Bài giảng giải tích hàm nhiều biến

TS NGUYỄN ĐỨC HẬU GIẢI TÍCH HÀM NHIỀU BIẾN SỐ (Dùng cho sinh viên Trường Đại học Thủy lợi) SECTION 12.1 DOUBLE INTEGRALS OVER RECTANGLES ◆ 843 and the corresponding approximations become more accurate when we use 16, 64, and 256 squares In the next section we will be able to show that the exact volume is 48 (a) m=n=4, VÅ41.5 (b) m=n=8, VÅ44.875 (c) m=n=16, VÅ46.46875 FIGURE The Riemann sum approximations to the volume under z=16-≈-2¥ become more accurate as m and n increase Խ EXAMPLE If R ෇ ͕͑x, y͒ Ϫ1 ഛ x ഛ 1, Ϫ2 ഛ y ഛ 2͖, evaluate the integral yy s1 Ϫ x dA R z SOLUTION It would be very difficult to evaluate this integral directly from Definition but, because s1 Ϫ x ജ 0, we can compute the integral by interpreting it as a volume If z ෇ s1 Ϫ x 2, then x ϩ z ෇ and z ജ 0, so the given double integral represents the volume of the solid S that below the circular cylinder x ϩ z ෇ Hàlies nội 2013 and above the rectangle R (See Figure 9.) The volume of S is the area of a semicircle with radius times the length of the cylinder Thus (0, 0, 1) S x FIGURE (1, 0, 0) (0, 2, 0) y yy s1 Ϫ x dA ෇ 12 ␲ ͑1͒2 ϫ ෇ 2␲ R The Midpoint Rule The methods that we used for approximating single integrals (the Midpoint Rule, the Trapezoidal Rule, Simpson’s Rule) all have counterparts for double integrals Here we consider only the Midpoint Rule for double integrals This means that we use a double Mục lục Chương HÀM NHIỀU BIẾN SỐ 1.1 Hệ tọa độ Oxyz 1.2 Mặt trụ mặt tròn xoay 1.3 Mặt bậc hai 1.4 Hệ tọa độ trụ 1.5 Hệ tọa độ cầu 1.6 Hàm véc tơ 11 1.7 Hàm nhiều biến số 13 1.8 Giới hạn hàm nhiều biến 16 1.9 Sự liên tục hàm nhiều biến số 17 1.10 Đạo hàm riêng 17 1.11 Vi phân toàn phần 19 1.12 Ứng dụng hình học 20 1.13 Đạo hàm theo hướng 20 1.14 Gradient 21 1.15 Đạo hàm hàm hợp 23 1.16 Đạo hàm hàm ẩn 25 1.17 Cực trị tự hàm nhiều biến số 27 1.18 Cực trị có điều kiện hàm nhiều biến số 29 1.19 Giá trị lớn giá trị nhỏ hàm nhiều biến 33 Chương TÍCH PHÂN BỘI 35 2.1 Tích phân bội hai 35 2.2 Tích phân lặp 37 2.3 Cách tính tích phân bội hai toạ độ Đề-các 38 2.4 Phép đổi biến tích phân bội hai 45 2.5 Các ứng dụng tích phân bội hai 48 2.6 Tích phân bội ba 55 2.7 Cách tính tích phân tọa độ đề 57 2.8 Phép đổi biến tích phân bội ba 60 2.9 Các ứng dụng tích phân bội ba 64 Chương TÍCH PHÂN ĐƯỜNG, TÍCH PHÂN MẶT 71 3.1 Trường véc tơ 71 3.2 Tích phân đường trường véc tơ 73 3.3 Định lý tích phân đường 80 3.4 Định lý Green 83 3.5 Curl Divergence 88 3.6 Tích phân mặt loại 89 3.7 Ứng dụng tích phân mặt loại 93 3.8 Mặt định hướng 94 3.9 Tích phân mặt trường véc tơ 95 3.10 Định lý Stokes 98 3.11 Định lý Gauss 101 TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com Lời nói đầu Bài giảng dành cho sinh viên Trường Đại học Thủy lợi học mơn Tốn (Giải tích hàm nhiều biến số) TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com systems for three-dimensional space This is the setting for the study of functions of two variables because the graph of such a function is a surface in space Vectors planes in space as well as velocities and acceleration of objects that move in space Chương 9.1 Three-Dimensional Coordinate Systems ● ● ● ● ● ● ● HÀM NHIỀU BIẾN SỐ z To locate a point in a plane, two numbers are necessary We know that any in the plane can be represented as an ordered pair ͑a, b͒ of real numbers, whe the x-coordinate and b is the y-coordinate For this reason, a plane is calle dimensional To locate a point in space, three numbers are required We represe O 648 CHAPTER VECTORS AND THE GEOMETRY OF SPACE point■in space by9 an ordered triple ͑a, b, c͒ of real numbers In order to represent points in space, we first choose a fixed point O (the 1.1 Hệ tọa độ Oxyz y anythat bottom corner of a room the corn and three directed lines through O are perpendicular to and eachcall other, call wall on your right is the y xz-plane, coordinate axes and labeled the xthe -axis, and Usually weinthink y-axis,the z-axis x Quy tắc bàn tay phải : Ngón tay hướng theo chiều Oz hướng quay Thethe runs along vertical, the intersection of thet x-axis as beingtrục horizontal and as being and we draw x- and y-axesdương z-axis FIGURE intersection floor and the righ of the axeschiều as in Figure Thethe direction of theofzthe -axis is determined ngược chiều kim Coordinate đồng hồaxes từ chiều dương củaentation trục Ox đến dương1.along trục Oy toward2: theIfceiling along intersection th right-hand rule as illustrated in Figure you curl the the fingers of your of righ you can now imagine around the zz-axis in the direction ofoctant, a 90Њ and counterclockwise rotationseven fromother the p z on the points same floor four ondirection the floo your thumb in theand positive x-axis to the positive y-axis, then (three corner point O z-axis P(a, b, c) Now any point inplanes space, let P is coordinate a be the The three coordinate axes determine theifthree illustrated , let contains distance from the the to Pthat b be the the xz-plane ure 3(a) The xy-plane is the plane and -axes; x y yz-plan c O -plane tothe represent point th P xWe P by tains theay- and z-axes; the xz-planexycontains - and These three coor z-axes the and we call a, b, and c the coordinates of P; planes divide space into eight 650 ■ CHAPTER VECTORS AND y parts, called octants The first octant, in the y THE GEOMETRY OF SPACE nate,axes and c is the z-coordinate Thus, to locate x ground, is determined by the positive b x gin O and move a units along the x-axis, the A Խ2 ϩ Խ to ABthe FIGURE and Խ P1 B Խ2c ෇units Խ P1parallel Խ2 z-axis as in Figure FIGURE z Hình 1.1: Quy tắc bàn tay phải tọa độ điểm Right-hand rule The point P͑a, b, c͒ determines a rectangu z Combining these equations, we get pendicular from P to the xy-plane, we get a p of P on the xy-plane Similarly Tọa độ điểm: Một điểm P có tọa độ (a, b, c) a, the b, projection c xác P1 P2 Խ2 ෇ Խ P1 A Խ2tions ϩ Խ AB BP2yz of ԽP ϩ onԽthe Խ Խ2-plane and xz-plane, respe y z -plane định cách chiếu lên trục tọa độ (Hình 1.1) plane ght wall͑Ϫ4 As numerical2 illustrations, 2theripoints l xz෇ Խ x Ϫ xFigure Խ ϩ Խ y2 Ϫ y1 Խ ϩ Խ z2 a wϪl Oz1 Խ t Khoảng cách hai điểm P1 (x1 , y1 , z1 ) P2 (x2 , y2 , zO2 ): f le 2 xy ෇ ͑x Ϫ x ͒ ϩ ͑y2 Ϫ y1z͒ xϩ ͑z2 Ϫ z1 ͒ fl P1 P2 = -pla (x1 − x2 )2 + (y1 − xy2 )2z + (z1 − z2n)e2 Therefore Խ P P Խ ෇ s͑x (0, 0, c) oor y Ϫ x ͒2 ϩ ͑y2 Ϫ y1 ͒2 ϩ ͑z2 Ϫ z1 ͒32 _4 R(0, b, c) (b) (a)k, Coordinate planes r: Phương trình mặt cầu (Hình 1.2 FIGURE bên trái) tâm 3C(h, l) bán EXAMPLE The distance fromkính the point P͑2, Ϫ1, 7͒ to the point Q͑1, Ϫ3, 5͒ is S(a, 0, c) (x − h)2 + (y Because many people P(a, b, c) have some difficulty visualizing diagrams of three-d Խ PQ Խ ෇ s͑1 Ϫ 2͒2 ϩ ͑Ϫ3 ϩ 1͒2 ϩ ͑5 Ϫ 7͒2 y _53(b)] L sional figures, − k) + (z − l)2 =your2may find it helpful to dox the following [see Figure ෇ s1 ϩ ϩ ෇ (0, b, 0) (a, 0, 0) EXAMPLE z P(x, y, z) r (_4, 3, _5) SECTION C͑h, 9.1 THREE-DIMEN Find an equation of a ysphere with radius r and center k, l͒ x SOLUTION By definition, a sphere is the set of all points P͑x, y, z͒ whose distance f Q(a, b, 0) z SOLUTION The inequalities Խ Խ C is r (See Figure 10.) Thus, P is on the sphere if and only if PC ෇ r Squarin FIGURE FIGURE ഛ x2 ϩ y2 ϩ both sides, we have PC ෇ r or Խ Խ The 2Cartesianasproduct bek͒rewritten ͑x Ϫ h͒2 ϩ can ͑y Ϫ ϩ ͑z Ϫ l͒2 ෇ r ‫ ޒ‬ϫ ‫ ޒ‬ϫ ‫͕͑ ෇ ޒ‬x dered triples of real numbers and is denoted ഛ sx ϩ y respondence between points in space and o P The result of Example is wortha remembering rectangular sothree-dimensional they represent the points whose di ͑x, y, z͒coordinat dinates, the first octant can be described th and at most But we are also given thatasz ഛ x y x positive xy-plane Thus, the given inequalities represen Equation of a Sphere An equation of a sphere with center and radius C͑h, k, l͒ r y two-dimensional geometry, the In spheres x ϩ y ϩ zanalytic ෇ and x ϩ y 2the ϩ is is a curve Itinis‫ޒ‬sketched In three-dimensional xy-plane in Figure 11 analytic FIGURE 10 FIGURE 11 Hình 1.2: Mặt cầu biểu diễn phần nằm hai nửa mặt ͑x Ϫ h͒2 cầu ϩ resents ͑y Ϫ k͒a2 surface ϩ ͑z Ϫ in l͒2‫෇ޒ‬ r C(h, k, l) In particular, if the center , 1then an surfaces equation in of ‫ޒ‬theare sphere is ≤O4, represent Phần không gian xác định bất đẳng thức ≤ x2 + yis2 the + origin zEXAMPLE zWhat ≤ (a) (b z ෇ 9.1 Exercises 2 2 biểu diễn vùng (tính phần nằm trên) hai mặt cầu x + xy2 ϩ+y 2zϩ z= ෇1r 2và ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● SOLUTION TS NGUYỄN ĐỨC HẬU The theequation set ͕͑x z ෇ represents Suppose you start at the origin, move (a) along theequation 10 Find an x-axis a dis3 tance of 54 units the positive then radius Descr in ‫ ޒ‬move whose isequation 3s7 This is EXAMPLE Showinthat of x ϩ y 2direction, ϩ z points ϩ and 4x Ϫ 6y ϩ 2z ϩz-coordinate ෇ is the http://nguyenduchau.wordpress.com downward distance of units Whatthe are xy the-plane coordinates nate planes and three units above it as in Fig sphere, and afind its center and radius of your position? 11 Find an equation SOLUTION We can rewrite the given equation in the form of an equation of a sphere ͑4, 3, Ϫ1͒ and 2.we Sketch the points (3, 0, 1), ͑Ϫ1, 0, 3͒, ͑0, 4, Ϫ2͒, and complete squares: (1, 1, 0) on a single set of coordinate axes 12 Find an equation 2 ϩ 4͒P͑6, ϩ 2, ͑y3͒ 6y ϩϪ1, 9͒ ϩ ϩ 9whose ϩ ce Which͑xof ϩ the4x points ,ϪQ͑Ϫ5, andϩ 2z ϩ 1͒ ෇ Ϫ6 ϩgin4 and 4͒, ͑z Which point lies in the R͑0, 3, 8͒ is closest to the xz-plane? ͑x ϩ 2͒2 ϩ ͑y Ϫ 3͒2 ϩ ͑z ϩ 1͒2 ෇ 13–14 yz-plane? ■ Show that t 676 tròn ■ CHAPTER 1.2 Mặt trụ mặt xoay VECTORS AND THE GEOMETRY OF SPACE x2 + y + z = đồng thời nằm phía (tính phần nằm trên) mặt Oxy Suppose the tetrahedron in the figurephẳng has a trirectangular vertex S (This means t three angles at are all right angles.) Let , , and C be the areas of the three fa S A B (Hình 1.2 bên phải) that meet at S, and let D be the area of the opposite face PQR Using the result o Phương trình đường thẳng qua điểm P0 (x0 ,lem y0 ,1,zor0 )otherwise, có có that véc tơ phương show v(a, b, c): D ෇ A2 ϩ B ϩ C x = x0 + at, y = y0 + bt, z = z0 + ct SECTION 9.5 EQU (This is a three-dimensional version of the Pythagorean Theorem.) Nếu ba số a, b, c khác phương trình đường thẳng có But thểif viết we solve the first two equations, we ge dạng: don’t satisfy the third equation Therefore, th the three equations Thus, L and L not in lines x − x0 y − y0 z − z0 = = a b c 9.5 Equations of Lines and Planes ● ● ● Planes ● ● ● ● ● ● ● Although a line in space is determined by a morewhen difficult to describe A single vector pa A line in the xy-plane is determined a point on the line and the directi the “direction” theequation plane, but vector line (its slope or anglenof inclination) are given.of The ofathe line per ca specify its direction Thus, a plane in space written using the P(x, y, z) point-slope form the plane space and a is vector orthogonal n that iswhen Likewise, a line L in three-dimensional determined we knot called a normal vector Let P͑x, y, z͒ be an a P0͑x , yr0 , zr-r¸ 0͒ on L and the direction of L In three dimensions the direction o r be P0 and to P LThen conveniently described by a vector, sothe weposition let v be vectors a vectorofparallel Let r¸ Figure normal vectorofn P is0 ortho be an arbitrary point on L and let(See theThe position vectors and P r0 and r be 6.) P¸(x¸, y¸, z¸) In particular, orthogonal to r Ϫ r0 and so A0 and OP A) If anisisthe vector with representa x they have representations OP y as in Figure 1, then the Triangle Law for vector addition gives r ෇ r0 ϩ a B a and6 v are parallel vectors, there is a4 scalar t such that a ෇ tv Thus n ؒ ͑r Ϫ r0 FIGURE z z P¸(x¸, y¸, z¸) a P(x, y, z) L r¸ O r v x y FIGURE Hình 1.3: Đường thẳng mặt phẳng r ෇can r0 be ϩ rewritten tv which as Phương trình mặt phẳng qua P0 (x0 , y0 , z0 ) có véc tơ pháp tuyến n(A, B, C): ؒr෇ value of the parameter t n z a vector of L Each gives then A(x − x0 ) + B(y − y0 ) which + C(zis − z0 ) =equation t>0 vector r of a point on L In other words, as t varies, the line is traced out by the vector r As Figure indicates, positive values of Equation to points t correspond Either Equation or is called av 1.2 Mặt trụ mặtr¸ trịn xoay lie on one side of P0 , whereas negative valuesaof to points thatw t correspond To obtain scalar equation for the plane, other side of P0 r0 ෇ ͗x , y0 , z0 ͘ Then the vector equation (4 the vectorsinh) the direction of the line in componen v that gives Mặt trụ mặt tạo đường thẳng LIf(đường giữ nguyên phương L is written b, c͘write ؒ ͗x rϪ෇x 0͗x, ,y v ෇ ͗a, b, c͘ , then we have tv ෇ ͗ta, tb, tc͘ We can͗a,also di chuyển chox ln ln song song với đường cong C (đường y , so the vector equation (1) becomes r0 ෇ nó, ͗x , ytựa , z ͘ 0 or t=0 t 1) y Comparing this equation đường ellipse (x−3)2 +2z = k −1, y = k Giao tuyến sinh loid Here, however, the a been shifted so that its ve cắt mặt mặt phẳng Oxy parabol có phương (3, 1, 0) ͑k Ͼ 1͒ are the ellipses 9.7 Cylindrical and Spherical Coordinates x trình y = + (x − 3) , z = Paraboloid minh họa ͑ hình vẽ bên ● ● ● ● In plane system FIGURE 13 geometry the polar coordinate The trace in the is -planeto xyused regions (See H.) In th The Appendix paraboloid is sketche coordinate systems that are similar to polar coordinate tions of some commonly occurring surfaces and solids whenbiểu we compute Trong hệ tọa độ trụ, điểm P (x, y, z) không gian inbaChapter chiều12 diễn volumes and triple int 1.4 of certain curves and ≈+2z@-6x-y+10=0 Hệ tọa độ trụ 9.6 Exercises ● ● ● ● ● ● ● ● ● ba tọa độ thứ tự P (r, θ, z), r θ tọa độ cực hình chiếu P mặt phẳng Oxy Hình 1.6 Cylindrical Coordinates In Example we considered the function h ෇ f ͑v, t͒, where h is the height of waves produced by wind at speed v for a z Intime thet.cylindrical system,questions a point P in thr Use Table tocoordinate answer the following (a) What the ordered value of ftriple ? What its meaning? ͑40, 15͒͑r, sented by isthe ␪, z͒is, where r and ␪ are (b) What meaning the function f ͑30, t͒? jection of isP the onto the xyof-plane and zhis෇the directed distan Describe the behavior of this function Figure 1) (c) What is the meaning of the function h ෇ f ͑v, 30͒? ToDescribe convertthefrom cylindrical to rectangular coordinat behavior of this function P(r,ă,z) z O ă x The figure shows vertical traces for a function z ෇ f ͑x, y͒ r y (r,ă,0) Which one of the graphs IIV has these traces? Explain x ෇ r cos ␪ z y ෇ r sin z ␪ k=_1 k=1 FIGURE Hình 1.6: Tọa độ trụ điểm whereas0to convert from rectangular to cylindrical coor _2 The cylindrical coordinates of a point y x Phép đổi biến tọa độ trụ _2 x = r cos θ, y = r sin θ, z=z r ෇ x 1ϩ y _1 (1.1) Traces in x=k tan ␪ ෇ y x Traces in y=k z Để tìm tọa độ trụ từ tọa độ vng góc ta sử dụng cácI đẳng z II Equations These thức equations follow from and in App y r2 = x2 + y , tan θ = , z = z EXAMPLE (1.2) x (a) Plot the point withy cylindrical coordinates ͑2,y 2␲͞3 x x coordinates (b) Find cylindrical coordinates of the point with recta http://nguyenduchau.wordpress.com z z III IV ͑3, Ϫ3, Ϫ7͒ TS NGUYỄN ĐỨC HẬU SOLUTION y x The point with cylindrical coordinates ͑2, 2␲y͞3, 1͒ (a) x Equations 1, its rectangular coordinates are z 2π ”2,       , 1’ These equations fo EXAMPLE 1.5 Hệ tọa độ cầu Ví dụ 1.4 (a) Vẽ điểm (2, 2π/3, 1) tọa độ trụ tìm tọa độ hệ tọa độ vng góc (b) Tìm tọa độ trụ điểm có tọa độ (3, −3, −7) hệ tọa độ vng góc (a) Plot the point w coordinates (b) Find cylindrica ͑3, Ϫ3, Ϫ7͒ SOLUTION (a) TheSECTION point 9.7 with C Equations 1, its rec z 2π (b) From Equations we have ”2,       , 1’ Giải √ (a) x = cos 2π y = sin 2π 3, z = Vậy tọa = −1, = √ độ vng góc (−1, 3, 1).√ (b) r = 32 + (−3)2 = 2, tan θ = −3 3√ = −1, chọn 7π θ = Vậy tọa độ trụ điểm cho (3 2, 7π/4, −7) x tan ␪ ෇ 2π ◆ 695 (b) Giải From Equations we have (a) Phương trình tọa2 ෇ độ3s2 trụ mặt cầu x2 + y + 2z = ϩ ͑Ϫ3͒ r ෇ s3 tan ␪ ෇ Ϫ3 ෇ Ϫ1 so y Ϫ3 ෇ z ෇ Ϫ7 Thus, the point isco ( Therefore, one set of cylindrical (3 s2, Ϫ␲͞4, Ϫ7) As with polar c FIGURE Ví dụ 1.5 Tìm phương trình tọa độ trụ (a) Mặt cầu x2 + y + 2z = −SPHERICAL 9.7 CYLINDRICAL COORDINATES (b) Paraboloid SECTION hyperbolic z = xAND y2 r ෇ s3 ϩ Cylindrical coordinates are use axis, and the z-axis is chosen to co axis of the circular cylinder with cylindrical coordinates this cylinder This is the reason for the name “cy 7␲2z = r2 + ϩ 2n␲ ␪෇ (b) Ta có x2 − y = r2 cos2 θ − r2 sin2 θ = r2 cos 2θ, phương trình mặt z ෇ Ϫ7 paraboloid hyperbolic z = x2 − y tọa độ trụ Therefore, one set of cylindrical coordinates is (3s2, 7␲͞4, Ϫ7) Another is r2 cos many 2θ choices (3s2, Ϫ␲͞4, Ϫ7) As with polar coordinates, there zare=infinitely (c, 0, 0) Cylindrical coordinates are trụ useful in problems symmetry aboutcó an mơt trục đối xứng (đặc Chú ý 1.1 Tọa độ thường đượcthat áp involve dụng với mặt x FIGURE axis, and the z-axis is chosen to coincide with this axis of symmetry For instance, the2 2 cácwith mặtCartesian trụ) xmặt = In c r=c, (r a= c) mặt nón 2 cylinder axisbiệt of thelàcircular cylinder equation is + theyz-axis ϩ y trụ ෇ cx + y (z = r) (xem Hình 1.7) z2 = x cylindrical coordinates this cylinder has the very simple equation r ෇ c (See Figure 3.) This is the reason for the name “cylindrical” coordinates EXAMPLE Describe the surface wh z z y y (c, 0, 0) FIGURE (0, c, 0) x x r=c, a cylinder nón z = r Hình 1.7: Mặt trụ r = FIGURE c mặt EXAMPLE Describe the surface whose equation in cylindricalz=r, coordinates a cone is z ෇ r z RE a cone y SOLUTION The equation says that the the same as r, the distance from th can vary So any horizontal trace in These traces suggest that the surfa converting the equation to rectangu have z We recognize the equation z ෇ x as being a circular cone whose axi SOLUTION The equation says that the z-value, or height, of each point on the surface is the same the distance fromcầu the point to the z-axis Because ␪ doesn’t appear, it EXAMPLE Find an equation in cyl 1.5.as r,Hệ tọa độ can vary So any horizontal trace in the plane z ෇ k ͑k Ͼ 0͒ is a circle of radius k 4x ϩ 4y ϩ z ෇ These traces suggest that the surface is a cone This prediction can be confirmed by Trongthehệequation tọa độto cầu, coordinates điểm P (x, y, z) không gian biểu diễn converting rectangular From thetrong first equation in (2) we ba chiều SOLUTION Since r ෇ x ϩ y from havebởi ba tọa độ thứ tự P (ρ, θ, φ), ρ khoảng cách từ gốc tọa độ đến P , θ z2 ෇ r ෇ x ϩ y TS NGUYỄN ĐỨC HẬU z2 ෇ Ϫ So an equation of the ellipsoid in c We recognize the equation z ෇ x ϩ y (by comparison with Tablehttp://nguyenduchau.wordpress.com in Section 9.6) as being a circular cone whose axis is the z-axis (see Figure 4) EXAMPLE Find an equation in cylindrical coordinates for the ellipsoid 4x ϩ 4y ϩ z ෇ SOLUTION Since r ෇ x ϩ y from Equations 2, we have 93 3.7 Ứng dụng tích phân mặt loại Giải Khử z từ hai phương trình hai mặt ta suy giao tuyến chúng đường có phương trình x2 + y = nằm mặt z = Vậy hình chiếu D mặt phẳng Oxy hình trịn x2 + y ≤ Trong tọa độ cực D = {0 ≤ r ≤ 1, ≤ θ ≤ 2π} Ta có z x = x x2 + y2 , z y y = x2 (x + z) dS = S y2 + , 1+z +z y =2 2dxdy D = = √ √ 2π dθ 2π (r cos θ + r)rdr r2 dr (cos θ + 1)dθ 0 √ 2π = 3.7 √ x2 + y x+ x Ứng dụng tích phân mặt loại Diện tích mặt cong A(S) = dS S Khối lượng mặt cong m= ρ(x, y, z)dS S ρ(x, y, z) khối lượng riêng mặt S điểm (x, y, z) Ví dụ 3.23 Tính khối lượng phần mặt paraboloid S: z = x2 + y , ≤ z ≤ biết tỉ khối điểm (x, y, z) ∈ S ρ(x, y, z) = z Giải Hình chiếu D mặt S lên mặt phẳng Oxy hình trịn x2 + y ≤ Khối lượng mặt TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com taking t͑x, y͒ ෇ ϩ x in Formula and converting to polar coordinates, we have ͱ ͩ ͪ ͩ ͪ yy z dS ෇ yy ͑1 ϩ x͒ S3 D ෇y 3.8 Mặt định hướng 2␲ y ෇ s2 y 964 964 ■ Ѩz Ѩx 1ϩ ϩ 2␲ y dA 2␲ 94 ͑r ϩ r cos ␪ ͒ dr d␪ ͫ ͱ ͩ ͪͬ ͩ ͪ 2␲ ␪ and sinconverting ␪ to polar coordinates, we have t͑x, y͒ ෇ ϩ x in Formula (212 ϩ 13 taking cos ␪) d␪ ෇ s2 ෇ s2 ␲ ϩ VECTOR CALCULUS m■ = CHAPTER 13zdS = (x +y ) 3Ѩz 24y dxdy 1yy+ z dS4x ෇ yy ͑1+ ϩ x͒ ϩ ϩ Ѩx Ѩz Ѩy S3 D dA 2␲ D z taking to polar coordinates, we have ෇ 1ϩϩ x inz෇Formula dS ෇ t͑x, zy͒dS dS zcos dSconverting ␪ ͒s1 ϩ ϩ r dr d␪ y0 ϩ y0 ͑1 ϩ4 rand 2 S S1 S 21 S 2␲ S Therefore yy 2π ͑1 ϩ r cos ␪ ͒s1 ϩ ϩ r dr d␪ 0CALCULUS CHAPTER 13 VECTOR ෇ s2 y Ѩz Ѩy yy yy yy ͱ ͩ ͪ ͩ ͪ Ѩz Ѩz ϩs2x͒y y 1͑rϩϩ r cos ␪ ͒ dr ϩd␪ dA yy z dS ෇ yy ͑1෇√ 3␲ 4tdtѨx (ĐổiѨybiến t = r ) + 4r෇2 rdr = π t + S D ϩ ϩ s2 ␲ ෇ ( ϩ s2 )␲ ␪ sin ␪ = r2 dθ √ 2␲ ෇y ෇1 s2 y y 2␲ ( 12 ϩ 13 cos ␪) d␪ ෇ s2 2␲ ෇ s2 ␲ yy1z dS ෇ yy z dS ϩ yy z dS ϩ yy z dS S1 S2 S3 ͫ P 3.8 ͬ S S SYou can construct S geometerto August Möbius (1790–1868).] one for yourself together as in Figure If an ant were crawl along the Möbius strip starting at by a taking a long rectangular strip of3paper, giving it a half-twist, and taping the short edges ␲ point P, it would end up on the “other of the strip is,␲ ෇ with upper togetherside” as in Figure 5.෇If an ant crawl along Möbius starting at a ϩwere 0(that ϩtos2 ( the ϩitss2 )␲ stripside point P, it would end up on2the “other side” of the strip (that is, with its upper side Mặt định hướng pointing in the opposite direction).pointing Then, if opposite the ant continued to crawl in the same in the direction) Then, if the ant continued to crawl in the same direction, FIGURE itMöbius would at the Surfaces same crossed P without direction, it would point end up back at the same ever point Phaving without ever having crossed A strip end up back Oriented an edge (If you have constructed a Möbius strip, try drawing a pencil line down the an edge (If you have constructed amiddle.) Möbius strip, try drawing a pencil line down the Therefore, a Möbius strip really has only one side In order to define integrals vector fields, we need to rule out nonorientable middle.) Therefore, a Möbius strip reallysurface has only oneofside B D P A Möbius strip ͬ 2␲ ϩ Oriented Surfaces ෇ s2 y y√͑r ϩ r cos ␪ ͒ dr d␪ π 0 3␲ = (u − u )du (Đổi biến ufields, = we 1෇+need 4t) ϩ ϩ s2 ␲ ෇ ( ϩ s2 )␲ 2␲ In order to define surface integrals of vector to rule out 2␲ ␪ nonorientable sin ␪ ෇ ( ϩ cos ␪ ) dnamed ␪ ෇ s2 after ෇ s2 ␲ ϩ the German s2 y [It is surfaces 1such as the Möbius strip shown in Figure Oriented Surfaces √ Möbius (1790–1868).] π August geometer You can construct one for yourself by taking order to define surface integrals of vector fields, we need to rule out nonorientable = rectangular + 2)ofTherefore (đvkl) Ingiving (50 strip P paper, surfaces dSMöbius ෇ yy strip z dSshown ϩ yyintaping zFigure dS ϩ4.yy a long ityyasazthehalf-twist, and the short edges such [ItzisdS named after the German 120 S FIGURE ͫ ͑1 ϩ r cos ␪ ͒s1 ϩ ϩ 02r dr d3␪ 0 Therefore C surfaces such as the Möbius strip shown in Figure [It is named after the German geometer August Möbius (1790–1868).] You can yourself by taking B D one for D construct A C a long rectangular strip Cof paper, giving it a half-twist, and taping the short edges together as in Figure If an ant were to crawl along the Möbius strip starting at a FIGURE D Constructing a Möbius strip A stripC(that is, with its upper side point P, it would end up on the “other side” of the pointing in the opposite direction) Then, if the ant continued to start crawl the same From now on we consider only orientable (two-sided) surfaces We within a surdirection, would enda up back at atthe same Hình it3.21: Mobius face has tangent plane every pointpoint at anyhaving boundarycrossed S thatMặt ͑x, y, z͒Ponwithout S (exceptever two unit normal vectors nstrip, (See Figure 6.) Ϫn1 at ͑x, y, and ntry ෇ drawing an edge (Ifpoint) you There have are constructed a Möbius a z͒pencil line down the middle.) Therefore, a Möbius strip really has only one side z B A B A FIGURE A Möbius strip FIGURE Constructing a Möbius strip n¡ Để định nghĩa tích phân mặt trường véc tơ ta cần phải định hướng mặt B D From nowB on we consider only orientable (two-sided) surfaces C We start with a sur1 face Smặt giống mặt Mobius (Hình 3.21) Ta coi mặt gồm có hai that has a tangent plane at every point ͑x, y, z͒ on S (except at any boundary A D A C hướng xácThere địnharebởi tơ pháp tuyến Hình 3.22 point) twovéc unit normal vectors n1 and n2 ෇ Ϫn1 at ͑x, y, z͒ (See Figure 6.) n™ FIGURE z Constructing a Möbius strip n¡ FIGURE y x From now on we consider only orientable (two-sided) surfaces We start with a surface S that has a tangent plane at every point ͑x, y, z͒ on S (except at any boundary point) There are two unit normal vectors n1 and n ෇ Ϫn1 at ͑x, y, z͒ (See Figure 6.) z n¡ n™ FIGURE y x n™ Hình 3.22: Hướng mặt FIGURE x y Thông thường mặt khơng kín hướng dương hướng có véc tơ pháp tuyến tạo với chiều dương trục Oz góc nhọn mặt kín hướng dương hướng có véc tơ pháp tuyến hướng phía ngồi (Hình 3.23) TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com Since the k-component is positive, gives for thethis sphere x 2thϩ If S is a smooth orientable surface given in p the r͑u,0v͒, then it is automatically supplied rwith ␾ ϫ r␪ ෇ y ru ϫ rv u ϫ rv FIGURE So the orientation Positiveand orientation the opposite 95 orientation is given by Ϫn F tion 10.5 we found the parametric representation r␾ ϫ n෇ z ϫ r͑␾, ␪ ͒ ෇ a sin ␾ cos ␪ i ϩԽ ra␾sin x ◆ SECTION 13.6 SURFACE INTEGRALS 965 3.9 Tích phân mặt trường véc tơ If it is possible to choose a unit normal vector n at every such point ͑x, y, z͒ so that n varies continuously over S, then S is called an oriented surface and the given choice of n provides S with an orientation There are two possible orientations for any orientable surface (see Figure 7) n n n z n n n n FIGURE n n n The two orientations of an orientable surface x FIGURE For a surface graph of t, we kín use Equation todương associate z ෇ 3.23: t͑x, y͒ given Hình Đốias the mặt khơng hướng hướng có véc 966 ■ CHAPTER 13 VECTOR CALCULUS with the surface a natural orientation given by the unit normal vector Positive orientation and n ෇ Խr n po for the sphere x ϩ y ϩ z ෇Observe in Exam a Thenthat from the sphere (s r␾ ϫ r␪ ෇ a sin 2␾ obtained cos ␪ i ϩ (see a sinFigu ␾ r␪ ϫ r␾ ෇ Ϫr␾ ϫ y y For a closed sur x a2 s and Խ r␾ ϫ r␪ Խis෇that convention outward from FIGURE pháp tuyến với induced by point So thetạo orientation r͑␾, ␪ ͒ is defined Figures and 9) Negative orientation tơ chiều dương Oz góc nhọn, mặt kín hướng dương hướng có véc Ѩt Ѩt r␾ ϫ r␪ i Ϫ j ϩphía k tơ pháp tuyếnϪhướng ngồi n ෇ of Vector ෇ sin ␾ cos ␪ i ϩ sin ␾ s z Surface Integrals Fields Ѩx Ѩy n෇ Խr ͱ ͩ ͪ ͩ ͪ Ѩt Ѩx 1ϩ ϩ Ѩt Ѩy ␾ ϫ r␪ Խ z Suppose that S isObserve an oriented surface with unit normal vector that n points in the same direction as with density ␳ ͑x,from andsphere velocity field flowing throug y, z͒ the v͑x, 8) y, z͒The (see Figure opposite (i Since the k-component is positive, this gives the upward orientation of the surface F=∏v imaginary surface that doesn’t impede the fluid flow, like obtained (see Figure 9) if we had reversed tha If S is a smooth orientable surface given in parametric form by a vector functionn stream.) Then ther␪rate flow (mass per unit time) per unit are r͑u, v͒, then it is automatically supplied with the orientation of the unit normal ϫ rof ␾ ෇ Ϫr␾ ϫ r␪ S ij vector y into small patches SFor in Figure 10 (compare Figure ij , as a closed surface, that is, awith surface that 1) is x S ru ϫ rv nar and so we can approximate of fluid crossing isSijth convention is thatthethemass positive orientation n෇ Խ ru ϫ rv Խ normal n per unitpoint timeoutward by the quantity from E, and inward-pointing norm FIGURE and the opposite orientation is given by Ϫn For instance, in Example in SecFigures and 9) Negative orientation y tion 10.5 we found the parametric representation x ͑ ␳ v ؒ n͒A͑Sij ͒ z r͑␾, ␪ ͒ ෇ a sin ␾ cos ␪ i ϩ a sin ␾ sin ␪ j FIGURE ϩ a cos ␾10 k Hình 3.24: Thơng lượng dịng qua mặt whereS␳., v, and n are evaluated at some point on Sij (Recall tha for the sphere x ϩ y ϩ z ෇ a Then in Example in Section 12.6 we found that r␾ ϫ r␪ ෇ a sin 2␾ cos ␪ i ϩ a sin 2␾ sin ␪ j ϩ a sin ␾ cos ␾ k 3.9 y Tích phân mặt trường véc tơ r ϫ r ෇ a sin ␾ Խ Խ x and FIGURE So the orientation induced by r͑␾, ␪ ͒ is defined by the unit normal vector Positive orientation z y x FIGURE Negative orientation ␾ ␪ vector ␳ v in the direction of the unit vector n is ␳ v ؒ n.) By sum and taking the limit we get, according to Definition 1, the surfac tion ␳ v ؒ n over S : n dS ෇ yy ␳ ͑x, y, z͒v͑x, y, z͒ ؒ n͑x, y, yy ␳ v ؒrằng Giả sử S mặt định hướng với véc tơ pháp tuyến đơn 7vị n, giả thiết S S r ϫr n ෇ với mật ෇ sin ␾độ cos ρ(x, ␪ i ϩ siny, ␾ z) sin ␪và j ϩ cos ␾ k ෇ véc r͑␾, ␪ ͒ tơ v(x, y, z) chảy qua S Tốc độ dòng dòng trường r ϫ r a Խ Խ and thistích is interpreted physically (khối lượng đơn vị thời gian) đơn vị diện ρv Nếu chiaasSthe rate of flow through S Observe that n points in the same direction as the position vector, that is, outward If we write F ෇ ␳ v, then F is also a vector field on ‫ ޒ‬3 and the nhỏ(see thành mảnh(inward) Sij Hình coibecomes Sij mặt phẳng fromđủ the sphere Figure 8) The opposite orientation would 3.24 have been obtained (see Figure 9) if we had reversed the order of the parameters because ta xấp xỉ khối lượng dòng chảy qua S ij theo hướng véc tơ n r ϫ r ෇ Ϫr ϫ r yy F ؒ n dS For a closed surface, that is, a surface that is the boundary of a solid region , the E đơn vị thời gian đại lượng S convention is that the positive orientation is the one for which the normal vectors ␪ ␾ ␾ ␪ ␾ ␪ ␾ ␪ point outward from E, and inward-pointing normals give the negative orientation (see Figures and 9) ρv.nA(Sij ) A surface integral of this form occurs frequently in physics, ev and is called the surface integral (or flux integral ) of F over S ρ, v n tính điểm Sij Lập tổng sau lấy giới hạn ta Definition If F is a continuous vector field defined on an tích phân mặt hàm ρv.n S with unit normal vector n, then the surface integral of F ov ρv.ndS = S ρ(x, y, z)v(x, y, z).n(x, y, z)dS yy F ؒ dS ෇ yy F ؒ n dS S S S This integral is also called the flux of F across S biểu diễn tốc độ dòng qua mặt S Nếu ta đặt F = ρv F trường véc tơ R tích phân trở thành F.ndS S August Mobius (1790-1868) nhà toán học người Đức In words, Definition says that the surface integral of a vect to the surface integral of its normal component over S (as previ If S is given by a vector function r͑u, v͒, then n is given by Definition and Equation we have yy F ؒ dS ෇ yy F ؒ S TS NGUYỄN ĐỨC HẬU S ෇ yy D ͫ Խ ru ϫ rv dS ru ϫ rv Խ F͑r͑u, v͒͒ ؒ http://nguyenduchau.wordpress.com Խ ru ϫ rv ru ϫ rv where D is the parameter domain Thus, we have ͬ r Խ Խ u ϫ 96 3.9 Tích phân mặt trường véc tơ Định nghĩa 3.5 Cho F = P i + Qj + Rk trường véc tơ xác định mặt định hướng S với véc tơ pháp tuyến đơn vị n tích phân mặt (loại hai) F S F.ndS F.dS = (3.9) S S P dydz + Qdzdx + Rdxdy = (3.10) S Tích phân cịn gọi thơng lượng F qua mặt S Các tính chất Tích phân mặt loại hai có tính chất tuyến tính cộng tính Nói chung khơng có tính chất bảo tồn thứ ta đổi hướng mặt tích phân đổi dấu Cách tính Tách tích phân mặt loại hai 3.10 thành tổng ba tích phân Xét tích phân (các tích phân khác tính tương tự) R(x, y, z)dydx = S R(x, y, z) cos(n, k)dS S Giả sử S có phương trình z = z(x, y) đường thẳng song song với trục Oz cắt S không điểm D hình chiếu S lên mặt phẳng Oxy miền xác định hàm z = z(x, y) Vì cos(n, k)dS = ±dxdy (dấu dương góc (n, k) góc nhọn, mặt kí hiệu S + , dấu âm trường hợp góc (n, k) góc tù, mặt kí hiệu S − ), tích phân mặt loại hai đưa tích phân hai lớp R(x, y, z)dydx = S+ R(x, y, z(x, y))dxdy D R(x, y, z)dydx = − S− R(x, y, z(x, y))dxdy D x2 y zdxdy, S mặt nửa mặt cầu x2 + y + z = Ví dụ 3.24 Tính S R2 , z ≤ Giải Hình chiếu D S mặt phẳng Oxy hình trịn x2 + y ≤ R2 Véc tơ pháp tuyến n tương ứng với mặt nửa mặt cầu S tạo với chiều k góc nhọn Phương trình mặt S, z ≤ 0, z = − R2 − x2 − y Do x2 y zdxdy = − S x2 y R2 − x2 − y dxdy D TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com 97 3.9 Tích phân mặt trường véc tơ Do tính đối xứng miền lấy tích phân hàm dấu tích phân nên tích phân cho bốn lần tích phân miền D1 với D1 phần D nằm góc phần tư thứ Tính tích phân tọa độ cực ta có π/2 2 x y zdxdy = −4 R S r5 cos2 θ sin2 θ dθ π/2 =− R R2 π/2 √ t2 (1 − cos 4θ)dθ Dùng phép đổi biến u = R2 − r2 d(r2 ) r4 sin 2θdθ =− R2 − r2 dr R2 − tdt R2 − t Sau vài phép biến đổi ta nhận x2 y zdxdy = − 2πR7 105 S zdxdy, S mặt ngồi ellipsoid Ví dụ 3.25 Tính S x2 a2 + y2 b2 + z2 c2 = Giải Hình chiếu D S mặt phẳng Oxy hình ellipse x2 y + ≤1 a2 b Do đường thẳng qua điểm thuộc D song song với trục Oz cắt S hai điểm nên ta chia S thành hai phần S1 , S2 với S1 xác định phương trình z = c 1− x2 a2 − y2 b2 véc tơ pháp tuyến S1 tạo với chiều dương trục Oz góc nhọn, S2 xác định phương trình z = −c − x2 a2 − y2 b2 có véc tơ pháp tuyến tạo với chiều dương trục Oz góc tù Dùng tính chất cộng tính tích phân mặt loại hai ta có zdxdy = S zdxdy + S1 zdxdy S2 1− =c x2 a2 − y2 dxdy − c b2 D x2 y − dxdy a2 b D 1− = 2c − 1− x2 a2 − y2 b2 dxdy D TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com normal that points toward the left E 32 Find a formula for xxS F ؒ dS similar to Formula 10 for the 40 U case where S is given by x ෇ k͑ y, z͒ and n is the unit normal that points forward (that is, toward the viewer when the axes are drawn in the usual way) 98 3.10 Định lý Stokes w E 33 Find the center of mass of the hemisphere 41 T Dùng phép đổi biến tọa độ 2cực ta2 được2 x ϩ y ϩ z ෇ a , z ജ 0, if it has constant density 2π c r y 34 Find the mass of a thin funnel in4πabc the shape of a cone zdxdy = 2abc S 3.10 − r rdr = dθ function is z ෇ sx ϩ y 2, ഛ z ഛ 4, if its density 0 ␳ ͑x, y, z͒ ෇ 10 Ϫ z 42 T 35 (a) Give an integral expression for the moment of inertia Iz Định lý Stokes i b w about the z -axis of a thin sheet in the shape of a surface Định lý 3.7 Cho mặt S có định hướng trơn function mảnh density is ␳có biên đường cong C S if the đơn, đóng trơn mảnh định hướng dương (một người đứng S hướng pháp tuyến dương từ chân đến đầu nhìn thấy hướng C ngược chiều kim đồng hồ) Trường véc tơ F có hàm thành phần liên tục có đạo hàm riêng miền mở R3 chứa S Khi 13.7 F.dr = curlF.dS Stokes’ Theorem C n n S C y x FIGURE C (3.11)● ● ● ● S z Ví dụ 3.26 Tính ● Stokes’ Theorem can be regard rem Whereas Green’s Theorem line integral around its plane bo gral over a surface S to a line int curve) Figure shows an orien of S induces the positive orien This means that if you walk in ing in the direction of n, then th F.dr, F(x, y, z) = −y i + xj + z k, C Stokes’ giao mặt Theorem phẳng y + z = mặt trụ x2 + y = (C hồ nhìn từ xuống) Giải Let S be an o by a kim simple, định hướng ngược chiều đồng closed, piecewi tion Let F be a vector field tives on an open region in ‫ޒ‬ C đường ellipse Hình 3.25 Ta tính trực tiếp ta áp dụng định lý Stokes Trước tiên ta có curlF = i j k ∂ ∂x −y ∂ ∂y ∂ ∂z z2 x = (1 + 2y)k C biên nhiều mặt, nhiên ta chọn C biên mặt S miền ellipse phía nằm mặt phẳng y + z = Như C có chiều dương TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com y C arranged to coincide with the right sid yC F ؒ dr 99 EXAMPLE Evaluate 3.10 Định lý Stokes xC F ؒ dr, where F curve of intersection of the plane y ϩ C to be counterclockwise when viewed z S SOLUTION The curve C (an ellipse) is sho C evaluated directly, it’s easier to use Sto y+z=2 Խ i Ѩ curl F ෇ Ѩx Ϫy D y Although there are many surfaces with the elliptical region S in the plane y ϩ 2 Hình 3.25: Giao củaFIGURE mặt trụ z = x + y = mặt phẳng y +upward, then C has the induced positiv x Hình chiếu D S mặt phẳng Oxy hình tròn x2 + y ≤ Véc tơ pháp tuyến S tạo với chiều dương Oz góc nhọn Áp dụng cơng thức 3.11 ta có F.dr = C curlF.dS = S (1 + 2y)dxdy S 2π = (1 + 2y)dxdy = D 2π = (1 + 2r sin θ)rdrdθ r3 r2 + sin θ 0 2π 1 + sin θ dθ dθ = 0 = (2π) + = π curlF.dS, F(x, y, z) = yzi + Ví dụ 3.27 Sử dụng cơng thức Stokes tính S y2 xzj + xyk S phần mặt cầu x2 + + z = nằm bên mặt trụ x2 + y = nằm phía mặt phẳng Oxy Giải Để tìm đường biên C S ta giải hệ gồm hai phương trình x2 + y + √ z2 = 2 2 x + y = Ta C đường tròn xác định x +√y = z = (vì z ≥ 0) Phương trình véc tơ C r(t) = cos ti + sin tj + 3k, ≤ t ≤ 2π √ √ r (t) = − sin ti + cos tj Do F (r(t)) = sin ti + cos tj + cos t sin tk Áp dụng công thức Stokes ta TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com ෇y 2␲ ෇ y 2␲ y ͫ ͑1 ϩ 2r sin ␪ ͒ r dr d␪ r2 r3 ϩ2 sin ␪ ͬ d␪ ෇ y 2␲ 0 ( 12 ϩ ෇ 12 ͑2␲͒ ϩ 100 0෇␲ 3.10 Định lý Stokes z xxS cu F͑x, y, z͒ ෇ yz i ϩ xz j ϩ xy k and S is the part of the sphere x lies inside the cylinder x ϩ y ෇ and above the xy-plane (S EXAMPLE Use Stokes’ Theorem to compute the integral ≈+¥+z@ =4 S C SOLUTION To find the boundary curve C we solve the equations x x ϩ y ෇ Subtracting, we get z ෇ and so z ෇ s3 (since the circle given by the equations x ϩ y ෇ 1, z ෇ s3 A vecto y r͑t͒ ෇ cos t i ϩ sin t j ϩ s3 k ≈+¥=1 x 0ഛtഛ so Hình 3.26: Phần mặt cầu x2 +4 y + z = nằm bên mặt trụ xrЈ͑t͒ +෇ y Ϫsin = t i ϩ cos t j FIGURE nằm phía mặt phẳng Oxy Also, we have F͑r͑t͒͒ ෇ s3 sin t i ϩ s3 cos t j ϩ cos t sin Therefore, by Stokes’ Theorem, 2π curlF.dS = F.dr = F (r(t)) r (t)dt yy curl F ؒ dS ෇ y C F ؒ dr ෇ y 2␲ F͑r͑t͒͒ ؒ rЈ͑ S S C 2π = ෇y √ √ − sin2 t + cos2 t dt = (Ϫs3 sin 2t ϩ s3 cos 2t) ෇ s3 y 2␲ 0 √ 2␲ 2π cos 2t dt ෇ Note that in Example we computed a surface integral simp ues of F on the boundary curve C This means that if we have an with the same boundary curve C, then we get exactly the same integral! In general, if S1 and S2 are oriented surfaces with the same or and satisfy theđường hypotheses of Stokes’ Theorem, then C ydx + zdy + xdz,both C trịn cos 2tdt = 0 Ví dụ 3.28 Sử dụng cơng thức Stokes tính C curl Fhướng ؒ dS ෇ y F ؒ dr ෇ yy curl F ؒ d giao mặt cầu x2 + y + z = R2 mặt phẳng x + y + 3z = hướngyytheo C S S dương This fact is useful when it is difficult to integrate over one surfac over the other Giải Ta coi C biên hình trịn tâm O bán kính R nằm √ √ mặt √phẳng x + y + z = S có véc tơ pháp tuyến đơn vị n = (1/ 3, 1/ 3, 1/ 3) = (cos α, cos β, cos γ) Áp dụng cơng thức Stokes ta có ydx + zdy + xdz = − C dydz + dzdx + dxdy S =− (cos α + cos β + cos γ)dS S √ dS = − 3πR2 √ =− S TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com 101 3.11 Định lý Gauss 3.11 Định lý Gauss Ta viết lại định lý Green mục 3.4 dạng véc tơ F.nds = C divF(x, y)dxdy D C biên định hướng dương D Nếu ta mở rộng định lý trường véc tơ R3 ta thu định lý Gauss (Divergence) Định lý 3.8 Cho E vật thể đơn mặt S biên phía ngồi E Trường véc tơ F có hàm thành phần liên tục có đạo hàm riêng miền mở R3 chứa E Khi F.ndS = S divF(x, y, z)dxdydz (3.12) E Ví dụ 3.29 Tìm thơng lượng trường véc tơ F(x, y, z) = zi + yj + xk qua phía mặt cầu đơn vị x2 + y + z = Giải Ta có divF = Mặt cầu đơn vị biên hình cầu B xác định x2 + y + z ≤ Theo định lý Gauss, thơng lượng cần tính xác định F.dS = S divFdxdydz = B dxdydz B 4π = V (B) = π(1)3 = 3 Ví dụ 3.30 Tìm F.dS S (y 2 exz )j F(x, y, z) = xyi + + + sin(xy)k S mặt bao phía ngồi vật thể E giới hạn trụ parabolic z = − x2 mặt phẳng z = 0, y = 0, y + z = Giải Để tính trực tiếp ta phải chia mặt S thành bốn mặt nhỏ Ở ta sử dụng định lý Gauss Ta có divF = y + 2y + = 3y Vật thể E xác định E = (x, y, z) | − ≤ x ≤ 1, ≤ z ≤ − x2 , ≤ y ≤ − z Ta có TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com ෇ V͑B͒ ෇ ␲ ͑1͒3 ෇ EXAMPLE Evaluate yy F ؒ dS, where S F͑x, y, z͒ ෇ xy i ϩ ( y ϩ e xz ) j ϩ sin͑xy͒ k S is the surface of the region E bounded by the parabolic cylinder z ෇ Ϫ x 3.11 Định lýand Gauss and the planes z ෇ 0, y ෇ 0, and y ϩ z ෇ (See Figure 2.) 102 z (0, 0, 1) y=2-z (1, 0, 0) (0, 2, 0) y x FIGURE z=1-≈ Hình 3.27: Vật thể giới hạn trụ parabolic z = − x2 mặt phẳng z = 0, y = 0, y + z = F.dS = divFdxdydz = S E 3ydxdydz E 1−x2 2−z ydydzdx =3 −1 0 1−x2 =3 −1 (2 − z)2 dzdx z=1−x2 (2 − z)3 − 3 = dx z=0 −1 1 =− x2 + − dx −1 184 = 35 Ví dụ 3.31 Tìm x2 dydz + y dzdx + z dxdy S S phía ngồi mặt cầu x2 + y + z = R2 Giải S biên hình cầu E : x2 + y + z ≤ R2 , P = x2 , Q = y , R = z Px = 2x, Qy = 3y , Rz = 3z Áp dụng công thức Gauss dùng phép đổi biến tọa độ cầu ta x2 dydz + y dzdx + z dxdy = S (2x + 3y + 3z )dxdydz E TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com 103 3.11 Định lý Gauss 2π = π dθ R (2ρ sin φ cos θ + 3ρ2 sin2 φ sin2 θ + 3ρ2 cos2 φ)ρ2 sin φdρ dφ 0 8πR5 = TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com cos ␪ s ෇ r␪ ͑␪ in radians͒ RIGHT ANGLE TRIGONOMETRY opp sin ␪ ෇ hyp hyp csc ␪ ෇ opp adj cos ␪ ෇ hyp hyp sec ␪ ෇ adj tan opp adj cot hyp opp ă adj cos ␪ ෇ x r sec ␪ ෇ r x tan ␪ ෇ y x cot ␪ ෇ x y ă y y x x x π y y=sec x π 2π x s3͞2 y=cot x π 1͞2 b sec x =A sin x cos x + tan2 x = sec2 x + cot2 x = csc2 x ADDITION SUBTRACTION FORMULAS sin(x + y)AND = sin x cos y + cos x sin y sin͑x ϩ− y͒ ෇ cos yxϩ cosyx − sincos y x sin y sin(x y)sin=x sin cos cos(x +෇y)sin=x cos sin͑x Ϫ y͒ cos yxϪcos cos yx − sin sin y x sin y cos(x − y) = cos x cos y + sin x sin y cos͑x ϩ y͒ ෇ cos x cos y Ϫ sin x sin y tan x + tan y tan(x +෇y)cos=x cos y ϩ sin x sin y cos͑x Ϫ y͒ − tan x tan y tan x ϩ tanxy − tan y tan tan͑x ϩ y͒ tan(x −෇y)1 = Ϫ tan x tan y + tan x tan y sin 2x = sin tan xxϪcos tanxy tan͑x Ϫ y͒ ෇ ϩ2 tan x tan cos 2x = cos x− siny2 x = cos2 x − = − sin2 x tan x tan 2x = − tan2 x − cos 2x + cos 2x DOUBLE-ANGLE FORMULAS sin2 x = cos2 x = sin 2x ෇ sin x cos2x Cơng thức biến đổi tổng thành tích 2 cos 2x ෇ cos 2x Ϫ sin 2x ෇ cos x Ϫ ෇ Ϫ sin x x+y x−y cos x + cos cos tanyx = cos 2 tan 2x ෇ Ϫ tan x x+y x−y cos x − cos y = −2 sin sin 2 x−y x+y cos sin x + sin y = sin HALF-ANGLE FORMULAS 2π x _1 ␲͞3 ␲͞2 a csc x = Công thức biến đổi tích thành tổng FUNCTIONS OF IMPORTANT ANGLES TRIGONOMETRIC − y) + cos(x + y)] cos x cos y = [cos(x ␪ sin ␪ cos ␪ tan ␪ radians sin x sin y = [cos(x − y) − cos(x + y)] 0Њ 02 1 30Њ ␲ ͞6 1͞2 s3͞2 s3͞3 sin x cos y = [sin(x − y) + sin(x + y)] 45Њ ␲͞4 s2͞2 s2͞2 60Њ 90Њ B c2 ෇ a2 ϩ b2 Ϫ 2ab cos C y=tan x _1 ␲ Ϫ ␪ ෇ cot ␪ b ෇ a ϩ c Ϫ 2ac cos B _1 y tan ␲ Ϫ ␪ ෇ sin ␪ a ෇ b ϩ c Ϫ 2bc cos A 2π y=csc x ␲ Ϫ ␪ ෇ cos ␪ ͩ ͪ ͩ ͪ THE LAW OF COSINES x 2π _1 sin C y=cos x x tan͑Ϫ␪͒ ෇ Ϫtan ␪ c π cos͑Ϫ␪͒ ෇ cos ␪ r y y=sin x sin͑Ϫ␪͒ ෇ Ϫsin ␪ sin A sin B sin C ෇ ෇ a b c (x, y) GRAPHS OF THE TRIGONOMETRIC FUNCTIONS 1 ϩ cot 2␪ ෇ csc 2␪ THE LAW OF SINES y LƯỢNG GIÁC y ϩ tan 2␪ ෇ sec 2␪ cos TRIGONOMETRIC FUNCTIONS r csc ␪ ෇ y sin ␪ ϩ cos 2␪ ෇ ͩ ͪ adj opp PHỤ LỤC y sin ␪ ෇ r sin ␪ cot ␪ ෇ tan ␪ s3 — sin 2x ෇ Ϫ cos 2x cos 2x ෇ ϩ cos 2x 2 TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com Phụ lục 105 GIỚI HẠN lim x→0 +∞, a > , 0, < a < lim ax = x→+∞ lim ax = x→−∞ lim loga x = +∞, a > , −∞, < a < lim arctan x = π , x→+∞ x→+∞ lim arccotx = 0, x→+∞ lim x→∞ lim x→0 1 = ∞, lim =0 x→∞ x x lim x→∞ lim x→0 lim arctan x = − x→−∞ 1+ x −∞, a > +∞, < a < π lim arccotx = π lim x→∞ an bm m = n, n < m ∞ n > m sin x =0 x x = e, lim (1 + x) x = e x→0 loga (1 + x) = , x ln a ax − = ln a, x 11 lim (1 + x)α − = α, x x→0 a>1 0 a 13 x2 ± a2 dx = x x2 ± a2 ± a2 ln |x + d (cot x) = − csc2 x = − dx sin x d d (arcsin x) = (sin−1 x) = √ 14 dx dx − x2 15 d d (arccos x) = (cos−1 x) = − √ dx dx − x2 16 d d (arctan x) = (tan−1 x) = dx dx + x2 17 d d (arccotx) = (cot−1 x) = − dx dx + x2 x2 ± a2 | + C, a > TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com x2 ± a2 | + C ෇ b ϩ b b c d R E F E R E N C E PA G E S ෇ natural logarithms, so Formula 10 enables calculators have a key for ϫ Scientific ෇ C ෇ 2␲ r s ෇ r ␪ ͑␪ in radians͒ ab sin ␪ b usc to use bc a calculator to compute a෇logarithm with any base (as shown in the next example) Similarly, Formula 10 allows us to graph any logarithmic function on a ␲ ␲ a graphing calculator or computer (see Exercises 43 and 44) tanrϪ1x ෇ y s&? tan y ෇ x and Ϫ Ͻ y Ͻ h EXPONENTS AND RADICALS ● ● ● ● ● ● ● ● ● ● ● ● ● G E O M E T RY xm ● x m x n ෇ x mϩn IONS Phụ lục n n na xy ෇ s xs y s n x y m͞n ͱ n ෇ ● ● ● ● ● ● b ෇ x mϪnEXAMPLE 10 Evaluate log correct to six decimal places ෇ r ෇ sx ෇ (sx ) m n m n x x s ෇ n y sy POLYNOMIALS xy ϩ y 2͒ xy ϩ y 2͒ ͑x Ϫ y͒2 ෇ x Ϫ 2xy ϩ y 3xy Ϫ y n͑n Ϫ 1͒ nϪ2 x y ͩͪ n nϪk k x y ϩ и и и ϩ nxy nϪ1 ϩ y n k и и ͑n Ϫ k ϩ 1͒ ؒ иии ؒ k r ͩͪ y nx෇ Ϫb Ϯ sb Ϫ 4ac 2a ABSOLUTE VALUE a Ͻ c ϩ c m APPENDIX B COORDINATE GEOMETRY ln͑2Ϫt͞25 ͒ ෇ ln APPENDIX B COORDINATE GEOMETRYure 20 ◆ 24 ■ A16 Խ Խ ͩ ca Ͼ cb ͩͪ ͩ or ■ or Խ Խ Խ Խ Խ Խ x ෇ a means x ෇ a or x ෇ Ϫa Point-slope equation of line through P1͑x 1, y1͒ with slope m: Exercises x Ͻ a means Ϫa Ͻ x Ͻ a B Hyperbolas x Ͼ a means yx Ϫ Ͼ ya1 ෇orm͑xxϪϽxϪa ͒ ● ● ● ● ● ● ● ● ● ● ● ● ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ● ● ● ● ● ● tended by a central angle of 72Њ ● ● ● ■ ● ● ■ ● ■ ● ■ ● ● ■ ● ■ ● ■ ● ■ ● ■ ● ● ■ ● ͑x Ϫ h͒2 ϩ ͑ y Ϫ k͒2 ෇ r ■ ● A circle has radius 1.5 m What angle is sl center of the circle by an arc m long? ● ● x Find the radius of a circular sector with an length cm EXAMPLE Evaluate ■ ● ● ● ● ● ● ● SOLUTION To evaluate the lim 11–24 ■distances Find an equation ■ Find theisdistance the points A1–2 hyperbola the set between of all points in a plane the difference of whose from of the line that satisfies thethegiven numerator and denomin conditions two fixed points and (the foci) is a constant This definition is illustrated in F F ͑1, m ͑1, Ϫ3͒, ͑5, 7͒ 1͒,and͑4,y-intercept 5͒ nator (We may assume tha Slope-intercept equation of line with slope b: P(x, y) Figure 23 11 Through ͑2, Ϫ3͒, slope Notice a hyperbola ellipse; the only the the slopedefinition of the lineof through P and Q.is similar to that of12.anThrough y ෇ mx3–4 ϩ b■ Findthat ͑Ϫ3, Ϫ5͒, slope Ϫ2 change is that the sum of distances has become a difference of distances It is left as F¡(_c, 0) F™(c, 0) x P͑Ϫ3, 3͒, Q͑Ϫ1, Ϫ6͒ P͑Ϫ1, Ϫ4͒, Q͑6, 0͒ 13 Through and ͑2, 1͒ ͑1, 6͒ Exercise 51 to show that when the foci are on the x-axis at ͑Ϯc, 0͒ and the difference CIRCLES of distances is Խ PF1 Խ Ϫ Խ PF2 Խ ෇ Ϯ2a, then the equation of the 14 hyperbola Throughis͑Ϫ1, Ϫ2͒ and ͑4, 3͒ y ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Show that the points ͑Ϫ2, 9͒, ͑4, 6͒, ͑1, 0͒, and ͑Ϫ5, 3͒ are Equation of the circle with center ͑h, k͒ and radius r: the vertices of a square FIGURE 23 x Ͻ Ϫa ● Likea slope parabolas, reflection property that has practical with of ellipses have an interesting 3–4 ■ Convert from radians to degrees (_4, 0) (4, 0) consequences If a source of light or sound is placed at one focuswith of slope a surface with Slope-intercept equation of line m and y-intercept b: (0, _3) (_5, 0) 3␲ (5, 0) x LINES Ϫ surface (a)or4␲sound is reflected off(b)the If a Ͻ b, then a ϩ c Ͻ b ϩ c elliptical cross-sections, then all the light to the Slope of cline P1͑x y1.͒ and other P2͑x 2,focus y2͒: (see Exercise 55) This principle is used in lithotripsy, a treatment y෇ mx8␲ϩ b for kidIfFIGURE and aϽ b 22 Ͼ through 0, then ca Ͻ1,cb 7␲ FIGURE 25 ney stones A reflector with elliptical is placed in such 4.cross-section (a) Ϫ (b) a way that the If9≈+16¥=144 a Ͻ b and c Ͻ 0, then ca Ͼ cb 9≈-16¥=144 ykidney Ϫ y1 stone is at one focus High-intensity sound waves generated at the other focus m෇ CIRCLES are reflected to the stone and destroy it without damaging surrounding tissue The x Ϫ x1 If a Ͼ 0, then patient is spared the trauma of surgery and recovers within few days Equation of the circle witha center ͑h, k͒ and radius r: 13 b Ͻ c, then a Ͻ c If FIGURE a Ͻ b and x ෇ Ϫa ns Ϫa Ͻ x Ͻ a ● ͪ ■ xϾa ͪ ͩͪ ■ ca Ͻ cb x෇a SOLUTION Instead of using trigonometric identities1as in Solution 1, 0.01 easier to use a diagram If y ෇ tanϪ1x, then tan y 100 and we can re ෇ x,෇ A15 illustrates the case y Ͼ 0) that (which FIGURE 20 Cone y In fact, by taking x large en y t1 b x ഛ Ϫa.according ෇ bln m Ϫ lnThis 24 shows thaty x ജ a 2, so x ෇ sx ജ a Therefore, we have x ജ a or r 2h2y= Therefore, to De y=_ a x VϪ෇ ␲ln DISTANCE AND MIDPOINT FORMULAS Ϫ1 ax 25 cos͑tan x͒ ෇ cos y ෇ 2This means π hyperbola consists of two parts, called its branches that the y x s1 ϩ x y=∆ ϩ ෇ (0, b) P125 ͑xa1,hyperbola y1͒ and P2͑x , yuseful 25 Distance we draw it 2is to first draw its asymptotes, which are the 2between 2͒: (_a, 0) a mWhen t ෇(a, 0) Ϫ ͑ln Ϫb ln 24͒ ෇ ͑ln 24 Ϫ ln m͒ (_a, 0) in Figure 24 Both branches of the hyperln lines y ෇ ͑b͞a͒xlnand y ෇ Ϫ͑b͞a͒x shown ͑x ϩ y͒2 ෇ x ϩb 2xy ϩay ͑x Ϫ ry͒2 ෇ x Ϫ 2xy ϩ y The inverse tangent function, tanϪ1 ෇ arctan, has domain ‫ޒ‬ (a, 0) is,͑ y bola approach 0the dasymptotes; they come arbitrarily close to the asymptotes ෇ s͑x Ϫ xthat 1͒ ϩ Ϫ y 1͒ (c, 0) x (_c, 0) Its graph is shown in Figure 21 We know that the lin ͑Ϫ ␲ ͞2, ␲ ͞2͒ x ͑x ϩ y͒(_c, 0) ෇ x ϩ0 3x ycϩ (c, 0) 3xy ϩ y 3x the Figure foci of 21.) a hyperbola are on find its the roles y-axis, weare equation by reversing x So thec 2inverse Similar reasoning shows th where 49Ifand Notice that thethe ෇ a Ϫfunction b (SeeisExercise x-intercepts r h vertical asymptotes of the graph of tan Since the graph of tanϪ1 is ob x and also have with Ϯa, theh y-intercepts are Ϯb, the fociofare ͑Ϯc,y.0͒, and the ellipse is symmetric ͑x Ϫ y͒3 ෇ x Ϫ 3x y ϩ 3xy Ϫ y ing the graph of the restricted tangent function about the line y ෇ x, i 25 are locatedxπ1onϩthe (0, _b) x y1 ϩ yat2 ͑0, Ϯc͒, then r foci respect to both axes If the an ellipse _ 2m͒ ,y-axis f Ϫ1of͑m͒ ෇ ͑ln Midpoint of P24 : ln lines y ෇ ␲͞2 and y ෇ Ϫ␲ asymptotes of the graph P2Ϫ n͑n Ϫ 1͒ nϪ2 ͞2 are horizontal 2 n y2 n 22 y2 nϪ1 EXAMPLE Find the foci and asymptotes of the hyperbola and 9x Ϫ 16y ෇ 144 ln 2 x x ͑x ϩ y͒ ෇ x ϩ nx 2y ϩ by interchanging x and y in (1) x weycan find its equation −1 + = c = a −b FIGURE − 24 = c2 = a2 +b2 sketch y = arctan x Of the six inverse trigonometric functions, arcsin and arctan are its graph.x = tan most useful for the purposes of calculus The inverse cosine function a2 21 b aThis b2 FIGURE 21 FIGURE function gives the time required for 2the mass to decay to m milligrams In par¥ ≈ y FIGURE 13 Sketch the graph of 9x ϩy=tan–! x=arctan x locate the foci.of the 16y ෇If144 n nϪk k EXAMPLE Exercise 46 by The144, remaining inverse trigonometric functions don’t ar   nxy =1 SOLUTION weand divide both sides equation it becomes nϪ1 n ≈ ¥ ϩ и и и ϩ x y ϩ и и и ϩ ϩ y ticular, the time required for the mass to be reduced to mg is It follows that the line y ෇ b@ a@ +   =1 1 y=´ k lim lim =0 a@ b@ SOLUTION Divide both sides of the equation LINES by 144: ` x =0, y ෇ 1͞x (This is an equilat x y=x x x _` x y DISTANCE AND MIDPOINT FORMULAS Ϫ ෇1 25 n͑n Ϫ 1͒ и и и ͑n Ϫ k ϩ 1͒ n Ϫ1 Slope line P56.58 16 1͑x1, yyears 1͒ and P2͑x 2, y2͒: ͑lnof Ϫ through t ෇ f ͑5͒ ෇ 24 ln 5͒ Ϸ where ෇ Distance between P ͑x , y ͒ and P ͑x , y ͒: Most of the Limit Laws k lnx ϩ y ෇ y ؒ 21ؒ 31 ؒ 1и и и ؒ k 2 16which9is of the form given in (2) with a ෇ and b ෇ Since c ෇ 16 ϩity can , be proved that th ෇It 25 y Ϫ y 1 (0, 3) Exercises d෇ ϩ ͑ yanswer that in ෇ Example in s͑x Ϫ x1͒2This y=ln x Laws and 10) are also va m theCestimate foci are The asymptotes are theSeclines y ෇ 34 x and y ෇ Ϫ 34 x The graph ͑Ϯ5,we 0͒.made Ϫ y1͒ agrees with the graphical QUADRATIC FORMULA , The is now in the standard form for an so we have ax 22 Ϫ ෇ x16 tionequation 1.5 particular, if we combine L is shown inellipse, Figure 25 Ϫ 4ac Ϫbx Ϯ sb are Ϯ4from anddegrees the y-intercepts b2 ෇ 9, a ෇ 4, and b ෇ The x-intercepts ing important calcu 1–2 ■ Convert to radians are Ϯ3 Find the length of a circularrule arc for subtended If ax ϩ(_4, 0) bx ϩ c ෇ 0, then x ෇ 2 y x1 ϩ x(4, 0) y2a y1 ϩ Also, x Point-slope equation of line through with slope m: P ͑x , y ͒ 2 3 1 , so and the foci are The graph is c ෇ a Ϫ b ෇ c ෇ ( Ϯs7, ) y ෇ e and its inverse (b) The graphs of the exponentials7 function function, ␲͞12 rad if the radius of the circle is 36 c y=_ y= x Midpoint of P1 P2 : , (a) 210Њ 9Њ x the natural x 2 sketched in Figure 22.are shown in Figure 13 Because the curve y ෇ e crosses the logarithm function, x {_œ„ 7, 0} {œ„ 7, 0} (a) Ϫ315Њ (b) 36Њ If a circle has radius the lengt m͑x Ϫ xthe INEQUALITIES AND ABSOLUTE VALUE ෇ 1͒ x-axis y-axis with a slope of 1, it follows that the reflected curve y Ϫ ෇ yln x crosses is a find positive int n cm, If10 x ϩ y ෇ ͑x ϩ y͒͑x Ϫ xy ϩ y 2͒ Sphere Cylinder x Ϫ y ෇4 ͑x 3Ϫ y͒͑x ϩ xy ϩ y 2͒ V ෇ ␲r y V ෇ ␲ r 2h A ෇ 4␲ r THEOREM BINOMIAL ■ A As x becomes large, we appro ͑x͒␲͞2 SOLUTION Let y ෇ tanϪ1x Then tan y ෇ negative W x and Ϫ␲values, ͞2 Ͻ y fϽ ● 3xy ϩ y ă CHAPTER LIMITS AND DERIVATIVES 107 ln Ϸ 0.773976 ln 24 x Ϫ y ෇ ͑x ϩ y͒͑x Ϫ y͒ ■ r EXAMPLE Simplify the expression cos͑tanϪ1x͒ SOLUTION Formula 10 gives log ෇ r 136 ● ͱ h b ͒ ns ● FACTORING SPECIAL POLYNOMIALS xn yn n ● ͩͪ ¨ ͩͪ ● cos y but, since tan y is known, it is easier to find sec y first: Circle Sector of Circle Sphere Cylinder Cone xn x n lim f ෇2 n 2 x lϱ 90 A෇ ␲ r A ෇ r ␪ V ෇ ␲ r V ෇ ␲ r h V ෇sec y y Sr that EXAMPLE 11 In Example in Section 1.5 we showed that the mass of ␲2ry h ෇ ϩ tan2 y ෇ ϩ x Ϫt͞25 C m͞n ෇ 2␲ rn remains r ␪ ͑␪ in radians͒ froms a෇24-mg sample after t years Find the A෇ 4␲ r is m ෇ f ͑t͒ ෇ 24 и n This means that both y ෇ x ෇s x m ෇ (s x )m inverse of this function and interpret it sec y ෇ s1 ϩ x ͑since sec y Ͼ for Ϫ␲͞2 Ͻ y Ͻ ␲͞ r n Ϫt͞25 sxSOLUTION We need to solve the equation m ෇ 24 и for t We start by isolating n x r EXAMPLE Find lim and ෇ n s x lϱ x y sythe exponential and taking natural logarithms of both sides: 1 r h r Ϫ1 œ„„„„„ 1+≈ h Thus cos͑tan x͒ ෇ cos y ă m SOLUTION sec Observe y x 2when s1 that x Ϫt͞25 Triangle ͑xy͒n ෇ x1n y n A ෇ bh ෇n 12 ab sin ␪ 1͞n x ෇s x x ෇ x mϪn xn xϪn ෇ n x ns ● Ϫn mn m ϩ ● ͑x ͒ ෇ x x ෇ n Formulas for area A, circumference x C, and volume V: m n DICALS ă GEOMETRIC FORMULAS a c ad ϩ bc ϩ ෇ b d bd a b a d ad ෇ ϫ ෇ c b c bc d x xn APPENDIX C TRIGONOM 15 Slope 3, y-intercept Ϫ2 x 11͒y, and C͑5, 15͒ (a)2 Show Ϫ ෇1 ෇ r that the points A͑Ϫ1, 3͒, B͑3, P is on the hyperbola when ͑x Ϫ h͒ ϩ ͑ y Ϫ2 k͒ are a by showing b collinear (lie on the same line) that Խ Խ Խ Խ Խ 16 Slope 5, y-intercept Խ 17 x-intercept 1, y-intercept Ϫ3 AB ϩ BC ෇ AC (b) Use slopes to show that A, B, and C are collinear | PF¡|-| PF™ |=Ϯ2a 2 y-intercept where c ෇ a ϩ b Notice that the x-intercepts are again Ϯa,18 Butx-intercept if we putϪ8x, ෇ ■ Sketch the graph of the equation in7–10 Equation we get y ෇ Ϫb 2, which is impossible, so there is19.noThrough y-intercept parallel to the x-axis ͑4, 5͒, The x ෇ is symmetric with respect y ෇ hyperbola toϪ2 both axes Through ͑4, 5͒, parallel to the y-axis the hyperbola further, at Equation and 20 obtain xyanalyze ෇0 ෇ look 9.To 10 y we Խ Խ ■ ■ ■ ■ ■ ■ ■ x2 y2 ෇1ϩ ജ1 a2 b ■ ■ ■ ■ ■ ■ 21 Through ͑1, Ϫ6͒, parallel to the line x ϩ 2y ෇ TS NGUYỄN ĐỨC HẬU http://nguyenduchau.wordpress.com ... 1.6 Hàm véc tơ 11 1.7 Hàm nhiều biến số 13 1.8 Giới hạn hàm nhiều biến 16 1.9 Sự liên tục hàm nhiều biến số... 1.15 Đạo hàm hàm hợp 23 1.16 Đạo hàm hàm ẩn 25 1.17 Cực trị tự hàm nhiều biến số 27 1.18 Cực trị có điều kiện hàm nhiều biến. .. ෇ Ϫ14? 17 1.9 Sự liên tục hàm nhiều biến số Chú ý 1.3 Các khái niệm khác, định lý phép tính giới hạn hàm nhiều biến định nghĩa cách tương tự lý thuyết giới hạn hàm số biến Ví dụ 1.21 Tính giới