More from Suneena PD

Page 1 :
848, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , In Exercises 109–114, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., , 112. The asymptotes of the hyperbola x 2>a 2  y 2>b 2  1 are, perpendicular to each other if and only if a  b., 113. If A and C are both positive constants, then, , 109. The graph of 2x 2  y 2  F  0 is a hyperbola, provided, that F 0., , Ax 2  Cy 2  Dx  Ey  F  0, is an ellipse., , 110. The graph of y 4  16ax 2, where a  0, is a parabola., 111. The ellipse b x  a y  a b , where a  b  0, is, contained in the circle x 2  y 2  a 2 and contains the, circle x 2  y 2  b 2., 2 2, , 10.2, , 2 2, , 2 2, , 114. If A and C have opposite signs, then, Ax 2  Cy 2  Dx  Ey  F  0, is a hyperbola., , Plane Curves and Parametric Equations, Why We Use Parametric Equations, Figure 1a gives a bird’s-eye view of a proposed training course for a yacht. In Figure 1b, we have introduced an xy-coordinate system in the plane to describe the position of, the yacht. With respect to this coordinate system the position of the yacht is given, by the point P(x, y), and the course itself is the graph of the rectangular equation, 4x 4  4x 2  y 2  0, which is called a lemniscate. But representing the lemniscate in, terms of a rectangular equation in this instance has three major drawbacks., y (mi), 1, P(x, y), 1, , 1, , x (mi), , 1, , FIGURE 1, , (a) The dots give the position of markers., , (b) An equation of the curve C is 4x4  4x 2  y 2  0., , First, the equation does not define y explicitly as a function of x or x as a function, of y. You can also convince yourself that this is not the graph of a function by applying the vertical and horizontal line tests to the curve in Figure 1b (see Section 0.2)., Because of this, we cannot make direct use of many of the results for functions developed earlier. Second, the equation does not tell us when the yacht is at a given point, (x, y). Third, the equation gives no inkling as to the direction of motion of the yacht., To overcome these drawbacks when we consider the motion of an object in the, plane or plane curves that are not graphs of functions, we turn to the following representation. If (x, y) is a point on a curve in the xy-plane, we write, x  f(t), , y  t(t), , where f and t are functions of an auxiliary variable t with (common) domain some, interval I. These equations are called parametric equations, t is called a parameter,, and the interval I is called a parameter interval.

Page 2 :
849, , 10.2 Plane Curves and Parametric Equations, , If we think of t on the closed interval [a, b] as representing time, then we can interpret the parametric equations in terms of the motion of a particle as follows: At t  a, the particle is at the initial point ( f(a), t(a)) of the curve or trajectory C. As t increases, from t  a to t  b, the particle traverses the curve in a specific direction called the, orientation of the curve, eventually ending up at the terminal point ( f(b), t(b)) of, the curve. (See Figure 2.), y, , ( f(b), g(b)), , C, ( f(a), g(a)), ( f(t), g(t)), , FIGURE 2, As t increases from a to b, the particle, traces the curve from ( f(a), t(a)) to, ( f(b), t(b)) in a specific direction., , a, , t, , t, , b, , x, , 0, , Parameter interval is [a, b]., , We can also interpret the parametric equations in geometric terms as follows: We, take the line segment [a, b] and, by a process of stretching, bending, and twisting, make, it conform geometrically to the curve C., , Sketching Curves Defined by Parametric Equations, Before looking at some examples, let’s define the following term., , DEFINITION Plane Curve, A plane curve is a set C of ordered pairs (x, y) defined by the parametric equations, x  f(t), , and, , y  t(t), , where f and t are continuous functions on a parameter interval I., , EXAMPLE 1 Sketch the curve described by the parametric equations, x  t2  4, , and, , y  2t, , 1, , t, , 2, , Solution By plotting and connecting the points (x, y) for selected values of t (Table 1),, we obtain the curve shown in Figure 3., TABLE 1, t, (x, y), , 1, , (3, 2), , 1, , 12, , 154,, , 1 2, , 0, , (4, 0), , 1, , 1, 2, , 154,, , 12, , 1, , 2, , (3, 2), , (0, 4)

Page 3 :
850, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, y, , C, , 1 0 1 2, , FIGURE 3, As t increases from 1 to 2, the curve, C is traced from the initial point, (3, 2) to the terminal point (0, 4)., , (0, 4), , 4, 3, 2, 1, , 32, , t, , x, , 1, 2, (3, 2), 3, , Alternative Solution We eliminate the parameter t by solving the second of the two, given parametric equations for t, obtaining t  12 y. We then substitute this value of t, into the first equation to obtain, 1 2, x  a yb  4, 2, , x, , or, , 1 2, y 4, 4, , This is an equation of a parabola that has the x-axis as its axis of symmetry and its, vertex at (4, 0). Now observe that t  1 gives (3, 2) as the initial point of the, curve and that t  2 gives (0, 4) as the terminal point of the curve. So tracing the graph, from the initial point to the terminal point gives the desired curve, as obtained earlier., , We will adopt the convention here, just as we did with the domain of a function,, that the parameter interval for x  f(t) and y  t(t) will consist of all values of t for, which f(t) and t(t) are real numbers, unless otherwise noted., , EXAMPLE 2 Sketch the curves represented by, a. x  1t and y  t, b. x  t and y  t 2, Solution, a. We eliminate the parameter t by squaring the first equation to obtain x 2  t. Substituting this value of t into the second equation, we obtain y  x 2, which is an, equation of a parabola. But note that the first parametric equation implies that, t  0, so x  0. Therefore, the desired curve is the right portion of the parabola, shown in Figure 4. Finally, note that the parameter interval is [0, ⬁) , and as t, increases from 0, the desired curve starts at the initial point (0, 0) and moves, away from it along the parabola., y, , FIGURE 4, As t increases from 0, the curve, starts out at (0, 0) and follows, the right portion of the parabola, with indicated orientation., , 4, 3, 2, 1, 0, , t, , t, , Parameter interval is [0, )., , (t  0), , (t  4), , (t  1), 1 2, , x, , t, , (x, y), , 0, 1, 2, 4, , (0, 0), (1, 1), ( 12, 2), (2, 4)

Page 4 :
10.2 Plane Curves and Parametric Equations, , 851, , b. Substituting the first equation into the second yields y  x 2. Although the rectangular equation is the same as that in part (a), the curve described by the parametric equations here is different from that of part (a), as we will now see. In this, instance the parameter interval is (⬁, ⬁). Furthermore, as t increases from ⬁, to ⬁ , the curve runs along the parabola y  x 2 from left to right, as you can see, by plotting the points corresponding to, say, t  1, 0, and 1. You can also see, this by examining the parametric equation x  t, which tells us that x increases as, t increases. (See Figure 5.), , y, , FIGURE 5, As t increases from ⬁ to ⬁ ,, the entire parabola is traced, out, from left to right., , 4, 3, 2, (t  1) 1, t, , 0, , (t  0), , (t  1), 1 2, , x, , t, , (x, y), , 1, 0, 1, , (1, 1), (0, 0), (1, 1), , Parameter interval is (, )., , For problems involving motion, it is natural to use the parameter t to represent time., But other situations call for different representations or interpretations of the parameters, as the next two examples show. Here, we use an angle as a parameter., , EXAMPLE 3 Describe the curves represented by the parametric equations, x  a cos u, , and, , y  a sin u, , a0, , with parameter intervals, a. [0, p], b. [0, 2p], c. [0, 4p], Solution, , We have cos u  x>a and sin u  y>a. So, y 2, x 2, 1  cos2 u  sin2 u  a b  a b, a, a, , giving us, x 2  y2  a2, This tells us that each of the curves under consideration is contained in a circle of, radius a, centered at the origin., a. If u  0, then x  a and y  0, giving (a, 0) as the initial point on the curve. As, u increases from 0 to p, the required curve is traced out in a counterclockwise, direction, terminating at the point (a, 0). (See Figure 6a.), b. Here, the curve is a complete circle that is traced out in a counterclockwise direction, starting at (a, 0) and terminating at the same point (see Figure 6b)., c. The curve here is a circle that is traced out twice in a counterclockwise direction, starting at (a, 0) and terminating at the same point (see Figure 6c).

Page 5 :
852, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, y, , (¨  π), 0, , Historical Biography, , ¨, , π, , ¨, , a, , (¨  0), a, , x, , (a), , Science Source/, Photo Researchers, Inc., , y, , ¨, 0, , ¨, , 2π, , a, , (¨  0, 2π), x, , a, , (b), y, , MARIA GAËTANA AGNESI, (1718–1799), Maria Gaetana Agnesi’s exceptional academic talents surfaced at an early age, and, her wealthy father provided her with the, best tutors. By the age of nine, she had, learned many languages in addition to her, native Italian, including Greek and Hebrew., At that same age, she translated into Latin, an article her tutor had written in Italian, defending higher education for women. She, then delivered it, from memory, to one of, the gatherings of intellectuals her father, hosted in their home. Agnesi developed a, deep interest in Newtonian physics, but her, primary interests became religion and, mathematics. After Agnesi wrote a book on, differential calculus, her talents attracted, the attention of Pope Benedict XIV, who, appointed her to a position at the University of Bologna. Despite being offered the, chair of mathematics at Bologna, Agnesi, left the academic world in 1752 so that she, could fulfill a desire to serve others. She, devoted the rest of her life to religious, charitable projects, including running a, home for the poor., Later, in Exercises 10.2, you will learn, that the curve in Problem 41 is called the, witch of Agnesi. Why was this given such, a peculiar name? It is actually because of a, mistranslation of Maria Agnesi’s 1748 work, Instituzione analitiche ad uso della, gioventu italiana. In her discussion of the, curve represented by the rectangular, equation y(x2  a2)  a3, Agnesi used, the Italian word versiera, which is derived, from the Latin vertere, meaning “to turn.”, However, this word was confused with, avversiera, meaning “witch” or “devil’s, wife,” and the curve became known as the, witch of Agnesi., , ¨, 4π ¨, , 0, , a, , (¨  0, 2π, 4π), x, , a, , (c), Parameter interval, , FIGURE 6, The curve is (a) a semicircle, (b) a complete circle, and (c) a complete circle traced out twice., All curves are traced in a counterclockwise direction., , EXAMPLE 4 Describe the curve represented by, x  4 cos u, Solution, , and, , y  3 sin u, , 0, , u, , 2p, , Solving the first equation for cos u and the second equation for sin u gives, cos u , , x, 4, , sin u , , y, 3, , and, , Squaring each equation and adding the resulting equations, we obtain, y 2, x 2, cos2 u  sin2 u  a b  a b, 4, 3, Since cos2 u  sin2 u  1, we end up with the rectangular equation, y2, x2, , 1, 16, 9, From this we see that the curve is contained in an ellipse centered at the origin. If, u  0, then x  4 and y  0, giving (4, 0) as the initial point of the curve. As u increases, from 0 to 2p, the elliptical curve is traced out in a counterclockwise direction, terminating at (4, 0) . (See Figure 7.)

Page 6 :
10.2 Plane Curves and Parametric Equations, y, , 853, , (¨  π2 ), , 2, 1, , (¨  π), , FIGURE 7, As u increases from 0 to 2p,, the curve that is traced out in, a counterclockwise direction, beginning and ending at, (4, 0) is an ellipse., , 4 3 2 1 0, 1, , 1, , 2, , (¨  0, 2π), x, 4, , 3, , 2, , (¨  3π2 ), EXAMPLE 5 A proposed training course for a yacht is represented by the parametric equations, x  sin t, , y  sin 2t, , and, , 0, , t, , 2p, , where x and y are measured in miles., a. Show that the rectangular equation of the course is 4x 4  4x 2  y 2  0., b. Describe the course., Solution, a. Using the trigonometric identity sin 2t  2 sin t cos t, we rewrite the second of, the parametric equations in the form, y  2 sin t cos t  2x cos t, , Since x  sin t, , Solving for cos t, we have, cos t , , y, 2x, , Then, using the identity sin2 t  cos2 t  1, we obtain, x2  a, , y 2, b 1, 2x, , x2 , , y2, 4x 2, , 1, , or, 4x 4  4x 2  y 2  0, b. From the results of part (a) we see that the required curve is symmetric with, respect to the x-axis, the y-axis, and the origin. Therefore, it suffices to concentrate first on drawing the part of the curve that lies in the first quadrant and then, make use of symmetry to complete the curve. Since both sin t and sin 2t are nonnegative only for 0 t p2 , we first sketch the curve corresponding to values of t, in C0, p2 D . With the help of the following table we obtain the curve shown in Figure 8. The direction of the yacht is indicated by the arrows., t, , 0, , (x, y), , (0, 0), , p, 6, , 1 12, 132 2, , p, 4, , 1 122, 1 2, , p, 3, , 1 132, 132 2, , p, 2, , (1, 0)

Page 7 :
854, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, y (mi), 1, , 1, , π, t, (t  π6 ) ( 4 ) π, (t  3 ), (t  π2 ), (t  0), 1, , x (mi), , 1, , FIGURE 8, The training course for the yacht, , EXAMPLE 6 Cycloids Let P be a fixed point on the rim of a wheel. If the wheel is, allowed to roll along a straight line without slipping, then the point P traces out a curve, called a cycloid (see Figure 9). Suppose that the wheel has radius a and rolls along the, x-axis. Find parametric equations for the cycloid., FIGURE 9, The cycloid is the curve, traced out by a fixed point P, on the rim of a rolling wheel., y, , P, a, , Solution Suppose that the wheel rolls in a positive direction with the point P initially, at the origin of the coordinate system. Figure 10 shows the position of the wheel after, it has rotated through u radians. Because there is no slippage, the distance the wheel, has rolled from the origin is, , C(a¨, a), ¨ a cos ¨, a sin ¨, a, , P, y, , d(O, M)  length of arc PM  au, , O, M, , x, , x, , giving its center as C(au, a). Also, from Figure 10 we see that the coordinates of P(x, y), satisfy, x  d(O, M)  a sin u  au  a sin u  a(u  sin u), , a¨, , FIGURE 10, The position of the wheel after it has, rotated through u radians, , and, y  d(C, M)  a cos u  a  a cos u  a(1  cos u), Although these results are derived under the tacit assumption that 0  u  p2 , it can, be demonstrated that they are valid for other values of u. Therefore, the required parametric equations of the cycloid are, x  a(u  sin u), , and, , y  a(1  cos u), , ⬁  u  ⬁, , The cycloid provides the solution to two famous problems in mathematics:, 1. The brachistochrone problem: Find the curve along which a moving particle, (under the influence of gravity) will slide from a point A to another point B,, not directly beneath A, in the shortest time (see Figure 11a)., 2. The tautochrone problem: Find the curve having the property that it takes the, same time for a particle to slide to the bottom of the curve no matter where the, particle is placed on the curve (see Figure 11b)., The brachistochrone problem—the problem of finding the curve of quickest, descent—was advanced in 1696 by the Swiss mathematician Johann Bernoulli. Offhand, one might conjecture that such a curve should be a straight line, since it yields

Page 8 :
10.2 Plane Curves and Parametric Equations, , 855, , A, , FIGURE 11, The cycloid provides the solution, to both the brachistochrone, and the tautochrone problem., , P, P, , B, , (b) The tautochrone problem, , (a) The brachistochrone problem, , the shortest distance between the two points. But the velocity of the particle moving, on the straight line will build up comparatively slowly, whereas if we take a curve that, is steeper near A, even though the path becomes longer, the particle will cover a large, portion of the distance at a greater speed. The problem was solved by Johann Bernoulli,, his older brother Jacob Bernoulli, Leibniz, Newton, and l’Hôpital. They found that the, curve of quickest descent is an inverted arc of a cycloid (Figure 11a). As it turns out,, this same curve is also the solution to the tautochrone problem., , 10.2, , CONCEPT QUESTIONS, , 1. What is a plane curve? Give an example of a plane curve, that is not the graph of a function., 2. What is the difference between the curve C1 with parametric, representation x  cos t and y  sin t, where 0 t 2p,, and the curve C2 with parametric representation x  sin t, and y  cos t, where 0 t 2p?, , 10.2, , EXERCISES, , In Exercises 1–28, (a) find a rectangular equation whose graph, contains the curve C with the given parametric equations, and, (b) sketch the curve C and indicate its orientation., 1. x  2t  1,, 2. x  t  2,, 3. x  1t,, 4. x  t 2,, , 3. Describe the relationship between the curve C1 with parametric equations x  f(t) and y  t(t), where 0 t 1,, and the curve C2 with parametric equations x  f(1  t) and, y  t(1  t), where 0 t 1., , yt3, y  2t  1; 1, , t, , 5. x  t 2  1,, , 0, , t, , y  2t  1, , 7. x  t 2,, , y  t 3; 2, , 22. x  t ,, , p, 2, , 1, , t, , y  e2t, y  3 ln t, , 23. x  ln 2t, y  t 2, , yt1, , 9. x  2 sin u,, , y  2 cos u;, , 0, , u, , 2p, , 10. x  cos 2u,, , y  3 sin u;, , 0, , u, , 2p, , 11. x  2 sin u,, , y  3 cos u;, , 0, , 13. x  2 cos u  2,, , 21. x  et,, , 2, , 24. x  et,, , y  sin u  2;, , u, , 20. x  et, y  et, , 2, , 1, 8. x  1  ,, t, , 12. x  cos u  1,, , 16. x  sec u, y  cos u, , 3, , t, , u, , y  cos 2u, , 19. x  sin2 u, y  sin4 u; 0, t, , 0, , 18. x  cos3 u, y  sin3 u, , 3, , y  2t 2  1; 2, , 6. x  t 3,, , 15. x  cos u,, , y  3 cos u  1;, , 17. x  sec u, y  tan u; p2  u  p2, , 5, , y9t, y  t  1;, , 14. x  sin u  3,, , u, 0, , y  3 sin u  1;, , 25. x  cosh t, y  sinh t, , 2p, u, 0, , y  ln t, y  2 cosh t, , 2, , 27. x  (t  1) ,, , y  (t  1)3;, , ,, , 1t, 1  t2, , 28. x , , 2p, u, , 26. x  3 sinh t,, , 2p, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 2t, 1  t2, , 2, , y, , 2, , 2p

Page 9 :
856, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , In Exercises 29–34, the position of a particle at time t is (x, y)., Describe the motion of the particle as t varies over the time, interval [a, b]., 29. x  t  1,, , y  1t;, , 30. x  sin pt,, , y  cos pt;, , 31. x  1  cos t,, , [0, 4], [0, 6], , y  2  sin t;, , 32. x  1  2 sin 2t,, , [0, 2p], , y  2  4 sin 2t;, , x  ru  d sin u, , [0, 2p], , y  e2t1; [0, ⬁), , 35. Flight Path of an Aircraft The position (x, y) of an aircraft flying in a fixed direction t seconds after takeoff is given by, x  tan(0.025pt) and y  sec(0.025pt)  1, where x and y, are measured in miles. Sketch the flight path of the aircraft, for 0 t 403., 36. Trajectory of a Shell A shell is fired from a howitzer with a, muzzle speed of √0 ft/sec. If the angle of elevation of the, howitzer is a, then the position of the shell after t sec is, described by the parametric equations, x  (√0 cos a)t, , y  (√0 sin a)t , , and, , x  2a cot u, , and, , y  2a sin2 u, , y, y  2a, P(x, y), , (0, a), ¨, , x, , 0, , 42. If a string is unwound from a circle of radius a in such a, way that it is held taut in the plane of the circle, then its end, P will trace a curve called the involute of the circle. Referring to the following figure, show that the parametric equations of the involute are, , 37. Let P1(x 1, y1) and P2 (x 2, y2) be two distinct points in the, plane. Show that the parametric equations, and, , 41. The witch of Agnesi is the curve shown in the following, figure. Show that the parametric equations of this curve are, , 1 2, tt, 2, , where t is the acceleration due to gravity (32 ft/sec2)., a. Find the range of the shell., b. Find the maximum height attained by the shell., c. Show that the trajectory of the shell is a parabola by, eliminating the parameter t., , x  x 1  (x 2  x 1)t, , y  r  d cos u, , and, , where u is the same parameter as that for the cycloid., Sketch the trochoid for the cases in which d  r and d  r., , 33. x  sin t, y  sin2 t; [0, 3p], 34. x  et,, , rolls without slipping along a straight line is called a trochoid. (The cycloid is the special case of a trochoid with, d  r.) Suppose that the circle rolls along the x-axis in the, positive direction with u  0 when the point P is at one of, the lowest points on the trochoid. Show that the parametric, equations of the trochoid are, , x  a(cos u  u sin u), , y  a(sin u  u cos u), , and, , y  y1  (y2  y1)t, , describe (a) the line passing through P1 and P2 if, ⬁  t  ⬁ and (b) the line segment joining P1, and P2 if 0 t 1., , y, String, , 38. Show that, x  a cos t  h, , and, , y  b sin t  k, , 0, , t, , P(x, y), , ¨, , 2p, 0, , are parametric equations of an ellipse with center at (h, k), and axes of lengths 2a and 2b., , (a, 0), , x, , 39. Show that, x  a sec t  h, , and, , y  b tan t  k, t 僆 1 p2 , p2 2 傼, , 1 p2 , 3p2 2, , are parametric equations of a hyperbola with center at (h, k), and transverse and conjugate axes of lengths 2a and 2b,, respectively., 40. Let P be a point located a distance d from the center of a, circle of radius r. The curve traced out by P as the circle, , In Exercises 43–46, use a graphing utility to plot the curve with, the given parametric equations., 43. x  2 sin 3t,, , y  3 sin 1.5t;, , t0, , 44. x  cos t  5 cos 3t,, , y  6 cos t  5 sin 3t;, , 45. x  2 cos t  cos 2t,, , y  2 sin t  sin 2t;, , 0, , t, , 2p, , 46. x  3 cos t  cos 3t,, , y  3 sin t  sin 3t;, , 0, , t, , 2p, , 0, , t, , 2p

Page 10 :
10.3 The Calculus of Parametric Equations, 47. The butterfly catastrophe curve, which is described by the, parametric equations, x  c(8at 3  24t 5), , and, , is the radius of the circle. The prolate cycloid is described, by the parametric equations, , y  c(6at 2  15t 4), , x  a(t  b sin t), , y  c(at 2  3t 4), , 51. The parametric equations x  cos2 t and y  sin2 t, where, ⬁  t  ⬁ , have the same graph as x  y  1., , occurs in the study of catastrophe theory. Plot the curve with, a  2 and c  0.5 for t in the parameter interval, [1.25, 1.25]., , 52. The graph of a function y  f(x) can always be represented, by a pair of parametric equations., 53. The curve with parametric equations x  f(t)  a and, y  t(t)  b is obtained from the curve C with parametric, equations x  f(t) and y  t(t) by shifting the latter horizontally and vertically., , 49. The Lissajous curves, also known as Bowditch curves,, have applications in physics, astronomy, and other sciences., They are described by the parametric equations, x  sin(at  bp),, , y  sin t, , a a rational number, and, , 54. The ellipse with center at the origin and major and minor, axes a and b, respectively, can be obtained from the circle, with equations x  f(t)  cos t and y  t(t)  sin t by multiplying f(t) and t(t) by appropriate nonzero constants., , Plot the curve with a  0.75 and b  0 for t in the parameter interval [0, 8p]., 50. The prolate cycloid is the path traced out by a fixed point at, a distance b  a from the center of a rolling circle, where a, , 10.3, , y  c(1  d cos t), , In Exercises 51–54, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., , 48. The swallowtail catastrophe curve, which is described by, the parametric equations, and, , and, , Plot the curve with a  0.1, b  2, c  0.25, and d  2 for, t in the parameter interval [10, 10]., , occurs in the study of catastrophe theory. Plot the curve with, a  7 and c  0.03 for t in the parameter interval, [1.629, 1.629]., , x  c(2at  4t 3), , 857, , The Calculus of Parametric Equations, Tangent Lines to Curves Defined by Parametric Equations, Suppose that C is a smooth curve that is parametrized by the equations x  f(t) and, y  t(t) with parameter interval I and we wish to find the slope of the tangent line, to the curve at the point P. (See Figure 1.) Let t 0 be the point in I that corresponds to, P, and let (a, b) be the subinterval of I containing t 0 corresponding to the highlighted, portion of the curve C in Figure 1. This subset of C is the graph of a function of x, as, you can verify using the Vertical Line Test. (The general conditions that f and t must, satisfy for this to be true are given in Exercise 66.), y, , C, ( f(b), g(b)), , ( f(a), g(a)), , FIGURE 1, We want to find the slope of the tangent, line to the curve at the point P., , a, Parameter interval, , to, , b, , t, , f(a), , P( f(to), g(to)), , f(b), , x

Page 11 :
858, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , Let’s denote this function by F so that y  F(x), where f(a)  x  f(b). Since, x  f(t) and y  t(t), we may rewrite this equation in the form, t(t)  F[ f(t)], Using the Chain Rule, we obtain, t¿(t)  F¿[f(t)] f ¿(t),  F¿(x)f ¿(t), If f ¿(t), , Replace f(t) by x., , 0, we can solve for F¿(x), obtaining, F¿(x) , , t¿(t), f ¿(t), , which can also be written, dy, dy, dt, , dx, dx, dt, , if, , dx, dt, , 0, , (1), , The required slope of the tangent line at P is then found by evaluating Equation (1) at, t 0. Observe that Equation (1) enables us to solve the problem without eliminating t., y, 2, 1, 0, , EXAMPLE 1 Find an equation of the tangent line to the curve, , (t  π4 ), , x  sec t, , FIGURE 2, The tangent line to the curve, at ( 12, 1), , p2  t  p2, , at the point where t  p>4. (See Figure 2.), , (√2, 1), 2, , y  tan t, , x, , Solution, , The slope of the tangent line at any point (x, y) on the curve is, dy, dy, dt, , dx, dx, dt, , , sec 2 t, sec t, , sec t tan t, tan t, , In particular, the slope of the tangent line at the point where t  p>4 is, p, dy, 4, 12, `, , ,  12, dx tp>4, p, 1, tan, 4, sec, , Also, when t  p>4, we have x  sec(p>4)  12 and y  tan(p>4)  1 giving, ( 12, 1) as the point of tangency. Finally, using the point-slope form of the equation, of a line, we obtain the required equation:, y  1  12(x  12), , or, , y  12x  1

Page 12 :
10.3 The Calculus of Parametric Equations, , 859, , Horizontal and Vertical Tangents, A curve C represented by the parametric equations x  f(t) and y  t(t) has a horizontal tangent at a point (x, y) on C where dy>dt  0 and dx>dt 0 and a vertical, tangent where dx>dt  0 and dy>dt 0, so that dy>dx is undefined there. Points where, both dy>dt and dx>dt are equal to zero are candidates for horizontal or vertical tangents, and may be investigated by using l’Hôpital’s rule., , EXAMPLE 2 A curve C is defined by the parametric equations x  t 2 and, y  t 3  3t., a. Find the points on C where the tangent lines are horizontal or vertical., b. Find the x- and y-intercepts of C., c. Sketch the graph of C., y, 2, C, 0, , 1, , 2, , 3, , x, , 2, , FIGURE 3, The graph of x  t 2, y  t 3  3t,, and the tangent lines at t  1, , Solution, a. Setting dy>dt  0 gives 3t 2  3  0, or t  1. Since dx>dt  2t 0 at these, values of t, we conclude that C has horizontal tangents at the points on C corresponding to t  1, that is, at (1, 2) and (1, 2). Next, setting dx>dt  0 gives, 2t  0, or t  0. Since dy>dt 0 for this value of t, we conclude that C has a, vertical tangent at the point corresponding to t  0, or at (0, 0)., b. To find the x-intercepts, we set y  0, which gives t 3  3t  t(t 2  3)  0, or, t   13, 0, and 13. Substituting these values of t into the expression for x, gives 0 and 3 as the x-intercepts. Next, setting x  0 gives t  0, which, when, substituted into the expression for y, gives 0 as the y-intercept., c. Using the information obtained in parts (a) and (b), we obtain the graph of C, shown in Figure 3., , Finding d 2y>dx 2 from Parametric Equations, Suppose that the parametric equations x  f(t) and y  t(t) define y as a twicedifferentiable function of x over some suitable interval. Then d 2y>dx 2 may be found, from Equation (1) with another application of the Chain Rule., d dy, a b, d y, d dy, dt dx, , a b, dx dx, dx, dx 2, dt, 2, , if, , Higher-order derivatives are found in a similar manner., , EXAMPLE 3 Find, Solution, , d 2y, dx 2, , if x  t 2  4 and y  t 3  3t., , First, we use Equation (1) to compute, dy, dy, 3t 2  3, dt, , , dx, dx, 2t, dt, , dx, dt, , 0, , (2)

Page 13 :
860, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , Then, using Equation (2), we obtain, d dy, d 3t 2  3, a b, a, b, d y, dt dx, dt, 2t, , , dx, 2t, dx 2, dt, 2, , (2t)(6t)  (3t 2  3)(2), 4t 2, 2t, , , , , 6t 2  6, 8t 3, , , , Use the Quotient Rule., , 3(t 2  1), 4t 3, , The Length of a Smooth Curve, In Section 5.4 we showed that the length L of the graph of a smooth function f on an, interval [a, b] can be found by using the formula, b, , L, , 冮 21  [f ¿(x)] dx, 2, , (3), , a, , We now generalize this result to include curves defined by parametric equations., We begin by explaining what is meant by a smooth curve defined parametrically. Suppose that C is represented by x  f(t) and y  t(t) on a parameter interval I. Then C, is smooth if f ¿ and t¿ are continuous on I and are not simultaneously zero, except possibly at the endpoints of I. A smooth curve is devoid of corners or cusps. For example, the cycloid that we discussed in Section 10.2 (see Figure 9 in that section) has, sharp corners at the values x  2npa and, therefore, is not smooth. However, it is, smooth between these points., Now let P  {t 0, t 1, p , t n} be a regular partition of the parameter interval [a, b]., Then the point Pk( f(t k), t(t k)) lies on C, and the length of C is approximated by the, length of the polygonal curve with vertices P0, P1, p , Pn. (See Figure 4.) Thus,, n, , L ⬇ a d(Pk1, Pk), , (4), , k1, , where, d(Pk1, Pk)  2[ f(t k)  f(t k1)]2  [t(t k)  t(t k1)]2, y, , Pn ( f(tn), g(tn)), , P2 ( f(t2), g(t2)), , Pn1 ( f(tn1), g(tn1)), C, Pk ( f(tk), g(tk)), , P1 ( f(t1), g(t1)), , Pk1 ( f(tk1), g(tk1)), P0 ( f(t0), g(t0)), a  t0 t1, , t2, , tk1 tk, , b  tn t, , 0, , Parameter interval, , FIGURE 4, The length of C is approximated by the length of the polygonal curve (the red lines)., , x

Page 14 :
10.3 The Calculus of Parametric Equations, , 861, , Now, since f and t both have continuous derivatives, we can use the Mean Value Theorem to write, f(t k)  f(t k1)  f ¿(t *k )(t k  t k1), and, t(t k)  t(t k1)  t¿(t **, k )(t k  t k1), where t *k and t **, k are numbers in (t k1, t k) . Substituting these expressions into Equation (4) gives, n, , n, , 2, L ⬇ a d(Pk1, Pk)  a 2[ f ¿(t *k )]2  [t¿(t **, k )] ⌬t, k1, , (5), , k1, , As in Section 5.4, we define, n, , L  lim a d(Pk1, Pk), n→⬁, k1, n, 2,  lim a 2[ f ¿(t *k )]2  [t¿(t **, k )] ⌬t, n→⬁, , (6), , k1, , The sum in Equation (6) looks like a Riemann sum of the function 2[ f ¿]2  [t¿]2, but, it is not, because t *k is not necessarily equal to t **, k . But it can be shown that the limit, in Equation (6) is the same as that of an expression in which t *k  t **, k . Therefore,, b, , L, , 冮 2[ f ¿(t)], ,  [t¿(t)]2 dt, , 2, , a, , and we have the following result., , THEOREM 1 Length of a Smooth Curve, Let C be a smooth curve represented by the parametric equations x  f(t) and, y  t(t) with parameter interval [a, b]. If C does not intersect itself, except possibly for t  a and t  b, then the length of C is, L, , 冮, , b, , 2[ f ¿(t)]2[t¿(t)]2 dt , , a, , 冮, , a, , b, , dy 2, dx 2, b  a b dt, B dt, dt, a, , (7), , Note Equation (7) is consistent with Equation (4) of Section 5.4. Both have the form, L  兰 ds, where (ds)2  (dx)2  (dy)2., , EXAMPLE 4 Find the length of one arch of the cycloid, x  a(u  sin u), , y  a(1  cos u), , (See Example 6 in Section 10.2.), Solution, , One arch of the cycloid is traced out by letting u run from 0 to 2p. Now, dx,  a(1  cos u), du, , and, , dy,  a sin u, du

Page 15 :
862, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , Therefore, using Equation (7), we find the required length to be, L, , 冮, , 2p, , 冮, , 2p, , dy 2, dx 2, b  a b du , B du, du, , 0, , , , 冮, , a, , 2p, , 2a 2 (1  cos u)2  a 2 sin2 u du, , 0, , 2a 2  2a 2 cos u  a 2 cos2 u  a 2 sin2 u du, , 0, , a, , 冮, , 2p, , 12(1  cos u) du, , sin2 u  cos2 u  1, , 0, , To evaluate this integral, we use the identity sin2 x  12 (1  cos 2x) with u  2x. This, gives 1  cos u  2 sin2(u>2), so, La, , 冮, , 2p, , B, , 0, ,  2a, , 冮, , 4 sin2, , 0, , 2p, , sin, , u, du, 2, , u, du, 2, , sin, , u,  0 on [0, 2p], 2, , u 2p,  4accos d, 2 0,  4a(1  1)  8a, , The Area of a Surface of Revolution, Recall that the formulas S  2p 兰 y ds and S  2p 兰 x ds (Formulas 11 and 12 of Section 5.4) give the area of the surface of revolution that is obtained by revolving the graph, of a function about the x- and y-axes, respectively. These formulas are valid for finding, the area of the surface of revolution that is obtained by revolving a curve described by, parametric equations about the x- and the y-axes, provided that we replace the element, of arc length ds by the appropriate expression. These results, which may be derived by, using the method used to derive Equation (7), are stated in the next theorem., , THEOREM 2 Area of a Surface of Revolution, Let C be a smooth curve represented by the parametric equations x  f(t) and, y  t(t) with parameter interval [a, b], and suppose that C does not intersect, itself, except possibly for t  a and t  b. If t(t)  0 for all t in [a, b], then the, area S of the surface obtained by revolving C about the x-axis is, S  2p, , 冮, , b, , y2[ f ¿(t)]2  [t¿(t)]2 dt  2p, , a, , b, , 冮, , dy 2, dx 2, y a b  a b dt, B dt, dt, , a, , (8), , If f(t)  0 for all t in [a, b], then the area S of the surface that is obtained by, revolving C about the y-axis is, b, , S  2p, , 冮 x2[ f ¿(t)], , 2, , b, ,  [t¿(t)]2 dt  2p, , a, , 冮 xB a dt b, dx, , 2, , a, , a, , dy 2, b dt, dt, , EXAMPLE 5 Show that the surface area of a sphere of radius r is 4pr 2., Solution, , We obtain this sphere by revolving the semicircle, x  r cos t, , y  r sin t, , 0, , t, , p, , (9)

Page 16 :
10.3 The Calculus of Parametric Equations, , 863, , about the x-axis. Using Equation (8), the surface area of the sphere is, S  2p, , 冮, , p, , r sin t2(r sin t)2  (r cos t)2 dt, , 0, ,  2pr, , 冮, , p, , 冮, , p, , sin t2r 2 (sin2 t  cos2 t) dt, , 0, ,  2pr, , sin2 t  cos2 t  1, , r sin t dt, , 0, ,  2pr 2 Ccos tD 0  2pr 2[(1)  1]  4pr 2, p, , 10.3, , CONCEPT QUESTIONS, , 1. Suppose that C is a smooth curve with parametric equations, x  f(t) and y  t(t) and parameter interval I. Write an, expression for the slope of the tangent line to C at the point, (x 0, y0) corresponding to t 0 in I., 2. Suppose that C is a smooth curve with parametric equations, x  f(t) and y  t(t) and parameter interval [a, b]. Furthermore, suppose that C does not cross itself, except possibly, for t  a. Write an expression giving the length of C., 3. Suppose that C is a smooth curve with parametric equations, x  f(t) and y  t(t) and parameter interval [a, b]. Suppose,, , 10.3, , further, that C does not intersect itself, except possibly for, t  a and t  b., a. Write an integral giving the area of the surface obtained, by revolving C about the x-axis assuming that t(t)  0, for all t in [a, b]., b. Write an integral giving the area of the surface obtained, by revolving C about the y-axis assuming that f(t)  0, for all t in [a, b]., , EXERCISES, , In Exercises 1–6, find the slope of the tangent line to the curve, at the point corresponding to the value of the parameter., 1. x  t 2  1,, , y  t 2  t;, , 2. x  t  t,, , y  t  2t  2; t  2, , 3, , 9. x  t 2  t,, , 2, , 3. x  1t,, , 1, y ; t1, t, , 4. x  e2t,, , y  ln t;, , 5. x  2 sin u,, , t1, , 10. x  e ,, t, , u, , p, 4, , t, , ye ;, , (0, 2), , (1, 1), , y  2(1  cos u);, , y  t 3  t 2;, , t1, , y  u sin u;, , p, u, 2, , 11. x  2t 2  1,, , y  t 3;, , 12. x  t ,, , 2, , 3, , u, , p, 6, , In Exercises 7 and 8, find an equation of the tangent line to the, curve at the point corresponding to the value of the parameter., , 8. x  u cos u,, , y  t 2  t 3;, , In Exercises 11 and 12, find the points on the curve at which the, slope of the tangent line is m., , t1, , y  3 cos u;, , 6. x  2(u  sin u),, , 7. x  2t  1,, , In Exercises 9 and 10, find an equation of the tangent line to the, curve at the given point. Then sketch the curve and the tangent, line(s)., , y  t  t;, , m3, m1, , In Exercises 13–16, find the points on the curve at which the, tangent line is either horizontal or vertical. Sketch the curve., 13. x  t 2  4,, , y  t 3  3t, , 14. x  t 3  3t,, , y  t2, , 15. x  1  3 cos t,, 16. x  sin t,, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , y  2  2 sin t, , y  sin 2t

Page 17 :
864, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, 33. x  sin2 t,, , In Exercises 17–24, find dy>dx and d 2y>dx 2., 17. x  3t  1,, , y  2t, , 2, , 19. x  1t,, , 18. x  t  t, y  t  2t, , 3, , 3, , 1, y, t, , 3, , 20. x  sin 2t,, , 2, , y  cos 2t, , 34. x  e cos t,, t, , y  cos 2t; 0, y  e sin t;, t, , t, 0, , p, t, , p, , 35. x  a(cos t  t sin t), y  a(sin t  t cos t);, , 0, , p, 2, , t, , y  (2  t 2)cos t  2t sin t;, , 21. x  u  cos u, y  u  sin u, , 36. x  (t 2  2)sin t  2t cos t,, 0 t p, , 22. x  et, y  e2t, , 37. Find the length of the cardioid with parametric equations, , 23. x  cosh t, y  sinh t, , x  a(2 cos t  cos 2t), , 24. x  2t 2  1, y  t ln t, , and y  a(2 sin t  sin 2t), , 38. Find the length of the astroid with parametric equations, , 25. Let C be the curve defined by the parametric equations, x  t 2 and y  t 3  3t (see Example 2). Find d 2y>dx 2, and, use this result to determine the intervals where C is concave, upward and where it is concave downward., , x  a cos3 t, , y  a sin3 t, , and, , (See the figure for Exercise 27. Compare with Exercise 25, in Section 5.4.), , 26. Show that the curve defined by the parametric equations, x  t 2 and y  t 3  3t crosses itself. Find equations of the, tangent lines to the curve at that point (see Example 2)., , 39. The position of an object at any time t is (x, y), where, x  cos2 t and y  sin2 t, 0 t 2p. Find the distance, covered by the object as t runs from t  0 to t  2p., , 27. The parametric equations of the astroid x 2>3  y 2>3  a 2>3, are x  a cos3 t and y  a sin3 t. (Verify this!) Find an, expression for the slope of the tangent line to the astroid in, terms of t. At what points on the astroid is the slope of the, tangent line equal to 1? Equal to 1?, , 40. The following figure shows the course taken by a yacht during a practice run. The parametric equations of the course are, , y, , x  412 sin t, , y  sin 2t, , 0, , t, , 2p, , where x and y are measured in miles. Find the length of the, course., y (mi), , a, , 1, , a, , x, , a, , 4 2, , 28. Find dy>dx and d 2y>dx 2 if, , 冮, , 1, , t, , sin u, du, u, , y, , and, , 冮, , ln t, , eu du, , 2, , 29. The function y  f(x) is defined by the parametric equations, x  t 5  5t 3  10t  2, , 4, , x (mi), , 1, , a, , x, , 2, , and y  2t 3  3t 2  12t  1, 2 t 2, , Find the absolute maximum and the absolute minimum, values of f., 30. Find the points on the curve with parametric equations, x  t 3  t and y  t 2 at which the tangent line is parallel to, the line with parametric equations x  2t and y  2t  4., , 41. Path of a Boat Two towns, A and B, are located directly opposite each other on the banks of a river that is 1600 ft wide, and flows east with a constant speed of 4 ft/sec. A boat leaving Town A travels with a constant speed of 18 ft/sec always, aimed toward Town B. It can be shown that the path of the, boat is given by the parametric equations, x  800(t 7>9  t 11>9), , y  1600t, , 0, , t, , 1, , Find the distance covered by the boat in traveling from, A to B., y, B, , In Exercises 31–36, find the length of the curve defined by the, parametric equations., 31. x  2t 2, y  3t 3; 0, , t, , 1, , 32. x  2t 3>2,, , 0, , t, , y  3t  1;, , 4, , A, 0, , x

Page 18 :
10.3 The Calculus of Parametric Equations, 42. Trajectory of an Electron An electron initially located at the, origin of a coordinate system is projected horizontally into, a uniform electric field with magnitude E and directed, upward. If the initial speed of the electron is √0, then its, trajectory is, , 865, , 47. Use the result of Exercise 46 to find the area of the region, under one arch of the cycloid x  a(u  sin u),, y  a(1  cos u)., y, , 1 eE, y   a bt 2, 2 m, , x  √0t, , where e is the charge of the electron and m is its mass., Show that the trajectory of the electron is a parabola., 0, , y, Screen, , 0, , x, , 2πa, , 48. Use the result of Exercise 46 to find the area of the region, enclosed by the ellipse with parametric equations, x  a cos u, y  b sin u, where 0 u 2p., 49. Use the result of Exercise 46 to find the area of the region, enclosed by the astroid x  a cos3 u, y  a sin3 u. (See the, figure for Exercise 27.), , x, , 50. Use the result of Exercise 46 to find the area of the region, enclosed by the curve x  a sin t, y  b sin 2t., Note: The deflection of electrons by an electric field is used to con-, , 51. Use the result of Exercise 46 to find the area of the region, lying inside the course taken by the yacht of Exercise 40., , trol the direction of an electron beam in an electron gun., , 43. Refer to Exercise 42. If a screen is placed along the vertical, line x  a, at what point will the electron beam hit the, screen?, 44. Find the point that is located one quarter of the way along, the arch of the cycloid, x  a(t  sin t), , y  a(1  cos t), , 0, , t, , 2p, , as measured from the origin. What is the slope of the tangent line to the cycloid at that point? Plot the arch of the, cycloid and the tangent line on the same set of axes., 45. The cornu spiral is a curve defined by the parametric equations, x  C(t) , , 冮, , t, , 0, , t, , cos(pu 2>2) du, , y  S(t) , , 冮 sin(pu >2) du, 2, , 0, , where C and S are the Fresnel functions discussed in Section 6.7., a. Plot the spiral. Describe the behavior of the curve as, t → ⬁ and as t → ⬁ ., b. Find the length of the spiral from t  0 to t  a., 46. Suppose that the graph of a nonnegative function F on an, interval [a, b] is represented by the parametric equations, x  f(t) and y  t(t) for t in [a, b]. Show that the area of, the region under the graph of F is given by, b, , 冮 t(t)f ¿(t) dt, a, , a, , or, , 冮 t(t)f ¿(t) dt, , In Exercises 52–57, find the area of the surface obtained by, revolving the curve about the x-axis., 52. x  t,, , y  2  t;, , 53. x  t 3,, , y  t 2;, , 54. x  13t 2,, 55. x , , 0, , t, , 0, , 1, , y  t  t 3;, , 1 3, t ,, 3, , y4, , 56. x  et sin t,, , 2, , t, , 0, , 1 2, t ;, 2, , 0, , y  et cos t;, , 57. x  t  sin t,, , t, , 1, t, , 212, , 0, , t, , p, 2, , y  1  cos t;, , 0, , t, , 2p, , In Exercises 58–61, find the area of the surface obtained by, rotating the curve about the y-axis., 58. x  t,, , y  2t;, , 59. x  3t ,, 2, , 0, , t, , y  2t ;, , 0, , 3, , 4, t, , 60. x  a cos t,, , y  b sin t;, , 61. x  et  t,, , y  4et>2;, , 0, , 1, p2, , t, , t, , 1, , p, 2, , 62. Find the area of the surface obtained by revolving the cardioid, x  a(2 cos t  cos 2t), , y  a(2 sin t  sin 2t), , about the x-axis., 63. Find the area of the surface obtained by revolving the astroid, x  a cos3 t, , y  a sin3 t, , about the x-axis., , b, , 64. Find the areas of the surface obtained by revolving one arch, of the cycloid x  a(u  sin u), y  a(1  cos u) about the, x- and y-axes.

Page 19 :
866, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , 65. Find the surface area of the torus obtained by revolving the, circle x 2  (y  b)2  r 2 (0  r  b) about the x-axis., Hint: Represent the equation of the circle in parametric form:, x  r cos t, y  b  r sin t, 0, , 2p., , t, , 66. Show that if f ¿ is continuous and f ¿(t) 0 for a t b,, then the parametric curve defined by x  f(t) and y  t(t), for a t b can be put in the form y  F(x)., cas In Exercises 67–70, (a) plot the curve defined by the parametric, , 74. Use the parametric representation of a circle in Exercise 73, to show that the circumference of a circle of radius a is, 2pa., 75. Find parametric equations for the Folium of Descartes,, x 3  y 3  3axy with parameter t  y>x., cas 76. Use the parametric representation of the Folium of Descartes, , to estimate the length of the loop., cas 77. Show that the length of the ellipse x  a cos t, y  b sin t,, , 0, , equations and (b) estimate the arc length of the curve accurate, to four decimal places., 67. x  2t ,, 2, , ytt ;, 3, , 0, , 68. x  sin(0.5t  0.4p),, , 0t, , 4p, , e, 3 2, t b;, 2, , 2  t  2, , y  2 cos 3t, , 0, , t, , is the eccentricity of the ellipse., cas 78. Use a computer or calculator and the result of Exercise 77, , to estimate the circumference of the ellipse, y2, x2, , 1, 100, 36, , p, 6, , about the x-axis., 72. a. Find an expression for the arc length of the curve defined, by the parametric equations, y  f ⬙(t)sin t  f ¿(t)cos t, , where a t b and f has continuous third-order derivatives., b. Use the result of part (a) to find the arc length, of the curve x  6t cos t  3t 2 sin t and, y  6t sin t  3t 2 cos t, where 0 t 1., 73. Show that, 1  t2, , y, , c, 2a 2  b 2, , a, a, , Note: The integral is called an elliptical integral of the second kind., , surface obtained by revolving the curve, , x, , 21  e2 sin2 t dt, , where, , (swallowtail castastrophe), , 2at, , p>2, , 0, , cas 71. Use a calculator or computer to approximate the area of the, , x  f ⬙(t)cos t  f ¿(t)sin t, , 冮, , y  0.2(6 sin t  sin 6t);, , y  t 2 a1 , , x  4 sin 2t, , 2p, where a  b  0, is given by, L  4a, , 1, , y  sin t;, , 69. x  0.2(6 cos t  cos 6t),, 0 t p, 70. x  2t(1  t 2),, , t, , t, , a(1  t 2), , accurate to three decimal places., In Exercises 79–80, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 79. If x  f(t) and y  t(t), f and t have second-order derivatives, and f ¿(t) 0, then, d 2y, dx, , 2, , , , f ¿(t)t⬙(t)  t¿(t)f ⬙(t), [ f ¿(t)]2, , 80. The curve with parametric equations x  f(t) and y  t(t) is, a line if and only if f and t are both linear functions of t., , 1  t2, , where a  0 and ⬁  t  ⬁ , are parametric equations of, a circle. What are its center and radius?, , 10.4, , Polar Coordinates, The curve shown in Figure 1a is a lemniscate, and the one shown in Figure 1b is called, a cardioid. The rectangular equations of these curves are, (x 2  y 2)2  4(x 2  y 2), , and, , x 4  2x 3  2x 2y 2  2xy 2  y 2  y 4  0, , respectively. As you can see, these equations are somewhat complicated. For example,, they will not prove very helpful if we want to calculate the area enclosed by the two, loops of the lemniscate shown in Figure 1a or the length of the cardioid shown in Figure 1b.

Page 20 :
10.4 Polar Coordinates, , 867, , y, , y, , 1, , 2 x, , 2, , 2 x, 1, , (a) A lemniscate, , (b) A cardioid, , FIGURE 1, A rectangular equation of the lemniscate in part (a) is (x 2  y 2)2  4(x 2  y 2), and an equation, of the cardioid in part (b) is x 4  2x 3  2x 2y 2  2xy 2  y 2  y 4  0., , A question that arises naturally is: Is there a coordinate system other than the rectangular system that we can use to give a simpler representation for curves such as the, lemniscate and cardioid? One such system is the polar coordinate system., , The Polar Coordinate System, P (r, ¨), r, ¨, O Pole, , Polar axis, , FIGURE 2, , To construct the polar coordinate system, we fix a point O called the pole (or origin), and draw a ray (half-line) emanating from O called the polar axis. Suppose that P is, any point in the plane, let r denote the distance from O to P, and let u denote the angle, (in degrees or radians) between the polar axis and the line segment OP. (See Figure 2.), Then the point P is represented by the ordered pair (r, u), also written P(r, u), where, the numbers r and u are called the polar coordinates of P., The angular coordinate u is positive if it is measured in the counterclockwise, direction from the polar axis and negative if it is measured in the clockwise direction., The radial coordinate r may assume positive as well as negative values. If r  0, then, P(r, u) is on the terminal side of u and at a distance r from the origin. If r  0, then, P(r, u) lies on the ray that is opposite the terminal side of u and at a distance of 冟 r 冟  r, from the pole. (See Figure 3.) Also, by convention the pole O is represented by the, ordered pair (0, u) for any value of u. Finally, a plane that is endowed with a polar, coordinate system is referred to as an ru-plane., (r, ¨), , P (r, ¨), r, , r, ¨, , ¨, , O, , O, |r|, P (r, ¨), , FIGURE 3, , (a) r > 0, , (b) r < 0, , EXAMPLE 1 Plot the following points in the ru-plane., , a. 1 1, 2p, 3 2, , b. 1 2, p4 2, , c. 1 2, p3 2, , d. (2, 3p)

Page 21 :
868, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , Solution, , The points are plotted in Figure 4., , (1, 2π3 ), , π, 3, , 2π, 3, 0, , 1, , 0, , 0, , 1, π, 4, , (2, 3π), , (2,  π4 ), (a), , 3π, , 1, , 1, , (2, π3 ), , (b), , (c), , (d), , FIGURE 4, The points in Example 1, , Unlike the representation of points in the rectangular system, the representation of, points using polar coordinates is not unique. For example, the point (r, u) can also be, written as (r, u  2np) or (r, u  (2n  1)p), where n is any integer. Figures 5a, and 5b illustrate this for the case n  1 and n  0, respectively., (r, ¨)  (r, ¨  π), , (r, ¨)  (r, ¨  2π), ¨π, , r, , ¨, , ¨, ¨  2π, , FIGURE 5, Representation of points using, polar coordinates is not unique., , (a) n  1, , (b) n  0, , Relationship Between Polar and Rectangular Coordinates, To establish the relationship between polar and rectangular coordinates, let’s superimpose an xy-plane on an ru-plane in such a way that the origins coincide and the positive x-axis coincides with the polar axis. Let P be any point in the plane other than the, origin with rectangular representation (x, y) and polar representation (r, u). Figure 6a, shows a situation in which r  0, and Figure 6b shows a situation in which r  0. If, r  0, we see immediately from the figure that, cos u , y, , x, r, , sin u , , y, r, y, , P (r, ¨), P (x, y), , P (x, y), |r|, , r, , ¨, , ¨, 0, , FIGURE 6, The relationship between polar, and rectangular coordinates, , x, , 0, P (r, ¨), P (x, y), , (a) r > 0, , (b) r < 0, , x

Page 22 :
10.4 Polar Coordinates, , 869, , so x  r cos u and y  r sin u. If r  0, we see by referring to Figure 6b that, cos u , , x, x, x, , , 冟r冟, r, r, , sin u , , y, y, y, , , 冟r冟, r, r, , so again x  r cos u and y  r sin u. Finally, in either case we have, x 2  y2  r 2, , tan u , , and, , y, x, , if x, , 0, , Relationship Between Rectangular and Polar Coordinates, Suppose that a point P (other than the origin) has representation (r, u) in polar, coordinates and (x, y) in rectangular coordinates. Then, x  r cos u, r 2  x 2  y2, , and, , y  r sin u, tan u , , and, , y, x, , if x, , (1), 0, , (2), , EXAMPLE 2 The point 1 4, p6 2 is given in polar coordinates. Find its representation, , in rectangular coordinates., Solution, , Here, r  4 and u  p>6. Using Equation (1), we obtain, x  r cos u  4 cos, y  r sin u  4 sin, , p, 13, 4ⴢ,  213, 6, 2, , p, 1, 4ⴢ 2, 6, 2, , Therefore, the given point has rectangular representation (2 13, 2)., , EXAMPLE 3 The point (1, 1) is given in rectangular coordinates. Find its representation in polar coordinates., Solution, , Here, x  1 and y  1. Using Equation (2), we have, r 2  x 2  y 2  (1)2  12  2, , and, tan u , , y,  1, x, , Let’s choose r to be positive; that is, r  12. Next, observe that the point (1, 1), lies in the second quadrant and so we choose u  3p>4 (other choices are, u  (3p>4)  2np, where n is an integer). Therefore, one representation of the given, point is 1 12, 3p, 4 2., , Graphs of Polar Equations, The graph of a polar equation r  f(u) or, more generally, F(r, u)  0 is the set of, all points (r, u) whose coordinates satisfy the equation.

Page 23 :
870, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , EXAMPLE 4 Sketch the graphs of the polar equations, and reconcile your results by, finding the corresponding rectangular equations., a. r  2, , b. u , , 2p, 3, , Solution, a. The graph of r  2 consists of all points P(r, u) where r  2 and u can assume, any value. Since r gives the distance between P and the pole O, we see that the, graph consists of all points that are located a distance of 2 units from the pole;, in other words, the graph of r  2 is the circle of radius 2 centered at the pole., (See Figure 7a.) To find the corresponding rectangular equation, square both sides, of the given equation obtaining r 2  4. But by Equation (2), r 2  x 2  y 2, and, this gives the desired equation x 2  y 2  4. Since this is a rectangular equation, of a circle with center at the origin and radius 2, the result obtained earlier has, been confirmed., y, , y, P (r, ¨), 2π, ¨ 3, , r2, 2, x, , O, , FIGURE 7, , 2π, 3, O, , x, , 2π, (b) The graph of ¨  3, , (a) The graph of r  2, , b. The graph of u  2p>3 consists of all points P(r, u) where u  2p>3 and r can, assume any value. Since u measures the angle the line segment OP makes with, the polar axis, we see that the graph consists of all points that are located on the, straight line passing through the pole O and making an angle of 2p>3 radians, with the polar axis. (See Figure 7b.) Observe that the half-line in the second, quadrant consists of points for which r  0, whereas the half-line in the fourth, quadrant consists of points for which r  0. To find the corresponding rectangular equation, we use Equation (2), tan u  y>x, to obtain, tan, , y, 2p, , x, 3, , or, , y,   13, x, , or y   13x. This equation confirms that the graph of u  2p>3 is a straight, line with slope  13., As in the case with rectangular equations, we can often obtain a sketch of the graph, of a simple polar equation by plotting and connecting some points that lie on the graph., , EXAMPLE 5 Sketch the graph of the polar equation r  2 sin u. Find a corresponding rectangular equation and reconcile your results., Solution The following table shows the values of r corresponding to some convenient, values of u. It suffices to restrict the values of u to those lying between 0 and p, since, values of u beyond p will give the same points (r, u) again.

Page 24 :
10.4 Polar Coordinates, , 0, , p, 6, , p, 4, , p, 3, , p, 2, , 2p, 3, , 3p, 4, , 5p, 6, , p, , r, , 0, , 1, , 12 ⬇ 1.4, , 13 ⬇ 1.7, , 2, , 13 ⬇ 1.7, , 12 ⬇ 1.4, , 1, , 0, , The graph of r  2 sin u is sketched in Figure 8. To find a corresponding rectangular equation, we multiply both sides of r  2 sin u by r to obtain r 2  2r sin u and, then use the relationships r 2  x 2  y 2 (Equation (2)) and y  r sin u (Equation (1)),, to obtain the desired equation, , (2, π2 ), (√ 3, π3 ), (√ 2, π4 ), (1, π6 ), , (√ 3, 2π3 ), (√ 2, 3π4 ), (1, 5π6 ), , U, , 871, , x 2  y 2  2y, , x 2  y 2  2y  0, , or, , Finally, completing the square in y, we have, , O, , FIGURE 8, The graph of r  2 sin u is a circle. To, plot the points, first draw the ray with, the desired angle, then locate the point, by measuring off the required distance, from the pole., , x 2  y 2  2y  (1)2  1, or, x 2  (y  1)2  1, which is an equation of the circle with center (0, 1) and radius 1, as obtained earlier., , It might have occurred to you that in the last several examples we could have, obtained the graphs of the polar equations by first converting them to the corresponding rectangular equations. But as you will see, some curves are easier to graph using, polar coordinates., , Symmetry, Just as the use of symmetry is helpful in graphing rectangular equations, its use is, equally helpful in graphing polar equations. Three types of symmetry are illustrated in, Figure 9. The test for each type of symmetry follows., , (r, ¨), , O, , ¨=π, 2, (r, π  ¨), π¨, , ¨, ¨, , (r, ¨), , ¨, O, , O, (r, ¨), , (r, ¨), , FIGURE 9, Symmetries of graphs, of polar equations, , (r, ¨), ¨, , (a) Symmetry with respect to, the polar axis, , (b) Symmetry with respect to, the line ¨ = π, 2, , (c) Symmetry with respect to, the pole, , Tests for Symmetry, a. The graph of r  f(u) is symmetric with respect to the polar axis if the, equation is unchanged when u is replaced by u., b. The graph of r  f(u) is symmetric with respect to the vertical line, u  p>2 if the equation is unchanged when u is replaced by p  u., c. The graph of r  f(u) is symmetric with respect to the pole if the equation is unchanged when r is replaced by r or when u is replaced by, u  p.

Page 25 :
872, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , To illustrate the use of the tests for symmetry, consider the equation r  2 sin u of, Example 5. Here, f(u)  2 sin u, and since, f(p  u)  2 sin(p  u)  2(sin p cos u  cos p sin u)  2 sin u  f(u), we conclude that the graph of r  2 sin u is symmetric with respect to the vertical line, u  p>2 (Figure 8)., , EXAMPLE 6 Sketch the graph of the polar equation r  1  cos u. This is the polar, form of the rectangular equation x 4  2x 3  2x 2y 2  2xy 2  y 2  y 4  0 of the cardioid that was mentioned at the beginning of this section (Figure 1b)., Solution, , Writing f(u)  1  cos u and observing that, f(u)  1  cos(u)  1  cos u  f(u), , we conclude that the graph of r  1  cos u is symmetric with respect to the polar, axis. In view of this, we need only to obtain that part of the graph between u  0 and, u  p. We can then complete the graph using symmetry., To sketch the graph of r  1  cos u for 0 u p, we can proceed as we did in, Example 5 by first plotting some points lying on that part of the graph, or we may proceed as follows: Treat r and u as rectangular coordinates, and make use of our knowledge of graphing rectangular equations to obtain the graph of r  f(u)  1  cos u on, the interval [0, p]. (See Figure 10a.) Then recalling that u is the angular coordinate, and r is the radial coordinate, we see that as u increases from 0 to p, the points on, the respective rays shrink to 0. (See Figure 10b, where the corresponding points are, shown.), r, 2, , a, , (π4 , 1  √22 ), b, , 1, , c, , π, 2, , ¨ π, 4, 1, , b, 1, , (π2 , 1), d, , π, 4, , ¨ π, 2, , ¨  3π, 4, , ( 3π4 , 1  √22 ), 3π, 4, , e, π, , c, 2, a, , d, e O, , ¨, , (a) r  f(¨), treating r and ¨ as rectangular coordinates, , 2, , ¨0, , (b) r  f(¨), treating r and ¨ as polar coordinates, , FIGURE 10, Two steps in sketching the graph of the polar equation r  1  cos u, , 1, , FIGURE 11, The graph of r  1  cos u, is a cardioid., , Finally, using symmetry, we complete the graph of r  1  cos u, as shown in Figure 11. It is called a cardioid because it is heart-shaped., , EXAMPLE 7 Sketch the graph of the polar equation r  2 cos 2u., Solution, , Write f(u)  2 cos 2u, and observe that, f(u)  2 cos 2(u)  2 cos 2u  f(u), , and, f(p  u)  2 cos 2(p  u)  2 cos(2p  2u),  2[cos 2p cos 2u  sin 2p sin 2u]  2 cos 2u  f(u)

Page 26 :
10.4 Polar Coordinates, , 873, , Therefore, the graph of the given equation is symmetric with respect to both the polar, axis and the vertical line u  p>2. It suffices, therefore, to obtain an accurate sketch, of that part of the graph for 0 u p2 and then complete the sketch of the graph using, symmetry. Proceeding as in Example 6, we first sketch the graph of r  2 cos 2u for, 0 u p2 treating r and u as rectangular coordinates (Figure 12a), and then transcribe, the information contained in this graph onto the graph in the ru-plane for 0 u p2 ., (See Figure 12b.), ¨ π, 2, , r, 2, , a, , (π8 , √ 2 ), , b, 1, , 0.5, c, , c, π, 8, , π, 4, , d, , ¨, , π, 2, , 3π, 8, , 1, , ¨ π, 4, ¨ π, 8, , b, a, 2, , ( 3π8 , √ 2 ), , 2, , FIGURE 12, Two steps in sketching the, graph of r  2 cos 2u, , ¨  3π, 8, , d, , e, , e 2, , (a) r  f(¨) treating r and ¨ as, rectangular coordinates, , (b) r  f(¨), treating r and ¨ as, polar coordinates, , Finally, using the symmetry that was established earlier (Figure 13a), we complete, the graph of r  2 cos 2u as shown in Figure 13b. This graph is called a four-leaved, rose., 2, , 2, 2, , FIGURE 13, The graph of r  2 cos 2u, is a four-leaved rose., , 2, (a), , (b), , The next example shows how the graph of a rectangular equation can be sketched, more easily by first converting it to polar form., , EXAMPLE 8 Sketch the graph of the equation (x 2  y 2)2  4(x 2  y 2) by first converting it to polar form. This is an equation of the lemniscate that was mentioned at, the beginning of this section., Solution To convert the given equation to polar form, we use Equations (1) and (2),, obtaining, (r 2)2  4(r 2 cos2 u  r 2 sin2 u),  4r 2(cos2 u  sin2 u), r 4  4r 2 cos 2u

Page 27 :
874, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , or, r 2  4 cos 2u, Observe that f(u)  21cos 2u is defined for p4, u p4 and 3p, u 5p, 4, 4 . Also,, observe that f(u)  f(u) and f(p  u)  f(u). (These computations are similar to, those in Example 7.) So the graph of r  21cos 2u is symmetric with respect to the, polar axis and the line u  p>2. The graph of r  f(u) for 0 u p4 , where r and u, are treated as rectangular coordinates, is shown in Figure 14a. This leads to the part of, the required graph for 0 u p4 shown in Figure 14b. Then, using symmetry, we, obtain the graph of r  21cos 2u and, therefore, that of (x 2  y 2)2  4(x 2  y 2), as, shown in Figure 15., r, ¨ π, 4, , 2, , π, 4, , (a), , 2, , ¨, , 2, , 2, , FIGURE 15, The graph of r  21cos 2u is a, lemniscate., , (b), , FIGURE 14, Two steps in sketching the graph of r  21cos 2u, , Tangent Lines to Graphs of Polar Equations, To find the slope of the tangent line to the graph of r  f(u) at the point P(r, u), let, P(x, y) be the rectangular representation of P. Then, x  r cos u  f(u) cos u, y  r sin u  f(u) sin u, We can view these equations as parametric equations for the graph of r  f(u) with, parameter u. Then, using Equation (1) of Section 10.3, we have, dy, dr, sin u  r cos u, dy, du, du, , , dx, dx, dr, cos u  r sin u, du, du, , if, , dx, du, , 0, , (3), , and this gives the slope of the tangent line to the graph of r  f(u) at any point P(r, u)., The horizontal tangent lines to the graph of r  f(u) are located at the points where, dy>du  0 and dx>du 0. The vertical tangent lines are located at the points where, dx>du  0 and dy>du 0 (so that dy>dx is undefined). Also, points where both dy>du, and dx>du are equal to zero are candidates for horizontal or vertical tangent lines,, respectively, and may be investigated using l’Hôpital’s Rule., Equation (3) can be used to help us find the tangent lines to the graph of r  f(u), at the pole. To see this, suppose that the graph of f passes through the pole when, u  u0. Then f(u0)  0. If f ¿(u0) 0, then Equation (3) reduces to, f ¿(u0) sin u0  f(u0) cos u0, dy, sin u0, , ,  tan u0, dx, f ¿(u0) cos u0  f(u0) sin u0, cos u0

Page 28 :
10.4 Polar Coordinates, , 875, , This shows that u  u0 is a tangent line to the graph of r  f(u) at the pole (0, u0) ., The following summarizes this discussion., u  u0 is a tangent line to the graph of r  f(u) at the pole if f(u0)  0 and, f ¿(u0) 0., , EXAMPLE 9 Consider the cardioid r  1  cos u of Example 6., a. Find the slope of the tangent line to the cardioid at the point where u  p>6., b. Find the points on the cardioid where the tangent lines are horizontal and where, the tangent lines are vertical., Solution, a. The slope of the tangent line to the cardioid r  1  cos u at any point P(r, u) is, given by, dr, sin u  r cos u, dy, (sin u)(sin u)  (1  cos u)cos u, du, , , dx, dr, (sin u)(cos u)  (1  cos u)sin u, cos u  r sin u, du, , , (cos2 u  sin2 u)  cos u, cos 2u  cos u, , 2 sin u cos u  sin u, sin 2u  sin u, , At the point on the cardioid where u  p>6, the slope of the tangent line is, dy, `, dx up>6, , p, p, 1, 13, cosa b  cosa b, , 3, 6, 2, 2, , ,  1, p, p, 13, 1, sina b  sina b, , 3, 6, 2, 2, , b. Observe that dy>du  0 if, cos 2u  cos u  0, 2 cos u  cos u  1  0, 2, , (2 cos u  1)(cos u  1)  0, that is, if cos u  or cos u  1. This gives, 1, 2, , u, , p, , p, or, 3, , 5p, 3, , Next, dx>du  0 if, sin 2u  sin u  0, 2 sin u cos u  sin u  0, sin u (2 cos u  1)  0, that is, if sin u  0 or cos u  12. This gives, u  0,, , p,, , 2p, ,, 3, , or, , 4p, 3

Page 29 :
876, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , In view of the remarks following Equation (3), we see that u  p>3 and, u  5p>3 give rise to horizontal tangents. To investigate the candidate u  p,, where both dy>du and dx>du are equal to zero, we use l’Hôpital’s Rule. Thus,, lim, , u→p, , ( 32 , π3 ), , dy, cos 2u  cos u,   lim, dx, u→p sin 2u  sin u,   lim, , 1, , (, , 1 2π, 2, 3, , ), , u→p, , 2 sin 2u  sin u, 0, 2 cos 2u  cos u, , Similarly, we see that, , (0, π), 0, , (2, 0), 2, , ( 12 , 4π3 ), , lim, , u→p, , Therefore, u  p also gives rise to a horizontal tangent. Thus, the horizontal tangent lines occur at, , 1, , 1 32, p3 2 ,, , ( 32 , 5π3 ), FIGURE 16, The horizontal and vertical tangents to, the graph of r  1  cos u, , ¨ π, 4, (y  x), , ¨  3π, 4, (y  x), , dy, 0, dx, , (0, p) , and, , 1 32, 5p3 2, , The vertical tangent lines occur at u  0, 2p>3, and 4p>3. The points are (2, 0),, 1 12, 2p3 2 , and 1 12, 4p3 2 . These tangent lines are shown in Figure 16., , EXAMPLE 10 Find the tangent lines of r  cos 2u at the origin., Solution, , Setting f(u)  cos 2u  0, we find that, 2u , , p, ,, 2, , 3p, ,, 2, , 5p, , or, 2, , 7p, 2, , u, , p, ,, 4, , 3p, ,, 4, , 5p, ,, 4, , 7p, 4, , or, , FIGURE 17, The tangent lines to the graph of, r  cos 2u at the origin, , 10.4, , or, , Next, we compute f ¿(u)  2 sin 2u. Since f ¿(u) 0 for each of these values of u, we, see that u  p>4 and u  3p>4 (that is, y  x and y  x) are tangent lines to the, graph of r  cos 2u at the pole (see Figure 17)., , CONCEPT QUESTIONS, , 1. Let P(r, u) be a point in the plane with polar coordinates r, and u. Find all possible representations of P(r, u)., 2. Suppose that P has representation (r, u) in polar coordinates, and (x, y) in rectangular coordinates. Express (a) x and y in, terms of r and u and (b) r and u in terms of x and y., 3. Explain how you would determine whether the graph of, r  f(u) is symmetric with respect to (a) the polar axis,, (b) the vertical line u  p>2, and (c) the pole., , 4. Suppose that r  f(u), where f is differentiable., a. Write an expression for dy>dx., b. How do you find the points on the graph of r  f(u), where the tangent lines are horizontal and where the tangent lines are vertical?, c. How do you find the tangent lines to the graph of, r  f(u) (if they exist) at the pole?

Page 30 :
10.4 Polar Coordinates, , 10.4, , 877, , EXERCISES, , In Exercises 1–8, plot the point with the polar coordinates. Then, find the rectangular coordinates of the point., , 49. r  4(1  sin u), , 50. r  3  3 cos u, 52. r  3 sec u, , 1. 1 4, p4 2, , 2. 1 2, p6 2, , 51. r  2 csc u, , 5. 1 12, p4 2, , 6. 1 1, p3 2, , 54. r , , 3. 1 4, 3p, 2 2, , 53. r  u,, , 4. (6, 3p), , 7. 1 4, 3p, 4 2, , 8. 1 5, 5p, 6 2, , In Exercises 9–16, plot the point with the rectangular coordinates. Then find the polar coordinates of the point taking r  0, and 0 u  2p., , 1, u, , u  0 (spiral), (spiral), , 55. r  eu,, , u  0 (logarithmic spiral), , 1, 56. r 2 , u, , (lituus), , 57. r 2  4 sin 2u (lemniscate), , 9. (2, 2), , 10. (1, 1), , 58. r  1  2 cos u (limaçon), , 11. (0, 5), , 12. (3, 4), , 59. r  3  2 sin u, , 13. (13, 13), , 14. (2 13, 2), , 60. r  sin 2u (four-leaved rose), , 15. (5, 12), , 16. (3, 1), , 61. r  sin 3u (three-leaved rose), , In Exercises 17–24, sketch the region comprising points whose, polar coordinates satisfy the given conditions., 17. r  1, , 18. r  1, , 19. 0, , r, , 2, , 20. 1, , r2, , 21. 0, , u, , p, 4, , 22. 0, , r, , 23. 1, , r, , 3,, , 24. 2  r  4,, , p6, , u, , 3,, , 0, , u, , p, 3, , p, 6, , p2  u  p2, , (limaçon), , 62. r  2 cos 4u (eight-leaved rose), 63. r  4 sin 4u, , (eight-leaved rose), , 64. r  2 sin 5u, , (five-leaved rose), , In Exercises 65–72, find the slope of the tangent line to the, curve with the polar equation at the point corresponding to the, given value of u., 65. r  4 cos u, u , , In Exercises 25–32, convert the polar equation to a rectangular, equation., 25. r cos u  2, , 26. r sin u  3, , 27. 2r cos u  3r sin u  6, , 28. r sin u  2r cos u, , 29. r 2  4r cos u, , 30. r 2  sin 2u, , 1, 31. r , 1  sin u, , 3, 32. r , 4  5 cos u, , In Exercises 33–38, convert the rectangular equation to a polar, equation., , 67. r  sin u  cos u,, 68. r  1  3 cos u,, , p, 3, u, u, , 69. r  u, u  p, 71. r 2  4 cos 2u, u , , u, , p, 4, , 70. r  sin 3u, u , , p, 3, , 66. r  3 sin u,, , p, 6, , p, 4, , p, 2, , 72. r  2 sec u,, , u, , p, 4, , 33. x  4, , 34. x  2y  3, , In Exercises 73–78, find the points on the curve with the given, polar equation where the tangent line is horizontal or vertical., , 35. x 2  y 2  9, , 36. x 2  y 2  1, , 73. r  4 cos u, , 74. r  sin u  cos u, , 37. xy  4, , 38. y 2  x 2  42x 2  y 2, , 75. r  sin 2u, , 76. r 2  4 cos 2u, , 77. r  1  2 cos u, , 78. r  1  sin u, , In Exercises 39–64, sketch the curve with the polar equation., 39. r  3, , 40. r  2, , p, 41. u , 3, , p, 42. u  , 6, , 43. r  3 cos u, , 44. r  4 sin u, , 45. r  3 cos u  2 sin u, , 46. r  2 sin u  4 cos u, , 47. r  1  cos u, , 48. r  1  sin u, , 79. Show that the rectangular equation, x 4  2x 3  2x 2y 2  2xy 2  y 2  y 4  0, is an equation of the cardioid with polar equation, r  1  cos u., 80. Show that the polar equation r  a sin u  b cos u, where a, and b are nonzero, represents a circle. What are the center, and radius of the circle?, , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 31 :
878, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , 81. a. Show that the distance between the points with polar, coordinates (r1, u1) and (r2, u2) is given by, , In Exercises 85–92, use a graphing utility to plot the curve with, the polar equation., 85. r  cos u(4 sin2 u  1), 0, , d  2r 21  r 22  2r1r2 cos(u1  u2), , 86. r  3 sin u cos u, 0, , b. Find the distance between the points with polar coordip, nates 1 4, 2p, 3 2 and 1 2, 3 2 ., , 82. Show that the curves with polar equations r  a sin u and, r  a cos u intersect at right angles., , u, 87. r  0.3c1  2 sina b d , 0, 2, (nephroid of Freeth), 1  10 cos u, , 0, 1  10 cos u, , 83. a. Plot the graphs of the cardioids r  a(1  cos u) and, r  a(1  cos u)., b. Show that the cardioids intersect at right angles except at, the pole., , 88. r , , 84. Let c be the angle between the radial line OP and the tangent line to the curve with polar equation r  f(u) at P (see, the figure). Show that, , 90. r 2 , , du, tan c  r, dr, Hint: Observe that c  f  u. Then use the trigonometric identity, tan(a  b) , , sin u  3.6 cos u, 2, , 0, , u  2p (hippopede curve), , 2, , sin2 u  cos2 u, , 91. r , , 0.1, , 0, cos 3u, , 92. r , , sin u, , 6p, u, , up, , , 0, , u  2p, , (devil’s curve), , (epi-spiral), , u  6p, , (cochleoid), , 93. If P(r1, u1) and P(r2, u2) represent the same point in polar, coordinates, then r1  r2., , r  f(¨), , 94. If P(r1, u1) and P(r2, u2) represent the same point in polar, coordinates, then u1  u2., , , , 95. The graph of r  f(u) has a horizontal tangent line at a, point on the graph if dy>du  0, where y  f(u)sin u., , P, ƒ, x, , O, , 10.5, , u  2p, , 89. r 2  0.8(1  0.8 sin2 u),, 1, 4, , u  4p, , In Exercises 93–95, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., , tan a  tan b, 1  tan a tan b, , y, , ¨, , u  2p, , u  2p, , 2, , Areas and Arc Lengths in Polar Coordinates, In this section we see how the use of polar equations to represent curves such as lemniscates and cardioids will simplify the task of finding the areas of the regions enclosed, by these curves as well as the lengths of these curves., , Areas in Polar Coordinates, r, , ¨, , A, , To develop a formula for finding the area of a region bounded by a curve defined by, a polar equation, we need the formula for the area of a sector of a circle, A, , 2π, , FIGURE 1, The area of a sector of a, circle is A  12 r 2u., , 1 2, r u, 2, , (1), , where r is the radius of the circle and u is the central angle measured in radians. (See, Figure 1.) This formula follows by observing that the area of a sector is u>(2p) times, that of the area of a circle; that is,, A, , u, 1, ⴢ pr 2  r 2u, 2p, 2

Page 32 :
10.5 Areas and Arc Lengths in Polar Coordinates, , 879, , Now let R be a region bounded by the graph of the polar equation r  f(u) and, the rays u  a and u  b, where f is a nonnegative continuous function and, 0 b  a  2p, as shown in Figure 2a. Let P be a regular partition of the interval, [a, b]:, a  u0  u1  u2  p  un  b, ¨∫, , ¨∫, r  f(¨), , ¨  ¨k*, ¨  ¨k 1, , Ψ, , ¨å, , ¨å, , R, , FIGURE 2, , rk*  f (¨k* ), ¨  ¨k, , 0, , 0, , (a) The region R, , (b) The k th subregion, , The rays u  uk divide R into n subregions R1, R2, p , Rn of area ⌬A1, ⌬A2, p ,, ⌬An, respectively. If we choose u*k in the interval [uk1, uk], then the area of ⌬Ak of the, kth subregion bounded by the rays u  uk1 and u  uk is approximated by the sector, of a circle with central angle, ⌬u , , ba, n, , and radius f(u*k ) (highlighted in Figure 2b). Using Equation (1), we have, ⌬Ak ⬇, , 1, [f(u*k )]2 ⌬u, 2, , Therefore, an approximation of the area A of R is, n, n, 1, A  a ⌬Ak ⬇ a [ f(u*k )]2 ⌬u, k1, k1 2, , (2), , But the sum in Equation (2) is a Riemann sum of the continuous function 12 f 2 over the, interval [a, b]. Therefore, it is true, although we will not prove it here, that, n, 1, A  lim a [ f(u*k )]2 ⌬u , n→⬁ k1 2, , 冮, , b, , a, , 1, [ f(u)]2 du, 2, , THEOREM 1 Area Bounded by a Polar Curve, Let f be a continuous, nonnegative function on [a, b] where 0 b  a  2p., Then the area A of the region bounded by the graphs of r  f(u), u  a, and, u  b is given by, A, , 冮, , b, , a, , 1, [f(u)]2 du , 2, , 冮, , b, , a, , 1 2, r du, 2, , Note When you determine the limits of integration, keep in mind that the region R, is swept out in a counterclockwise direction by the ray emanating from the origin, starting at the angle a and terminating at the angle b.

Page 33 :
880, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, π, ¨4, , 2, , 2, , EXAMPLE 1 Find the area of the region enclosed by the lemniscate r 2  4 cos 2u., , This lemniscate has rectangular equation x 4  2x 2y 2  4x 2  4y 2  y 4  0, as you, can verify., Solution The lemniscate is shown in Figure 3. Making use of symmetry, we see that, the required area A is four times that of the area swept out by the ray emanating from, the origin as u increases from 0 to p>4. In other words,, , FIGURE 3, The region enclosed by the lemniscate, r 2  4 cos 2u, , A4, , 冮, , 0, , p>4, , 1 2, r du  8, 2, ,  C4 sin 2uD 0, , p>4, , 冮, , p>4, , cos 2u du, , 0, , 4, , EXAMPLE 2 Find the area of the region enclosed by the cardioid r  1  cos u., Solution The graph of the cardioid r  1  cos u, sketched previously in Example 6, in Section 10.4, is reproduced in Figure 4. Observe that the ray emanating from the, origin sweeps out the required region exactly once as u increases from 0 to 2p. Therefore, the required area A is, , 1, , 2, , A, , 1, , 冮, , 2p, , 0, , 1 2, r du , 2, , , , FIGURE 4, The region enclosed by the, cardioid r  1  cos u, , , , , , 1, 2, 1, 2, 1, 2, , 冮, , 2p, , 冮, , 2p, , 冮, , 2p, , 冮, , 2p, , 0, , 1, (1  cos u)2 du, 2, , (1  2 cos u  cos2 u) du, , 0, , 0, , 0, , a1  2 cos u , , 1  cos 2u, b du, 2, , 3, 1, a  2 cos u  cos 2ub du, 2, 2, , 2p, 1 3, 1, 3, c u  2 sin u  sin 2ud  p, 2 2, 4, 2, 0, , EXAMPLE 3 Find the area inside the smaller loop of the limaçon r  1  2 cos u., ¨  2π, 3, , r  1  2 cos ¨, , ¨0, , Solution We first sketch the limaçon r  1  2 cos u (Figure 5). Observe that the, region of interest is swept out by the ray emanating from the origin as u runs from, 2p>3 to 4p>3. We can also take advantage of symmetry by observing that the required, area is double the area of the smaller loop lying below the polar axis. Since this region, is swept out by the ray emanating from the origin as u runs from 2p>3 to p, we see, that the required area is, A2, , 冮, , p, , 2p>3, , ¨  4π, 3, , FIGURE 5, The limaçon r  1  2 cos u, , , , 冮, , p, , 冮, , p, , 冮, , p, , 1 2, r du , 2, , 冮, , p, , r 2 du, , 2p>3, , (1  2 cos u)2 du, , 2p>3, , , , (1  4 cos u  4 cos2 u) du, , 2p>3, , , , 2p>3, , c1  4 cos u  4a, , 1  cos 2u, b d du, 2

Page 34 :
10.5 Areas and Arc Lengths in Polar Coordinates, , , , 冮, , 881, , p, , (3  4 cos u  2 cos 2u) du, , 2p>3, ,  C3u  4 sin u  sin 2uD 2p>3, p, ,  3p  a2p  4 ⴢ, , 13, 13, 313, , bp, 2, 2, 2, , Area Bounded by Two Graphs, ¨∫, , R, , ¨å, , Consider the region R bounded by the graphs of the polar equations r  f(u) and, r  t(u), and the rays u  a and u  b, where f(u)  t(u)  0 and 0 b  a  2p., (See Figure 6.) From the figure we can see that the area A of R is found by subtracting the area of the region inside r  t(u) from the area of the region inside r  f(u)., Using Theorem 1, we obtain the following theorem., , r  f(¨), , THEOREM 2 Area Bounded by Two Polar Curves, , r  g(¨), 0, , FIGURE 6, R is the region bounded by the, graphs of r  f(u) and r  t(u) for, a u b., , Let f and t be continuous on [a, b], where 0 t(u) f(u) and, 0 b  a  2p. Then the area A of the region bounded by the graphs of, r  t(u), r  f(u), u  a, and u  b is given by, 1, 2, , A, , b, , 冮 {[ f(u)], , 2, ,  [t(u)]2} du, , a, , EXAMPLE 4 Find the area of the region that lies outside the circle r  3 and inside, the cardioid r  2  2 cos u., ∫π, 3, 3, r  g(¨), , r  f(¨), , 2, , R, 3, , 3, , 4, , Solution We first sketch the circle r  3 and the cardioid r  2  2 cos u. The, required region is shown shaded in Figure 7., To find the points of intersection of the two curves, we solve the two equations, simultaneously. We have 2  2 cos u  3 or cos u  12, which gives u  p>3. Since, the region of interest is swept out by the ray emanating from the origin as u varies from, p>3 to p>3, we see that the required area is, by Theorem 2,, , 2, 3, , 1, 2, , A, ¨  π, 3, , FIGURE 7, R is the region outside the circle r  3, and inside the cardioid r  2  2 cos u., , b, , 冮 {[f(u)], , 2, ,  [t(u]2} du, , a, , where f(u)  2  2 cos u  2(1  cos u), t(u)  3, a  p>3, and b  p>3. If we, take advantage of symmetry, we can write, 1, A  2a b, 2, , 冮, 冮, 冮, , 冮, , p>3, , {[2(1  cos u)]2  32} du, , 0, , p>3, , , , (4  8 cos u  4 cos2 u  9) du, , 0, p>3, , 0, , a5  8 cos u  4 ⴢ, , 1  cos 2u, b du, 2, , p>3, , (3  8 cos u  2 cos 2u) du, , 0, ,  C3u  8 sin u  sin 2uD 0, , p>3, ,  ap  8a, , 13, 13, 913, b, b, p, 2, 2, 2

Page 35 :
882, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , Arc Length in Polar Coordinates, To find the length of a curve C defined by a polar equation r  f(u) for a, we use Equation (1) in Section 10.4 to write the parametric equations, x  r cos u  f(u) cos u, , y  r sin u  f(u) sin u, , and, , a, , u, , u, , b,, , b, , for the curve, regarding u as the parameter. Then, dx,  f ¿(u) cos u  f(u) sin u, du, , dy,  f ¿(u) sin u  f(u) cos u, du, , and, , Therefore,, a, , dy 2, dx 2, b  a b  [ f ¿(u)]2 cos2 u  2f ¿(u) f(u) cos u sin u  [ f(u)]2 sin2 u, du, du,  [ f ¿(u)]2 sin2 u  2f ¿(u) f(u) cos u sin u  [ f(u)]2 cos2 u,  [ f ¿(u)]2  [ f(u)]2, , sin2 u  cos2 u  1, , Consequently, if f ¿ is continuous, then Theorem 1 in Section 10.3 gives the arc length, of C as, L, , 冮, , b, , a, , dy 2, dx 2, b  a b du , B du, du, a, , b, , 冮 2[ f ¿(u)], , 2, ,  [ f(u)]2 du, , a, , THEOREM 3 Arc Length, Let f be a function with a continuous derivative on an interval [a, b]. If the graph, C of r  f(u) is traced exactly once as u increases from a to b, then the length, L of C is given by, L, , 冮, , b, , 2[ f ¿(u)]2  [ f(u)]2 du , , a, , 冮, , b, , a, , dr 2, b  r 2 du, B du, a, , EXAMPLE 5 Find the length of the cardioid r  1  cos u., Solution The cardioid is shown in Figure 8. Observe that the cardioid is traced exactly, once as u runs from u to 2p. However, we can also take advantage of symmetry to see, that the required length is twice that of the length of the cardioid lying above the polar, axis. Thus,, , 1, , 2, , L2, , 1, , 冮, , 0, , p, , dr 2, b  r 2 du, B du, a, , But r  1  cos u, so, FIGURE 8, The cardioid r  1  cos u, , dr,  sin u, du, Therefore,, L2, , 冮, , p, , 冮, , p, , 2(sin u)2  (1  cos u)2 du, , 0, , 2, , 0, , 2sin2 u  1  2 cos u  cos2 u du

Page 36 :
10.5 Areas and Arc Lengths in Polar Coordinates, , 2, , 883, , 冮, , p, , 12  2 cos u du, , sin2 u  cos2 u  1, , 0, ,  212, , p, , 冮, , 11  cos u du  212, , 0, , 4, , 冮, , 0, , p, , 冮, , p, , 0, , ` cos, , u, ` du  4, 2, , 冮, , p, , cos, , 0, , u, du, 2, , B, , 2 cos2, , cos, , u, du, 2, , u,  0 on [0, p], 2, , u p,  c4(2) sin d  8, 2 0, , Area of a Surface of Revolution, The formulas for finding the area of a surface obtained by revolving a curve defined, by a polar equation about the polar axis or about the line u  p>2 can be derived by, using Equations (8) and (9) of Section 10.3 and the equations x  r cos u and, y  r sin u., , THEOREM 4 Area of a Surface of a Revolution, Let f be a function with a continuous derivative on an interval [a, b]. If the graph, C of r  f(u) is traced exactly once as u increases from a to b, then the area of, the surface obtained by revolving C about the indicated line is given by, a. S  2p, , 冮, , b, , 冮, , b, , a, , b. S  2p, , a, , dr 2, r sin u a b  r 2 du, B du, dr 2, r cos u a b  r 2 du, B du, , (about the polar axis), (about the line u  p>2), , r cos ¨, , Note In using Theorem 4, we must choose [a, b] so that the surface is only traced, once when C is revolved about the line., , 1, , EXAMPLE 6 Find the area S of the surface obtained by revolving the circle r  cos u, about the line u  p>2. (See Figure 9.), , (a), π, __, 2, , Solution Observe that the circle is traced exactly once as u increases from 0 to p., Therefore, using Theorem 4 with r  cos u, a  0, and b  p, we obtain, S  2p, , 冮, , b, , 冮, , p, , 冮, , p, , f(u) cos u, , a, , 0, ,  2p, , dr 2, b  r 2 du, B du, a, , cos u(cos u)2(sin u)2  (cos u)2 du, , 0, , (b), , FIGURE 9, The solid obtained by revolving, the circle r  cos u (a) about, the line u  p>2 is a torus (b)., ,  2p, , cos2 u du  p, , 0, ,  pcu , , 冮, , p, , (1  cos 2u) du, , 0, , sin 2u p, d  p2, 2, 0, , Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

Page 37 :
884, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, , Points of Intersection of Graphs in Polar Coordinates, In Example 4 we were able to find the points of intersection of two curves with representations in polar coordinates by solving a system of two equations simultaneously., This is not always the case. Consider for example, the graphs of the cardioid, r  1  cos u and the circle r  3 cos u shown in Figure 10. Solving the two equations simultaneously, we obtain, , (, ), 3_ π, _, 2 3, , 1, , 1, , 2, , 3, , 3 cos u  1  cos u, , 1, , cos u , , ( ), 3_ 5π, __, 2, 3, , (3), , 1, 2, , or u  p>3 and 5p>3. Therefore, the points of intersection are 1 32, p3 2 and 1 32, 5p, 3 2 . But, one glance at Figure 10 shows the pole as a third point of intersection that is not revealed, in our calculation. To see how this can happen, think of the cardioid as being traced, by the point (r, u) satisfying, , FIGURE 10, The graphs of the cardioid, r  1  cos u and the circle, r  3 cos u, , r  f(u)  1  cos u, , 0, , u, , 2p, , with u as a parameter. If we think of u as representing time, then as u runs from u  0, through u  2p, the point (r, u) starts at (2, 0) and traverses the cardioid in a counterclockwise direction, eventually returning to the point (2, 0). (See Figure 11a.) Similarly, the circle is traced twice in the counterclockwise direction, by the point (r, u),, where, r  t(u)  3 cos u, , 0, , u, , 2p, , and the parameter u, once again representing time, runs from u  0 through u  2p, (see Figure 11b)., , ( 32 , π3 ), (0, π), , (2, 0), , (0, π2 ), , ( 32 , π3 ), (3, 0), , ( 32 , 23π ), ( 32 , 53π ), FIGURE 11, , (a), , (b), , Observe that the point tracing the cardioid arrives at the point 1 32, p3 2 on the cardioid, at precisely the same time that the point tracing the circle arrives at the point 1 32, p3 2 on, the circle. A similar observation holds at the point 1 32, 5p, 3 2 on each of the two curves., These are the points of intersection found earlier., Next, observe that the point tracing the cardioid arrives at the origin when u  p., But the point tracing the circle first arrives at the origin when u  p>2 and then again, when u  3p>2. In other words, these two points arrive at the origin at different times,, so there is no (common) value of u corresponding to the origin that satisfies both Equations (3) simultaneously. Thus, although the origin is a point of intersection of the two, curves, this fact will not show up in the solution of the system of equations. For this, reason it is recommended that we sketch the graphs of polar equations when finding, their points of intersection.

Page 38 :
10.5 Areas and Arc Lengths in Polar Coordinates, , 885, , EXAMPLE 7 Find the points of intersection of r  cos u and r  cos 2u., Solution, , We solve the system of equations, r  cos u, r  cos 2u, , We set cos u  cos 2u and use the identity cos 2u  2 cos2 u  1. We obtain, 2 cos2 u  cos u  1  0, (2 cos u  1)(cos u  1)  0, So, 1, , cos u  , , ( 12 , 43π ), (1, 0), 1, , 1, , FIGURE 12, , 10.5, , u, , 2p, ,, 3, , 4p, ,, 3, , or, , 0, , CONCEPT QUESTIONS, 2. Suppose that f has a continuous derivative on an interval, [a, b]. If the graph C of r  f(u) is traced exactly once, as u increases from a to b, write an integral giving the, length of C., 3. Suppose that f is a function with a continuous derivative on, [a, b] and the graph C of r  f(u) is traced exactly once as, u increases from a to b. Write an integral giving the area of, the surface obtained by revolving C about (a) the polar axis,, y  0, and (b) the line u  p>2, x  0., , EXERCISES, , 1. a. Find a rectangular equation of the circle r  4 cos u, and, use it to find its area., b. Find the area of the circle of part (a) by integration., 2. a. By finding a rectangular equation, show that the polar, equation r  2 cos u  2 sin u represents a circle. Then, find the area of the circle., b. Find the area of the circle of part (a) by integration., In Exercises 3–8, find the area of the region bounded by the, curve and the rays., 3. r  u,, , cos u  1, , 1 4p, These values of u give 1 12, 2p, 3 2 , 1 2 , 3 2 , and (1, 0) as the points of intersection. Since, both graphs also pass through the pole, we conclude that the pole is also a point of, intersection. (See Figure 12.), , 1. a. Let f be nonnegative and continuous on [a, b], where, 0 b  a  2p. Write an integral giving the area of, the region bounded by the graphs of r  f(u), u  a,, and u  b. Make a sketch of the region., b. If f and t are continuous on [a, b] and 0 t(u) f(u),, where 0 a b 2p, write an integral giving the area, of the region bounded by the graphs of r  t(u),, r  f(u), u  a, and u  b. Make a sketch of the region., , 10.5, , or, , that is,, , 1, , ( 12 , 23π ), , 1, 2, , u  0,, , up, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 4. r , , 1, p, p, , u , u, u, 6, 3, , p, 5. r  eu, u   , u  0, 2, 6. r  e2u, u  0, u , , p, 4, , 7. r  1cos u, u  0, u , 8. r  cos 2u, u  0,, , u, , p, 2, , p, 16

Page 39 :
886, , Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates, 27., , In Exercises 9–12, find the area of the shaded region., 9., , 28., 1, , 10., , 1, 1, , ⫺a, , ⫺2⫺1, , 2, , 4, , a, , 6, , ⫺2, ⫺3, ⫺4, , ⫺1, , 1, , 2, , ⫺1, , r ⫽ a √cos 2¨, , ⫺1, , 1, , 2, , ⫺1, , r ⫽ 1 ⫹ cos ¨, r ⫽ √cos 2¨, r ⫽ 2 cos 3¨, r ⫽ 2 cos ¨, , r⫽¨, , 11., , 12., , In Exercises 29–34, find all points of intersection of the given, curves., , 1.5, 1, , ⫺1, , 2, , 1, , ⫺1, , 2, , ⫺1.5, r ⫽ 2 cos 3¨, , r ⫽ 1 ⫺ cos ¨, , 29. r ⫽ 1 and, , r ⫽ 1 ⫹ cos u, , 30. r ⫽ 3 and, , r ⫽ 2 ⫹ 2 cos u, , 31. r ⫽ 2 and, , r ⫽ 4 cos 2u, , 32. r ⫽ 1 and, , r 2 ⫽ 2 cos 2u, , 33. r ⫽ sin u and, , r ⫽ sin 2u, , 34. r ⫽ cos u and, , r ⫽ cos 2u, , In Exercises 13–18, sketch the curve, and find the area of the, region enclosed by it., , In Exercises 35–40, find the area of the region that lies outside, the first curve and inside the second curve., , 13. r ⫽ 3 sin u, , 14. r ⫽ 2(1 ⫺ cos u), , 35. r ⫽ 1 ⫹ cos u,, , r ⫽ 3 cos u, , 15. r ⫽ sin u, , 16. r ⫽ 3 sin 2u, , 36. r ⫽ 1 ⫺ sin u,, , r⫽1, , 17. r ⫽ 2 sin 2u, , 18. r ⫽ 2 sin 3u, , 37. r ⫽ 4 cos u,, , r⫽2, , 38. r ⫽ 3 sin u,, , r ⫽ 2 ⫺ sin u, , 2, , 2, , In Exercises 19–22, find the area of the region enclosed by one, loop of the curve., 19. r ⫽ cos 2u, , 20. r ⫽ 2 cos 3u, , 21. r ⫽ sin 4u, , 22. r ⫽ 2 cos 4u, , 39. r ⫽ 1 ⫺ cos u, r ⫽, , 3, 2, , 40. r ⫽ 2 cos 3u, r ⫽ 1, In Exercises 41–46, find the area of the region that is enclosed, by both of the curves., , In Exercises 23–24, find the area of the region described., 23. The inner loop of the limaçon r ⫽ 1 ⫹ 2 cos u, , 41. r ⫽ 1,, , 24. Between the loops of the limaçon r ⫽ 1 ⫹ 2 sin u, , r ⫽ 2 sin u, , 42. r ⫽ cos u,, , r ⫽ 13 sin u, , In Exercises 25–28, find the area of the shaded region., , 43. r ⫽ sin u,, , r ⫽ 1 ⫺ sin u, , 25., , 44. r ⫽ cos u, r ⫽ 1 ⫺ cos u, , 26., , 1, , 2, , 45. r 2 ⫽ 4 cos 2u,, , r ⫽ 12, , 46. r ⫽ 13 sin u,, , r ⫽ 1 ⫹ cos u, , 1, , In Exercises 47–54, find the length of the given curve., ⫺0.5, , 47. r ⫽ 5 sin u, , 1, , ⫺0.5, , ⫺1, , 1, , 48. r ⫽ 2u;, ⫺u, , 49. r ⫽ e ;, r ⫽ sin ¨, r ⫽ cos ¨, , ⫺1, r ⫽ 1, r ⫽ 1 ⫹ sin ¨, , 0 ⱕ u ⱕ 2p, 0 ⱕ u ⱕ 4p, , 50. r ⫽ 1 ⫹ sin u; 0 ⱕ u ⱕ 2p, u, 51. r ⫽ sin3 ; 0 ⱕ u ⱕ p, 3

Page 40 :
10.5 Areas and Arc Lengths in Polar Coordinates, , 52. r  cos2, , u, 2, , 54. r  sec u;, , 53. r  a sin4, 0, , cas 66. Plot the curve r  sin(3 cos u) , and find an approximation, , u, 4, , of the area enclosed by the curve accurate to four decimal, places., , p, 3, , u, , In Exercises 55–60, find the area of the surface obtained by, revolving the given curve about the given line., 55. r  4 cos u about the polar axis, 56. r  2 cos u about the line u , , 60. r  e ,, , 0, , p, 2, , u, , 67. r  21  u2, where 0, , 2p (involute of a circle), , u, , 6p, , u, , (parabolic spiral), , p (bifolia), , u, , 70. a. Let f be a function with a continuous derivative in an, interval [a, b]. If the graph C of r  f(u) is traced, exactly once as u increases from a to b, show that the, rectangular coordinates of the centroid of C are, , p, 2, , b, , p, about the line u , 2, , x, , In Exercises 61 and 62, find the area of the region enclosed by, the given curve. (Hint: Convert the rectangular equation to a, polar equation.), 61. (x 2  y 2)3  16x 2y 2, , mation of its length accurate to two decimal places., , 69. r  3 sin u cos2 u, where 0, , 58. r 2  cos 2u about the polar axis, , au, , cas In Exercises 67–69, (a) plot the curve, and (b) find an approxi-, , 68. r  0.21u  1, where 0, , p, 2, , 57. r  2  2 cos u about the polar axis, , 59. r 2  cos 2u about the line u , , 887, , 62. x 4  y 4  4(x 2  y 2), , 63. Let P be a point other than the origin lying on the curve, r  f(u). If c is the angle between the tangent line to the, r, curve at P and the radial line OP, then tan c , . (See, dr>du, Section 10.4, Exercise 84.), , 冮 r cos u2(r¿), , 2, ,  r 2 du, , a, , b, , 冮 2(r¿), , 2, ,  r 2 du, , a, , and, b, , y, , 冮 r sin u2(r¿), , 2, ,  r 2 du, , a, , b, , 冮 2(r¿), , 2, ,  r 2 du, , a, , Hint: See the directions for Exercises 45 and 46 in Exercises 5.7., , b. Use the result of part (a) to find the centroid of the upper, semicircle r  a, where a  0 and 0 u p., cas 71. a. Plot the curve with polar equation r  2 cos3 u where, , u p2 ., p2, b. Find the Cartesian coordinates of the centroid of the, region bounded by the curve of part (a)., , , , r  f(¨), P, ¨, , cas 72. a. Plot the graphs of r  1  cos u and r  3 cos u for, ƒ, , 0 u 2p, treating r and u as rectangular coordinates., b. Refer to page 884. Reconcile your results with the discussion of finding the points of intersection of graphs in, polar coordinates., , O, , a. Show that the angle between the tangent line to the logarithmic spiral r  emu and the radial line at the point of, tangency is a constant., b. Suppose the curve with polar equation r  f(u) has the, property that at any point on the curve, the angle c, between the tangent line to the curve at that point and, the radial line from the origin to that point is a constant., Show that f(u)  Cemu, where C and m are constants., 64. Find the length of the logarithmic spiral r  aemu between, the point (r0, u0) and the point (r, u), and use this result to, deduce that the length of a logarithmic spiral is proportional, to the difference between the radial coordinates of the, points., 65. Show that the length of the parabola y  (1>2p)x on the, interval [0, a] is the same as the length of the spiral r  pu, for 0 r a., 2, , cas In Exercises 73 and 74, (a) find the polar representation of the, , curve given in rectangular coordinates, (b) plot the curve, and, (c) find the area of the region enclosed by a loop (or loops) of, the curve., 73. x 3  3xy  y 3  0 (folium of Descartes), y, 74. (x 2  y 2)1>2  cosc4 tan1 a b d  0 (rhodenea), x, In Exercises 75 and 76, determine whether the statement is true, or false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 75. If there exists a u0 such that f(u0)  t(u0), then the graphs of, r  f(u) and r  t(u) have at least one point of intersection., 76. If P is a point of intersection of the graphs of r  f(u) and, r  t(u), then there must exist a u0 such that f(u0)  t(u0).

Page 41 :
11.5, , 945, , Lines and Planes in Space, , 55. Find a and b if a  具1, a, 3典 and b  具2, 3, b典 are parallel., 56. Find s and t such that (a  b)  a  s a  t b, where, a  具2, 1, 3典 and b  具1, 3, 4典., , ␻, v, , In Exercises 57–62, 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 shows it is false., , R, 0, , 57. a  b  b  a  0, , Suppose that the axis of rotation is parallel to the vector, 2i  2j  k. What is the speed of a particle at the instant it, passes through the point (3, 5, 2)?, 54. Find a unit vector in the plane that contains the vectors, a  i  3j  2k and b  i  2j  4k, and is perpendicular to the vector c  3i  j  2k., , 11.5, , 58. [(a  b)  c]  [(a  b)  c]  0, 59. a ⴢ (a  b)  0, 60. a  b  a  c if and only if, , bc, , 61. (a  b)  (a  b)  2a  b, 62. If a  0, a ⴢ b  a ⴢ c, and a  b  a  c then b  c., , Lines and Planes in Space, z, , Equations of Lines in Space, Figure 1 depicts an airplane flying in a straight line above a ground radar station. How, fast is the distance between the airplane and the radar station changing at any time?, How close to the radar station does the airplane get? To answer questions such as these,, we need to be able to describe the path of the airplane. More specifically, we want to, be able to represent a line in space algebraically., In this section we will see how lines as well as planes in space can be described, in algebraic terms. We begin by considering a line in space. Such a line is uniquely, determined by specifying its direction and a point through which it passes. The direction may be specified by a vector that has the same direction as the line. So suppose, that the line L passes through the point P0(x 0, y0, z 0) and has the same direction as the, vector v  具a, b, c典. (See Figure 2.), !, Let P(x, y, z) be any point on L. Then the vector P0P is parallel to v. But two vectors are parallel if and only if one is a scalar multiple of the other. Therefore, there, exists some number t, called a parameter, such that, !, P0P  tv, !, or, since P0P  具x  x 0, y  y0, z  z 0典, we have, , 0, y, x, , FIGURE 1, The path of the airplane is a straight, line., z, L, P0(x0, y0, z0), , P(x, y, z), , v, , 具x  x 0, y  y0, z  z 0典  t具a, b, c典  具ta, tb, tc典, , 0, y, x, , FIGURE 2, The line L passes through P0 and is, parallel to the vector v., , Equating the corresponding components of the two vectors then yields, x  x 0  ta,, , y  y0  tb,, , and, , z  z 0  tc, , Solving these equations for x, y, and z, respectively, gives the following standard parametric equations of the line L., , DEFINITION Parametric Equations of a Line, The parametric equations of the line passing through the point P0(x 0, y0, z 0) and, parallel to the vector v  具a, b, c典 are, x  x 0  at,, , y  y0  bt,, , and, , z  z 0  ct, , (1)

Page 42 :
946, , Chapter 11 Vectors and the Geometry of Space, , Each value of the parameter t corresponds to a point P(x, y, z) on L. As t takes on, all values in the parameter interval (⬁, ⬁), the line L is traced out. (See Figure 3.), z, L, P(x, y, z), , FIGURE 3, As t runs through all values, in the parameter interval, (⬁, ⬁), L is traced out., z, P0 (2, 1, 3) (t  0), 3, , (t  2), , L, , y, x, , EXAMPLE 1 Find parametric equations for the line passing through the point, P0 (2, 1, 3) and parallel to the vector v  具1, 2, 2典., , x  2  t,, , 5, x, , 0, , Solution We use Equation (1) with x 0  2, y0  1, z 0  3, a  1, b  2, and, c  2, obtaining, , (t  1), , 3, , t, t, Parameter interval, , y, , v  1, 2, 2, , FIGURE 4, The line L and some points on L corresponding to selected values of t. Note, the orientation of the line., , y  1  2t,, , z  3  2t, , and, , The line L in question is sketched in Figure 4., Suppose that the vector v  具a, b, c典 defines the direction of a line L. Then the, numbers a, b, and c are called the direction numbers of L. Observe that if a line L is, described by a set of parametric equations (1), then the direction numbers of L are precisely the coefficients of t in each of the parametric equations., There is another way of describing a line in space. We start with the parametric, equations of the line L,, x  x 0  at,, , y  y0  bt,, , z  z 0  ct, , and, , If the direction numbers a, b, and c are all nonzero, then we can solve each of these, equations for t. Thus,, t, , x  x0, ,, a, , t, , y  y0, ,, b, , t, , and, , z  z0, c, , which gives the following symmetric equations of L., , DEFINITION Symmetric Equations of a Line, The symmetric equations of the line L passing through the point P0 (x 0, y0, z 0), and parallel to the vector v  具a, b, c典 are, y  y0, x  x0, z  z0, , , a, c, b, , (2), , Note Suppose a  0 and both b and c are not equal to zero, then the parametric equations of the line take the form, x  x 0,, , y  y0  bt,, , and, , z  z 0  ct

Page 43 :
11.5, , Lines and Planes in Space, , 947, , and the line lies in the plane x  x 0 (parallel to the yz-plane). Solving the second and, third equations for t leads to, y  y0, z  z0, x  x 0,, , c, b, which are the symmetric equations of the line. We leave it to you to consider and interpret the other cases., , EXAMPLE 2, a. Find parametric equations and symmetric equations for the line L passing through, the points P(3, 3, 2) and Q(2, 1, 4)., b. At what point does L intersect the xy-plane?, Solution, !, a. The direction of L is the same as that of the vector PQ  具5, 4, 6典. Since L, passes through P(3, 3, 2), we can use Equation (1) with a  5, b  4,, c  6, x 0  3, y0  3, and z 0  2, to obtain the parametric equations, x  3  5t,, , y  3  4t,, , z  2  6t, , and, , Next, using Equation (2), we obtain the following symmetric equations for L:, y3, x3, z2, , , 5, 4, 6, b. At the point where the line intersects the xy-plane, we have z  0. So setting, z  0 in the third parametric equation, we obtain t  13. Substituting this value, of t into the other parametric equations gives the required point as 1 43, 53, 0 2 ., (See Figure 5.), z, 5, Q(2, 1, 4), , ( , , 0), 4, 3, , FIGURE 5, The line L intersects the, xy-plane at the point 1 43, 53, 0 2 ., , 5, 3, , P(3, 3, 2), , 0, 4, , 3, , L, , y, , x, , Suppose that L 1 and L 2 are lines having the same directions as the vectors v1 and, v2, respectively. Then L 1 is parallel to L 2 if v1 is parallel to v2., , EXAMPLE 3 Let L 1 be the line with parametric equations, x  1  2t,, , y  2  3t,, , and, , z2t, , and let L 2 be the line with parametric equations, x  3  4t,, , y  1  4t,, , and, , z  3  4t, , a. Show that the lines L 1 and L 2 are not parallel to each other., b. Do the lines L 1 and L 2 intersect? If so, find their point of intersection.

Page 44 :
948, , Chapter 11 Vectors and the Geometry of Space, , Solution, a. By inspection the direction numbers of L 1 are 2,3, and 1. Therefore, L 1 has the, same direction as the vector v1  具2, 3, 1典. Similarly, we see that L 2 has direction given by the vector v2  具4, 4, 4典  4具1, 1, 1典. Since v1 is not a, scalar multiple of v2, the vectors are not parallel, so L 1 and L 2 are not parallel as, well., b. Suppose that L 1 and L 2 intersect at the point P0(x 0, y0, z 0). Then there must exist, parameter values t 1 and t 2 such that, x 0  1  2t 1,, , y0  2  3t 1,, , and, , z, , z 0  2  t1, t 1 corresponds to P0 on L 1., , 5, , and, x 0  3  4t 2,, (1, 5, 1), , L1, , 0, 5, 4, , L1, , L2, , FIGURE 7, The lines L 1 and L 2 are skew lines., , and, , z 0  3  4t 2, t 2 corresponds to P0 on L 2., , This leads to the system of three linear equations, , y, , x, , FIGURE 6, The lines L 1 and L 2 intersect at the, point (1, 5, 1)., , y0  1  4t 2,, , L2, , 1  2t 1 , , 3  4t 2, , 2  3t 1 , , 1  4t 2, , 2  t 1  3  4t 2, that must be satisfied by t 1 and t 2. Adding the first two equations gives, 3  t 1  4, or t 1  1. Substituting this value of t 1 into either the first or, the second equation then gives t 2  1. Finally, substituting these values of t 1, and t 2 into the third equation gives 2 1  3  4(1)  1, which shows that, the third equation is also satisfied by these values. We conclude that L 1 and L 2, do indeed intersect at a point., To find the point of intersection, substitute t 1  1 into the parametric equations defining L 1, or, equivalently substitute t 2  1 into the parametric equations, defining L 2. In both cases we find that x 0  1, y0  5, and z 0  1, so the point, of intersection is (1, 5, 1). (See Figure 6.), Two lines in space are said to be skew if they do not intersect and are not parallel. (See Figure 7.), , EXAMPLE 4 Flight Path of Two Airplanes As two planes fly by each other, their flight, paths are given by the straight lines, L 1:, , x1t, , y  2  3t, , z4t, , x  2  2t, , y  4  3t, , z  1  4t, , and, L 2:, , Show that the lines are skew and, therefore, that there is no danger of the planes colliding., Solution The directions of L 1 and L 2 are given by the directions of the vectors, v1  具1, 3, 1典 and v2  具2, 3, 4典, respectively. Since one vector is not a scalar, multiple of the other, the lines L 1 and L 2 are not parallel. Next, suppose that the two, lines do intersect at some point P0 (x 0, y0, z 0) . Then, x 0  1  t1, , y0  2  3t 1, , z 0  4  t1

Page 45 :
11.5, , Lines and Planes in Space, , 949, , and, x 0  2  2t 2, , y0  4  3t 2, , z 0  1  4t 2, , for some t 1 and t 2. Equating the values of x 0, y0, and z 0 then gives, 1  t1 , , 2  2t 2, , 2  3t 1  4  3t 2, 4  t1 , , 1  4t 2, , Solving the first two equations for t 1 and t 2 yields t 1  19 and t 2  59. Substituting these, values of t 1 and t 2 into the third equation gives 379  299, a contradiction. This shows, that there are no values of t 1 and t 2 that satisfy the three equations simultaneously., Thus, L 1 and L 2 do not intersect. We have shown that L 1 and L 2 are skew lines, so there, is no possibility of the planes colliding., , Equations of Planes in Space, A plane in space is uniquely determined by specifying a point P0 (x 0, y0, z 0) lying in, the plane and a vector n  具a, b, c典 that is normal (perpendicular) to it. (See Figure 8.), To find! an equation of the plane, let P(x, y, z) be any point in the plane. Then the vector P0P must be orthogonal to n. But two vectors are orthogonal if and only if their, dot product is equal to zero. Therefore, we must have, !, (3), n ⴢ P0P  0, !, Since P0P  具x  x 0, y  y0, z  z 0典, we can also write Equation (3) as, 具a, b, c典 ⴢ 具x  x 0, y  y0, z  z 0典  0, or, a(x  x 0)  b(y  y0)  c(z  z 0)  0, , n, n, , ! FIGURE 8, The vector P0P lying in the, plane must be orthogonal !to the, normal n so that n ⴢ P0P  0., , P(x, y, z), P0 (x0 , y0 , z0 ), , DEFINITION The Standard Form of the Equation of a Plane, The standard form of the equation of a plane containing the point P0 (x 0, y0, z 0), and having the normal vector n  具a, b, c典 is, a(x  x 0)  b(y  y0)  c(z  z 0)  0, , (4), , EXAMPLE 5 Find an equation of the plane containing the point P0(3, 3, 2) and having a normal vector n  具4, 2, 3典. Find the x-, y-, and z-intercepts, and make a sketch, of the plane.

Page 46 :
950, , Chapter 11 Vectors and the Geometry of Space, , Solution We use Equation (4) with a  4, b  2, c  3, x 0  3, y0  3, and, z 0  2, obtaining, , z, (0, 0, 4), , 4(x  3)  2(y  3)  3(z  2)  0, or, 4x  2y  3z  12, (0, 6, 0), , (3, 0, 0), , y, , x, , FIGURE 9, The portion of the plane, 4x  2y  3z  12 in the, first octant, , To find the x-intercept, we note that any point on the x-axis must have both its yand z-coordinates equal to zero. Setting y  z  0 in the equation of the plane, we, find that x  3. Therefore, 3 is the x-intercept. Similarly, we find that the y- and, z-intercepts are 6 and 4, respectively. By connecting the points (3, 0, 0) , (0, 6, 0) , and, (0, 0, 4) with straight line segments, we obtain a sketch of that portion of the plane, lying in the first octant. (See Figure 9.), , EXAMPLE 6 Find an equation of the plane containing the points P(3, 1, 1),, Q(1, 4, 2), and R(0, 1, 4)., Solution To use Equation (4), we need, to find a vector normal, to the plane in question., !, !, Observe that both! of the! vectors PQ  具2, 5, 1典 and PR  具3, 2, 3典 lie in the plane,, so the vector PQ  PR is normal to the plane. Denoting this vector by n, we have, i, !, !, n  PQ  PR  † 2, 3, , j k, 5 1 †  13i  3j  11k, 2 3, , Finally, using the point P(3, 1, 1) in the plane (any of the other two points will, also do) and the normal vector n just found, with a  13, b  3, c  11, x 0  3,, y0  1, and z 0  1, Equation (4) gives, 13(x  3)  3(y  1)  11(z  1)  0, or, upon simplification,, 13x  3y  11z  47, The plane is sketched in Figure 10., z, 4, , R(0, 1, 4), , n, PR, Q(1, 4, 2), PQ, , P(3, –1, 1), , FIGURE 10, The normal to the, ! plane!, is n  PQ  PR ., , 4, y, 4, , x, , By expanding Equation (4) and regrouping the terms, as we did in Examples 5 and, 6, we obtain the general form of the equation of a plane in space,, ax  by  cz  d, , (5)

Page 47 :
11.5, , Lines and Planes in Space, , 951, , where d  ax 0  by0  cz 0. Conversely, given ax  by  cz  d with a, b, and c, not all equal to zero, we can choose numbers x 0, y0, and z 0 such that, ax 0  by0  cz 0  d. For example, if c  0, we can pick x 0 and y0 arbitrarily and, solve the equation ax 0  by0  cz 0  d for z 0, obtaining z 0  (d  ax 0  by0)>c., Therefore, with these choices of x 0, y0, and z 0, Equation (5) takes the form, ax  by  cz  ax 0  by0  cz 0, or, a(x  x 0)  b(y  y0)  c(z  z 0)  0, which we recognize to be an equation of the plane containing the point (x 0, y0, z 0) and, having a normal vector n  具a, b, c典. (See Equation (4).) An equation of the form, ax  by  cz  d, with a, b, and c not all zero, is called a linear equation in the, three variables x, y, and z., , THEOREM 1, Every plane in space can be represented by a linear equation ax  by  cz  d,, where a, b, and c are not all equal to zero. Conversely, every linear equation, ax  by  cz  d represents a plane in space having a normal vector 具a, b, c典., , Note Notice that the coefficients of x, y, and z are precisely the components of the, normal vector n  具a, b, c典. Thus, we can write a normal vector to a plane by simply, inspecting its equation., , Parallel and Orthogonal Planes, Two planes with normal vectors m and n are parallel to each other if m and n are parallel; the planes are orthogonal if m and n are orthogonal. (See Figure 11.), , m, m, n, , FIGURE 11, Two planes are parallel if m and, n are parallel and orthogonal, if m and n are orthogonal., , n, , (a) Parallel planes, , (b) Orthogonal planes, , EXAMPLE 7 Find an equation of the plane containing P(2, 1, 3) and parallel to the, plane defined by 2x  3y  4z  6., Solution By Theorem 1 the normal vector of the given plane is n  具2, 3, 4典. Since, the required plane is parallel to the given plane, it also has n as a normal vector. Therefore, using Equation (4), we obtain, 2(x  2)  3(y  1)  4(z  3)  0

Page 48 :
952, , Chapter 11 Vectors and the Geometry of Space, , or, , n1, , 2x  3y  4z  19, , P2, n2, , as an equation of the plane., , ¨, P1, , ¨, , The Angle Between Two Planes, Two distinct planes in space are either parallel to each other or intersect in a straight, line. If they do intersect, then the angle between the two planes is defined to be the, acute angle between their normal vectors (see Figure 12)., , FIGURE 12, The angle between two planes is the, angle between their normal vectors., , EXAMPLE 8 Find the angle between the two planes defined by 3x  y  2z  1 and, 2x  3y  z  4., Solution, , The normal vectors of these planes are, n 1  具3, 1, 2典, , and, , n 2  具2, 3, 1典, , Therefore, the angle u between the planes is given by, cos u , , , n1 ⴢ n2,  n 1  n 2 , , Use Equation (2) of Section 11.3., , 具3, 1, 2典 ⴢ 具2, 3, 1典, 3(2)  (1)(3)  2(1), 1, , , 14, 19  1  4 14  9  1, 114 114, , or, u  cos1 a, , 1, b  86°, 14, , EXAMPLE 9 Find parametric equations for the line of intersection of the planes defined by 3x  y  2z  1 and 2x  3y  z  4., n1, P2, , n1  n2, n2, , L, , P1, , Solution We need the direction of the line of intersection L as well as a point on L., To find the direction of L, we observe that a vector v is parallel to L if and only if it, is orthogonal to the normal vectors of both planes. (See Figure 13 for the general case.), In other words, v  n 1  n 2, where n 1 and n 2 are the normal vectors of the two planes., Here, the normal vectors are n 1  具3, 1, 2典 and n 2  具2, 3, 1典, so the vector v is, given by, i, v  n1  n2  † 3, 2, , FIGURE 13, The vector n 1  n 2 has the same, direction as L, the line of intersection, of the two planes., , j, 1, 3, , k, 2 †  5i  7j  11k, 1, , To find a point on L, let’s set z  0 in both of the equations defining the planes. (This, will give us the point where L intersects the xy-plane.) We obtain, 3x  y  1, , and, , 2x  3y  4, , Solving these equations simultaneously gives x  117 and y  10, 11 . Finally, by using Equation (1), the required parametric equations are, x, , 7,  5t,, 11, , y, , 10,  7t,, 11, , and, , z  11t

Page 49 :
11.5, , Lines and Planes in Space, , 953, , The Distance Between a Point and a Plane, To find a formula for the distance between a point and a plane, suppose that P1 (x 1, y1, z 1), is a point not lying in the plane ax  by  cz  d. Let P0(x 0, y0, z 0) be any point lying, in the plane. Then, as you can see in Figure 14, the distance, ! D between P1 and the, plane is given by the length of the vector projection of P0P1 onto the normal vector, n  具a, b, c典 of, ! the plane. Equivalently, D is the absolute value of the scalar component of P0P1 along n. Using Equation (6) of Section 11.3 (and taking the absolute, value), we obtain, !,  P0P1 ⴢ n , D, n, , P1(x1, y1, z1), projn P0P1, , D, , n, ax  by  cz  d, , P0 (x0 , y0 , z0 ), , FIGURE 14, !, The distance from P1 to the plane is the length of proj n P0P1 ., , !, But P0P1  具x 1  x 0, y1  y0, z 1  z 0典, so we have, D, , , ,  具x 1  x 0, y1  y0, z 1  z 0典 ⴢ 具a, b, c典 , 2a 2  b 2  c2,  a(x 1  x 0)  b(y1  y0)  c(z 1  z 0) , 2a 2  b 2  c2,  ax 1  by1  cz 1  (ax 0  by0  cz 0) , 2a 2  b 2  c2, , Since P0(x 0, y0, z 0) lies in the plane, its coordinates must satisfy the equation of the, plane, that is, ax 0  by0  cz 0  d; so we can write D in the following form:, D, ,  ax 1  by1  cz 1  d , 2a 2  b 2  c2, , (6), , EXAMPLE 10 Find the distance between the point (2, 1, 3) and the plane 2x , , 3y  z  1., , Solution Using Equation (6) with x 1  2, y1  1, z 1  3, a  2, b  3, c  1, and d  1, we obtain, D, ,  2(2)  3(1)  1(3)  1 , 22  (3)  1, 2, , 2, , 2, , , , 5, 5114, , 14, 114

Page 50 :
954, , Chapter 11 Vectors and the Geometry of Space, , 11.5, , CONCEPT QUESTIONS, 3. a. Write the standard form of an equation of the plane, containing the point P0(x 0, y0, z 0) and having the normal, vector n  具a, b, c典., b. What is the general form of the equation of a plane in, space?, 4. a. What is the angle between two planes in space? How do, you find it?, b. Write the formula giving the distance between a point, and a plane in space., , 1. a. Write the parametric equations of the line passing, through the point P0 (x 0, y0, z 0) and having the same, direction as the vector v  具a, b, c典., b. What are the symmetric equations of the line of part (a)?, c. Write the parametric and symmetric equations of the line, that passes through the point P(x 0, y0, z 0) and has the, same direction as the vector v  具a, b, 0典., 2. If you are given two lines L 1 and L 2 in space, how do, you determine whether they are (a) parallel to each other,, (b) perpendicular to each other, or (c) skew?, , 11.5, , EXERCISES, , In Exercises 1 and 2 describe in your own words the strategy, you might adopt to solve the problem. For example, to find the, parametric equations of the line passing through two distinct, points, you might use this strategy:, Let P and Q denote the two, !, points, and write the vector PQ that gives the direction numbers, of the line. Then using this information and either P or Q, write, the desired equations using Equation (1)., 1. Find parametric equations of a line, given that the line, a. Passes through a given point and is parallel to a given, line., b. Passes through a given point and is perpendicular to two, distinct lines passing through that point., c. Passes through a given point lying in a given plane and, is perpendicular to the plane., d. Is the intersection of two given nonparallel planes., 2. Find an equation of a plane, given that the plane, a. Contains three distinct points., b. Contains a given line and a point not lying on the line., c. Contains a given point and is parallel to a given plane., d. Contains two intersecting nonparallel lines., e. Contains two parallel and distinct lines., , 9. 1 1, 2, 12 2 and 1 1, 32, 3 2, , 10., , 11. Find parametric and symmetric equations of the line passing, through the point (1, 2, 1) and parallel to the line with, parametric equations x  1  t, y  2  2t, and, z  2  3t. At what points does the line intersect the, coordinate planes?, 12. Find parametric equations of the line passing through the, point (1, 3, 2) and parallel to the line with symmetric, equation, y1, x2, , z2, 3, 3, At what point does the line intersect the yz-plane?, 13. Determine whether the point (3, 6, 1) lies on the line L, passing through the point (1, 4, 3) and parallel to the, vector v  i  j  k., 14. Find parametric equations of the line that is parallel to the, line with equation, y4, x1, z1, , , 4, 5, 2, , In Exercises 3–6, find parametric and symmetric equations for, the line passing through the point P that is parallel to the, vector v., 3. P(1, 3, 2);, 4. P(1, 4, 2);, , and contains the point of intersection of the lines, , v  具2, 4, 5典, v  2i  3j  k, , 5. P(3, 0, 2); v  2i  j  3k, 6. P(0, 1, 3) ; v  具2, 3, 4典, In Exercises 7–10, find parametric and symmetric equations for, the line passing through the given points., 7. (2, 1, 4) and (1, 3, 7), 8. (3, 2, 1) and (3, 4, 4), , 1 12, 13, 14 2 and 1 12, 13, 34 2, , L 1:, , x4t, , L 2:, , x  6  2t, , y5t, y  11  4t, , z  1  2t, z  3  t, , In Exercises 15–18, determine whether the lines L 1 and L 2 are, parallel, are skew, or intersect each other. If they intersect, find, the point of intersection., 15. L 1: x  1  3t, y  2  3t, z  3  t, L 2: x  1  4t, y  2  6t, z  4  t, 16. L 1: x  1  2t, y  1  3t, z  2  t, L 2: x  3  t, y  2  2t, z  3  t, , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 51 :
11.5, z1, x2, , ,y3, 4, 1, y3, x2, L 2:, , z1, 2, 2, , 35. (3, 4, 5);, , 17. L 1:, , 36. (1, 3, 0);, , y1, x4, , z4, 1, 6, y1, z5, x1, L 2:, , , 2, 4, 1, , 18. L 1:, , Lines and Planes in Space, , y1, x2, z3, , , 2, 3, 5, y3, x4, , ,, 3, 5, , z2, , In Exercises 37 and 38, find an equation of the plane passing, through the given points and perpendicular to the given plane., 2x  3y  4z  3, , 37. (2, 1, 1) and (1, 3, 2);, , In Exercises 19–22, determine whether the lines L 1 and L 2 intersect. If they do intersect, find the angle between them., 19. L 1: x  1  t, y  3  2t, z  t, L 2: x  2  3t, y  3  2t, z  1  t, , 38. (1, 3, 0) and (2, 1, 4);, , 3x  4y  5z  1, , In Exercises 39–42, determine whether the planes are parallel,, orthogonal, or neither. If they are neither parallel nor orthogonal, find the angle between them., , 20. L 1: x  2  3t, y  2  2t, z  3  t, L 2: x  3  t, y  1  t, z  4  5t, , 39. x  2y  z  1,, 40. 2x  y  4z  7,, , 6x  3y  12z  1, , y2, x1, z1, , , 21. L 1:, 3, 2, 4, y4, x2, L 2:, , z3, 2, 4, , 41. 3x  y  2z  2,, , 2x  3y  z  4, , 43. x  y  2z  6;, , n  具1, 2, 4典, n  2i  4k, , 26. (3, 0, 3);, , n  具0, 0, 1典, , 46. 3x  y  2z  4,, , 27. (3, 6, 2);, , 2x  3y  z  4, , 28. (2, 1, 0);, , x  2y 3z  1, x  3z  1, , 48. Find an equation of the plane that passes through the point, (3, 2, 4) and is perpendicular to the line, , 49. Find an equation of the plane that contains the lines given by, x  1  2t, x2t, , In Exercises 31 and 32, find an equation of the plane that passes, through the three given points., 31. (1, 0, 2), (1, 3, 2), (2, 3, 0), 32. (2, 3, 1) , (1, 2, 3), (1, 2, 4), In Exercises 33–36, find an equation of the plane that passes, through the given point and contains the given line., 34. (1, 2, 3);, , y  1  2t,, , x  1  2t,, , x  4y  2z  7, 2x  y  3z  6, , y2, z4, x1, , , 2, 3, 4, , 1, 1, 1, x y z2, 2, 3, 4, , x  1  t,, , z  1  t, , y1, x1, z, , , 2, 3, 2, , 47. Find parametric equations of the line that passes through the, point (2, 3, 1) and is perpendicular to the plane, 2x  4y  3z  4., , In Exercises 27–30, find an equation of the plane that passes, through the given point and is parallel to the given plane., , 33. (1, 3, 2);, , y  2  t,, , In Exercises 45 and 46, find parametric equations for the line of, intersection of the planes., 45. 2x  3y  4z  3,, , n  i  2j  k, , 25. (1, 3, 0);, , 30. (0, 2, 1);, , x  1  t,, , 44. 2x  3y  4z  12;, , In Exercises 23–26, find an equation of the plane that has the, normal vector n and passes through the given point., , 29. (1, 2, 3);, , 3x  2y  2z  5, , In Exercises 43 and 44, find the angle between the plane and, the line., , y1, z3, x4, , , 3, 2, 3, y8, x  32, ,z4, L 2:, , 6, 2, , 24. (1, 3, 2);, , 2x  3y  4z  3, , 42. 4x  4y  2z  7,, , 22. L 1:, , 23. (2, 1, 5);, , 955, , z  3  2t, , y  2  3t,, , z3t, , y  2  3t, y  1  2t, , z1t, z  5  3t, , 50. Find an equation of the plane that passes through the point, (3, 4, 1) and contains the line of intersection of the planes, x  y  2z  1 and 2x  3y  z  2., 51. Find an equation of the plane that is orthogonal to the plane, 3x  2y  4z  7 and contains the line of intersection of, the planes 2x  3y  z  3 and x  2y  3z  5., 52. Find an equation of the plane that is parallel to the line of intersection of the planes x  y  2z  3 and 2x  3y  z  4, and contains the points (2, 3, 5) and (3, 4, 1).

Page 52 :
956, , Chapter 11 Vectors and the Geometry of Space, , In Exercises 53 and 54, find the point of intersection, if any, of, the plane and the line., 53. 2x  3y  z  9;, z  3  2t, 54. x  y  2z  13;, , x  2  3t,, , y  1  t,, , y2, x1, , z1, 3, 4, , In Exercises 55 and 56, find the distance between the point and, the plane., 2x  3y  4z  7, , 55. (3, 1, 2);, , 56. (1, 3, 2);, , 63. Show that the distance D between the parallel planes, ax  by  cz  d1 and ax  by  cz  d2 is, D, , In Exercises 57 and 58, show that the two planes are parallel, and find the distance between them., 57. x  2y  4z  1,, , x  2y  4z  7, , 58. 2x  3y  z  2,, , 4x  6y  2z  8, , 2a 2  b 2  c2, , In Exercises 64 and 65, find the distance between the skew lines., (Hint: Use the result of Exercise 63.), 64. x  1  5t, y  1  2t, z  2  3t, x  1  t, y  1  2t, z  1  t, 65., , 3x  y  z  2, ,  d1  d2 , , y4, x1, z3, , , 2, 6, 2, , and, , and x  2 , , y2, z1, , 5, 3, , 66. Find the distance between the line given by, x  1  3t,, , y  2  6t,, , and, , z  1  2t, , and the plane that passes through the point (2, 3, 1) and is, perpendicular to the line containing the points (2, 1, 4) and, (4, 4, 10) ., , 59. Let P be a point that is not on the line L. Show that the distance D between the point P and the line L is, !,  QP  u , D, u, , In Exercises 67–72, 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 shows it is false., , where u is a vector having the same direction as L, and Q is, any point on L., , 67. If the lines L 1 and L 2 are both perpendicular to the line L 3,, then L 1 must be parallel to L 2., , In Exercises 60 and 61, use the result of Exercise 59 to find the, distance between the point and the line., x  2  t, y  1  2t, z  3  t, , 60. (3, 4, 6);, , 68. If the lines L 1 and L 2 do not intersect, then they must be, parallel to each other., 69. If the planes P1 and P2 are both perpendicular to the plane, P3, then P1 must be perpendicular to P2., , y1, z3, x2, , , 3, 1, 2, , 70. If the planes P1 and P2 are both parallel to a line L, then P1, and P2 must be parallel to each other., , 62. Find the distance between the point (1, 4, 2) and the line, passing through the points (1, 3, 1) and (1, 2, 3)., , 71. There always exists a unique plane passing through a given, point and a given line., , 61. (1, 2, 3);, , 72. Given any two lines that are not coincident, there is a plane, containing the two lines., , 11.6, , Surfaces in Space, In Section 11.5 we saw that the graph of a linear equation in three variables is a plane, in space. In general, the graph of an equation in three variables, F(x, y, z)  0, is a surface in 3-space. In this section we will study surfaces called cylinders and quadric surfaces., The paraboloidal surface shown in Figure 1a is an example of a quadric surface. A, uniformly rotating liquid acquires this shape as a result of the interaction between the, force of gravity and centrifugal force. As was explained in Section 10.1, this surface, is ideal for radio and optical telescope mirrors. (See Figure 1b.) Mathematically, a, paraboloid is obtained by revolving a parabola about its axis of symmetry. (See Figure 1c.)

Page 53 :
11.6, , Surfaces in Space, , 957, , Axis of, symmetry, , Parabola, (a) Surface of rotating liquid, , (b) Surface of a radio telescope, , (c) Surface obtained by revolving a, parabola about its axis, , FIGURE 1, , Traces, Just as we can use the x- and y-intercepts of a plane curve to help us sketch the graph, of a plane curve, so can we use the traces of a surface in the coordinate planes to help, us sketch the surface itself. The trace of a surface S in a plane is the intersection of, the surface and the plane. In particular, the traces of S in the xy-plane, the yz-plane,, and the xz-plane are called the xy-trace, the yz-trace and the xz-trace, respectively., To find the xy-traces, we set z  0 and sketch the graph of the resulting equation, in the xy-plane. The other traces are obtained in a similar manner. Of course, if the surface does not intersect the plane, there is no trace in that plane., , EXAMPLE 1 Consider the plane with equation 4x  2y  3z  12. (See Example 5, in Section 11.5.) Find the traces of the plane in the coordinate planes, and sketch the, plane., Solution To find the xy-trace, we first set z  0 in the given equation to obtain, the equation 4x  2y  12. Then we sketch the graph of this equation in the, xy-plane. (See Figure 2a.) To find the yz-trace, we set x  0 to obtain the equation, 2y  3z  12, whose graph in the yz-plane gives the required trace. (See Figure 2b.), The xz-trace is obtained in a similar manner. (See Figure 2c.) The graph of the plane, in the first octant is sketched in Figure 2d., z, , z, , z, 4, , 0, , x, , 3, , (a) xy-trace, V, , 0, 6, , y, x, , y, x, , (b) yz-trace, , 4, , 4, , 0, 6, , z, , 3, , (c) xz-trace, , 0, y, x, , 3, , 6, , y, , (d) The plane in the first octant, , FIGURE 2, The traces of the plane 4x  2y  3z  12 in the coordinate planes are shown in parts (a)–(c)., , Sometimes it is useful to obtain the traces of a surface in planes that are parallel, to the coordinate planes, as illustrated in the next example.

Page 54 :
958, , Chapter 11 Vectors and the Geometry of Space, , EXAMPLE 2 Let S be the surface defined by z  x 2  y 2., , Historical Biography, , SPL/Photo Researchers, Inc., , a. Find the traces of S in the coordinate planes., b. Find the traces of S in the plane z  k, where k is a constant., c. Sketch the surface S., Solution, a. Setting z  0 gives x 2  y 2  0, from which we see that the xy-trace is the, origin (0, 0). (See Figure 3a.) Next, setting x  0 gives z  y 2, from which we, see that the yz-trace is a parabola. (See Figure 3b.) Finally, setting y  0 gives, z  x 2, so the xz-trace is also a parabola. (See Figure 3c.), b. Setting z  k, we obtain x 2  y 2  k, from which we see that the trace, of S in the plane z  k is a circle of radius 1k centered at the point of, intersection of the plane and the z-axis, provided that k  0. (See Figure 3d.), Observe that if k  0, the trace is the point (0, 0) (degenerate circle) obtained, in part (a)., c. The graph of z  x 2  y 2 sketched in Figure 3e is called a circular paraboloid, because its traces in planes parallel to the coordinate planes are either circles or, parabolas., , HERMANN MINKOWSKI, (1864–1909), Hermann Minkowski spent much of his, childhood in Königsberg, Germany, and, eventually attended the University of, Königsberg. While he was at the university,, his exceptional mathematical talent surfaced. In 1881 he submitted a proof for a, problem that had been posed by the Paris, Académie Royale des Sciences. Although, his 140-page proof was not the only one, submitted, it was deemed better formulated than that of British mathematician, H. J. Smith. The prize was awarded to, Minkowski when he was just 19 years old., Minkowski received his doctorate from, the University of Königsberg in 1885 and, began a career in teaching. He taught at, the University of Zurich, the University of, Bonn, and the University of Königsberg., Some of his mathematical achievements were the geometry of numbers, his, geometrical concept of volume, his investigation of ternary quadratic forms, and a, generalization of this technique to ellipsoids and other convex shapes such as, cylinders. Minkowski had an exceptional, grasp of geometrical concepts, and in, 1905, when participating in a seminar on, electron theory, he linked the theory with, that of subatomic particles proposed by, Einstein and Hendrik Lorentz. Minkowski, identified the need to visualize space as a, four-dimensional, non-Euclidian space-time, continuum. This 4-space concept became, the basis of the theory that Einstein used, in his general theory of relativity., Minkowski died of a ruptured appendix, when he was just 44 years old., , z, , z, , z, , 0, y, x, , y, x, , (a) xy-trace is a point., , y, x, , (b) yz-trace is a parabola., z, , (c) xz-trace is a parabola., z, xz-trace, , yz-trace, , zk, , y, x, (d) If z  k, the trace is a circle., V, , FIGURE 3, The traces of the surface S, , xy-trace, , y, , x, (e) The surface z  x2  y2

Page 55 :
11.6, , Surfaces in Space, , 959, , Cylinders, We now turn our attention to a class of surfaces called cylinders., , DEFINITION Cylinder, Let C be a curve in a plane, and let l be a line that is not parallel to that plane., Then the set of all points generated by letting a line traverse C while parallel to, l at all times is called a cylinder. The curve C is called the directrix of the cylinder, and each line through C parallel to l is called a ruling of the cylinder. (See, Figure 4.), , l, , l, Rulings, C (directrix), , FIGURE 4, Two cylinders. The curve C is the, directrix. The rulings are parallel to l., , C, , Cylinders in which the directrix lies in a coordinate plane and the rulings are perpendicular to that plane have relatively simple algebraic representations. Consider, for, example, the surface S with equation f(x, y)  0. The xy-trace of S is the graph C of, the equation f(x, y)  0 in the xy-plane. (See Figure 5a.) Next, observe that if (x, y, 0), is any point on C, then the point (x, y, z) must satisfy the equation f(x, y)  0 for any, value of z (since z is not present in the equation). But all such points lie on the line, perpendicular to the xy-plane and pass through the point (x, y, 0). This shows that the, surface S is a cylinder with directrix f(x, y)  0 and rulings that are parallel to the, direction of the axis of the missing variable z. (See Figure 5b.), z, , z, (x, y, z), S, , FIGURE 5, The surface f(x, y)  0 is a, cylinder with directrix C defined, by f(x, y)  0 in the xy-plane, and with rulings parallel to, the direction of the axis, of the missing variable z., , 0, y, x, , C, , (x, y, 0), , (a) C is the xy-trace., , y, x, , C, , (b) C is the directrix of the cylinder S., The rulings are parallel to the z-axis., , EXAMPLE 3 Sketch the graph of y  x 2  4., Solution The given equation has the form f(x, y)  0, where f(x, y)  x 2  y  4., Therefore, its graph is a cylinder with directrix given by the graph of y  x 2  4 in, the xy-plane and rulings parallel to the z-axis (corresponding to the variable missing in, the equation). The graph of y  x 2  4 in the xy-plane is the parabola shown in Figure 6a. The required cylinder is shown in Figure 6b. It is called a parabolic cylinder.

Page 56 :
960, , Chapter 11 Vectors and the Geometry of Space, , y, , z, , 4, , y  x2  4, 2, , x, , 2, , 4, 2, , 4, , x, (b) The parabolic cylinder y  x2  4, , (a) The directrix shown in the xy-plane, V, , y, , FIGURE 6, The graph of y  x 2  4 is sketched in two steps., , EXAMPLE 4 Sketch the graph of, Solution, , y2, z2,   1., 4, 9, , The given equation has the form f(y, z)  0, where, f(y, z) , , y2, z2,  1, 4, 9, , Its graph is a cylinder with directrix given by, y2, z2,  1, 4, 9, and rulings parallel to the x-axis. The graph of, y2, z2,  1, 4, 9, in the yz-plane is the ellipse shown in Figure 7a. The required cylinder is shown in, Figure 7b. It is called an elliptic cylinder., z, 3, , z, 3, , 2, , 2, , 2 y, 3, 3, x, 2, , 2, , y, z, (a) The directrix ––  ––  1 shown in the yz-plane, 4, 9, V, , FIGURE 7, , y2 z2, (b) The cylinder ––  ––  1, 4, 9, , y

Page 57 :
11.6, , Surfaces in Space, , 961, , EXAMPLE 5 Sketch the graph of z  cos x., Solution The given equation has the form f(x, z)  0, where f(x, z)  z  cos x., Therefore, its graph is a cylinder with directrix given by the graph of z  cos x in the, xz-plane and rulings parallel to the y-axis. The graph of the directrix in the xz-plane is, shown in Figure 8a, and the graph of the cylinder is sketched in Figure 8b., z, , z, , 1, π_2 1, , __, 3π, 2, , π, _, 2, , 3π, __, 2, , x, y, x, (b) The cylinder z  cos x, , (a) The directrix shown in the xz-plane, , FIGURE 8, , !, , z, , Note that, while an equation in two variables represents a curve in 2-space, the, same equation represents a cylinder when we are working in 3-space. For example, the equation x 2  y 2  1 represents a circle in the plane, but the same equation represents a right circular cylinder in 3-space., , Quadric Surfaces, Z, X, , The equation of a sphere given in Section 11.2 and the equations in Examples 1, 2,, and 3 in this section are special cases of the second-degree equation in x, y, and z, Ax 2  By 2  Cz 2  Dxy  Exz  Fyz  Gx  Hy  Iz  J  0, , Y, y, , x, , FIGURE 9, By translating and rotating the xyzsystem, we have the XYZ-system in, which the paraboloid is in standard, position with respect to the latter., , where A, B, C, p , J are constants. The graph of this equation is a quadric surface., By making a suitable translation and/or rotation of the coordinate system, a quadric, surface can always be put in standard position with respect to a new coordinate system. (See Figure 9.) With respect to the new system the equation will assume one of, the two standard forms, AX 2  BY 2  CZ 2  J  0, , or, , AX 2  BY 2  IZ  0, , For this reason we will restrict our study of quadric surfaces to those represented by, the equations, AX 2  BY 2  CZ 2  J  0, , or, , AX 2  BY 2  IZ  0, , When we sketch the following quadric surfaces, we will find it useful to look at their, traces in the coordinate planes as well as planes that are parallel to the coordinate planes., In the remainder of this section, unless otherwise noted, a, b, and c denote positive real numbers., , Ellipsoids, , The graph of the equation, x2, a2, , , , y2, b2, , , , z2, c2, , 1

Page 58 :
962, , Chapter 11 Vectors and the Geometry of Space, , is an ellipsoid because its traces in the planes parallel to the coordinate planes are, ellipses. In fact, its trace in the plane z  k, where c  k  c, is the ellipse, x2, a2, , , , y2, b2, , 1, , k2, c2, , and, in particular, its trace in the xy-plane is the ellipse, x2, a2, , , , y2, b2, , 1, , shown in Figure 10a., Similarly, you may verify that its traces in the planes x  k (a  k  a) and, y  k (b  k  b) are ellipses and, in particular, that its yz- and xz-traces are the, ellipses, y2, b2, , , , z2, c2, , 1, , x2, , and, , a2, , , , z2, c2, , 1, , respectively. (See Figures 10b–c.) The ellipsoid is sketched in Figure 10d., , z, , z, , z, , z, , c, , c, , c, , a, b, a, , x, , (a) xy-trace, , b, , b, , y, , a, y, , x, , y, , x, , (b) yz-trace, , a, , b, , y, , x, , (c) xz-trace, , (d) The ellipsoid, , FIGURE 10, y2, x2, z2, The traces in the coordinate planes and the ellipsoid 2  2  2  1, a, b, c, , Note that if a  b  c, then the ellipsoid is in fact a sphere of radius a with center at, the origin., , Hyperboloids of One Sheet, , The graph of the equation, x, , 2, , a2, , , , y2, b2, , , , z2, c2, , 1, , is a hyperboloid of one sheet. The xy-trace of this surface is the ellipse, x2, a2, , , , y2, b2, , 1

Page 59 :
11.6, , Surfaces in Space, , 963, , (Figure 11a) whereas both the yz- and xz-traces are hyperbolas (Figures 11b–c), as you, may verify. The trace of the surface in the plane z  k is an ellipse, x2, a2, , , , y2, b2, , 1, , k2, c2, , As  k  increases, the ellipses grow larger and larger. The hyperboloid is sketched in, Figure 11d., z, , b, , x, (a) xy-trace, , z, , a, a, , b, , b, y, , z, , z, , b, , a, , b, y, , b, , y, , y, , x, , x, (b) yz-trace, , a, , x, , (c) xz-trace, , (d) A hyperboloid of one sheet, , FIGURE 11, y2, x2, z2, The traces in the coordinate plane and the hyperboloid of one sheet 2  2  2  1, a, b, c, , The z-axis is called the axis of the hyperboloid. Note that the orientation of the, axis of the hyperboloid is associated with the term that has a minus sign in front of it., Thus, if the minus sign had been in front of the term involving x, then the surface would, have been a hyperboloid of one sheet with the x-axis as its axis., , Hyperboloids of Two Sheets, , , The graph of the equation, x2, a2, , , , y2, b2, , , , z2, c2, , 1, , is a hyperboloid of two sheets. The xz- and yz-traces are the hyperbolas, , , x2, a, , 2, , , , z2, c, , 2, , 1, , , , and, , y2, b, , 2, , , , z2, c2, , 1, , sketched in Figures 12a–b. The trace of the surface in the plane z  k is an ellipse, x2, a, , 2, , , , y2, b, , 2, , , , k2, c2, , 1, , provided that  k   c. There are no values of x and y that satisfy the equation if,  k   c, so the surface is made up of two parts, as shown in Figure 12c: one part lying, on or above the plane z  c and the other part lying on or below the plane z  c., The axis of the hyperboloid is the z-axis. Observe that the sign associated with the, variable z is positive. Had the positive sign been in front of one of the other variables,, then the surface would have been a hyperboloid of two sheets with its axis along the, axis associated with that variable.

Page 60 :
964, , Chapter 11 Vectors and the Geometry of Space, , z, , z, , z, , c, , c, , c, , c, , y, , y, , c, , c, x, , x, , x, (a) xz-trace, , y, , (b) yz-trace, , (c) A hyperboloid of two sheets, , FIGURE 12, y2, x2, z2, The traces in the xz- and yz-planes and the hyperboloid of two sheets  2  2  2  1, a, b, c, , Cones, , The graph of the equation, x2, a2, , , , y2, b2, , , , z2, c2, , 0, , is a double-napped cone. The xz- and yz-traces are the lines z  (c>a)x and, z  (c>b)y, respectively. (See Figures 13a–b.) The trace in the plane z  k is an ellipse,, x2, a2, , , , y2, b2, , , , k2, c2, , As  k  increases, so do the lengths of the axes of the resulting ellipses. The traces in, planes parallel to the other two coordinate planes are hyperbolas. The cone is sketched, in Figure 13c. The axis of the cone is the z-axis., z, , z, , z, , y, , x, (a) xz-trace, , y, , x, , y, , x, , (b) yz-trace, , (c) A cone, , FIGURE 13, y2, x2, z2, The traces in the xz- and yz-planes and the cone 2  2  2  0, a, b, c, , Paraboloids, , The graph of the equation, x2, a2, , , , y2, b2, ,  cz

Page 61 :
11.6, , Surfaces in Space, , 965, , where c is a real number, is called an elliptic paraboloid because its traces in planes, parallel to the xy-coordinate plane are ellipses and its traces in planes parallel to the, other two coordinate planes are parabolas. If a  b, the surface is called a circular, paraboloid. We will let you verify these statements. The graph of an elliptic paraboloid with c  0 is sketched in Figure 14a. The axis of the paraboloid is the z-axis, and, its vertex is the origin., z, , z, , y, y, x, , x, , (a) An elliptic paraboloid, 2, , x, y, ––2  ––2  cz,, a, b, , FIGURE 14, , (b) A hyperbolic paraboloid, , 2, , Hyperbolic Paraboloids, , x2 y2, ––2  ––2  cz,, a, b, , c0, , c0, , The graph of the equation, x2, a2, , , , y2, b2, ,  cz, , where c is a real number, is called a hyperbolic paraboloid because the xz- and, yz-traces are parabolas and the traces in planes parallel to the xy-plane are hyperbolas., The graph of a hyperbolic paraboloid with c  0 is shown in Figure 14b., , EXAMPLE 6 Identify and sketch the surface 12x 2  3y 2  4z 2  12  0., Solution, , Rewriting the equation in the standard form, , , y2, x2, z2, ,  1, 1, 4, 3, , we see that it represents a hyperboloid of two sheets with the y-axis as its axis., To sketch the surface, observe that the surface intersects the y-axis at the points, (0, 2, 0) and (0, 2, 0), as you can verify by setting x  z  0 in the given equation., Next, let’s find the trace in the plane y  k. We obtain, x2, z2, k2,  , 1, 1, 3, 4, In particular, the trace in the plane y  6 is the ellipse, x2, z2,  8, 1, 3, , or, , x2, z2, , 1, 8, 24, , A sketch of this trace is shown in Figure 15a. The completed sketch of the hyperboloid, of two sheets is shown in Figure 15b.

Page 62 :
966, , Chapter 11 Vectors and the Geometry of Space, , z, , z, , y6, , y, , y, , x, V, FIGURE 15, Steps in sketching a, hyperboloid of two sheets, , x, , (a) The trace in the plane y  6, , (b) The hyperboloid 12x2  3y 2  4z2  12  0, , EXAMPLE 7 Identify and sketch the surface 4x  3y 2  12z 2  0., Solution, , Rewriting the equation in the standard form,, 4x  3y 2  12z 2, , we see that it represents a paraboloid with the x-axis as its axis. To sketch the surface,, let’s find the trace in the plane x  k. We obtain, 4k  3y 2  12z 2, Letting k  3, we see that the trace in the plane x  3 is the ellipse with equation, 12  3y 2  12z 2 or, in standard form,, y2, z2,  1, 4, 1, A sketch of this trace is shown in Figure 16a. The completed sketch of the paraboloid, is shown in Figure 16b., z, , z, , y, , y, , x3, , x, (a) The trace in the plane x  3, V, , x, (b) The paraboloid 4x  3y2  12z2  0, , FIGURE 16, , We now give a summary of the quadric surfaces and their general shapes, and we, also suggest an aid for sketching these surfaces. Note that in many instances, finding, the intercepts and using a judiciously chosen trace will be sufficient to help you obtain, a good sketch of the surface.

Page 63 :
11.6, , Equation, , Surface (computer generated), , Aid for Sketching the Figure, Find x-, y-, and z-intercepts, and then, sketch., , Ellipsoid, x, , 2, , a, , 2, , , , y, , 2, , b, , 2, , , , z, , 2, , c2, , 1, z, , z, , c, , Note: All signs are positive., , a, , b, , x, , x, , x, , a2, , , , y, , b2, , , , z, , 2, , c2, , 1, , c, , z, , z, , Notes:, 1. One sign is negative., 2. The axis lies along the coordinate axis, associated with the variable with the, negative coefficient., , zk, , y, , y, , x, , x, , Sketch the trace on the plane z  k (in, this case) for an appropriate value of k., Find the z-intercept (in this case) and use, symmetry., , Hyperboloid of Two Sheets, , , x, , 2, , a2, , , , y, , 2, , b2, , , , z, , 2, , c2, , y, , Sketch the trace on the plane z  k (in, this case) for an appropriate value of k, and for z  0. Then use symmetry., , Hyperboloid of One Sheet, 2, , b, , a, , y, , 2, , 967, , Surfaces in Space, , 1, , Notes:, 1. Two signs are negative., 2. The axis lies along the coordinate axis, associated with the variable with the, positive coefficient., , z, , z, , zk, , y, , x, , y, , x, , (continued)

Page 64 :
968, , Chapter 11 Vectors and the Geometry of Space, , Equation, , Surface (computer generated), , Aid for Sketching the Figure, Sketch the trace on the plane z  k (in, this case) for an appropriate value of k., Then use symmetry., , Cone, x, , 2, , a, , 2, , , , y, , 2, , b, , 2, , , , z, , 2, , c2, , 0, z, , Notes:, 1. One sign is negative., 2. The constant term is zero., 3. The axis lies along the coordinate axis, associated with the variable with the, negative coefficient., , z, , zk, , y, , x, , x, , Sketch the trace on the plane z  k (in, this case) for an appropriate value of k., , Paraboloids, x, , 2, , a2, , , , y, , 2, , b2, , y, ,  cz, z, , z, , Notes:, 1. There are two positive signs., 2. The axis lies along the coordinate, axis associated with the variable of, degree 1., 3. It opens upward if c  0 and opens, downward if c  0., , zk, , y, , y, x, , x

Page 65 :
11.6, , Equation, , Surface (computer generated), , Aid for Sketching the Figure, a. For the case c  0, sketch the parabolas, , Hyperbolic Paraboloid, x, , 2, , a2, , , , y, , 2, , b2, , 969, , Surfaces in Space, ,  cz, , z, , Note: There is one positive and one negative sign., , x2, , z, , and, , ca 2, , y2, cb 2, , z, , z, , y, , y, x, , x, , b. Sketch the hyperbola, x2, a, , 2, , , , y2, b2, ,  ck, , for an appropriate value of k., z, , y, x, , c. Complete your sketch., z, , y, x

Page 66 :
970, , Chapter 11 Vectors and the Geometry of Space, , 11.6, , CONCEPT QUESTIONS, 3. a. What is a quadric surface?, b. Write a standard equation for (1) an ellipsoid, (2) a, hyperboloid of one sheet, (3) a hyperboloid of two, sheets, (4) a cone, (5) an elliptic paraboloid, and (6) a, hyperbolic paraboloid., , 1. What is the trace of a surface in a plane? Illustrate by showing the trace of the surface z  x 2  y 2 in the plane z  4., 2. What is a cylinder? Illustrate by sketching the cylinder, x  y 2  4., , 11.6, , EXERCISES, , In Exercises 1–12, sketch the graph of the cylinder with the, given equation., 1. x 2  y 2  4, , 2. y 2  z 2  9, , 3. x  z  16, , 4. y  4x, , 5. z  4  x, , 6. y  z 2  9, , 2, , 2, , 2, , 7. 9x  4y  36, 2, , (c), , (d), , z, , z, , 3, , 2, , 3, , 3, , y, , 8. x  4z  16, , 2, , 2, , 9. yz  1, , y, , 2, , 10. y  x  1, 2, , 11. z  cos y, , 2, , 12. y  sec x,, , x, , p2, , x, , p, 2, , (e), , x, (f), , z, , z, , In Exercises 13–20, match each equation with one of the graphs, labeled (a)–(h)., 13. x 2 , , y2, z2,  1, 16, 4, , 14. x 2 , , 15. 2x  2y  z  1, 2, , 2, , 2, , 19., , 2, , 2, , 3, , 2, , z, 17. x 2   y, 4, 2, , 1, , 16. x  y  z  0, , 2, , 2, , 2, , y2,  z2  9, 9, , 3, , y, , y, , y,  z, 18. x 2 , 4, , x, , 20. x 2  z 2  y, , (g), , x, , 2, , y, x, z, , , 1, 4, 9, 25, (a), , (b), , z, , (h), , z, , z, , z, 5, , 2, , 1, , y, 4, , y, , y, x, , x, 2, , x, , 3, , y, , x, , In Exercises 21–44, write the given equation in standard form, and sketch the surface represented by the equation., 21. 4x 2  y 2  z 2  4, 22. 4x 2  4y 2  z 2  16, 23. 9x 2  4y 2  z 2  36, 24. 36x 2  100y 2  225z 2  900, 25. 4x 2  4y 2  z 2  4, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

Page 67 :
11.7, 26. 9x 2  9z 2  4y 2  36, , 27. x 2  4y 2  z 2  4, , 28. 9x  9y  4z  36, , 29. z  x  y  1, , 30. y  x  9z  9, , 31. 4x 2  y 2  2z 2  4  0, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 32. 4x  3y  12z  12  0, 2, , 2, , 2, , Cylindrical and Spherical Coordinates, , 54. Show that the straight lines, L 1: x  a  t, y  b  t, z  b 2  a 2  2(b  a)t and, L 2: x  a  t, y  b  t, z  b 2  a 2  2(b  a)t,, passing through each point (a, b, b 2  a 2) on the hyperbolic, paraboloid z  y 2  x 2, both lie entirely on the surface., Note: This shows that the hyperbolic paraboloid is a ruled surface,, that is, a surface that can be swept out by a line moving in space., The only other quadric surfaces that are ruled surfaces are cylinders,, cones, and hyperboloids of one sheet., , 33. x 2  y 2  z 2  0, , 34. y 2  z 2  x 2, , 35. 9x 2  4y 2  z 2  0, , 36. x 2  4y 2  16z 2  0, , 37. x 2  y 2  z, , 38. y 2  z 2  x, , 39. x  9y  z, , 40. x 2  z 2  y  1, , 41. z  x 2  y 2  4, , 42. z  x 2  4y 2  4, , the surface with the given equation., , 43. y  x  z, , 44. x  y  z, , 55. 2x 2  3y 2  6z 2  36, , 56. x 2  4y 2  z 2  2, , 57. 2x  9y  z  1, , 58. x 2  3y 2  z 2  0, , 2, , 2, , 2, , 2, , 2, , cas In Exercises 55–60, use a computer algebra system (CAS) to plot, , 2, , 2, , In Exercises 45–50, sketch the region bounded by the surfaces, with the given equations., 45. x  3y  2z  6,, 46. z  2x 2  y 2, , x  0,, , and, , x  z  2,, , 48. y 2  z 2  1,, , x 2  z 2  1,, , 2, , 49. z  2x  y, 2, , 50. z  x 2  y 2, , 2, , and, , x  0,, x  0,, , y  0, and, y  0, and, , z9x y, 2, , z 0, z 0, , 2, , z  2  x 2  y2, , and, , 2, , 2, , 59. x 2  y 2  z  0, 60. x 2  2x  4y 2  16y  z  0, , z 0, , z 2, , 47. x  y  4,, 2, , y  0, and, , In Exercises 51 and 52, find an equation of the surface satisfying, the conditions. Identify the surface., , In Exercises 61–64, 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 shows it is false., 61. The graph of y  x  3 in 3-space is a line lying in the, xy-plane., 62. The graph of 9x 2  4z 2  1 and y  0 is the graph of an, ellipse lying in the xz-plane., 63. The surface, , 51. The set of all points equidistant from the point (3, 0, 0), and the plane x  3., , 53. Show that the curve of intersection of the surfaces, 2x 2  y 2  3z 2  2y  6 and 4x 2  2y 2  6z 2  4x  4, lies in a plane., , x2, a2, , b2, , , , z2, c2, ,  4 is an ellipsoid obtained, x2, , 64. The surface, , x2, a2, , , , y2, b2, , 2, , , , y2, , 2, , , , z2, ,  c(z  z 0) is obtained by translating, , y2, x2,  2  cz vertically., 2, a, b, , Cylindrical and Spherical Coordinates, Just as certain curves in the plane are described more easily by using polar coordinates, than by using rectangular coordinates, there are some surfaces in space that can be, described more conveniently by using coordinates other than rectangular coordinates., In this section we will look at two such coordinate systems., , z, , P(r, ¨, z), , ¨, y, , The Cylindrical Coordinate System, , z, , 0, , x, , y2, ,  1 by a factor, a, b, c2, of 2 in each of the x-, y-, and z-directions., , the paraboloid, , x, , , , by stretching the ellipsoid, , 52. The set of all points whose distance from the y-axis is twice, its distance from the xz-plane., , 11.7, , 971, , r, , y, (r, ¨, 0), , FIGURE 1, The cylindrical coordinate system, , The cylindrical coordinate system is just an extension of the polar coordinate system, in the plane to a three-dimensional system in space obtained by adding the (perpendicular) z-axis to the system (see Figure 1)., A point P in this system is represented by the ordered triple (r, u, z), where r and, u are the polar coordinates of the projection of P onto the xy-plane and z is the directed, distance from (r, u, 0) to P.

Page 68 :
972, , Chapter 11 Vectors and the Geometry of Space, , The relationship between rectangular coordinates and cylindrical coordinates can, be seen by examining Figure 1. If P has representation (x, y, z) in terms of rectangular coordinates, then we have the following equations for converting cylindrical coordinates to rectangular coordinates and vice versa., Converting Cylindrical to Rectangular Coordinates, x  r cos u, , y  r sin u, , zz, , (1), , Converting Rectangular to Cylindrical Coordinates, r 2  x 2  y2, , tan u , , y, x, , zz, , (2), , EXAMPLE 1 The point 1 3, p4 , 3 2 is expressed in cylindrical coordinates. Find its rect-, , angular coordinates., z, , Solution, we have, , 3, 2, 1, , P, 3, π4 , 3, , (, , ) (√,, , 1, 1, 2, x, , 3√2, 2, , π, 4, , (3, , 0), π, 4, , 3 2, 2, , 3√2, ,, 2, , We are given that r  3, u  p>4, and z  3. Using the equations in (1),, x  r cos u  3 cos, , ), , 3, , y  r sin u  3 sin, , 2, 3√2, 2, , y, , FIGURE 2, The point P can be written as, 312, 1 312, 2 , 2 , 3 2 in rectangular coordinates., , p, 312, , 4, 2, , p, 312, , 4, 2, , and, z3, Therefore, the rectangular coordinates of the given point are, ure 2.), , 312, 1 312, 2 , 2 , 3 2 . (See Fig-, , EXAMPLE 2 The point ( 12, 12, 2) is expressed in rectangular coordinates. Find, its cylindrical coordinates., Solution We are given that x   12, y  12, and z  2. Using the equations in, (2), we have, r 2  x 2  y 2  ( 12)2  ( 12)2  2  2  4, and, tan u , So r , , y, 12, ,  1, x, 12, , 2, and, u  tan1(1)  np , , where n is an integer; and z  2., , 3, p  np, 4

Page 69 :
11.7, z, (√ 2, √ 2, 2), , a2,, , 2, , √ 2, , √2 2, , a2,, , y, , 2, , 3p, , 2b, 4, , If we pick r  0, , and, , r, ¨, , 973, , We have two choices for r and infinitely many choices for u. For example, we have, the representations, , P, , 2, , Cylindrical and Spherical Coordinates, , 7p, , 2b, 4, , If we pick r  0, , The point is shown in Figure 3. Note that neither the combination r  2 and u  7p>4, nor r  2 and u  3p>4 in Example 2 will do. (Why?), , x, , FIGURE 3, The point P can be written as, 1 2, 3p4 , 2 2 or 1 2, 7p4 , 2 2 , among, others, in cylindrical coordinates., , Cylindrical coordinates are especially useful in describing surfaces that are symmetric about the z-axis. For example, the circular cylinder with rectangular equation, x 2  y 2  c2 has the simple representation r  c in the cylindrical coordinate system., (See Figure 4.), , z, , EXAMPLE 3 Find an equation in cylindrical coordinates of the surface with the given, rectangular equation., , (0, c, 0), , a. x 2  y 2  9z, , y, , (c, 0, 0), , b. x 2  y 2  z 2, , c. 9x 2  9y 2  4z 2  36, , Solution In each case we use the relationship r 2  x 2  y 2., a. We obtain r 2  9z as the required equation., b. Here, r 2  z 2., c. Here we have 9(x 2  y 2)  4z 2  36,, , x, , FIGURE 4, The circular cylinder has the simple, representation r  c in cylindrical, coordinates., , 9r 2  4z 2  36, The surfaces are shown in Figure 5., , z, , z, , z, 3, , 2, y, , y, , x, , x, (a) x  y  9z; r  3 √z, 2, , 2, , 2, , (paraboloid), , y, , x, , (b) x  y  z ; r  z, 2, , 2, , 2, , (cone), , (c) 9x  9y 2  4z 2  36; z 2  9  _94_r 2, 2, , (ellipsoid), , FIGURE 5, , EXAMPLE 4 Find an equation in rectangular coordinates for the surface with the, given cylindrical equation., a. u , , p, 4, , b. r 2 cos 2u  z 2  4

Page 70 :
974, , Chapter 11 Vectors and the Geometry of Space, , Solution, a. Using the equations in (2), we have, y, p,  tan u  tan  1, x, 4, or, yx, b. First, we use the trigonometric identity cos 2u  cos2 u  sin2 u to rewrite the, given equation in the form, r 2(cos2 u  sin2 u)  z 2  4, r 2 cos2 u  r 2 sin2 u  z 2  4, Using the equations in (1), we then obtain, x 2  y2  z2  4, The surfaces are shown in Figure 6. Note that Figure 6a shows only the part of the, plane in the first octant., , z, , z, , 0, , 0, , π, _, 4, , y, , y, , x, , x, , π, (a) ¨  _ ; y  x, 4, , (vertical plane), , (b) r2 cos 2¨  z2  4; x2  y2  z2  4 (hyperboloid of two sheets), , FIGURE 6, , The Spherical Coordinate System, z, , ƒ ®, , O, x, , ¨, , P(®, ¨, ƒ), z, , r, , y, , In the spherical coordinate system a point P is represented by an ordered triple, (r, u, f), where r is the distance between P and the origin, u is the same angle as the, one used in the cylindrical coordinate system, and f is the angle between the positive, z-axis and the line segment OP. (See Figure 7.) Note that the spherical coordinates satisfy r 0, 0 u  2p, and 0 f p., The relationship between rectangular coordinates and spherical coordinates can be, seen by examining Figure 7. If P has representation (x, y, z) in terms of rectangular, coordinates, then, , y, x, , FIGURE 7, The spherical coordinate system, , x  r cos u, , and, , y  r sin u, , r  r sin f, , and, , z  r cos f, , Since, we have the following equations for converting spherical coordinates to rectangular, coordinates and vice versa.

Page 71 :
11.7, , Cylindrical and Spherical Coordinates, , 975, , Converting Spherical to Rectangular Coordinates, x  r sin f cos u, , y  r sin f sin u, , z  r cos f, , (3), , Converting Rectangular to Spherical Coordinates, r2  x 2  y 2  z 2, , tan u , , y, x, , cos f , , z, r, , (4), , EXAMPLE 5 The point 1 3, p3 , p4 2 is expressed in spherical coordinates. Find its rect-, , angular coordinates., Solution, , Using the equations in (3) with r  3, u  p>3, and f  p>4, we have, x  r sin f cos u  3 sin, , p, p, 12 1, 312, cos  3a, ba b , 4, 3, 2, 2, 4, , y  r sin f sin u  3 sin, , p, p, 12 13, 316, sin  3a, ba, b, 4, 3, 2, 2, 4, , and, z  r cos f  3 cos, , p, 12, 312,  3a, b, 4, 2, 2, , Thus, in terms of rectangular coordinates the given point is, , 316 312, 1 312, 4 , 4 , 2 2., , EXAMPLE 6 The point ( 13, 3, 2) is given in rectangular coordinates. Find its spherical coordinates., Solution, , We use the equations in (4). First, we have, r2  x 2  y 2  z 2  ( 13)2  32  22  3  9  4  16, , so r  4. (Remember that r, , 0.) Next, from, tan u , , y, 3, ,  13, x, 13, , we see that u  p>3. Finally, from, cos f , , z, 2, 1,  , r, 4, 2, , we see that f  p>3. Therefore, in terms of spherical coordinates, the given point is, 1 4, p3 , p3 2 ., Spherical coordinates are particularly useful in describing surfaces that are, symmetric about the origin. For example, the sphere with rectangular equation, x 2  y 2  z 2  c2 has the simple representation r  c in the spherical coordinate system. (See Figure 8a.) Also, shown in Figures 8b–8c are surfaces described by the equations u  c and f  c, where 0  c  p2 .

Page 72 :
976, , Chapter 11 Vectors and the Geometry of Space, , z, z, , z, ®c, , ƒc, 0, y, x, (a) ®  c, , ¨c, , y, , x, , x, , (b) ¨  c, , (sphere), , y, , (vertical plane), , (c) ƒ  c, , (half-cone), , FIGURE 8, , EXAMPLE 7 Find an equation in spherical coordinates for the paraboloid with rectangular equation 4z  x 2  y 2., Solution, , Using the equations in (3), we obtain, 4r cos f  r2 sin2 f cos2 u  r2 sin2 f sin2 u,  r2 sin2 f(cos2 u  sin2 u),  r2 sin2 f, , or, r sin2 f  4 cos f, , EXAMPLE 8 Find an equation in rectangular coordinates for the surface with spherical equation r  4 cos f., Solution, , Multiplying both sides of the given equation by r gives, r2  4r cos f, , Then, using the equations in (4), we have, x 2  y 2  z 2  4z, or, upon completing the square in z, we obtain, x 2  y 2  (z  2)2  4, which is an equation of the sphere with center (0, 0, 2) and radius 2., , 11.7, , CONCEPT QUESTIONS, , 1. Sketch the cylindrical coordinate system, and use it as an, aid to help you give the equations (a) for converting cylindrical coordinates to rectangular coordinates and (b) for converting rectangular coordinates to cylindrical coordinates., , 2. Sketch the spherical coordinate system, and use it as an aid, to help you give the equations (a) for converting spherical, coordinates to rectangular coordinates and (b) for converting, rectangular coordinates to spherical coordinates.

Page 73 :
11.7, , 11.7, , Cylindrical and Spherical Coordinates, , 977, , EXERCISES, , In Exercises 1–6, the point is expressed in cylindrical coordinates. Write it in terms of rectangular coordinates., , 43. f , , p, 4, , 44. z  4r 2, , 1. 1 3, p2 , 2 2, , 2. (4, 0, 3 2, , 45. z  4  r 2, , 46. r  6 sin u, , 47. r cos f  3, , 48. r sin f  3, , 5., , 6. (1, p, p), , 49. r sec u  4, , 50. r  csc u, , 51. z  r 2 sin2 u, , 52. r  4 cos f, , 53. r  z  16, , 54. 3r 2  4z 2  12, , 55. r  2 csc f sec u, , 56. r2 (sin2 f  2 cos2 f)  1, , 57. r 2  3r  2  0, , 58. r2  4r  3  0, , 3., , 1 12, 13 2, 1 3, p6 , 2 2, p, 4,, , 4. 1 2,, , p, 3,, , 52, , In Exercises 7–12, the point is expressed in rectangular coordinates. Write it in terms of cylindrical coordinates., 7. (2, 0, 3), , 8. (3, 3, 3), , 9. (1, 13, 5), 11. ( 13, 1, 2), , 10. ( 12,  12, 4), 12. ( 13, 1, 4), , In Exercises 13–18, the point is expressed in spherical coordinates. Write it in terms of rectangular coordinates., 14. 1 2, p2 , p6 2, , 13. (5, 0, 0), , 15. 1 2, 0, p4 2, , 16. 1 3, p4 , 3p, 4 2, , 17. 1 5, p6 , p4 2, , 18. 1 1, p, p2 2, , In Exercises 19–24, the point is expressed in rectangular coordinates. Write it in terms of spherical coordinates., 19. (2, 0, 0), , 20. (1, 1, 1), , 2, , 2, , In Exercises 59–66, write the given equation (a) in cylindrical, coordinates and (b) in spherical coordinates., 59. x 2  y 2  z 2  4, , 60. x 2  y 2  z 2  4, , 61. x 2  y 2  2z, , 62. x 2  y 2  9, , 63. 2x  3y  4z  12, , 64. x 2  y 2  4y, , 65. x 2  z 2  4, , 66. x 2  y 2  z 2  1, , In Exercises 67–70, sketch the region described by the inequalities., 67. r, 69. 0, , z, , 68. r 2, , 2, , u, , 2p,, , f, , p, 4,, , 0, , f, , p, 6,, , 0, , z, , r, , 4  r2, , a sec f, , 21. ( 13, 0, 1), , 22. (2, 213, 4), , 70. 0, , 23. (0, 2 13, 2), , 24. ( 13, 1, 213), , 71. Spherical Coordinate System for the Earth A spherical coordinate, system for the earth can be set up as follows. Let the origin, of the system be at the center of the earth, and choose the, positive x-axis to pass through the point of intersection of, the equator and the prime meridian and the positive z-axis to, pass through the North Pole. Recall that the parallels of latitude are measured from 0° to 90° degrees north and south of, the equator and the meridians of longitude are measured, from 0° to 180° east and west of the prime meridian., , In Exercises 25–30, the point is expressed in cylindrical coordinates. Write it in terms of spherical coordinates., 25. 1 2, p4 , 0 2, , 26. 1 2, p2 , 2 2, , 29. 1 4,, , 30. 1 12, p2 , 5 2, , 27. 1 4, p3 , 4 2, p, 6,, , 62, , 28. (12, p, 5), , In Exercises 31–36, the point is expressed in spherical coordinates. Write it in terms of cylindrical coordinates., 31. (3, 0, 0), 33. 1 2,, 35. 1 1,, , 3p p, 2, 2, p p, 4, 3, , 2, , 2, , 2, , 32. 1 5,, , p p, 6, 2, , 36. 1 5,, , p 3p, 4, 4, , 34. 1 4, p6 , p6 2, , 2, , z, Greenwich, (England), P( ®, ¨, ƒ), , 2, , 37. Find the distance between 1 2, 0 2 and (1, p, 2), where the, points are given in cylindrical coordinates., p, 3,, , r, , 0, , p, 38. Find the distance between 1 4, p2 , 2p, 3 2 and 1 3, p, 2 2 , where the, points are given in spherical coordinates., , y, Equator (0° latitude), x, , Prime meridian (0° longitude), , In Exercises 39–58, identify the surface whose equation is given., 39. r  2, , 40. z  2, , 41. r  2, , 42. u , , p, 6, , (spherical, coordinates), , V Videos for selected exercises are available online at www.academic.cengage.com/login., , a. Express the locations of Los Angeles (latitude 34.06°, North, longitude 118.25° West) and Paris (latitude 48.52°, North, longitude 2.20° East) in terms of spherical coordinates. Take the radius of the earth to be 3960 miles.

Page 74 :
978, , Chapter 11 Vectors and the Geometry of Space, b. Express the points found in part (a) in terms of rectangular coordinates., c. Find the great-circle distance between Los Angeles and, Paris. (A great circle is the circle obtained by intersecting a sphere with a plane passing through the center of, the sphere.), , 72. A geodesic on a surface is the curve that minimizes the distance between any two points on the surface. Suppose a, cylinder of radius r is oriented so that its axis coincides with, the z-axis of a cylindrical coordinate system. If P1 (r, u1, z 1), and P2(r, u2, z 2) are two points on the cylinder, show that the, length of the geodesic joining P1 to P2 is, 2r (u2  u1)  (z 2  z 1), 2, , CHAPTER, , 11, , 2, , 2, , In Exercises 73–76, 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 shows it is false., 73. The representation of a point in the cylindrical coordinate, system is not unique., 74. The equation u  p>4 in cylindrical coordinates represents, the plane with rectangular equation y  x., 75. The equation u  c, where c is a constant, in cylindrical, coordinates and the equation u  c in spherical coordinates, represent different surfaces., 76. The surface defined by the spherical equation f  p>4 is, the same as the surface defined by the rectangular equation, x 2  y 2  z 2., , REVIEW, , CONCEPT REVIEW, In Exercises 1–14, fill in the blanks., , c. If u is the angle that the unit vector u makes with the, positive x-axis, then u can be written in terms of u as, ., u, , 1. a. A vector is a quantity that possesses both, and, ., b. A vector can be represented by an, ; the, points in the, of the vector, and the, of the arrow, gives its magnitude., !, c. The vector v  AB has, point, and, point, ., d. Two vectors v and w are equal if they have the same, and, ., , 5. a. The standard equation of a sphere with center, and radius, is, (x  h)2  (y  k)2  (z  l)2  r 2., b. The midpoint of the line segment joining the points, ., P1(x 1, y1, z 1) and P2 (x 2, y2, z 2) is, , 2. a. The scalar multiple of a scalar c and a vector v is the, vector, whose magnitude is, times, that of v and the direction is the same as that of v if, and opposite that of v if, ., b. The vector v  w is represented by the arrow with tail at, the, point of v and head at the, point, of w., c. The vector v  w is defined to be the vector, ., 3. a. A vector in the plane is an ordered pair a , of real numbers, and, , called, the, components of a. The zero vector is, ., 0, b. If a  具a1, a2典 and b  具b1, b2典, then a  b , If a  具a1, a2典 and c is a scalar, then ca , , ., ., , 4. a. If a  0, then a unit vector having the same direction as, ., a is u , b. The vectors i , and j , are called, the, vectors. If a is any vector in the, plane, then a can be expressed in terms of i and j as, ., a, , 6. a. The vector with initial point, ! P1(x 1, y1, z 1) and terminal, point P2 (x 2, y2, z 2) is P1P2 , ., b. A vector a  具a1, a2, a3典 in 3-space can be written in, terms of the basis vectors i , ,j, ,, and k , as a , ., 7. a. The dot product of a  具a1, a2, a3典 and b  具b1, b2, b3典 is, and is a, ., aⴢb, b. The magnitude of a vector a can be written in terms of, the dot product as  a  , ., c. The angle between two nonzero vectors a and b is given, by cos u , ., 8. a. Two nonzero vectors a and b are orthogonal if and only, if, ., b. The angles a, b, and g that a nonzero vector a makes, with the positive x-, y-, and z-axes, respectively, are, called the, angles of a., c. The direction cosines of a nonzero vector, ., a  ai  bj  ck satisfy, 9. a. The vector obtained by projecting b onto the line containing the vector a is called the, of b onto a; it is also called the, of b along a.

Page 75 :
Review Exercises, b. The length of proj ab is called the, of b along a., c. proj ab , ., d. The work done by a constant force F in moving, an object along a straight line from P to Q is, ., W, 10. a. If a and b are nonzero vectors in 3-space, then a  b is, to both a and b., b. If 0 u p is the angle between a and b, then, a  b , ., c. Two nonzero vectors a and b are parallel if and only if, ., ab, 11. a. The scalar triple product of a, b, and c is, ., b. The volume V of the parallelepiped determined by a, b,, and c is V , ., 12. a. The parametric equations of the line passing through, ., P0 (x 0, y0, z 0) and parallel to v  具a, b, c典 are, , 979, , b. The symmetric equations of the line passing through, ., P0(x 0, y0, z 0) and parallel to v  具a, b, c典 are, 13. a. The standard form of the equation of the plane containing the points P0(x 0, y0, z 0) and having the normal vector, ., n  具a, b, c典 is, b. The linear equation ax  by  cz  d represents a, in space having the normal vector, ., The acute angle between two intersecting planes is the, angle between their, ., 14. a. If a point has rectangular coordinates (x, y, z) and, cylindrical coordinates (r, u, z), then x , ,, , and z , ; and r 2 , y, , and z , ., tan u , b. If a point has rectangular coordinates (x, y, z) and, spherical coordinates (r, u, f), then x , ,, , and z , ; and r2 , y, , cos f , ., tan u , , ,, , ,, , REVIEW EXERCISES, In Exercises 1–17, let a  2i  j  3k, b  i  2j  k, and, c  3i  2j  k. Find the given quantities., 1. 2a  3b, , 2. a ⴢ (b  c), , 3.  3a  2b , , 4.  a    c , , 5. a  c, , 6.  b  (c  a) , , 7. a ⴢ (b  c), , 8.  a  a , , 9. a  (b  c), 10. The angle between a and b, 11. Two unit vectors having the same direction as c, 12. A vector having twice the magnitude of a and direction, opposite to that of a, 13. The direction cosines of b, 14. The scalar projection of b onto a, 15. The vector projection of b onto a, 16. The scalar projection of b  c onto a, 17. The volume of the parallelepiped determined by a, b, and c, 18. Which of the following are legitimate operations?, a. a ⴢ (b  c), b. a  (b ⴢ c), c. ( a  b   b  a  c), , 23. Find the acute angle between two diagonals of a cube., 24. Find the volume of the parallelepiped with adjacent sides, AB, AC, and AD, where A(2, 1, 1) , B(2, 0, 1) , C(4, 1, 1) ,, and D(5, 2, 0) ., 25. a. Find a vector perpendicular to the plane passing through, the points P(1, 2, 2) , Q(2, 3, 1) , and R(3, 2, 1) ., b. What is the area of the triangle with vertices P, Q,, and R?, 26. A force F  i  2j  4k moves an object along the line, segment from (3, 1, 1) to (2, 1, 1) . Find the work done, by the force if the distance is measured in feet and the force, is measured in pounds., 27. A constant force has a magnitude of 20 newtons and acts in, the direction of the vector a  2i  j  3k. If this force, moves a particle along the line segment from (1, 2, 1) to, (2, 1, 4) and the distance is measured in meters, find the, work done by the force., 28. Two men wish to push a crate in the x-direction as shown in, the figure. If one man pushes with a force F1 of magnitude, 80 N in the direction indicated in the figure and the second, man pushes in the indicated direction, find the force F2 with, which he must push., F1, , 19. Show that a  2i  3j  4k and b  3i  6j  3k are, orthogonal., 20. Find the value of x such that 3i  xj  2k and, 2xi  3j  6k are orthogonal., 21. Find c such that a  2i  3j  k, b  i  2j  3k, and, c  i  3j  ck are coplanar., 22. Find two unit vectors that are orthogonal to 具3, 1, 2典 and, 具2, 3, 2典., , 60º, x, , 30º, , F2

Page 76 :
980, , Chapter 11 Vectors and the Geometry of Space, , 29. Two forces F1 and F2 with magnitude 10 N and 8 N, respectively, are applied to a bar as shown in the figure. Find the, resultant torque about the point 0., , F2, F1, 30º, 0, 4m, , 3m, , In Exercises 30–33, find (a) parametric equations and (b) symmetric equations for the line satisfying the given conditions., , In Exercises 43 and 44, find the distance between the parallel, planes., 43. x  2y  3z  2;, , 2x  4y  6z  6, , 44. 2x  3y  z  2;, , 6x  9y  3z  10, , 45. Find the distance between the point (3, 4, 5) and the plane, 2x  4y  3z  12., 46. Find the curve of intersection of the plane x  z  5 and, the cylinder x 2  y 2  9., In Exercises 47–50, describe and sketch the region in 3-space, defined by the inequality or inequalities., 47. x 2  y 2, , 4, , 30. Passes through (2, 3, 1) and has the direction of, v  i  2j  3k, , 48. 1, , x  z2, , 49. y, , x,, , 31. Passes through (1, 2, 4) and (2, 1, 3), , 50. 9x  4y  36,, , 32. Passes through (2, 1, 3) and is parallel to the line with, parametric equations x  1  2t, y  2  3t, z  1  t, , In Exercises 51–58, identify and sketch the surface represented, by the given equation., , 2, , 0, , 2, , 4, x, , 33. Passes through (1, 2, 4) and is perpendicular to, u  具1, 2, 1典 and v  具3, 2, 5典, , 51. 2x  y  6, , In Exercises 34–37, find an equation of the plane satisfying the, given conditions., , 55. 4x  9z  y, , 34. Passes through (1, 2, 3) and has a normal vector, 2i  3j  5k, , 57. x 2  z 2  y, , 35. Passes through (2, 4, 3) and is parallel to the plane with, equation 2x  4y  3z  12, 36. Passes through (2, 1, 1), (2, 2, 4), and (3, 1, 5), 37. Passes through (3, 2, 2) and is parallel to the xz-plane, 38. Find the point of intersection (if any) of the line with parametric equations x  1  2t, y  1  t, and z  2  3t, and the plane 2x  3y  4z  6, , 1,, , 2, , 0, , 0, , z, , 1,, , 0, , z, , 1, , 2, , 52. x  9  y 2, , 53. x  y 2  z 2, 2, , y, , 54. 9x 2  4y 2  z 2  36, , 2, , 2, , 56. 225x 2  100y 2  36z 2  900, 58. z  sin y, 59. The point (1, 1, 12) is expressed in rectangular coordinates., Write it in terms of cylindrical and spherical coordinates., 60. The point 1 2, p6 , 4 2 is expressed in cylindrical coordinates., Write it in terms of rectangular and spherical coordinates., , 61. The point 1 2, p4 , p3 2 is expressed in spherical coordinates., Write it in terms of rectangular and cylindrical coordinates., , 39. Find the distance between the point (2, 1, 4) and the plane, 2x  3y  4z  12, , In Exercises 62–66, identify the surface whose equation is given., , 40. Determine whether the lines with parametric equations, x  3  3t, y  1  2t, z  3  2t, and x  1  2t,, y  2  3t, z  2  t are parallel, are skew, or intersect., If they intersect, find the point of intersection., , 62. z  2, , 41. Show that the lines with symmetric equations, y3, z, x1, , , 2, 1, 2, and, y1, x2, z1, , , 3, 1, 3, intersect, and find the angle between the two lines., 42. Determine whether the planes 2x  3y  4z  12 and, 2x  3y  5z  8 are parallel, perpendicular, or neither., If they are neither parallel nor perpendicular, find the angle, between them., , 64. f , , 63. u , , p, 3, , p, 3, , (spherical, coordinates), , 65. r  2 sin u, , 66. r  2 sec f, In Exercises 67–70, write the given equation (a) in cylindrical, coordinates and (b) in spherical coordinates., 67. x 2  y 2  2, , 68. x 2  y 2  z 2  9, , 69. x 2  y 2  2z 2  1, , 70. x 2  y 2  z 2  2y, , In Exercises 71 and 72, sketch the region described by the given, inequalities., 71. 0, , r, , 72. 0, , f, , z,, p, 3,, , 0, r, , u, 2, , p, 2