2. Prove it.
3. Try to produce a Bézier curve with a loop by changing the second control point in Problem 1.
4. Some laser printers use Bézier curves to represent letters and other symbols. Experiment with control points until you ﬁnd a Bézier curve that gives a reasonable representation of the
letter C.
5. More complicated shapes can be represented by piecing together two or more Bézier curves. Suppose the ﬁrst Bézier curve has control points P0 , P1, P2 , P3 and the second one has control points P3 , P4 , P5 , P6. If we want these two pieces to join together smoothly, then the
tangents at P3 should match and so the points P2, P3, and P4 all have to lie on this common
tangent line. Using this principle, ﬁnd control points for a pair of Bézier curves that represent the letter S.  11.3 Polar Coordinates P (r, ¨ )
r O ¨ FIGURE 1 polar axis x A coordinate system represents a point in the plane by an ordered pair of numbers called
coordinates. Usually we use Cartesian coordinates, which are directed distances from two
perpendicular axes. Here we describe a coordinate system introduced by Newton, called
the polar coordinate system, which is more convenient for many purposes.
We choose a point in the plane that is called the pole (or origin) and is labeled O. Then
we draw a ray (halfline) starting at O called the polar axis. This axis is usually drawn horizontally to the right and corresponds to the positive xaxis in Cartesian coordinates.
If P is any other point in the plane, let r be the distance from O to P and let be the
angle (usually measured in radians) between the polar axis and the line OP as in Figure 1.
Then the point P is represented by the ordered pair r, and r, are called polar coordinates of P. We use the convention that an angle is positive if measured in the counterclockwise direction from the polar axis and negative in the clockwise direction. If P O,
then r 0 and we agree that 0, represents the pole for any value of . 5E11(pp 706715) 706 ❙❙❙❙ 1/18/06 9:48 AM Page 706 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES (r, ¨ ) ¨+π We extend the meaning of polar coordinates r, to the case in which r is negative by
agreeing that, as in Figure 2, the points r, and r, lie on the same line through O
and at the same distance r from O, but on opposite sides of O. If r 0, the point r,
lies in the same quadrant as ; if r 0, it lies in the quadrant on the opposite side of the
pole. Notice that r, represents the same point as r,
. ¨
O EXAMPLE 1 Plot the points whose polar coordinates are given. (_r, ¨) (a) 1, 5 FIGURE 2 4 (b) 2, 3 (c) 2, 2 3 (d) 3, 3 4 SOLUTION The points are plotted in Figure 3. In part (d) the point
3, 3 4 is located
three units from the pole in the fourth quadrant because the angle 3 4 is in the second
quadrant and r
3 is negative.
5π
4 3π
O (2, 3π) 3π
4 O
O O _ 2π
3 5π ”1, ’
4
2π
”2, _ ’
3 FIGURE 3 ”_3, 3π ’
4 In the Cartesian coordinate system every point has only one representation, but in the
polar coordinate system each point has many representations. For instance, the point
1, 5 4 in Example 1(a) could be written as 1, 3 4 or 1, 13 4 or
1, 4 .
(See Figure 4.)
5π
4 O π
4 13π
4 O O _ 3π
4 ”1, 5π ’
4 3π
”1, _ 4 ’ O 13π
”1, 4 ’ π ”_1, ’
4 FIGURE 4 In fact, since a complete counterclockwise rotation is given by an angle 2 , the point
represented by polar coordinates r, is also represented by
r,
y
P (r, ¨ )=P (x, y) r y ¨
O FIGURE 5 x x 2n and r, 2n 1 where n is any integer.
The connection between polar and Cartesian coordinates can be seen from Figure 5, in
which the pole corresponds to the origin and the polar axis coincides with the positive
xaxis. If the point P has Cartesian coordinates x, y and polar coordinates r, , then,
from the ﬁgure, we have
x
y
cos
sin
r
r
and so
1 x r cos y r sin Although Equations 1 were deduced from Figure 5, which illustrates the case where
r 0 and 0
2, these equations are valid for all values of r and . (See the general deﬁnition of sin and cos in Appendix D.) 5E11(pp 706715) 1/18/06 9:48 AM Page 707 SECTION 11.3 POLAR COORDINATES ❙❙❙❙ 707 Equations 1 allow us to ﬁnd the Cartesian coordinates of a point when the polar coordinates are known. To ﬁnd r and when x and y are known, we use the equations r2 2 x2 y2 y
x tan which can be deduced from Equations 1 or simply read from Figure 5.
EXAMPLE 2 Convert the point 2,
SOLUTION Since r 3 from polar to Cartesian coordinates. 2 and 3, Equations 1 give x r cos 2 cos y r sin 2 sin 3
3 2
2 1
2 1 s3
2 s3 Therefore, the point is (1, s3 ) in Cartesian coordinates.
EXAMPLE 3 Represent the point with Cartesian coordinates 1,
coordinates. 1 in terms of polar SOLUTION If we choose r to be positive, then Equations 2 give r
tan sx 2 y2 y
x s1 2 1 1 2 s2 Since the point 1, 1 lies in the fourth quadrant, we can choose
7 4. Thus, one possible answer is (s2,
4); another is s2, 7 4 or
4. NOTE Equations 2 do not uniquely determine when x and y are given because, as
increases through the interval 0
2 , each value of tan occurs twice. Therefore, in
converting from Cartesian to polar coordinates, it’s not good enough just to ﬁnd r and
that satisfy Equations 2. As in Example 3, we must choose so that the point r, lies in
the correct quadrant.
■ Polar Curves 1 r= 2 r=4 The graph of a polar equation r f , or more generally F r,
0, consists of all
points P that have at least one polar representation r, whose coordinates satisfy the
equation. r=2
r=1
x FIGURE 6 EXAMPLE 4 What curve is represented by the polar equation r 2? SOLUTION The curve consists of all points r,
with r 2. Since r represents the distance from the point to the pole, the curve r 2 represents the circle with center O and
radius 2. In general, the equation r a represents a circle with center O and radius a .
(See Figure 6.) 5E11(pp 706715) 708 ❙❙❙❙ 1/18/06 9:48 AM Page 708 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES EXAMPLE 5 Sketch the polar curve (3, 1) SOLUTION This curve consists of all points r,
such that the polar angle is 1 radian. It
is the straight line that passes through O and makes an angle of 1 radian with the polar
axis (see Figure 7). Notice that the points r, 1 on the line with r 0 are in the ﬁrst
quadrant, whereas those with r 0 are in the third quadrant. (2, 1) ¨=1
(1, 1)
O 1. 1
x EXAMPLE 6 (_1, 1) (a) Sketch the curve with polar equation r 2 cos .
(b) Find a Cartesian equation for this curve. (_2, 1) SOLUTION FIGURE 7 (a) In Figure 8 we ﬁnd the values of r for some convenient values of and plot the
corresponding points r, . Then we join these points to sketch the curve, which appears
to be a circle. We have used only values of between 0 and , since if we let increase
beyond , we obtain the same points again.
r
0 F IGURE 8 Table of values and
graph of r=2 cos ¨ 2 cos
2
s3
s2
1
0
1
s2
s3
2 6
4
3
2
23
34
56 ” œ2 , ’
„π
4 π
”1, ’
3 ”œ3 , ’
„π
6 (2, 0)
π
”0, ’
2 2π
”_1, ’
3 ”_ œ2 , ’
„ 3π
4 „6 ’
”_ œ3 , 5π (b) To convert the given equation into a Cartesian equation we use Equations 1 and 2.
From x r cos we have cos
x r, so the equation r 2 cos becomes r 2 x r,
which gives
2x r2 x2 y2 x2 or y2 2x Completing the square, we obtain
x 1 2 y2 1 which is an equation of a circle with center 1, 0 and radius 1.
 Figure 9 shows a geometrical illustration
that the circle in Example 6 has the equation
r 2 cos . The angle OPQ is a right angle
(Why?) and so r 2 cos . y P
r
¨ O FIGURE 9 2 Q x 0 5E11(pp 706715) 1/18/06 9:48 AM Page 709 S ECTION 11.3 POLAR COORDINATES r EXAMPLE 7 Sketch the curve r 2 π π
2 2π ¨ 3π
2 709 sin . SOLUTION Instead of plotting points as in Example 6, we ﬁrst sketch the graph of
r 1 sin in Cartesian coordinates in Figure 10 by shifting the sine curve up one
unit. This enables us to read at a glance the values of r that correspond to increasing
values of . For instance, we see that as increases from 0 to 2, r (the distance from O )
increases from 1 to 2, so we sketch the corresponding part of the polar curve in Figure
11(a). As increases from 2 to , Figure 10 shows that r decreases from 2 to 1, so we
sketch the next part of the curve as in Figure 11(b). As increases from to 3 2,
r decreases from 1 to 0 as shown in part (c). Finally, as increases from 3 2 to 2 ,
r increases from 0 to 1 as shown in part (d). If we let increase beyond 2 or decrease
beyond 0, we would simply retrace our path. Putting together the parts of the curve from
Figure 11(a)–(d), we sketch the complete curve in part (e). It is called a cardioid
because it’s shaped like a heart. 1
0 1 ❙❙❙❙ F IGURE 10 r=1+sin ¨ in Cartesian coordinates,
0¯¨¯2π π π ¨= 2 ¨= 2 2
O O
1 O ¨=0 ¨=π O ¨=π 3π O
3π ¨= 2 (a) ¨= 2 (c) (b) ¨=2π (d) (e) FIGURE 11 Stages in sketching the
cardioid r=1+sin ¨ EXAMPLE 8 Sketch the curve r Module 11.3 helps you see how polar
curves are traced out by showing animations similar to Figures 10–13. Tangents
to these polar curves can also be visualized as in Figure 15 (see page 712). cos 2 . SOLUTION As in Example 7, we ﬁrst sketch r
cos 2 , 0
2 , in Cartesian coordinates in Figure 12. As increases from 0 to 4, Figure 12 shows that r decreases from
1 to 0 and so we draw the corresponding portion of the polar curve in Figure 13 (indicated by !). As increases from 4 to 2, r goes from 0 to 1. This means that the
distance from O increases from 0 to 1, but instead of being in the ﬁrst quadrant this
portion of the polar curve (indicated by @) lies on the opposite side of the pole in the
third quadrant. The remainder of the curve is drawn in a similar fashion, with the arrows
and numbers indicating the order in which the portions are traced out. The resulting
curve has four loops and is called a fourleaved rose. r π ¨= 2 1 ¨= $ ! π
4 @ π
2 3π
4 # % π * 5π
4 3π
2 ^ 7π
4 π 3π
4 & ¨= 4 ^ $
2π ¨ ! % ⑧ ¨=π & ¨=0 @ # FIGURE 12 FIGURE 13 r=cos 2¨ in Cartesian coordinates Fourleaved rose r=cos 2¨ 5E11(pp 706715) 710 ❙❙❙❙ 1/18/06 9:49 AM Page 710 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES Symmetry
When we sketch polar curves it is sometimes helpful to take advantage of symmetry. The
following three rules are explained by Figure 14.
(a) If a polar equation is unchanged when is replaced by
, the curve is symmetric
about the polar axis.
(b) If the equation is unchanged when r is replaced by r, or when is replaced by
, the curve is symmetric about the pole. (This means that the curve remains
unchanged if we rotate it through 180° about the origin.)
(c) If the equation is unchanged when is replaced by
, the curve is symmetric
2.
about the vertical line
(r, π¨ ) (r, ¨ ) (r, ¨) π¨ (r, ¨ ) ¨
O ¨
O (_ r, ¨ ) _¨ O (r, _¨ ) (a) (b) (c) FIGURE 14 The curves sketched in Examples 6 and 8 are symmetric about the polar axis, since
cos
cos . The curves in Examples 7 and 8 are symmetric about
2 because
sin
sin and cos 2
cos 2 . The fourleaved rose is also symmetric
about the pole. These symmetry properties could have been used in sketching the curves.
2 and then
For instance, in Example 6 we need only have plotted points for 0
reﬂected about the polar axis to obtain the complete circle. Tangents to Polar Curves
To ﬁnd a tangent line to a polar curve r
metric equations as
x r cos f cos f we regard as a parameter and write its paray r sin f sin Then, using the method for ﬁnding slopes of parametric curves (Equation 11.2.2) and the
Product Rule, we have
dy
dr
sin
r cos
dy
d
d
3
dx
dx
dr
cos
r sin
d
d
0 (provided that
We locate horizontal tangents by ﬁnding the points where dy d
dx d
0 ). Likewise, we locate vertical tangents at the points where dx d
0 (pro0).
vided that dy d
Notice that if we are looking for tangent lines at the pole, then r 0 and Equation 3 simpliﬁes to
dy
dr
tan
if
0
dx
d 5E11(pp 706715) 1/18/06 9:50 AM Page 711 S ECTION 11.3 POLAR COORDINATES For instance, in Example 8 we found that r cos 2
0 when
means that the lines
4 and
3 4 (or y x and y
r cos 2 at the origin. ❙❙❙❙ 711 4 or 3 4. This
x) are tangent lines to EXAMPLE 9 (a) For the cardioid r 1 sin of Example 7, ﬁnd the slope of the tangent line
when
3.
(b) Find the points on the cardioid where the tangent line is horizontal or vertical.
SOLUTION Using Equation 3 with r dr
sin
d
dr
cos
d dy
dx 1 sin , we have r cos cos
cos r sin cos 1 2 sin
1 2 sin 2
sin sin
cos 1
1 3 1 cos
sin cos 1 2 sin
1 sin 1 2 sin (a) The slope of the tangent at the point where
dy
dx sin
sin 3 is cos 3 1 2 sin 3
sin 3 1 2 sin 3
s3 )
s3 2)(1 s3 ) 1
2 (1
1
1 (1 s3
s3 1 s3
s3 )(1 s3 ) (2 1 (b) Observe that
dy
d cos 1 2 sin dx
d 1 sin 1 when 2 sin 0 3
7
11
,
,
22
6
6 when 0 3
5
,,
266 , Therefore, there are horizontal tangents at the points 2, 2 , ( 1 , 7 6), ( 1 , 11 6) and
2
2
vertical tangents at ( 3 , 6) and ( 3 , 5 6). When
3 2, both d y d and d x d are
2
2
0, so we must be careful. Using l’Hospital’s Rule, we have
lim l3 2 dy
dx lim l3 2 1
3 l3 1
3 l3 lim lim 1
1 2 2 2 sin
2 sin 1 cos
sin sin
cos lim l3 2 1 cos
sin 5E11(pp 706715) ❙❙❙❙ 712 1/18/06 9:50 AM Page 712 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES π lim By symmetry, ”2, ’
2
3
”1+ œ„ , ’ π
23 m=_1 dy
dx 2 Thus, there is a vertical tangent line at the pole (see Figure 15).
NOTE Instead of having to remember Equation 3, we could employ the method used to
derive it. For instance, in Example 9 we could have written ”3 π , ’
26 3
” , 5π ’
26 l3 ■ (0, 0) x FIGURE 15 1 sin cos y 1
1
” , 7π ” , 11π ’ 2 ’
26
6 r cos
r sin 1 sin sin 1
2 cos
sin sin 2 sin 2 Then we have Tangent lines for r=1+sin ¨ dy
dx dy d
dx d cos
sin 2 sin cos
cos 2 cos
sin sin 2
cos 2 which is equivalent to our previous expression. Graphing Polar Curves with Graphing Devices
2 _1.4 1.4 Although it’s useful to be able to sketch simple polar curves by hand, we need to use a
graphing calculator or computer when we are faced with a curve as complicated as the one
shown in Figure 16.
Some graphing devices have commands that enable us to graph polar curves directly.
With other machines we need to convert to parametric equations ﬁrst. In this case we take
the polar equation r f
and write its parametric equations as _0.3 x FIGURE 16 r=sin ¨+sin#(5¨/2) r cos f cos y r sin f sin Some machines require that the parameter be called t rather than .
EXAMPLE 10 Graph the curve r sin 8 5 . SOLUTION Let’s assume that our graphing device doesn’t have a builtin polar graphing
command. In this case we need to work with the corresponding parametric equations,
which are
1 _1 1 _1 FIGURE 17 r=sin(8¨/5) x r cos sin 8 5 cos y r sin sin 8 5 sin In any case we need to determine the domain for . So we ask ourselves: How many
complete rotations are required until the curve starts to repeat itself? If the answer is
n, then
2n
8
8
8
16 n
sin
sin
sin
5
5
5
5
and so we require that 16 n 5 be an even multiple of . This will ﬁrst occur when
n 5. Therefore, we will graph the entire curve if we specify that 0
10 .
Switching from to t, we have the equations
x sin 8t 5 cos t y sin 8t 5 sin t 0 t 10 and Figure 17 shows the resulting curve. Notice that this rose has 16 loops. 5E11(pp 706715) 1/18/06 9:51 AM Page 713 S ECTION 11.3 POLAR COORDINATES ❙❙❙❙ 713 EXAMPLE 11 Investigate the family of polar curves given by r
1 c sin . How does
the shape change as c changes? (These curves are called limaçons, after a French word
for snail, because of the shape of the curves for certain values of c.)
SOLUTION Figure 18 shows computerdrawn graphs for various values of c. For c
1
there is a loop that decreases in size as c decreases. When c 1 the loop disappears and
the curve becomes the cardioid that we sketched in Example 7. For c between 1 and 1 the
2
cardioid’s cusp is smoothed out and becomes a “dimple.” When c decreases from 1 to 0,
2
the limaçon is shaped like an oval. This oval becomes more circular as c l 0, and when
c 0 the curve is just the circle r 1.  In Exercise 53 you are asked to prove analytically what we have discovered from the graphs
in Figure 18. c=1 c=1.7 c=0.7 c=0.5 c=0.2 c=2.5 c=_2
c=0 FIGURE 18 c=_ 0.8 Exercises 1–2  Plot the point whose polar coordinates are given. Then ﬁnd
two other pairs of polar coordinates of this point, one with r 0
and one with r 0. 1. (a) 1, ■ 3–4 2, (b) 2 2. (a) 3, 0
■ ■ ■ (c) 7 ■ ■ ■ 1,
■ 2
■ 6. (a)
■ 4. (a) 2, 2
■ ■ ■ ■ ■ 1, (c) (b) 4, 3 3
■ 4) (c)
■ ■ 5
■ The Cartesian coordinates of a point are given.
(i) Find polar coordinates r, of the point, where r
2.
0
(ii) Find polar coordinates r, of the point, where r
0
2. 6
■  5. (a) 1, 1 (b) (2 s3, 2) ■ ■ ■ r 8. r 0, 9. 0 r 4, r 5, 3 4 5 2, 3
■ ■ ■ ■ ■ 4 r 3, 3 7 3 4 3 4 2
3 2 3 2 5 6 ■ 12.
5–6 (b) ■ 11. 2 3 2,
■ ■ 7. 1 ■ Plot the point whose polar coordinates are given. Then ﬁnd
the Cartesian coordinates of the point.
(b) (2 s2, 3 s3 ) 1, 10. 2 ■  2 ( ■ 7–12  Sketch the region in the plane consisting of points whose
polar coordinates satisfy the given conditions. (c) 3, 2 4 (b) 2,
■ 3. (a) 3, c=_1 The remaining parts of Figure 18 show that as c becomes negative, the shapes change
in reverse order. In fact, these curves are reﬂections about the horizontal axis of the corresponding curves with positive c. Members of the family of
limaçons r=1+c sin ¨  11.3 c=_ 0.5 c=_ 0.2 ■ 1
■ r 1,
■ ■ ■ ■ ■ ■ ■ ■ ■ 0 and
13. Find the distance between the points with polar coordinates 0 and 1, 6 and 3, 3 4. 14. Find a formula for the distance between the points with polar coordinates r 1, 1 and r 2 , 2 . ■ 5E11(pp 706715) ❙❙❙❙ 714 15–20 1/18/06 9:52 AM Page 714 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES Identify the curve by ﬁnding a Cartesian equation for the  curve.
15. r 2 16. r cos 17. r 3 sin 18. r 19. r
■ ■ ■ ■ ■ ■ 4 2 sec (called a conchoid)
has the line x 2 as a vertical asymptote by showing that
lim r l x 2. Use this fact to help sketch the conchoid. 1 2 sin 20. r csc ■ 49. Show that the polar curve r 50. Show that the curve r 2 cos tan sec
■ ■ ■ ■ ■ 51. Show that the curve r sin tan (called a cissoid of
Diocles) has the line x 1 as a vertical asymptote. Show also
that the curve lies entirely within the vertical strip 0 x 1.
Use these facts to help sketch the cissoid. 21–26  Find a polar equation for the curve represented by the
given Cartesian equation. 21. x 22. x 2 3 23. x
25. x y
2 ■ y 2 24. x 2 ■ 26. x 2cx
■ ■ ■ ■ ■ y2 9 y
2 52. Sketch the curve x 2 9 y 2 ■ ■ ■ ■ 27. (a) A line through the origin that makes an angle of reasons for your choices. (Don’t use a graphing device.)
(a) r sin 2
(b) r sin 4
(c) r sec 3
(d) r
sin
(e) r 1 4 cos 5
(f) r 1 s 28. (a) A circle with radius 5 and center 2, 3 (b) A circle centered at the origin with radius 4 29–46 ■ ■ ■ ■ I ■ 30. r 2 6 31. r sin 33. r 21 35. r , 3r 32. r 2 2 34. r 1 0 36. r
38. r
40. r sin 5 1
■ 47–48 42. r 2
44. r ■ ■ ■ ■ 55–60  Find the slope of the tangent line to the given polar curve
at the point speciﬁed by the value of . 1
1 ■ 1 sin 2 2 46. r 2 cos 2
■ 3 cos 2 cos
■ ■ 2
■  48. r
2 1, ■ r 1
■ cos ,
■ ln , 60. r 6
■ 2 58. r ■ 56. r 6 ■ ■ 61. r
π ■ ■ ■ ■ 3 cos e sin 3 ,
■ 62. r 63. r 2π ¨ π _2
■ 3 ■ 6
■ ■ ■  Find the points on the given curve where the tangent line
is horizontal or vertical. 0 2π ¨ sin , 61–66 2 0 ■ 2 sin , 59. r 1 ■ 55. r
57. r ■ The ﬁgure shows the graph of r as a function of in
Cartesian coordinates. Use it to sketch the corresponding polar
curve. 47. VI 2 cos 3 2 cos 3 2 45. r V 0 ln , 4 cos 2 43. r ■ sin 2 cos 4 39. r III ■ 3 cos sin 2 37. r ■ II IV ■ Sketch the curve with the given polar equation.  29. 41. r ■ 4 x 2 y 2. 54. Match the polar equations with the graphs labeled I–VI. Give 6 with the positive xaxis
(b) A vertical line through the point 3, 3 ■ 3 r 1 c sin has an inner loop when c
1.
Prove that this is true, and ﬁnd the values of that
correspond to the inner loop.
(b) From Figure 18 it appears that the limaçon loses its dimple
when c 1 . Prove this.
2 ■  For each of the described curves, decide if the curve
would be more easily given by a polar equation or a Cartesian
equation. Then write an equation for the curve. ■ y2 53. (a) In Example 11 the graphs suggest that the limaçon 1 27–28 ■ 2 csc (also a conchoid) has the
1 as a horizontal asymptote by showing that
y
1. Use this fact to help sketch the conchoid. line y
lim r l 1 64. r 65. r
■ ■ ■ ■ ■ ■ cos 66. r cos 2
■ ■ ■ ■ ■ ■ cos sin e
2
■ sin 2
■ ■ ■ ■ 5E11(pp 706715) 1/18/06 9:52 AM Page 715 S ECTION 11.4 AREAS AND LENGTHS IN POLAR COORDINATES b cos , where
a sin
0, represents a circle, and ﬁnd its center and radius. 68. Show that the curves r a sin and r a cos intersect at ; 80. The astronomer Giovanni Cassini (1625–1712) studied the right angles.
69–74 family of curves with polar equations Use a graphing device to graph the polar curve. Choose
;
the parameter interval to make sure that you produce the entire
curve.
 69. r 1 2 sin 70. r s1 71. r sin e 2 0.8 sin sin 4 73. r 2 74. r
■ (butterﬂy curve) cos 4 2 cos
■ a4 3 ■ ■ ■ ■ ■ ■ ■ ■ tan 6 and
1 sin
r 1 sin
3 related to the graph of r 1 sin ?
In general, how is the graph of r f
related to the
graph of r f ? [Hint: Observe that r
dr d
in the ﬁgure.] ; 76. Use a graph to estimate the ycoordinate of the highest points r=f(¨ ) sin 2 . Then use calculus to ﬁnd the exact ÿ ; 77. (a) Investigate the family of curves deﬁned by the polar equa P tions r sin n , where n is a positive integer. How is the
number of loops related to n ?
(b) What happens if the equation in part (a) is replaced by
r
sin n ? ¨ 1 c sin n ,
where c is a real number and n is a positive integer. How does
the graph change as n increases? How does it change as c
changes? Illustrate by graphing enough members of the family
to support your conclusions. 82. (a) Use Exercise 81 to show that the angle between the tangent ; ; 79. A family of curves has polar equations
r 1
1 ˙ O ; 78. A family of curves is given by the equations r  11.4 0 f . If
is the angle between the tangent line at P and the radial line
OP, show that ; 75. How are the graphs of r on the curve r
value. c4 81. Let P be any point (except the origin) on the curve r 6 ■ 2c 2 r 2 cos 2 where a and c are positive real numbers. These curves are
called the ovals of Cassini even though they are oval shaped
only for certain values of a and c. (Cassini thought that these
curves might represent planetary orbits better than Kepler’s
ellipses.) Investigate the variety of shapes that these curves
may have. In particular, how are a and c related to each other
when the curve splits into two parts? (hippopede) 5 sin cos
■ r4 (nephroid of Freeth) 2 2 cos 4 2 72. r 715 Investigate how the graph changes as the number a changes. In
particular, you should identify the transitional values of a for
which the basic shape of the curve changes. 67. Show that the polar equation r ab ❙❙❙❙ a cos
a cos line and the radial line is
4 at every point on the
curve r e .
(b) Illustrate part (a) by graphing the curve and the tangent
lines at the points where
0 and
2.
(c) Prove that any polar curve r f
with the property that
the angle between the radial line and the tangent line is a
constant must be of the form r Ce k , where C and k are
constants. Areas and Lengths in Polar Coordinates
In this section we develop the formula for the area of a region whose boundary is given by
a polar equation. We need to use the formula for the area of a sector of a circle r
¨
FIGURE 1 1 A 1
2 r2 where, as in Figure 1, r is the radius and is the radian measure of the central angle.
Formula 1 follows from the fact that the area of a sector is proportional to its central angle:
A
2
r 2 1 r 2 . (See also Exercise 35 in Section 8.3.)
2 5E11(pp 716725) 716 ❙❙❙❙ 1/18/06 5:30 PM Page 716 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES r=f(¨) Let
be the region, illustrated in Figure 2, bounded by the polar curve r f
and by the rays
a and
b, where f is a positive continuous function and where
0 b a 2 . We divide the interval a, b into subintervals with endpoints 0 , 1 ,
. The rays
into n smaller regions with
2 , . . . , n and equal width
i then divide
*
central angle
i
i 1 . If we choose
i in the i th subinterval
i 1, i , then the
area Ai of the i th region is approximated by the area of the sector of a circle with central
angle
and radius f i* . (See Figure 3.)
Thus, from Formula 1 we have ¨=b
b
O ¨=a
a FIGURE 2
f(¨ i*) ¨=¨ i1 1
2 Ai ¨=¨ i 2 * f i and so an approximation to the total area A of is n ¨=b 1
2 A 2 2 * f i i1 Î¨
¨=a It appears from Figure 3 that the approximation in (2) improves as n l . But the sums
1
2
in (2) are Riemann sums for the function t
, so
2f O
FIGURE 3 n
1
2 lim nl f b1
2
a 2 * i y i1 2 f d It therefore appears plausible (and can in fact be proved) that the formula for the area A of
the polar region is
b1
2 f y y A 3 b1
2 a 2 d Formula 3 is often written as A 4 a r2 d with the understanding that r f . Note the similarity between Formulas 1 and 4.
When we apply Formula 3 or 4 it is helpful to think of the area as being swept out by
a rotating ray through O that starts with angle a and ends with angle b.
EXAMPLE 1 Find the area enclosed by one loop of the fourleaved rose r
r=cos 2¨ π ¨= 4 SOLUTION The curve r
cos 2 was sketched in Example 8 in Section 11.3. Notice from
Figure 4 that the region enclosed by the right loop is swept out by a ray that rotates from
4 to
4. Therefore, Formula 4 gives A π
¨=_ 4 FIGURE 4 cos 2 . y
y 0 1
2 [ 41
2
4
4 r2 d 1
2 y cos 2 2 d
1
4 sin 4 4
4 y 0 0 cos 2 2 d
41
2 4 8 1 cos 4 d 5E11(pp 716725) 1/18/06 5:31 PM Page 717 S ECTION 11.4 AREAS AND LENGTHS IN POLAR COORDINATES EXAMPLE 2 Find the area of the region that lies inside the circle r
the cardioid r 1 sin . π 5π ¨= 6
O r=1+sin ¨ 1
2 A F IGURE 5 y 5 6 2 3 sin 6 1
2 d y 5 6 1 6 sin Since the region is symmetric about the vertical axis
A 2 y
y
3
r=f(¨) 3 sin and outside 1
2 2
6
2
6 2 y 6 8 sin 2
3 1
2 9 sin 2 d
1 2 1 6 2 sin 4 cos 2 2 sin 2 y d 2, we can write
sin 2 2 sin d d 2 sin
2 cos 2 [because sin 2 d 1
2 1 cos 2 2
6 Example 2 illustrates the procedure for ﬁnding the area of the region bounded by two
polar curves. In general, let be a region, as illustrated in Figure 6, that is bounded by
curves with polar equations r f , r t ,
a, and
b, where f
t
0
and 0 b a 2 . The area A of is found by subtracting the area inside r t
from the area inside r f , so using Formula 3 we have ¨=b
¨=a 717 SOLUTION The cardioid (see Example 7 in Section 11.3) and the circle are sketched in
Figure 5 and the desired region is shaded. The values of a and b in Formula 4 are determined by ﬁnding the points of intersection of the two curves. They intersect when
1
3 sin
1 sin , which gives sin
6, 5 6. The desired area can be
2 , so
found by subtracting the area inside the cardioid between
6 and
5 6 from
the area inside the circle from 6 to 5 6. Thus r=3 sin ¨ ¨= 6 ❙❙❙❙ r=g(¨) A y b1
2 1
2 O y a 2 f d y b1
2 2 t a d FIGURE 6  CAUTION b a f 2 t 2 d The fact that a single point has many representations in polar coordinates
sometimes makes it difﬁcult to ﬁnd all the points of intersection of two polar curves. For
instance, it is obvious from Figure 5 that the circle and the cardioid have three points
of intersection; however, in Example 2 we solved the equations r 3 sin and
r 1 sin and found only two such points, ( 3, 6) and ( 3, 5 6). The origin is also
2
2
a point of intersection, but we can’t ﬁnd it by solving the equations of the curves because
the origin has no single representation in polar coordinates that satisﬁes both equations.
Notice that, when represented as 0, 0 or 0, , the origin satisﬁes r 3 sin and so it
lies on the circle; when represented as 0, 3 2 , it satisﬁes r 1 sin and so it lies on
the cardioid. Think of two points moving along the curves as the parameter value
increases from 0 to 2 . On one curve the origin is reached at
0 and
; on the
other curve it is reached at
3 2. The points don’t collide at the origin because they
reach the origin at different times, but the curves intersect there nonetheless.
Thus, to ﬁnd all points of intersection of two polar curves, it is recommended that you
draw the graphs of both curves. It is especially convenient to use a graphing calculator or
computer to help with this task.
■ 5E11(pp 716725) 718 ❙❙❙❙ 1/18/06 5:32 PM Page 718 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES E XAMPLE 3 Find all points of intersection of the curves r
1π ” , ’
23 1
r=2 1π ” 2 ’ , 6 r=cos 2¨ 1
2 cos 2 and r 1
2 . 1
2 SOLUTION If we solve the equations r
, we get cos 2
and, therecos 2 and r
fore, 2
3, 5 3, 7 3, 11 3. Thus, the values of between 0 and 2 that satisfy both equations are
6, 5 6, 7 6, 11 6. We have found four points of
intersection: ( 1, 6), ( 1, 5 6), ( 1, 7 6), and ( 1, 11 6).
2
2
2
2
However, you can see from Figure 7 that the curves have four other points of intersection—namely, ( 1, 3), ( 1, 2 3), ( 1, 4 3), and ( 1, 5 3). These can be found using
2
2
2
2
1
symmetry or by noticing that another equation of the circle is r
2 and then solving
1
the equations r cos 2 and r
.
2 Arc Length FIGURE 7 To ﬁnd the length of a polar curve r f , a
write the parametric equations of the curve as
x r cos f b, we regard cos y r sin f as a parameter and
sin Using the Product Rule and differentiating with respect to , we obtain
dx
d
so, using cos 2
dx
d dr
cos
d
sin 2 2 dy
d dy
d r sin dr
sin
d r cos 1, we have
2 2 dr
d cos 2
dr
d 2r dr
cos
d 2 sin 2 2r r 2 sin 2 sin
dr
sin
d cos r 2 cos 2 dr 2
r2
d
Assuming that f is continuous, we can use Theorem 11.2.6 to write the arc length as y L dx
d b a 2 2 dy
d d Therefore, the length of a curve with polar equation r y L 5 b dr
d r2 a EXAMPLE 4 Find the length of the cardioid r 1 f ,a b, is 2 d sin . SOLUTION The cardioid is shown in Figure 8. (We sketched it in Example 7 in
Section 11.3.) Its full length is given by the parameter interval 0
2 , so
Formula 5 gives
O L y 2 r2 0 FIGURE 8 r=1+sin ¨ y 2 0 s2 dr
d
2 sin 2 d
d y 2 0 s1 sin 2 cos 2 d 5E11(pp 716725) 1/18/06 5:33 PM Page 719 ❙❙❙❙ S ECTION 11.4 AREAS AND LENGTHS IN POLAR COORDINATES 719 We could evaluate this integral by multiplying and dividing the integrand by
s2 2 sin , or we could use a computer algebra system. In any event, we ﬁnd that
the length of the cardioid is L 8.  11.4 Exercises 1–4  Find the area of the region that is bounded by the given
curve and lies in the speciﬁed sector. 1. r s, 2. r e 3. r sin , 4. r 2 3 ■ ■ 2 ■ 3 (inner loop)
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5. 2 cos ■ ■ ■ ■ ■ ■ 23. r
25. r 4 sin ,
2 1 ■ 29–34 ■ sin , 33. r 3 2 sin , 3 sin r 3 cos
■ ■ ■ ■ ■ 30. r cos
r r r ■ 32. r cos 2 2 sin , a sin , ■ r r sin 2 , 34. r 1 cos ■ 31. r 8. 1 sin , r 26. r 2 1 ■ ■ ■ ■ Find the area of the region that lies inside both curves.  29. r
r=1+sin ¨ r cos , ■ 24. r 2 3 cos , r 28. r r=¨ r 8 cos 2 , 27. r 6. 7. sec . 23–28  Find the area of the region that lies inside the ﬁrst curve
and outside the second curve. Find the area of the shaded region.  ■ r 0
■ 2 sin ■ 22. Find the area enclosed by the loop of the strophoid 2 , 1 ■ 4 ssin , ■ 5–8 0 21. r sin 2 ,
2 r 2 sin 2 , sin
r 1 2 b cos , a ■ ■ ■ 0, b
■ 0
■ ■ ■ ■ ■ 35. Find the area inside the larger loop and outside the smaller loop
1
2 of the limaçon r cos . 36. Find the area between a large loop and the enclosed small loop of the curve r
r=sin 4¨ r=4+3 sin ¨
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 37–42
■   9. r Find all points of intersection of the given curves. 11. r 2 ; 15–16  15. r
■ ■ ■ ■ ■ 2
■ 16. r 2 sin 6
■ ■ ■ ■ ■ 2 sin
■ ■ 17–21 ■ ■ ■ ■ sin 2 18. r 4 sin 3 19. r 3 cos 5 20. r 2 cos 4 r sin ,
2 r 2 cos 2
r ■ 1
r r sin 2 ,
■ cos cos
sin 3 sin 2
r
■ 2 cos 2
■ ■ ■ ■ ■ ■ ■ ; 43. The points of intersection of the cardioid r 3 sin 9
■ 42. r
■  Find the area of the region enclosed by one loop of
the curve. 17. r cos 3 , 41. r cos 2
■ cos , 40. r Graph the curve and ﬁnd the area that it encloses.
1 ■ ■ 2, 39. r cos sin 2 14. r 2 cos 3
■ 31 12. r 2 4 cos 2 13. r
■ 10. r sin , 38. r Sketch the curve and ﬁnd the area that it encloses.
3 cos 2 cos 3 . ■ 37. r
9–14 1 ■ ■ 1 sin and the
2
2, can’t be found exactly.
spiral loop r 2 ,
Use a graphing device to ﬁnd the approximate values of at
which they intersect. Then use these values to estimate the area
that lies inside both curves. ; 44. Use a graph to estimate the values of for which the curves
r 3 sin 5 and r 6 sin intersect. Then estimate the
area that lies inside both curves. 5E11(pp 716725) 720 ❙❙❙❙ 45–48  Page 720 2 47. r 55. (a) Use Formula 11.2.7 to show that the area of the surface Find the exact length of the polar curve.
3 sin , 3
2 ■ ■ ■ 2 ■ 46. r e, 48. r 0 ,0 ■ 49–52 5:33 PM CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 45. r ■ 1/18/06 , ■ ■ 0 generated by rotating the polar curve 2 0 r 2 ■ ■ ■ ■ Use a calculator to ﬁnd the length of the curve correct to
four decimal places. 51. r
■ 50. r 3 sin 2
sin
■ 52. r 2
■ ■ ■ ■ ■ S 4 sin 3
1
■ cos
■ y b a 2 r sin r2 b
a b
dr
d ) about the
2 d (b) Use the formula in part (a) to ﬁnd the surface area
generated by rotating the lemniscate r 2 cos 2 about the
polar axis. 3
■ a (where f is continuous and 0
polar axis is  49. r f ■ ■ 56. (a) Find a formula for the area of the surface generated by ; 53–54  cos 4 53. r
■ Graph the curve and ﬁnd its length. ■ ■  11.5 cos 2 54. r 4
■ ■ ■ ■ ■ ■ 2
■ ■ ■ rotating the polar curve r f , a
b (where f is
continuous and 0 a b
), about the line
2.
(b) Find the surface area generated by rotating the lemniscate
r 2 cos 2 about the line
2. Conic Sections
In this section we give geometric deﬁnitions of parabolas, ellipses, and hyperbolas and
derive their standard equations. They are called conic sections, or conics, because they
result from intersecting a cone with a plane as shown in Figure 1. ellipse parabola hyperbola FIGURE 1 Conics parabola axis
focus vertex
FIGURE 2 F directrix Parabolas
A parabola is the set of points in a plane that are equidistant from a ﬁxed point F (called
the focus) and a ﬁxed line (called the directrix). This deﬁnition is illustrated by Figure 2.
Notice that the point halfway between the focus and the directrix lies on the parabola; it is
called the vertex. The line through the focus perpendicular to the directrix is called the
axis of the parabola.
In the 16th century Galileo showed that the path of a projectile that is shot into the
air at an angle to the ground is a parabola. Since then, parabolic shapes have been used
in designing automobile headlights, reﬂecting telescopes, and suspension bridges. 5E11(pp 716725) 1/18/06 5:34 PM Page 721 S ECTION 11.5 CONIC SECTIONS y P(x, y)
F(0, p) y y=_p 721 (See Problem 14 on page 220 for the reﬂection property of parabolas that makes them so
useful.)
We obtain a particularly simple equation for a parabola if we place its vertex at the origin O and its directrix parallel to the xaxis as in Figure 3. If the focus is the point 0, p ,
then the directrix has the equation y
p. If P x, y is any point on the parabola, then the
distance from P to the focus is
sx 2 PF px O ❙❙❙❙ y p 2 and the distance from P to the directrix is y p . (Figure 3 illustrates the case where
p 0.) The deﬁning property of a parabola is that these distances are equal: FIGURE 3 sx 2 y p 2 y p We get an equivalent equation by squaring and simplifying:
x2
y2 2py 2 y p2 y2 x2 x2 y 4py p p 2 2py y p 2 p2 An equation of the parabola with focus 0, p and directrix y 1 x 2 p is 4py If we write a 1 4p , then the standard equation of a parabola (1) becomes y ax 2.
It opens upward if p 0 and downward if p 0 [see Figure 4, parts (a) and (b)]. The
graph is symmetric with respect to the yaxis because (1) is unchanged when x is replaced
by x.
y y y y=_p (0, p)
x x (0, p) y=_p (a) ≈=4py, p>0 ( p, 0) ( p, 0) 0
0 y (b) ≈=4py, p<0 x 0 x 0 x=_p x=_p (c) ¥=4px, p>0 (d) ¥=4px, p<0 FIGURE 4 If we interchange x and y in (1), we obtain
2 y2 4px which is an equation of the parabola with focus p, 0 and directrix x
p. (Interchanging x and y amounts to reﬂecting about the diagonal line y x.) The parabola opens
to the right if p 0 and to the left if p 0 [see Figure 4, parts (c) and (d)]. In both cases
the graph is symmetric with respect to the xaxis, which is the axis of the parabola. 5E11(pp 716725) 722 ❙❙❙❙ 1/18/06 5:34 PM Page 722 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES y E XAMPLE 1 Find the focus and directrix of the parabola y 2 ¥+10x=0 10 x 0 and sketch the graph.
SOLUTION If we write the equation as y 2 10 x and compare it with Equation 2, we see
5
that 4p
10, so p
( 5, 0) and the directrix is
2 . Thus, the focus is p, 0
2
5
x 2 . The sketch is shown in Figure 5. ”_ 5 , 0’
2
x 0 Ellipses 5
x= 2 An ellipse is the set of points in a plane the sum of whose distances from two ﬁxed points
F1 and F2 is a constant (see Figure 6). These two ﬁxed points are called the foci (plural of
focus). One of Kepler’s laws is that the orbits of the planets in the solar system are ellipses
with the Sun at one focus.
In order to obtain the simplest equation for an ellipse, we place the foci on the xaxis at
the points c, 0 and c, 0 as in Figure 7 so that the origin is halfway between the foci.
Let the sum of the distances from a point on the ellipse to the foci be 2a 0. Then P x, y
is a point on the ellipse when
PF1
PF2
2a FIGURE 5
P F¡ F™ FIGURE 6 that is,
P(x, y) c 2 y2 sx c or y sx 2 sx c 2 y2 2a sx y2 2a 2 y2 c Squaring both sides, we have
F¡(_c, 0) 0 F™(c, 0) x2 x 2cx c2 y2 which simpliﬁes to 4a 2 4a s x as x c 2 c y2 2 y2 a2 x2 2cx c2 y2 cx We square again: FIGURE 7 a2 x2
which becomes 2cx
a2 c2 y2 c2 x2 a4 a2y2 2a 2cx a2 a2 c 2x 2 c2 From triangle F1 F2 P in Figure 7 we see that 2c 2a, so c a and, therefore,
a 2 c 2 0. For convenience, let b 2 a 2 c 2. Then the equation of the ellipse becomes
b 2x 2 a 2 y 2 a 2b 2 or, if both sides are divided by a 2b 2,
x2
a2 3 y
(0, b) (_a, 0) a b
(_c, 0) 0 c
(0, _b) (a, 0)
(c, 0) x 1 Since b 2 a 2 c 2 a 2, it follows that b a. The xintercepts are found by setting
y 0. Then x 2 a 2 1, or x 2 a 2, so x
a. The corresponding points a, 0 and
a, 0 are called the vertices of the ellipse and the line segment joining the vertices
is called the major axis. To ﬁnd the yintercepts we set x 0 and obtain y 2 b 2, so
y
b. Equation 3 is unchanged if x is replaced by x or y is replaced by y, so the
ellipse is symmetric about both axes. Notice that if the foci coincide, then c 0, so a b
and the ellipse becomes a circle with radius r a b.
We summarize this discussion as follows (see also Figure 8).
4 The ellipse FIGURE 8 ≈¥
+ =1
a@ b@ y2
b2 has foci c, 0 , where c 2 x2
a2
a2 y2
b2 1 a b 2, and vertices b 0 a, 0 . 5E11(pp 716725) 1/18/06 5:35 PM Page 723 S ECTION 11.5 CONIC SECTIONS y If the foci of an ellipse are located on the yaxis at 0,
tion by interchanging x and y in (4). (See Figure 9.) (0, a) ❙❙❙❙ 723 c , then we can ﬁnd its equa (0, c)
5
(_b, 0) The ellipse (b, 0)
0 x2
b2 x (0, _c) has foci 0, c , where c 2 y2
a2
a2 1 a b b 2, and vertices 0, 0
a. (0, _a) EXAMPLE 2 Sketch the graph of 9x 2 FIGURE 9 ≈¥
+ =1, a˘b
b@ a@ 16y 2 SOLUTION Divide both sides of the equation by 144: x2
16
y (_4, 0)
{_œ„, 0}
7 y2
9 1 The equation is now in the standard form for an ellipse, so we have a 2 16, b 2 9,
a 4, and b 3. The xintercepts are 4 and the yintercepts are 3. Also,
c 2 a 2 b 2 7, so c s7 and the foci are ( s7, 0). The graph is sketched in
Figure 10. (0, 3) (4, 0)
0 144 and locate the foci. {œ„, 0}
7 x EXAMPLE 3 Find an equation of the ellipse with foci 0, 2 and vertices 0, 3. SOLUTION Using the notation of (5), we have c b2 a2 c2 9 4 2 and a 3. Then we obtain
5, so an equation of the ellipse is (0, _3) x2
5 FIGURE 10 9≈+16¥=144 Another way of writing the equation is 9x 2 y2
9
5y 2 1
45. Like parabolas, ellipses have an interesting reﬂection property that has practical consequences. If a source of light or sound is placed at one focus of a surface with elliptical
crosssections, then all the light or sound is reﬂected off the surface to the other focus (see
Exercise 59). This principle is used in lithotripsy, a treatment for kidney stones. A reﬂector with elliptical crosssection is placed in such a way that the kidney stone is at one focus.
Highintensity sound waves generated at the other focus are reﬂected to the stone and
destroy it without damaging surrounding tissue. The patient is spared the trauma of surgery
and recovers within a few days.
y Hyperbolas
P(x, y) F¡(_c, 0) 0 FIGURE 11 P is on the hyperbola when
 PF¡ PF™ = 2a F™(c, 0) x A hyperbola is the set of all points in a plane the difference of whose distances from two
ﬁxed points F1 and F2 (the foci) is a constant. This deﬁnition is illustrated in Figure 11.
Hyperbolas occur frequently as graphs of equations in chemistry, physics, biology, and
economics (Boyle’s Law, Ohm’s Law, supply and demand curves). A particularly signiﬁcant application of hyperbolas is found in the navigation systems developed in World Wars
I and II (see Exercise 51).
Notice that the deﬁnition of a hyperbola is similar to that of an ellipse; the only change
is that the sum of distances has become a difference of distances. In fact, the derivation of
the equation of a hyperbola is also similar to the one given earlier for an ellipse. It is left 5E11(pp 716725) 724 ❙❙❙❙ 1/18/06 5:35 PM Page 724 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES as Exercise 52 to show that when the foci are on the xaxis at c, 0 and the difference of
distances is PF1
PF2
2a, then the equation of the hyperbola is
x2
a2 6 b y y= a x (_a, 0) (a, 0) (_c, 0) (c, 0) 0 x2
a2 x 7 1 The hyperbola
x2
a2 y (0, c)
a a y=_ b x y= b x has foci c, 0 , where c 2
y
b a x. (0, a)
(0, _a) 0 a2 1 b 2, vertices a, 0 , and asymptotes x 8 FIGURE 13
¥
≈
 =1
a@
b@ The hyperbola
y2
a2 has foci 0, c , where c 2
y
a b x.
y y2
b2 If the foci of a hyperbola are on the yaxis, then by reversing the roles of x and y we
obtain the following information, which is illustrated in Figure 13. (0, _c) 3 y2
b2 1 This shows that x 2 a 2, so x
a. Therefore, we have x a or x
a. This
sx 2
means that the hyperbola consists of two parts, called its branches.
When we draw a hyperbola it is useful to ﬁrst draw its asymptotes, which are the dashed
lines y
b a x and y
b a x shown in Figure 12. Both branches of the hyperbola
approach the asymptotes; that is, they come arbitrarily close to the asymptotes. [See
Exercise 55 in Section 4.5, where y
b a x is shown to be a slant asymptote.] FIGURE 12
≈
¥
 =1
a@
b@ y=_ 4 x 1 where c 2 a 2 b 2. Notice that the xintercepts are again a and the points a, 0 and
a, 0 are the vertices of the hyperbola. But if we put x 0 in Equation 6 we get
y2
b 2, which is impossible, so there is no yintercept. The hyperbola is symmetric
with respect to both axes.
To analyze the hyperbola further, we look at Equation 6 and obtain b y=_ a x y2
b2 a2 x2
b2 1 b 2, vertices 0, a , and asymptotes 3 y= 4 x EXAMPLE 4 Find the foci and asymptotes of the hyperbola 9x 2 16y 2 144 and sketch its graph.
(_4, 0)
(_5, 0) SOLUTION If we divide both sides of the equation by 144, it becomes (4, 0)
0 FIGURE 14 9≈16¥=144 (5, 0) x x2
16 y2
9 1 which is of the form given in (7) with a 4 and b 3. Since c 2 16 9 25, the
3
foci are 5, 0 . The asymptotes are the lines y 3 x and y
4
4 x. The graph is shown
in Figure 14. 5E11(pp 716725) 1/18/06 5:36 PM Page 725 S ECTION 11.5 CONIC SECTIONS EXAMPLE 5 Find the foci and equation of the hyperbola with vertices 0, tote y ❙❙❙❙ 725 1 and asymp 2 x. SOLUTION From (8) and the given information, we see that a
1
b a 2 2 and c 2
hyperbola is a2 5
4 b2 1 and a b 2. Thus,
s5 2) and the equation of the . The foci are (0,
y2 4x 2 1 Shifted Conics
As discussed in Appendix C, we shift conics by taking the standard equations (1), (2), (4),
(5), (7), and (8) and replacing x and y by x h and y k.
EXAMPLE 6 Find an equation of the ellipse with foci 2, 1, 2 , 5, 2 , 4, 2 and vertices 2. S OLUTION The major axis is the line segment that joins the vertices 1, 2 , 5, 2
and has length 4, so a 2. The distance between the foci is 2, so c 1. Thus,
b 2 a 2 c 2 3. Since the center of the ellipse is 3, 2 , we replace x and y in (4)
by x 3 and y 2 to obtain x 3 2 y 4 2 2 1 3 as the equation of the ellipse.
EXAMPLE 7 Sketch the conic 9x 2
y 4y 2 72 x 8y 176 0 and ﬁnd its foci. 3 y1=_ 2 (x4) SOLUTION We complete the squares as follows: 4 y2
(4, 4) 4 y2 (4, 1) (4, _2) 9 x2 1 3 FIGURE 15 9≈4¥72x+8y+176=0 176 8x 16 176 1 2 9x 4 2 y 1 2 x 4 4 144 2 9 y1= 2 (x4) 8x 4y
x 0 2y 9 x2 2y 4 36
1 This is in the form (8) except that x and y are replaced by x 4 and y 1. Thus,
a 2 9, b 2 4, and c 2 13. The hyperbola is shifted four units to the right and one
unit upward. The foci are (4, 1 s13 ) and (4, 1 s13 ) and the vertices are 4, 4 and
3
4, 2 . The asymptotes are y 1
4 . The hyperbola is sketched in
2x
Figure 15. 5E11(pp 726735) ❙❙❙❙ 726 1/18/06 Page 726 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES  11.5
1–8 9:35 AM Exercises
23. 2y 2 Find the vertex, focus, and directrix of the parabola and
sketch its graph.
 2y 2 1. x
3. 4 x 2 4. y y 5. x 2 2 8y 3 7. y 2 2y 12 x 25 ■ ■ x2 2. 4y ■ ■ 2 ■ 1 8. y
■ ■ y 5 10. y 4y 9y ■ 2 12 x 64 x ■ 8 90y ■ ■ 0
305 ■ ■ ■ 2x 2
25. x 2 16 ■ ■ 27. x 2 ■ y y 2 29. y ■ 2y 2y 2 4x 4y ■ 26. x 2 1 2 ■ 28. y ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ focus 0, 32. Parabola, x 2 vertex 0, 0 ,
vertex 1, 0 ,  focus 34. Parabola, focus 3, 6 , 4, 0 , Find the vertices and foci of the ellipse and sketch 36. Parabola, y2
5 13. 4 x 2 y2 15. 9x 2 18 x 16. x 2
■ 1 17–18 14. 4 x 2 25y 2 25 ■ ■ ■ ■ 18. vertex 3, 2
axis the xaxis, vertical axis, ■ ■ ■ foci
foci 0, foci 0, 2 , 0, 6 vertices 0, 0 , 0, 8 foci 0, 1, 41. Ellipse, center 2, 2 , 42. Ellipse, ■ foci 2, 0 ,
5, vertices 0, 1 , 8, 2, 0 , 5, 0 vertices focus 0, 2 , x 2 foci foci 1, 3 and 7, 3 , 46. Hyperbola, foci 2,
vertices 48. Hyperbola, x foci 0, 47. Hyperbola, 1 foci 2, 2 and 6, 2 ,
x 2 and y 6 asymptotes y
■
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 3, vertex 5, 2 vertices 0, 6, 0 , ■ 1 passing through 2, 1 45. Hyperbola, 1 13 vertex 9, 43. Hyperbola, y 1
0 2, 3 , 0, 3 , passing through 44. Hyperbola, y ■ 2 40. Ellipse, 0
■ ■ 5 39. Ellipse, 7 0 2 38. Ellipse, Find an equation of the ellipse. Then ﬁnd its foci. 17. ■ 37. Ellipse, 27
4y y 16
2 and 1, 9 1 6x ■  y2
100 16 2y 2
■ x2
12.
64 4y 2 4x
■ directrix x vertex 0, 0 ,
passing through (1, 4) its graph.
x2
11.
9 6x 2 directrix x 35. Parabola,
11–16 ■ 1 8y 2 ■ 33. Parabola,
■ y2 30. 4 x 3 ■ 31. Parabola, 0 ■ ■  Find an equation for the conic that satisﬁes the given
conditions. 1 x ■ 31–48 1
_2 ■  Identify the type of conic section whose equation is given
and ﬁnd the vertices and foci. 2 9–10  Find an equation of the parabola. Then ﬁnd the focus and
directrix. 9. 2 25–30 12 x ■ ■ 12 x 6. x 0 24. 16 x 0 3x 2 4, 0 vertices vertices 2, 3 and 6, 3 2 and 2, 8 ,
3, 0 , ■ 1 vertices 2, 0 and 2, 6 asymptotes y ■ 2x x
■ ■ ■ ■ ■ ■ 49. The point in a lunar orbit nearest the surface of the moon is
19–24 Find the vertices, foci, and asymptotes of the hyperbola
and sketch its graph. 19.  x2
144 21. y 2 y2
25
x2 1
4 20. y2
16 22. 9x 2 x2
36 1 4y 2 36 called perilune and the point farthest from the surface is called
apolune. The Apollo 11 spacecraft was placed in an elliptical
lunar orbit with perilune altitude 110 km and apolune altitude
314 km (above the moon). Find an equation of this ellipse if
the radius of the moon is 1728 km and the center of the moon
is at one focus. 5E11(pp 726735) 1/18/06 9:35 AM Page 727 S ECTION 11.5 CONIC SECTIONS 50. A crosssection of a parabolic reﬂector is shown in the ﬁgure. The bulb is located at the focus and the opening at the focus
is 10 cm.
(a) Find an equation of the parabola.
(b) Find the diameter of the opening CD , 11 cm from
the vertex.
C y2 4px at the point x 0 , y 0 can be written as
y0y 2p x x0 (b) What is the xintercept of this tangent line? Use this fact to
draw the tangent line.
ellipse x 2 4y 2 10 to estimate the length of the 4. 58. The planet Pluto travels in an elliptical orbit around the Sun 5 cm
11 cm
F
5 cm V (at one focus). The length of the major axis is 1.18 10 10 km
and the length of the minor axis is 1.14 10 10 km. Use Simpson’s Rule with n 10 to estimate the distance traveled by the
planet during one complete orbit around the Sun.
59. Let P x 1, y1 be a point on the ellipse x 2 a 2 B
D
51. In the LORAN (LOng RAnge Navigation) radio navigation system, two radio stations located at A and B transmit simultaneous signals to a ship or an aircraft located at P. The
onboard computer converts the time difference in receiving
these signals into a distance difference PA
PB , and
this, according to the deﬁnition of a hyperbola, locates the
ship or aircraft on one branch of a hyperbola (see the ﬁgure).
Suppose that station B is located 400 mi due east of station A
on a coastline. A ship received the signal from B 1200 microseconds ( s) before it received the signal from A.
(a) Assuming that radio signals travel at a speed of 980 ft s,
ﬁnd an equation of the hyperbola on which the ship lies.
(b) If the ship is due north of B, how far off the coastline is
the ship? y 2 b 2 1 with
foci F1 and F2 and let and be the angles between the lines
.
PF1, PF2 and the ellipse as in the ﬁgure. Prove that
This explains how whispering galleries and lithotripsy work.
Sound coming from one focus is reﬂected and passes through
the other focus. [Hint: Use the formula in Problem 13 on
page 220 to show that tan
tan .]
y
å P(⁄, ›)
∫ F¡ 0 x F™ ¥
≈
+ b@ =1
a@ 60. Let P x 1, y1 be a point on the hyperbola x 2 a 2 y2 b2 1
with foci F1 and F2 and let and be the angles between the
lines PF1 , PF2 and the hyperbola as shown in the ﬁgure. Prove
that
. (This is the reﬂection property of the hyperbola. It
shows that light aimed at a focus F2 of a hyperbolic mirror is
reﬂected toward the other focus F1 .) P A coastline B y 400 mi
sending stations
å 52. Use the deﬁnition of a hyperbola to derive Equation 6 for a hyperbola with foci c, 0 and vertices
x2 b2 P
∫ a, 0 .
F¡ 53. Show that the function deﬁned by the upper branch of the hyperbola y 2 a 2 0 F™ 1 is concave upward. 54. Find an equation for the ellipse with foci 1, 1 and 1, 1 and major axis of length 4.
55. Determine the type of curve represented by the equation x2
k y2
k 16 727 56. (a) Show that the equation of the tangent line to the parabola 57. Use Simpson’s Rule with n A ❙❙❙❙ P 1 in each of the following cases: (a) k 16, (b) 0 k 16,
and (c) k 0.
(d) Show that all the curves in parts (a) and (b) have the same
foci, no matter what the value of k is. F¡ F™ x 5E11(pp 726735) 728 ❙❙❙❙ 1/18/06 9:35 AM Page 728 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES  11.6 Conic Sections in Polar Coordinates
In the preceding section we deﬁned the parabola in terms of a focus and directrix, but we
deﬁned the ellipse and hyperbola in terms of two foci. In this section we give a more uniﬁed treatment of all three types of conic sections in terms of a focus and directrix. Furthermore, if we place the focus at the origin, then a conic section has a simple polar equation.
In Chapter 14 we will use the polar equation of an ellipse to derive Kepler’s laws of planetary motion.
1 Theorem Let F be a ﬁxed point (called the focus) and l be a ﬁxed line (called
the directrix) in a plane. Let e be a ﬁxed positive number (called the eccentricity).
The set of all points P in the plane such that PF
Pl (The ratio of the distance from F to
the distance from l is the constant e.) e is a conic section. The conic is
(a) an ellipse if e 1 (b) a parabola if e 1 (c) a hyperbola if e 1 Proof Notice that if the eccentricity is e
y l (directrix)
P
r 1, then PF
Pl and so the given condition simply becomes the deﬁnition of a parabola as given in Section 11.5.
Let us place the focus F at the origin and the directrix parallel to the yaxis and
d units to the right. Thus, the directrix has equation x d and is perpendicular to the
polar axis. If the point P has polar coordinates r, , we see from Figure 1 that
PF ¨ Thus, the condition PF F r Pl x=d Pl d e, or PF r cos e Pl , becomes x r cos ¨ r 2 ed r cos d
C FIGURE 1 If we square both sides of this polar equation and convert to rectangular coordinates,
we get
x 2 y 2 e 2 d x 2 e 2 d 2 2dx x 2
or 1 e2 x2 2de 2x y2 e 2d 2 After completing the square, we have
3 If e x e 2d
1 e2 2 y2
1 e2 e 2d 2
1 e2 2 1, we recognize Equation 3 as the equation of an ellipse. In fact, it is of the form
x h
a 2 2 y2
b2 1 5E11(pp 726735) 1/18/06 9:35 AM Page 729 SECTION 11.6 CONIC SECTIONS IN POLAR COORDINATES ❙❙❙❙ 729 where
h 4 x=d
directrix
x y This shows that e 2d
1 e2 c 1 where c 2 a2 h
a F x ed
1e cos ¨ c
a b2 By solving Equation 2 for r, we see that the polar equation of the conic shown in Figure 1 can be written as
ed
r
1 e cos y directrix x If the directrix is chosen to be to the left of the focus as x
d, or if the directrix is chosen to be parallel to the polar axis as y
d, then the polar equation of the conic is given
by the following theorem, which is illustrated by Figure 2. (See Exercises 21–23.)
6 (c) r= 2 2 and see that
e F h y2
b2 x y=d e 2d 2
1 e2 If e 1, then 1 e 2 0 and we see that Equation 3 represents a hyperbola. Just as we
did before, we could rewrite Equation 3 in the form x=_d
directrix (b) r= b2 2 and conﬁrms that the focus as deﬁned in Theorem 1 means the same as the focus deﬁned
in Section 11.5. It also follows from Equations 4 and 5 that the eccentricity is given by
c
e
a ed
1+e cos ¨ (a) r= e 2d 2
1 e2 a2 In Section 11.5 we found that the foci of an ellipse are at a distance c from the center,
where
e 4d 2
c2 a2 b2
5
1 e2 2 y F e 2d
1 e2 Theorem A polar equation of the form ed
1+e sin ¨ r
y 1 ed
e cos or r 1 ed
e sin represents a conic section with eccentricity e. The conic is an ellipse if e
a parabola if e 1, or a hyperbola if e 1.
F y=_d
(d) r= 1, x directrix
ed
1e sin ¨ FIGURE 2 Polar equations of conics EXAMPLE 1 Find a polar equation for a parabola that has its focus at the origin and whose
directrix is the line y
6.
SOLUTION Using Theorem 6 with e
1 and d
that the equation of the parabola is r 1 6, and using part (d) of Figure 2, we see
6
sin 5E11(pp 726735) 730 ❙❙❙❙ 1/18/06 9:36 AM Page 730 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES EXAMPLE 2 A conic is given by the polar equation r 3 10
2 cos Find the eccentricity, identify the conic, locate the directrix, and sketch the conic.
SOLUTION Dividing numerator and denominator by 3, we write the equation as r
y 10
3 d focus
0 e x (10, 0) F IGURE 3 EXAMPLE 3 Sketch the conic r . Since ed 10
3 , 5 12
.
4 sin 2 SOLUTION Writing the equation in the form y
π ”6, ’
2 r π 1 6
2 sin y=3 (directrix)
(6, π) 0 x (6, 0) focus
FIGURE 4 r= 10
3
2
3 2
3 so the directrix has Cartesian equation x
5. When
0, r 10; when
,
r 2. So the vertices have polar coordinates 10, 0 and 2, . The ellipse is sketched
in Figure 3. (2, π) ”2, ’
2 cos From Theorem 6 we see that this represents an ellipse with e
we have 10
r= 32 cos ¨ x=_5
(directrix) 1 10
3
2
3 12
2+4 sin ¨ we see that the eccentricity is e 2 and the equation therefore represents a hyperbola.
Since ed 6, d 3 and the directrix has equation y 3. The vertices occur when
2 and 3 2, so they are 2, 2 and 6, 3 2
6, 2 . It is also useful to
plot the xintercepts. These occur when
0, ; in both cases r 6. For additional
accuracy we could draw the asymptotes. Note that r l
when 2 4 sin l 0 or
1
0 and 2 4 sin
0 when sin
2 . Thus, the asymptotes are parallel to the rays
7 6 and
11 6. The hyperbola is sketched in Figure 4.
When rotating conic sections, we ﬁnd it much more convenient to use polar equations
than Cartesian equations. We just use the fact (see Exercise 75 in Section 11.3) that the
graph of r f
is the graph of r f
rotated counterclockwise about the origin
through an angle .
EXAMPLE 4 If the ellipse of Example 2 is rotated through an angle
ﬁnd a polar equation and graph the resulting ellipse. 11
10
r=32 cos (¨π/4) _5 15
10
r= 32 cos ¨ _6 FIGURE 5 SOLUTION We get the equation of the rotated ellipse by replacing
equation given in Example 2. So the new equation is r 3 10
2 cos 4 about the origin,
with 4 in the 4 We use this equation to graph the rotated ellipse in Figure 5. Notice that the ellipse has
been rotated about its left focus. 5E11(pp 726735) 1/18/06 9:36 AM Page 731 S ECTION 11.6 CONIC SECTIONS IN POLAR COORDINATES ❙❙❙❙ 731 In Figure 6 we use a computer to sketch a number of conics to demonstrate the effect
of varying the eccentricity e. Notice that when e is close to 0 the ellipse is nearly circular,
whereas it becomes more elongated as e l 1 . When e 1, of course, the conic is a
parabola. e=0.1 e=0.5 e=0.68 e=1 e=0.86 e=1.1 e=0.96 e=1.4 e=4 FIGURE 6  11.6 Exercises 1–8  Write a polar equation of a conic with the focus at the
origin and the given data. eccentricity 7,
4 1. Hyperbola,
2. Parabola, directrix x 4. Hyperbola,
5. Parabola, directrix x eccentricity 2,
vertex 4 , 3 6. Ellipse,
7. Ellipse, eccentricity 1,
2 8. Hyperbola,
■ ■ ■ ; 18. Graph the parabola r directrix y vertex 1, 5 2 2 sin and its directrix. Also
graph the curve obtained by rotating this parabola about its
focus through an angle 6. 2 ■ ; 19. Graph the conics r 4 sec directrix r
■ e 1 e cos with e 0.4, 0.6, 0.8,
and 1.0 on a common screen. How does the value of e affect
the shape of the curve? 2 directrix r eccentricity 3,
■ 5 2 eccentricity 0.8, ■ r 1 4 3 cos and graph the conic and its directrix.
(b) If this conic is rotated counterclockwise about the origin
through an angle 3, write the resulting equation and
graph its curve. 6 4 eccentricity 3,
4 3. Ellipse, directrix y ; 17. (a) Find the eccentricity and directrix of the conic 6 csc ■ ■ ■ ; 20. (a) Graph the conics r
■ ■ 9–16  (a) Find the eccentricity, (b) identify the conic, (c) give an
equation of the directrix, and (d) sketch the conic. 10. r
12. r 1 11. r 6 15. r
■ 9
2 cos
3
8 cos 3 16. r ■ ■ ■ ■ ■ 6
2 sin 2 14. r 4 13. r ■ 1
sin 12
4 sin 9. r 4
3 cos 2 21. Show that a conic with focus at the origin, eccentricity e, and directrix x 4
cos
■ d has polar equation
r 5
2 sin 2
■ ed 1 e sin for e 1 and various values of d. How does the value of d affect the shape of
the conic?
(b) Graph these conics for d 1 and various values of e. How
does the value of e affect the shape of the conic? ed
e cos 22. Show that a conic with focus at the origin, eccentricity e, and directrix y d has polar equation
r ■ 1 ■ ■ 1 ed
e sin 5E11(pp 726735) ❙❙❙❙ 732 1/18/06 9:36 AM Page 732 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 23. Show that a conic with focus at the origin, eccentricity e, and directrix y r 1 ed
e sin 24. Show that the parabolas r r aphelion, respectively. Use Exercise 25(a) to show that the
perihelion distance from a planet to the Sun is a 1 e and
the aphelion distance is a 1 e .
(b) Use the data of Exercise 25(b) to ﬁnd the distances from
Earth to the Sun at perihelion and at aphelion. d has polar equation d1 cos c 1 cos and
intersect at right angles. 27. The orbit of Halley’s comet, last seen in 1986 and due to return in 2062, is an ellipse with eccentricity 0.97 and one
focus at the Sun. The length of its major axis is 36.18 AU.
[An astronomical unit (AU) is the mean distance between Earth
and the Sun, about 93 million miles.] Find a polar equation for
the orbit of Halley’s comet. What is the maximum distance
from the comet to the Sun? 25. (a) Show that the polar equation of an ellipse with directrix x d can be written in the form
r a 1 e2
1 e cos (b) Find an approximate polar equation for the elliptical orbit
of Earth around the Sun (at one focus) given that the eccentricity is about 0.017 and the length of the major axis is
about 2.99 10 8 km.
26. (a) The planets move around the Sun in elliptical orbits with 28. The HaleBopp comet, discovered in 1995, has an elliptical orbit with eccentricity 0.9951 and the length of the major axis
is 356.5 AU. Find a polar equation for the orbit of this comet.
How close to the Sun does it come?
29. The planet Mercury travels in an elliptical orbit with eccentric ity 0.206. Its minimum distance from the Sun is 4.6 10 7 km.
Use the results of Exercise 26(a) to ﬁnd its maximum distance
from the Sun. the Sun at one focus. The positions of a planet that are closest to and farthest from the Sun are called its perihelion and
planet 10 9 km
at perihelion and 7.37 10 km at aphelion. Use Exercise 26
to ﬁnd the eccentricity of Pluto’s orbit. 30. The distance from the planet Pluto to the Sun is 4.43
9 r
aphelion  ¨
Sun 31. Using the data from Exercise 29, ﬁnd the distance traveled by perihelion 11 Review the planet Mercury during one complete orbit around the Sun.
(If your calculator or computer algebra system evaluates deﬁnite integrals, use it. Otherwise, use Simpson’s Rule.) ■ CONCEPT CHECK 1. (a) What is a parametric curve? (b) How do you sketch a parametric curve?
2. (a) How do you ﬁnd the slope of a tangent to a parametric curve?
(b) How do you ﬁnd the area under a parametric curve?
3. Write an expression for each of the following: (a) The length of a parametric curve
(b) The area of the surface obtained by rotating a parametric
curve about the xaxis
4. (a) Use a diagram to explain the meaning of the polar coordi nates r, of a point.
(b) Write equations that express the Cartesian coordinates
x, y of a point in terms of the polar coordinates.
(c) What equations would you use to ﬁnd the polar coordinates
of a point if you knew the Cartesian coordinates?
5. (a) How do you ﬁnd the slope of a tangent line to a polar curve?
(b) How do you ﬁnd the area of a region bounded by a polar
curve? ■ (c) How do you ﬁnd the length of a polar curve?
6. (a) Give a geometric deﬁnition of a parabola. (b) Write an equation of a parabola with focus 0, p and directrix y
p. What if the focus is p, 0 and the directrix
is x
p?
7. (a) Give a deﬁnition of an ellipse in terms of foci. (b) Write an equation for the ellipse with foci
vertices a, 0 . c, 0 and 8. (a) Give a deﬁnition of a hyperbola in terms of foci. (b) Write an equation for the hyperbola with foci c, 0 and
vertices a, 0 .
(c) Write equations for the asymptotes of the hyperbola in
part (b).
9. (a) What is the eccentricity of a conic section? (b) What can you say about the eccentricity if the conic section
is an ellipse? A hyperbola? A parabola?
(c) Write a polar equation for a conic section with eccentricity
e and directrix x d. What if the directrix is x
d?
y d? y
d? 5E11(pp 726735) 1/18/06 9:37 AM Page 733 C HAPTER 11 REVIEW ■ TRUEFALSE QUIZ Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
1. If the parametric curve x f t , y t t satisﬁes t 1
then it has a horizontal tangent when t 1. xs 2 ft 2 tt f t,y t t ,a t 0, y 2. x 1 2t 3. x tan , 4. x 2 cos , ■ ■ y e,
y ■ e t, EXERCISES 4 t ■ ■ ■ ■ ■ x ■ ■ ■ ■ 19. x 3 t 21. r t 1 cos 2 sin 4 3 1
cos 3 2 cos 14. r
■ ■ ■ ■ 2 1, y 6t ■ ■ 2
■ ■ ■ ■ 1
t 2; cos 3 ; ■ ■  2t t 1 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Find dy d x and d 2 y d x 2.
t cos t, y t sin t 1 t 2, y t ■ 8
3 sin
■ t 2; t 24. x cos 3 4 1 23. x 10. r
12. r ■ ■ e; 23–24 8. r sec 2 ■ ln t, y 20. x t cos ■ ■  Find the slope of the tangent line to the given curve at the
point corresponding to the speciﬁed value of the parameter. Sketch the polar curve. 1 ■ 19–22 ■ 1 ■ y2 2 4 3 cos and its directrix. Also
graph the ellipse obtained by rotation about the origin through
an angle 2 3. _1 13. r ■ 22. r 1 9. r 16. x 2 2 ; 18. Graph the ellipse r 1 11. r 2 y is called a
sin
cochleoid. Use a graph of r as a function of in Cartesian
coordinates to sketch the cochleoid by hand. Then graph it with
a machine to check your sketch. ■ y 1 3 x is a parabola. 2y ; 17. The curve with polar equation r
sin sx. 7. r t. ■ ■ f t and y t t to sketch the parametric curve x f t , y t t . Indicate with arrows the
direction in which the curve is traced as t increases.  t ,y 15. x 1 6. Use the graphs of x 7–14 t 4 have the same graph  Find a polar equation for the curve represented by the
given Cartesian equation. 5. Write three different sets of parametric equations for the curve y t 2, y 15–16 t 1 ■ y 2 4, and x 2 sin 3t,
all have the same graph. 10. A hyperbola never intersects its directrix. cot
y 1 have the 9. A tangent line to a parabola intersects the parabola only once. in polar coordinates, then 2 y sin 2 6 8. The graph of y 2 dt .  Sketch the parametric curve and eliminate the parameter to
ﬁnd the Cartesian equation of the curve. 4 t, 3 as x b, is 1–4 t2 2 cos 3t 0 7. The parametric equations x ■ 1. x 2, x 2
t2 6. The equations r 4. If a point is represented by x, y in Cartesian coordinates (where x 0) and r,
tan 1 y x . sin 2 and r 1 same graph. f t and y t t are twice differentiable, then
d 2 y dt 2 d 2x dt 2 .
d 2y dx 2
b
a 733 ■ 5. The polar curves r 2. If x 3. The length of the curve x ❙❙❙❙ ■ ■ ■ t3
■ ■ ; 25. Use a graph to estimate the coordinates of the lowest point on
■ ■ the curve x t 3 3 t, y t 2
ﬁnd the exact coordinates. t 1. Then use calculus to 5E11(pp 726735) ❙❙❙❙ 734 1/18/06 9:37 AM Page 734 CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 26. Find the area enclosed by the loop of the curve in Exercise 25. 2a cos t 43. a cos 2 t y 2a sin t a sin 2 t 29. Find the area enclosed by the curve r 2 46. 25x 9 cos 5 . ■ 30. Find the area enclosed by the inner loop of the curve cot and  36. x 2 37. r 1, 38. r 3 ■ sin
■ 2 t 3, y
3t, y 0 t cosh 3t, ■ ■ ■ ■ ■ 0 5 and
3, 0 and x.
2 and major axis 100 and that has its other focus at y t two lines of slope m that are tangent to the ellipse
x 2 a 2 y 2 b 2 1 and their equations are
y m x sa 2m 2 b 2. 1 53. Find a polar equation for the ellipse with focus at the origin, eccentricity 1 , and directrix with equation r
3
■ ■ ■ ■ ■ ■ ■ 4 sec . ■ 54. Show that the angles between the polar axis and the
39–40 Find the area of the surface obtained by rotating the given
curve about the xaxis.
 39. x 4 st, 40. x 2 ■ ■ t3
3 y
3t, ■ ■ 1
,
2t 2 1 t 0 t 1 ■ ■ ■ asymptotes of the hyperbola r
are given by cos 1 1 e . ed 1 e cos ,e 1, 55. In the ﬁgure the circle of radius a is stationary, and for every 4 cosh 3t, y , the point P is the midpoint of the segment QR. The curve
traced out by P for 0
is called the longbow curve.
Find parametric equations for this curve. ■ ■ ■ ■ ■ y ; 41. The curves deﬁned by the parametric equations
x t2
t2 c
1 y ■ 2. 52. Show that if m is any real number, then there are exactly 2 3, 0
■ 59
■ 2. with the parabola x 2
the origin. 2 ■ 16y
■ 51. Find an equation for the ellipse that shares a vertex and a focus sin . 2 Find the length of the curve.
3 t 2, ■ with length 8. cos 2 but outside the curve r 35. x 50 x ■ 0 50. Find an equation of the ellipse with foci 3, cos . sin 34. Find the area of the region that lies inside the curve 35–38 4y 55 asymptotes 2y 2 sin and r
2 36y
2 49. Find an equation of the hyperbola with foci 33. Find the area of the region that lies inside both of the circles r x
2 48. Find an equation of the hyperbola with foci 0, 2 cos . r 16 vertices 0, 32. Find the points of intersection of the curves r r y2 ■ y
2 and 4 cos . r 1 47. Find an equation of the parabola with focus 0, 6 and directrix 3 sin . 31. Find the points of intersection of the curves r y2
8 x
9 45. 6y 2 28. Find the area enclosed by the curve in Exercise 27. 1 Find the foci and vertices and sketch the graph. 44. 4 x 2 have vertical or horizontal tangents? Use this information to
help sketch the curve. r 
2 27. At what points does the curve x 43–46 t t2
t2 R 2a c
1 are called strophoids (from a Greek word meaning “to turn or
twist”). Investigate how these curves vary as c varies. P
a a
; 42. A family of curves has polar equations r sin 2 where
a is a positive number. Investigate how the curves change as
a changes. y=2 a Q ¨
0 x 5E11(pp 726735) 1/18/06 9:37 AM Page 735 PROBLEMS
PLUS 1. A curve is deﬁned by the parametric equations x y t 1 cos u
du
u y y t 1 sin u
du
u Find the length of the arc of the curve from the origin to the nearest point where there is a
vertical tangent line.
2. (a) Find the highest and lowest points on the curve x 4 C AS y 4 x 2 y 2.
(b) Sketch the curve. (Notice that it is symmetric with respect to both axes and both of the
lines y
x, so it sufﬁces to consider y x 0 initially.)
(c) Use polar coordinates and a computer algebra system to ﬁnd the area enclosed by the
curve. ; 3. What is the smallest viewing rectangle that contains every member of the family of polar
curves r 1 c sin , where 0 c 1? Illustrate your answer by graphing several members of the family in this viewing rectangle.
4. Four bugs are placed at the four corners of a square with side length a. The bugs crawl counterclockwise at the same speed and each bug crawls directly toward the next bug at all
times. They approach the center of the square along spiral paths.
(a) Find the polar equation of a bug’s path assuming the pole is at the center of the square.
(Use the fact that the line joining one bug to the next is tangent to the bug’s path.)
(b) Find the distance traveled by a bug by the time it meets the other bugs at the center.
a a a a
5. A curve called the folium of Descartes is deﬁned by the parametric equations x 3t
1 t 3 y 3t 2
1 t3 (a) Show that if a, b lies on the curve, then so does b, a ; that is, the curve is symmetric
with respect to the line y x. Where does the curve intersect this line?
(b) Find the points on the curve where the tangent lines are horizontal or vertical.
(c) Show that the line y
x 1 is a slant asymptote.
(d) Sketch the curve.
(e) Show that a Cartesian equation of this curve is x 3 y 3 3xy.
(f) Show that the polar equation can be written in the form
r CAS 3 sec tan
1 tan 3 (g) Find the area enclosed by the loop of this curve.
(h) Show that the area of the loop is the same as the area that lies between the asymptote
and the inﬁnite branches of the curve. (Use a computer algebra system to evaluate the
integral.) 735 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas.
 Fall '09
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