45.
46. x
y
1
−1x = sin t
y = sin 2t
x
y
1
−1
1−1
x = sin 2t y = sin 3t
47. Cycloid
a. Find the length of one arch of the cycloid
x =a (t -sin t ), y =a (1-cos t ).
b. Find the area of the surface generated by revolving one arch of the cycloid in part (a) about the x -axis for a =1. 48. Volume Find the volume swept out by revolving the region
bounded by the x -axis and one arch of the cycloid
x =t -sin t , y =1-cos t
about the x -axis.
49. Find the volume swept out by revolving the region bounded by the
x -axis and the graph of
x =2t , y =t (2-t )
about the x -axis.
50. Find the volume swept out by revolving the region bounded by the
y -axis and the graph of
x =t (1-t ), y =1+t 2
about the y -axis.
COMPUTER EXPLORATIONS
In Exercis 51–54, u a CAS to perform the following steps for the given curve over the clod interval.
a. Plot the curve together with the polygonal path approxima-tions for n =2, 4, 8 partition points over the interval. (See Figure 11.16.)
b. Find the corresponding approximation to the length of the curve by summing the lengths of the line gments.
c. Evaluate the length of the curve using an integral. Compare your approximations for n =2, 4, 8 with the actual length given by the integral. How does the actual length compare with the approximations as n increas? Explain your answer. 51. x =13 t 3, y =1
2 t 2, 0...t (1)
52. x =2t 3-16t 2+25t +5, y =t 2+t -3, 0…t …6 53. x =t -cos t , y =1+sin t , -p …t …p 54. x =e t cos t , y =e t sin t , 0…t …p
for the length of the curve x =g (y ), c …y …d (Section 6.3, Equation 4), is a special ca of the parametric length formula
L =L b
a B
a dx dt
b 2
+a dy dt b 2 dt .
U this result to find the length of each curve.b. x =y 3>2, 0…y …4>3c. x =3
2 y 2>3, 0...y (1)
43. The curve with parametric equations
x =(1+2 sin u ) cos u , y =(1+2 sin u ) sin u
is called a limaçon and is shown in the accompanying figure. Find
the points (x , y ) and the slopes of the tangent lines at the points for
a. u =0.
b. u =p >2 .
c. u =4p >3 .
x
y 1
−1
3
2
1
india44. The curve with parametric equations
2011美国电影排行榜x =t , y =1-cos t , 0…t …2p
changelingis called a sinusoid and is shown in the accompanying figure. Find
the point (x , y ) where the slope of the tangent line is
a. largest.
b. smallest.
x
y 2
02p
The curves in Exercis 45 and 46 are called Bowditch curves or
L issajous figures . In each ca, find the point in the interior of the first quadrant where the tangent to the curve is horizontal, and find the equations of the two tangents at the origin.
T 11.3 Polar Coordinates
In this ction we study polar coordinates and their relation to Cartesian coordinates. You will e that polar coordinates are very uful for calculating many multiple inte-grals studied in Chapter 15. They are also uful in describing the paths of planets and satellites.
Definition of Polar Coordinates
To define polar coordinates, we first fix an origin O (called the pole ) and an initial ray from O (Figure 11.20). Usually the positive x -axis is chon as the initial ray. Then each point P can be located by assigning to it a polar coordinate pair (r , u ) in which r gives the directed distance from O to P and u gives the directed angle from the initial ray to ray OP . So we label the point P as
P (r , u )
Directed angle from initial ray to OP
Directed distance from O to P
O
r
Initial ray
Origin (pole)
x
P (r , u )
u
FIGURE 11.20 To define polar coordi-
nates for the plane, we start with an origin, called the pole, and an initial ray.
As in trigonometry, u is positive when measured counterclockwi and negative when measured clockwi. The angle associated with a given point is not unique. While a point in the plane has just one pair of Cartesian coordinates, it has infinitely many pairs of polar coordinates. For instance, the point 2 units from the origin along the ray u =p >6 has polar coordinates r =2, u =p >6. It also has coordinates r =2, u =-11p >6 (Figure 11.21). In some situations we allow r to be negative. That is why we u directed distance in defining P (r , u ). The point P (2, 7p >6) can be reached by turning 7p >6 radians coun-terclockwi from the initial ray and going forward 2 units (Figure 11.22). It can also be reached by turning p >6 radians counterclockwi from the initial ray and going backward 2 units. So the point also has polar coordinates r =-2, u =p >6.EXAMPLE 1 Find all the polar coordinates of the point P (2, p >6).
Solution We sketch the initial ray of the coordinate system, draw the ray from the origin
that makes an angle of p >6 radians with the initial ray, and mark the point (2, p >6) (Figure 11.23). We then find the angles for the other coordinate pairs of P in which r =2 and r =-2.
For r =2, the complete list of angles is
p 6, p 6{2p , p 6{4p , p
6
{6p ,c .For r =-2, the angles are
-
5p 6, - 5p 6{2p , - 5p 6{4p , - 5p 6
{6p ,c .The corresponding coordinate pairs of P are
a 2,
p
6
+2n p b , n =0, {1, {2,c and
a -2, -
5p
6
+2n p b , n =0, {1, {2,c .When n =0, the formulas give (2, p >6) and (-2, -5p >6). When n =1, they give (2, 13p >6) and (-2, 7p >6), and so on.
Polar Equations and Graphs
If we hold r fixed at a constant value r =a ≠0, the point P (r , u ) will lie ͉a ͉ units from the origin O . As u varies over any interval of length 2p , P then traces a circle of radius ͉a ͉ centered at O (Figure 11.24).
If we hold u fixed at a constant value u =u 0 and let r vary between -q and q , the point P (r , u ) traces the line through O that makes an angle of measure u 0 with the initial ray. (See Figure 11.22 for an example.)
O
x
Initial ray u = 0
u = p ͞6
−11p
6
P 2, = P 2,−11p 6p 6a b a b
FIGURE 11.21 Polar coordinates are not
unique.
O
x
u = 0
u = p ͞6
p ͞6
7p ͞6
P 2, = P –2,p 67p 6
a b a b FIGURE 11.22 Polar coordinates can
have negative r -values.
O
7p ͞6–5p ͞6
Initial ray
x
6日语初级单词
2, = –2, –5p 6
p 6= –2, 7p etc.
p 6
a b a b a b FIGURE 11.23 The point P (2, p >6)
has infinitely many polar coordinate pairs (Example 1).
x
a r = a O
FIGURE 11.24 The polar equation for a
circle is r =a .
EXAMPLE 2 A circle or line can have more than one polar equation.
(a) r =1 and r =-1 are equations for the circle of radius 1 centered at O .
(b) u =p >6, u =7p >6, and u =-5p >6 are equations for the line in Figure 11.23. Equations of the form r =a and u =u 0 can be combined to define regions, s egments, and rays.
Equations Relating Polar and Cartesian Coordinates
x =r cos u , y =r sin u , r 2=x 2+y 2, tan u =y
x
The first two of the equations uniquely determine the Cartesian coordinates x and y given the polar coordinates r and u . On the other hand, if x and y are given, the third equa-tion gives two possible choices for r (a positive and a negative value). For each (x , y )≠(0, 0), there is a unique u ∊30, 2p ) satisfying the first two equations, each then giving a polar coordinate reprentation of the Cartesian point (x , y ). The other polar coor-dinate reprentations for the point can be determined from the two, as in Example 1.EXAMPLE 4 Here are some plane curves expresd in terms of both polar coordi-nate and Cartesian coordinate equations. Polar equation Cartesian equivalent
r cos u =2 x =2 r 2 cos u sin u =4 xy =4 r 2 cos 2 u -r 2 sin 2 u =1
x 2-y 2=1 r =1+2r cos u
y 2-3x 2-4x -1=0
r =1-cos u
x 4+y 4+2x 2y 2+2x 3+2xy 2-y 2=0
Some curves are more simply expresd with polar coordinates; others are not.
EXAMPLE 3 Graph the ts of points who polar coordinates satisfy the following conditions.
(a) 1…r …2 and 0…u …p 2
(b) -3…r …2 and u =
p 4(c)
2p 3…u …5p
6
(no restriction on r )Solution The graphs are shown in Figure 11.25.
Relating Polar and Cartesian Coordinates
When we u both polar and Cartesian coordinates in a plane, we place the two origins together and
let the initial polar ray be the positive x -axis. The ray u =p >2, r 70, becomes the positive y -axis (Figure 11.26). The two coordinate systems are then related by the following equations.
x
y
1(a)
2
joe什么意思
x
y
3
董事长英文(b)
2(c)
x
y
1 ≤ r ≤ 2, 0 ≤ u ≤p
2
u = ,p 4
−3 ≤ r ≤ 2p 4
2p 3
5p 6
2p 35p
中学生英语作文范文6≤ u ≤FIGURE 11.25 The graphs of typical
inequalities in r and u (Example 3).
x
y
Common origin
Initial ray
x
y
r P (x , y ) = P (r , u )
u = 0, r ≥ 0
u Ray u =p
2
FIGURE 11.26 The usual way to relate
polar and Cartesian coordinates.
EXAMPLE 5 Find a polar equation for the circle x 2+( y -3)2=9 (Figure 11.27).
Solution We apply the equations relating polar and Cartesian coordinates:
x
2+(y -3)2=9
x 2+y 2-6y +9=9 Expand ( y -3)2.
x 2+y 2-6y =0 Cancelation
r 2-6r sin u =0 x 2+y 2=r 2, y =r sin u
r =0 or r -6 sin u =0
r =6 sin u Includes both possibilities
EXAMPLE 6 Replace the following polar equations by equivalent Cartesian equa-tions and identify their graphs.(a) r cos u =-4(b) r 2=4r cos u (c) r =
4
2 cos u -sin u
Solution We u the substitutions r cos u =x , r sin u =y , and r 2=x 2+y 2.
(a) r cos u =-4
getudtoThe Cartesian equation: r cos u =-4
x =-4 The graph: Vertical line through x =-4 on the x @axis (b) r 2=4r cos u
The Cartesian equation: r 2=4r cos u x 2+y 2=4x Substitute. x 2-4x +y 2=0 x 2-4x +4+y 2=4 Complete the square. (x -2)2+y 2=4 Factor. The graph: Circle, radius 2, center (h , k )=(2, 0)
(c) r =
4
2 cos u -sin u
The Cartesian equation:
r (2 cos u -sin u )=42r cos u -r sin u =42x -y =4y =2x -4
Substitute.x
y
(0, 3)
x 2 + (y − 3)2 = 9
or r = 6 sin u
FIGURE 11.27 The circle in Example 5.
Multiply by r .Substitute.Solve for y .
The graph: Line, slope m =2, y @intercept b =-4
Polar Coordinates
1. Which polar coordinate pairs label the same point? a. (3, 0) b. (-3, 0) c. (2, 2p >3) d. (2, 7p >3) e. (-3, p ) f. (2, p >3)
g. (-3, 2p )
h. (-2, -p >3)
googol2. Which polar coordinate pairs label the same point? a. (-2, p >3) b. (2, -p >3) c. (r , u ) d. (r , u +p ) e. (-r , u ) f. (2, -2p >3)cri中国国际广播电台
g. (-r , u +p )
h. (-2, 2p >3)
EXERCISES
11.3
