CHAPTER 1 VECTOR AND SCALARS The sum of a number of vectors a  b  c  d  (i) Draw the vectors as a chain. A E d c D a B b C (ii) Then: a  b  AC AC c  AD bc  AD AD  d  AE a  b  c  d  AE Thus the sum of all vectors a, b, c, d is given by the single vector joining the start of the first to the end of the last – in this case, AE . If the vector diagram is a closed figure e.g. f F e ~ E d ~ D A ~ a ~ B b ~ C c ~ the end of the last vector coincides with the beginning of the first, the resultant sum is a vector with no magnitude.
CHAPTER 1 VECTOR AND SCALARS A P B Q C O A B D B C R OP  PQ  OQ  A  B and PQ  QR  PR  B  C OP  PR  OR  D i.e. A     D OQ  QR  OR  D i.e. A  B C  D then A  B  C   A  B C Thus the order of addition of any number of vectors is immaterial. COMPONENTS OF A VECTOR Any vector A in 3-dimensions can be represented with initial point at the origin O of a rectangular coordinate system. Z X A1iˆ O A2 ˆj A3kˆ Y Let (A1, A2, A3) be the rectangular coordinates of the terminal point of vector A with initial point at O. The vectors A1iˆ, A2 ˆj and A3kˆ are called the component vectors of
A in the x, y and z directions respectively. The resultant of A1iˆ, A2 ˆj and A3kˆ is the vector A so that A  A1iˆ  A2 ˆj  A3kˆ The magnitude of A is A  A12  A22  A32 . In particular, the position vector or radius vector r from O to the point (x, y, z) is written as r  xiˆ  yˆj  zkˆ 7
CHAPTER 1 VECTOR AND SCALARS and has magnitude r  r  x 2  y 2  z 2 Definition: A unit vector is a vector having unit magnitude. If A is a vector with magnitude A 0 then, A A  aˆ is a unit vector having the same direction as A . Example 1.3 Find the unit vector paralleled to the resultant of vectors: r1  2iˆ  4 ˆj 5kˆ , r2  iˆ  2 ˆj  3kˆ Resultant R  r1  r 2     3ˆ  6 ˆ 2k R  R  3iˆ  6 ˆj 2kˆ  32  62   7 Then a unit vector parallel to R is rˆ  R R  1 7  6 ˆj 2kˆ. Verify that rˆ has magnitude 1. Example 1.4 Determine the vector having initial point P (x1, x1, x1) and terminal point Q (x2, y2,z2) and find its magnitude. Z P(X,Y,Z X r1 O r 2 Q(X,Y,Z Y The position vector of P is r1  x1iˆ  y1 ˆj  z1kˆ The position vector of Q is r 2  x2iˆ  y2 ˆj  z2kˆ 2
CHAPTER 1 VECTOR AND SCALARS Magnitude of PQ  PQ  x2 x1  y2 y1 2  z2 z1 2 Which is the distance between P and Q. DIRECTION COSINES: The direction of a vector in three dimensions is determined by the angles that the vector makes with the three axes of reference. Z P Let X OP  r  aiˆ  bˆj  ckˆ  r b Y Then a b c r r r  cos a  r cos  cos  a  r cos   cos a  r cos Also If Then a2  b2  c2  r 2 r 2 cos2   r 2 cos2   r 2 cos2   r 2 cos2   cos2   cos2   1 l  cos , m  cos  , n  cos ; l 2  m2  n2  1 Note: [l,m,n] written in square brackets are called the direction cosines of the vector OP and are the values of the cosines of the angles which the vector makes with the three axes of reference. So for the vector r  aiˆ  bˆj  ckˆ l  a r ; m  b r ; n  c r and of course r  a2  b2  c2 Example 1.5 Find the direction cosines [l,m,n] of the vector r  3iˆ 2 ˆj  6kˆ Here a=3, b=-2, c=6; r  9  4  36  7 l  37 , m  , n  67
CHAPTER 1 VECTOR AND SCALARS SCALAR FIELD If to each point (x,y,z) of a region R in space there corresponds a number or scalar (x, y, z) , then  is called a scalar function of position and we say that a scalar field  has been defined in R. Example 1.6 The temperature at any point within or on the earth’s surface at a certain time defines a scalar field. Example 1.7  (x, y, z)  x 3 y z 2 defines a scalar field. Example 1.8 Given the scalar field defined by  (x, y, z)  3x 2 z xy 3  5 , find  at the point (-1,-2,-3).   2  5  VECTOR FIELD If to each point (x,y,z) of a region R in space there corresponds a vector V (x, y, z) , then V is called a vector function of position and we say that a vector field V has been defined in R. Example 1.9 If the velocity at any point (x,y,z) within a moving fluid is known at a certain time, then a vector field is defined. Example 1.10 V (x, y, z)  xy 2iˆ 2 yz 3 ˆj  x 2 zkˆ defines a vector field. A vector field that is independent of time is called a steady-state vector field. FURTHER EXAMPLES Example 1.11 Given two non-collinear vectors a and b , find an expression for any vector r lying in the plane determined by a and b . Non-collinear vectors are vectors that are not parallel to D R the same line. Thus when their initial points coincide, A r they determine a plane. Let r be any vector lying in the plane of a and b and having its initial point O a b C B
CHAPTER 1 VECTOR AND SCALARS Example 1.13 Find the equation of a straight line that passes through two given points A and B having position vectors a and b with respect to an origin O. A O a r b P B Let r be the position vector of any point P on the line through A and B. Then, OA OA   AB  AP  OB OP But Thus i .e . AP OP r    a t   t ( b t AB a ) ; t If the equation is written  A t a t b r  0 , the sum of the coefficients of a , b and r is 1-t+t-1=0. Alternatively, since AP and PB are collinear, we have for some scalars m and n m AP  n PB m a  n b r  m  n or so that m ( r a )  n ( b r ) which we call the symmetric form. Example 1.14 At noon, two boats P and Q are at points whose position vectors are 4 iˆ  8 ˆj and 4 iˆ  3 ˆj respectively. Both boats are moving at constant velocity; the velocity of P is 4 iˆ  ˆj and the velocity of Q is 2 iˆ  5 ˆj (all distances are in kilometers and the
CHAPTER 1 VECTOR AND SCALARS QUESTIONS FOR DISCUSSION 1. If a and b are non-collinear vectors and ~ ~ A  x  4 y a  2 x  y b and B  ~ ~  y 2 x  2 a  2 x 3 y 1 b ~ ~ find x and y such that 3 A  2 B Ans. x  56 43 , y  22 43 2. Prove that if a and b are non-collinear, then x a  y b  0 implies that ~ ~ ~ ~ x=y=0. ~ ~ ~ ~ ~ ~ and y1=y2. 4. If a , b , c are non-coplanar vectors determine whether the vectors ~ ~ ~ r 1  2 a 3 b  c , r ~ ~ ~ 2  3 a 5 b  2 c and ~ ~ ~ 3  4 a 5 b  c are ~ ~ ~ linearly independent or dependent. 5. Prove that the medians of a triangle meet in a common point that is a point of trisection of the medians. 6. Find the resultant of the following displacements: A, 20 km300 South of East; B, 50 km due West; C, 40 km Northeast; D, 30 km 600 South of West. Ans. Magnitude 20.9 km, direction 21039ˈSouth of West 7. A 100 kg mass is suspended from the center of a rope as shown in the adjoining figure. Determine the tension T in the rope. Ans. 100kg.
CHAPTER 2 THE DOT AND CROSS PRODUCTS In the case where two vectors are in the same direction, the angle between them is zero. a So a .b  ab cos 0  ab ~ ~ b On writing the two vectors A and B in their component form, i.e. A  A1iˆ  A2 ˆj  A3kˆ B  B1iˆ  B2 ˆj  B3kˆ Then Now Also A B      A1B1ˆ ˆ  A1B2 ˆ ˆ  A1B3 ˆ k  A2 B1 ˆj iˆ  A2 B2 ˆj ˆj  A2 B3 ˆj kˆ  A3 B1k ˆ  A3 B2k ˆ  A3 B3k k iˆ iˆ  ˆj ˆj  kˆ kˆ  0  1 iˆ ˆj  iˆ kˆ  ˆj kˆ  90  0 We are left with A B  A1B1  A2 B2  A3 B3 Example 2.3 If Then A  2iˆ  2 ˆj kˆ and B  6iˆ 3 ˆj  2kˆ A B  AB cos  A1B1  A2 B2  A3 B3  3.7 cos  12 6 2  4 Thus A B  4 and the angle between A and B is cos  214  790 Example 2.4 Determine the values of a so that A  aiˆ 2 ˆj  kˆ and B  2aiˆ  aˆj 4kˆ are perpendicular (⊥). A and B are ⊥if their dot product is zero. Thus aiˆ 2 ˆj  kˆ 2aiˆ  aˆj 4kˆ  2a 2 2a 4  0 or
CHAPTER 2 THE DOT AND CROSS PRODUCTS Example 2.5 Show that the vectors A  3iˆ 2 ˆj  kˆ , B  iˆ 3 ˆj  5kˆ, C  2iˆ  ˆj 4kˆ form a right- angled triangle. This can be shown in two ways. Either we show that the sum of two vectors equals the third or the sum of the three vectors is identically zero. By trial and error we find that A  B  C , so that the vectors do form a triangle. On determining the dot products of the vectors in turn we discover that A C  0 . Thus A and C are ⊥and the triangle is a right-angled triangle. Example 2.6 Find the angles that the vector A  3iˆ 6 ˆj  2kˆ makes with the coordinate axis. We let the angles be  ,  , that A makes with the positive x,y,z axes respectively. then and Thus 2 A ˆ   i  3 cos  73   73  64.60 Similarly and cos   76   76  1490 cos   72  72  73.40 Example 2.7 Find the projection of the vector A  iˆ 2 ˆj  kˆ on the vector B  4iˆ 4 ˆj  7kˆ . A P  Q R S B Projection of A on B  RS  PQ  A cos Q  A bˆ where bˆ is the unit vector in the direction of B .
CHAPTER 2 THE DOT AND CROSS PRODUCTS Thus bˆ  B B  4iˆ 4 ˆj  7kˆ 2  i j  9 k and 4 9 j  k  9 19 9 Example 2.8 Prove that C 2  A2  B 2 2 AB cos where A , B , C are the three sides of a triangle with  the angle between the vectors A and B . C B  A From the diagram we have A  B  C or C  A B Thus Hence C C      A A 2 A B  B B C 2  A2  B 2 2 AB cos We call C 2  A2  B 2 2 AB cos the Cosine rule for plane triangles. Example 2.9 Find an equation for the plane perpendicular to the vector A  2iˆ  3 ˆj  6kˆ and passing through the terminal point of the vector B  iˆ  5 ˆj  3kˆ . Also find the distance from the origin to the plane. Z P(x,y,z) Q A r 
CHAPTER 2 THE DOT AND CROSS PRODUCTS We let r be the position vector of point P and Q, the terminal point of B . Since PQ  B r is perpendicular to A,  A  0 or r A  B A gives the equation of the plane. In rectangular coordinates, we have xiˆ  yˆj  zkˆ 2iˆ  3 ˆj  6kˆ  iˆ  5 ˆj  2iˆ  3 ˆj  6kˆ or 2x  3y  6z  35 The distance from the origin to the plane is the projection of B on A . Thus, a unit vector aˆ in the direction of A is A A  72 iˆ  73 ˆj  76 kˆ . Then the projection of B on A  B a      5 . Example 2.10 Given that A  3iˆ  ˆj  2kˆ and B  iˆ 2 ˆj 4kˆ are position vectors of points P and Q respectively, a) Find an equation for the plane passing through Q and perpendicular to line PQ. Z P R Q A r B O Y X Let position vector of some point R on the plane be r . Then PQ  OQ OP  3 ˆj 6kˆ Since the plane is perpendicular to PQ Then
CHAPTER 2 Hence THE DOT AND CROSS PRODUCTS 2iˆ 3 ˆj x 1iˆ  y  2 ˆj  z  4kˆ  0 2x  3y  6z  28  0 b) What is the distance from the point (-1,1,1) to the plane? The distance from the point (x1,y1,z1) to the plane is given by: distance  ax1  by1  cz1 d a 2  b 2  c 2 Where ax+by+cz=d is the equation of the plane. Thus the distance from (-1,1,1) to the plane is 2 1  3  6  28 22  32  62  7  5 Z THE CROSS PRODUCT For vectors in space, we normally define axes of reference by the ‘right-hand rule’. OX, OY, OZ form a right-handed set if rotation from OX to OY takes a right-handed corkscrew action along the positive direction of OZ. Similarly, rotation from OY to OZ gives right-hand corkscrew action along the positive direction of OX etc. X y Definition: The cross or vector product of A and B written A  B and read as ‘ A cross B ’ is defined as a vector having the magnitude AB sin  , where  is the angle between the two vectors. The product vector acts in a direction perpendicular to A and B in such a sense that A , B , and A  B form a right-handed system-in that order. A  B A  B  AB sin   B
CHAPTER 2 THE DOT AND CROSS PRODUCTS To determine the angle  between the vectors A and B ; we have A B  AB sin  Now A B  162 142  52  477 A  A  32  22  42  29 B  B  22  32  22  17 477  129 17 sin  or 477 2917 790 Example 2.12 If A  3iˆ ˆj  2kˆ , B  2iˆ  ˆj kˆ and C  iˆ 2 ˆj  2kˆ Find (a) (b) A B C , A B  C . (a) (b) A B   7 ˆj  5kˆ A B C  iˆ  7 ˆj  5kˆ iˆ 2 ˆj  2kˆ 24iˆ  7 ˆj 5kˆ B  C     0ˆ 5 ˆ 5k A B  C 3ˆ ˆ  2k 0ˆ 5 ˆ 5k  15ˆ 15 ˆ Thus A B C A B  C , showing the need for parentheses in A B  C to avoid ambiguity. Example 2.13 The area of a parallelogram with sides A and B is A B i.e. area of parallelogram  h B A h where h  A sin  25 
CHAPTER 2 THE DOT AND CROSS PRODUCTS Thus area  A sin  B  A B We can also deduce from this that the area of a triangle is half the area of a parallelogram. Hence area of a triangle is 1 2 A B . Example 2.14 Find the area of the triangle having vertices at P(1,3,2), Q(2,-1,1), R(-1,2,3). We have PQ  iˆ 4 ˆj kˆ ; PR  ˆj  kˆ , and area of triangle  1 PQ  PR 2 2  1 107 2 TRIPLE PRODUCTS Scalar Triple Product: If A, B, C are three vectors, the scalar formed by the product A   is called the scalar triple product. If A  A1iˆ  A2 ˆj  A3kˆ , B  B1iˆ  B2 ˆj  B3kˆ and C  C1iˆ  C2 ˆj  C3kˆ . iˆ ˆj kˆ Then B  C  B1 C1 A1 A B  C  B1 C1 B2 C2 A2 B2 C2 B3 C3 A3 B3 C3
CHAPTER 2 THE DOT AND CROSS PRODUCTS Example 2.15 If A  2iˆ 3 ˆj  4kˆ , B  iˆ 2 ˆj 3kˆ and C  2iˆ  ˆj  2kˆ . iˆ ˆj kˆ Then B  C  1 2 3 2 1 2  8 ˆj  5kˆ  8 ˆj  5kˆ A    ˆ 8 ˆ  5k    24  20  42 Scalar triple products satisfy the following properties: (i) B B B A 1 2 3 1 A A A B A C B A A 3 1 B C A B C A B C Since interchanging two rows in a determinant reverses the sign. Then A       (ii) A 1 A A B A A 3 A since two rows are identical The scalar triple product in absolute represents the volume of a parallelepiped with sides A , B and C . COPLANAR VECTORS The scalar triple product provides a test to show whether three given vectors lie in the same plane. A BC
CHAPTER 2 THE DOT AND CROSS PRODUCTS If A happens to be in the same plane as B and C , then A will be perpendicular to B  C . Hence A   A B  C cos 90  0 . Therefore, three vectors A , B and C are coplanar if A   0 . Example 2.16 The vectors A  iˆ  2 ˆj 3kˆ , B  2iˆ ˆj  2kˆ and C  3iˆ  ˆj kˆ are coplanar since 1 2 A B  C  2 1 3 2  0 3 1 1 Example 2.17 Find the volume of a parallelepiped whose edges are represented by A  2iˆ 3 ˆj  4kˆ , B  iˆ  2 ˆj kˆ and C  3iˆ ˆj  2kˆ 2 3 Volume of parallelepiped is A B  C  1 2 3 1 4 1  7 2 Example 2.18 Find the constant a such that the vectors 2iˆ ˆj  kˆ , iˆ  2 ˆj 3kˆ and 3iˆ  aˆj  5kˆ are coplanar. Solution: If coplanar then      0 i.e. 2 1 3 a 5 2 3  0 from which we obtain a=-4. Example 2.19 Find an equation for the plane determined by the points P1(2,-1,1), P2(3,2,-1) and P3(-1,3,2). The position vectors of P1, P2, P3 and any point P(x,y,z) are respectively
CHAPTER 2 THE DOT AND CROSS PRODUCTS Then PP1  r r1, P2 P1  r2 r1, PP3  r3 r1 all lie in the required plane, so that r r1  r2 r1  r3 r1  0 i.e x 2ˆ y  1 ˆ  z 1k  ˆ  3 ˆ 2k  3ˆ  4 ˆ  k  0 or 11x  5 y  13z  30 VECTOR TRIPLE PRODUCTS OF THREE VECTORS If A , B and C are three vectors, then A    and   C are called vector triple products. iˆ A  B  C  A1iˆ  A2 ˆj  A3kˆ  B1 ˆj B2 kˆ B3 C1 C2 C3    B2C3 B3C2 ˆ B1C3 B3C1  ˆ  B1C2 k  iˆ ˆj kˆ  A1 B2C3 B3C2 A2 B3C1 B1C3 A3 B1C2 B2C1  A2 B1C2 B2C1  A3 B3C1 B1C3 iˆ A1 B1C2 B2C1  A3 B2C3 B3C2 ˆj  A1 B3C1 B1C3  A2 B2C3 B3C2 k Also we have B C  A2 B1C2  A3 B1C3 A2C1B2 A3C1B3 ˆ  B2 A1C1  B2 A3C3 C2 A1B1 C2 A3 B3  ˆj  B3 A1C1  B3 A2C2 C3 A1B1 C3 A2 B2 k and so we have shown that A    B C . Now since A  B C   A  B we interchange the roles of A, B and C in the above argument and have A  B C   A  B   BC  A  BA  C AB  C  and as such we use the above relations to easily show that
CHAPTER 2 THE DOT AND CROSS PRODUCTS QUESTIONS 1. If A  5iˆ 2 ˆj  3kˆ ; B  3iˆ  ˆj 2kˆ ; C  iˆ 3 ˆj  4kˆ ; determine a) The scalar triple product b) The vector triple products i.) ii.) A    A  B  C  A  B C Ans. -12 Ans. 62iˆ  44 ˆj 74kˆ Ans. 109iˆ  7 ˆj 22kˆ 2. Determine the value of p such that the vectors A, B and C are coplanar when A  2iˆ  ˆj  4kˆ ; B  3iˆ  2 ˆj  pkˆ ; C  iˆ  4 ˆj  2kˆ ; Ans. P=6 3. Find the area of a parallelogram having diagonals A  3iˆ  ˆj 2kˆ and B  iˆ 3 ˆj  4kˆ . Ans. 5 3 4. Find the shortest distance from (6,-4,4) to the line joining (2,1,2) and (3,-1,4). Ans. 3 5. A force given by F  3iˆ  2 ˆj 4kˆ is applied at the point (1,-1,2). Find the moment of F about the point (2,-1,3). Ans. 2iˆ 7 ˆj 2kˆ
CHAPTER 3 DIFFERENTIATION OF VECTORS If a vector A depends on a scalar variable t, then A can be represented as At  and A is then said to be a function of t and t is called a parameter. In the component form we write A  A1 t iˆ  A2 t  ˆj  A3 t kˆ; Differentiating w r t , we get d dt At ˆ d dt dt dt Since d dt At  is itself a vector depending on t, we can consider its derivative w r t t . If this derivative exists it is denoted by order derivatives are described. Example 3.1 If Au  3u 2  4iˆ  2u 5 ˆj  4u 3kˆ, find d 2 A d A du 2 d A  6uiˆ  2 ˆj  12u 2 kˆ d 2 A  6iˆ  24ukˆ du 2 d 2 dt 2 At  . In like manner higher If we wanted these derivatives at u=2, then we would have obtained d A and d 2 A du 2 u2  6iˆ  48kˆ Example 3.2 If A  iˆ  Int 2  1ˆj tan t kˆ , find d A dt , d 2 A dt 2 , d A dt , d 2 A dt 2 at t  0 d A dt 2t t t 0   0 ˆj kˆ u
18 CHAPTER 3 DIFFERENTIATION OF VECTORS d A dt  12  02  12 Differentiating the second time w r t t ; d 2 A 2  iˆ  2 2t 2 ˆ 2 sec2 t tan t k 2 t0  iˆ  2 ˆj  0kˆ and d 2 A 2  12  22  02  5 Example 3.3 A particle moves along a curve whose parametric equations are x  , y  2 cos 3t, , z  2 sin 3t , where t is the time. a) Determine its velocity and acceleration at any time. The position vector r of the particle is r  xiˆ  yˆj  zkˆ  iˆ  2 cos 3tˆj  2 sin 3tkˆ r  d r dt  iˆ 6 sin 3tˆj  6 cos 3tkˆ and a  d 2 r dt 2  iˆ 18 cos 3tˆj 18 sin 3tkˆ b) Find the magnitudes of the velocity and acceleration at t=0. At t=0, d r dt   6kˆ and d 2 r dt 2  iˆ 18 ˆj. Then and d r dt d 2 r dt 2 t0  t0   12   37  325 1 6
CHAPTER 3 DIFFERENTIATION OF VECTORS Example 3.6 Show that r  C1 cos 2t  C2 sin 2t  satisfies the differential equation d 2 r 2  2 d r dt  5r  0 where C1 and C2 are constants. d r dt  C1 cos 2t C2 sin 2t   2C1 sin 2t  2C2 cos 2t  d 2 r dt 2 On substituting these derivatives; on the left, the claim follows PARTIAL DIFFERENTIATION OF VECTORS: If a vector A is dependent on more than one scalar variable i.e. A=A(x1,y2,z3) say, then we differentiate A with respect to each independent variable partially denoted by i  1, 2, 3 For A  A1iˆ  A2 ˆj  A3kˆ , the partial derivative of A w.r.t. x1 is A  ˆ  ˆ  k etc Similarly, the second partial derivative w.r.t. x1 is 2 A 2  2 A1 ˆ  2 2 A2 ˆ  2 2 A3 k 2 Example 3.7 If A  x 2 yziˆ 2xz 3 ˆj  xz 2 kˆ and B  2ziˆ  yˆj x 2 kˆ , 2   at (1,0, -2). iˆ ˆj kˆ We have A  B  x 2 yz 2xz 3 xz 2 2z y x 2  2x3 z 3 xyz 2 ˆ  x 4 yz  2xz 3 ˆ  x 2 y 2 z  4xz 4 k
CHAPTER 3 DIFFERENTIATION OF VECTORS ELEMENTS OF DIFFERENTIAL GEOMETRY Differential Geometry involves a study of space curves and surfaces. If C is a space curve defined by the function rs where s is arc length, then we have d r ds which is a unit tangent vector to C denoted by Tˆ . Suppose r  rt  is an arbitrary representation of the curve C with the same orientation as r  rs , then we find r  d r dt  d r ds ds dt the quantity ds dt d r dt d r ds  d r d r dt dt  r r d r dt ds Example 3.8 Find the unit tangent vector along the helix r  a cos tiˆ  a sin tˆj  btkˆ where a and b are non-zero. Solution: d r dt  sin tiˆ  a cos tˆj  bkˆ d r dt  ds dt    1 2 Thus Tˆ  d r dt ds dt Definition: The rate at which Tˆ changes with respect to S is a Z measure of the curvature of C and is given by dTˆ ds . C Nˆ direction of d Tˆ
CHAPTER 3 DIFFERENTIATION OF VECTORS And d ds  dTˆ dt ds dt aw2 a w  b 2 cos wtˆ  sin  So that K  dTˆ ds aw2 a w  b 2 * Two limiting cases of (*) are worth noting. First if b=0, then z=0, and the helix reduces to a circle of radius a in the xy-plane, while equation (*) reduces to K  1a . Secondly, if a=0, then x=y=0 and z=bt. This tells us that the point moves along the z-axis. In this case the curvature k=0 i.e. a straight line (see following figure) Z y X Definition: We defined a vector Bˆ perpendicular to both Tˆ and Nˆ by the formula Bˆ  Tˆ  Nˆ This vector is called the binormal vector of the curve. The three vectors Tˆ Nˆ and Bˆ (in this order) define a moving right-handed coordinate system at any specified point of C. Definition: A plane to a curve at a point P containing the tangent and principal normal at P is called an osculating plane. Definition: The normal plane is the plane through P perpendicular to the tangent. Definition: The rectifying plane is the plane through P which is perpendicular to the principal normal. Definition: The arc of rotation of the binormal vector (i.e. d Bˆ ds ) the torsion vector of the curve and its magnitude is denoted by .   1 is called
CHAPTER 3 DIFFERENTIATION OF VECTORS A set of relations involving derivatives of the fundamental vectors Tˆ Nˆ and Bˆ is known collectively as the Frenet-Serret formulae given by dTˆ ds  kNˆ , dNˆ ds Bˆ KTˆ , dBˆ ds  Or in matrix form as K  0 dB ds and they obey the cyclic vector multiplication in an anticlockwise sense: Tˆ Nˆ Bˆ So that Tˆ  Nˆ  Bˆ ; Nˆ  Bˆ  Tˆ and Bˆ  Tˆ  Nˆ Clockwise move gives the negatives. Example 3.11 Prove that the curvature of the space curve r  r t  is given numerically by K  We have   r r  3 r , where dots denote differentiation with respect to  r  d r dt  d r ds ds dt   Tˆ r  r  d 2 r dt 2  T r  T  d dt  r  dTˆ ds 2 r  Tˆ d dt   r  KNˆ r 2      2    3  3 r r  T dt
CHAPTER 3 DIFFERENTIATION OF VECTORS Thus   r r  3 r K    r r  3 r Example 3.12 Prove that     r r r   2 r r for the space curve r  r t  . Given curve is r  r t  Tˆ  d r ds  d r dt dt ds   r dt ds dTˆ ds  KNˆ   r  r  r ds 2 d 2t ds 2 2 K  ds 2 ds3 ds   r ds 3   3 r d 2t ds 2   r d 3t dt ds3 ds where ds KN  K B KT  K 2 T  KB 5 2  r r  r r 2 ds ds ds Taking dot product with Tˆ ; we obtain    r r ds 2 ds ds ds 3 2 2  ds   6
CHAPTER 3 DIFFERENTIATION OF VECTORS ds 6    r r r  6 r ; dt ds  1 ds dt  1 d r dt 1 r Thus      r r r  6 2     r r   2  6 r r r  6  r    r r   2 r r QUESTIONS 1. Find Tˆ and k for the space curve x   sin  , y  cos , z  4 sin2 . Ans. K  1 8 6 2 cos , T  3  cos cos2  2 sin  sin2  12 cos 4 2. Find the torsion of the curve x  2t 1 t , y  t 2 t , z  t  2 . Ans. T=0. The curve lies on the plane x-3y+3z=5 3. Show that the equation of the tangent line, principal normal and bi-normal to the space curve r  rt at the point t=to can be written respectively r  r o  t Tˆo , r  r o  t Nˆ o , r  r o  t Bˆo , where t is a parameter. 4. Find equations for (a) tangent, (b) principal normal and (c) bi-normal to the our
curve x=3cos t, y=3sin t, z=4t at the point where  . Ans. (a) Tangent:  . (b) Normal: r   4 ˆj  tiˆ . (c) Bi-normal: r   4 ˆj  t  54 5  42
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL Example 4.1 The gradient of   2x2 yz3 is    4xyz 3iˆ  2x 2 z 3 ˆj  6x 2 yz 2 kˆ Now given that    x, y, z is continuously differentiable with respect to its variables x,y,z, then if we let x=x(t), y=y(t), z=z(t) where t is a parameter, we have  t    x(t), y(t), z(t) On differentiating with respect to t, we obtain d dt  dt  dt  dt or d  dx  dy  dz i  ˆ  ˆ where r  xiˆ  yˆj  zkˆ Hence total differential d of  is given by d  grad  d r If ds is a small element of an arc, then d ds  d r ds  grad Now if we take r  rˆt , then d r dt  d r ds dt  d r dr dt Meaning that d r ds d r dr dt dt If we denote this unit vector by ˆ i.e. d r ds  ˆ , the result becomes d ds  ˆ  grad
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL Example 4.2 Find the directional derivative of   x 2 yz  4xz 2 at (1,-2,-1) in the direction 2iˆ ˆj 2kˆ .  2 yz  4xz 2   2xyz  4z 2 iˆ  x 2 zˆj  x 2 y  8zxkˆ  8iˆ ˆj 10kˆ at Unit vector in the direction of 2iˆ ˆj 2kˆ is ˆ  2iˆ ˆj 2kˆ 3 Required directional derivative is 1 3  163  13  20 3  37 3 If  x, y, z  constant , this relationship represents a surface in space, depending on the value ascribed to the constant. If d r is the displacement in this surface, then d  0 since  is constant over the surface. Thus from grad  d r  d we have d  grad  d r  0 such small displacements d r in the surface. But grad  d r  grad d r cos  0    2 grad is perpendicular to d r The unit vector Nˆ in the direction of grad is called the unit normal vector at P. Unit Normal Vector Nˆ 
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL Example 4.3 Find the unit normal to the surface x 2 y  2xz  4 at the point (2,-2,3)   2xy  2ziˆ  x 2 ˆj  2xkˆ   4 ˆj  4kˆ at Unit normal to the surface   13 iˆ  23 ˆj  23 kˆ If  and are differentiable scalars, then we have; 1.) 2.)           Example 4.4 If   x 2 yz  xz 2 and  xy 2 z z 3 , evaluate  at the point (2,1,3).        y 2 zˆ  2xyzˆ   k  2 2 2 2  303ˆ 12 ˆ 25k 2121ˆ 12 ˆ 16k   3  Example 4.5 Find r 3 1 3 3 2 2 2 2 2 3 3 3 2 2 1 1 1 2 2 2 2 1 2
CHAPTER 4 Example 4.6 GRADIENT, DIVERGENCE AND CURL Prove that r   f  r r 1 2 Example 4.7 Find if   In r  12 In x 2  y 2  z 2 . 1 2  1 ˆ  ˆ  ˆ 2 1 r 2x 2  y 2  z 2  r 2 Example 4.8 Show that is a vector perpendicular to the surface  (x, y, z)  c where c is a constant. Let r  xiˆ  yˆj  zkˆ be the position vector to any point P(x,y,z) on the surface. Then d r  dxiˆ  dyˆj  dzkˆ lies in the tangent plane to the surface at P. But d  dx  dy  dz  0 or i  ˆ  i.e.  d r  0 so that is perpendicular to d r and therefore to the surface. Example 4.9 Let V  x  2 y  4ziˆ  2x 3y z ˆj  4x y  2zkˆ . Show that V can be expressed as the gradient of a scalar function. Assume V   ˆ 
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL So that  x  2 y  4z   1 2 2 holding y and z constant. Similarly,  2x 3y z   2xy 32 y 2 yz  gx, z  4x y  2z   4xz yz  z 2  hx, y Comparing the three equations, we have a common value of  if we choose f y, z  3y 2 2  z 2 , gx, z  x2 2  z 2 , hx, y  2 2 2 2 So that   2 2 2 2  z 2  2xy  4xz yz DIVERGENCE: Let Ax, y, z  A1iˆ  A2 ˆj  A3kˆ be defined and differentiable at every point (x,y,z) in a specified region. Then the divergence of A , written  A or div A is defined by i  ˆ  ˆ    We note that the gradient operator acts on a scalar and gives a vector while divergence operation acts on a vector and gives a scalar. Example 4.10 If A  2x 2 yiˆ 2xy 2  y 3 z ˆj  3 y 2 z 2 kˆ , determine  A . i  ˆ  ˆ  4xy 4xy 6 y 2 z  6 y 2 z  0
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL Example 4.11 Find   where    x, y, z  is a differentiable scalar function.  ˆj   ˆj  kˆ   kˆ    2 2  2 2  2 2  2 The operator 2  2 2  2 2  2 2 is Example 4.12 Evaluate  r 3 r . 3  ˆj  ˆ 2  y 2  z 2 3 2  3 2 2  y 2  z 2 3 2 2  y 2  z 2 3 2 2 2 3 2 3  3    6r 3 1 2 Example 4.13 r r c
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL Example 4.16 Evaluate    if A is a differentiable vector field.  A  iˆ A1 ˆj A2 kˆ A3 ˆ     iˆ ˆj kˆ  j             Example 4.17 Evaluate   where  is a scalar field.    iˆ ˆj kˆ 2 2 2  0 x
CHAPTER 4 Example 4.18 Evaluate    GRADIENT, DIVERGENCE AND CURL     iˆ A1 ˆj A2 kˆ A3    0 Thus    0 Definition: A vector F is called irrotational if curl F  0 Example 4.19 Show that the vector r r2 is irrotational. r i j k ˆj   kˆ  0iˆ  0 ˆj  0kˆ  0 Hence irrotational. x x x
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS LINE INTEGRALS Let rt   xt iˆ  yt  ˆj  zt kˆ where rt  is the position vector of (x,y,z,) define a curve C joining points P1 and P2 where t=t1 and t=t2 respectively. The curve C is composed of a finite number of curves (see figure) for each of which rt  has a continuous derivative. P2 rd r d r C O r P1 Let F x, y, z  F1iˆ  F2 ˆj  F3kˆ be a vector function of P2 d r n position defined and continuous along C. Then the d r integral of the tangential component of F along C from P1 to P2 dr2 dr1 P1 C P2 is an example of a line integral. If F is a force on a particle moving along C, this line integral represents the work done by the force. If C is a simple closed curve, the integral around C is often denoted by F  d r In general, any integral, which is to be evaluated along a curve is called a line integral. If  is a scalar field, then the line integral of V along the curve C from P1 to P2 is  d r
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Since y  2x 2 dy  4xdx and hence we have 1 2 1 3 5 Example 5.5 Find the work done in moving a particle once around a circle C in the xy plane, if the circle has center at the origin and radius 3 and if the force field is given by F  2x y  zˆ   ˆ  3x 2 y  4zk In the plane z  0 So that d r  dxˆ  dyˆ where x  3cos , y  3sin  where 0  2 Then F  d r       2x ydx  x  ydy 2 2 4 9 2 0 1 4 1 4  18 Theorem: If F  everywhere in a region R of space, where x, y, z is single- valued and has continuous derivatives in R, then: 1)  d r is independent of the path C in R joining P1 and P2. 2) F  d r  0 around any closed curve C in R. In such a case F is called a conservative vector field and  is its scalar potential. A vector field F is conservative if and only if  F  0 or F  so that F  d r  F1dx  F2dy  F3dz  d , an exact differential.
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Example 5.6 a) Show that F  2xy  z 3 iˆ  x 2 ˆj  3xz 2kˆ is a conservative force field. b) Find the scalar potential. c) Find the work done in moving an object in this field from (1,-2,1) to (3,1,4). a) F is conservative if  F  0 i j k  F   0 2xy  z 3 x 2 2 Hence F is conservative. b) F   ˆ  ˆ  k F  d r  dx  dy  3 2 2        const. 2 3 x, y, z  x 2 y  xz 3  const. Work done  202 P2 P1 3,1,4 59
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS We first define the area of S by dividing the region R into small rectangles . Directly above each there lies a patch of surface that we may approximate with a portion of of the tangent plane since F(x,y,z)=C may not be a plane. The vector u and v lie along the edges of the patch in the tangent plane. Both  v and are normal to . Now u  v n is the area of the orthogonal projection of the parallelogram determined by u and v onto a plane whose normal is nˆ . This translates into the statement u  v n  .....( A) Now u  v is the area . Hence (A) becomes u  v n cos   where  is the angle between u  v and nˆ . cos   so that  cos  provided that cos  0 . On summing; k or k R dA We therefore define the area of S to be the value of this integral whenever the integral exists. For any particular surface F(x,y,z)=C we have  nˆ  n cos  or 1 cos   nˆ  nˆ   nˆ which results into Surface area  R  uˆ dA provided  nˆ 0 .
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS In case of a smooth surface S given by an equation Z=f(x,y) defined over a region Rxy in the xy- plane, we write F(x,y,z)=f(x,y)-z=0 and consider S as the level surface F(x,y,z)=0 of the function F. Taking the normal to be nˆ  kˆ to Rxy we have  f xiˆ  f y ˆj kˆ   nˆ  kˆ  k  1 2 2 S  Rxy  nˆ dA  Rxy 2 2 Similarly, formulations are easily given for the area of smooth surface x=f(y,z) over a region Ryz and y=f(x,z) over region Rxz. Example 5.9 Find the area of the upper cap cut from the sphere x 2  y 2  z 2  2 by the cylinder x 2  y 2  1. Z x 2  y 2  z 2  2 2  y 2  1 1 1 Y X The cap part of the surface F x, y, z  x 2  y 2  z 2  2 , projects onto the disc R : 2  y 2  1in the xy- plane. At any point (x,y,z) in the space we have F  2  y 2  z 2  2xiˆ  2 yˆj  2zkˆ  2 x 2  y 2  z 2 On the level surface F  x 2  y 2  z 2  2  2 2 63
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Taking nˆ  kˆ ;  n    k  2z Then S  R  nˆ dA  R 2 2 2z  2 R dA z Now since we are in the xy- plane, we have from x 2  y 2  z 2  2 that z  2 x 2 y 2 Hence S  2 x2 dxdy 2 x 2 y 2 We let x  r cos , y  r sin  , then, dxdy  Jdrd  , where J is the Jacobian of (x,y) with respect to r, , i.e. r sin  sin  r cos  r Hence dxdy  rdrd  Then S  2 1 2 rdrd  2 r 2 2 0     Example 5.10 Find the area of the surface cut from the bottom of the paraboloid z  x 2  y 2 by the plane z  1. The surface projects onto the disk R : x 2  y 2  1 in the xy- plane. We apply z  f x, y  x 2  y 2 to get Area  fx2  fy 2 1dxdy  Rxy 2 1  4r 2 1   R0 x2  6 4x2  4 y 2 1 S Z z  1 z  x 2  y 2
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Example 5.11 Evaluate F  nˆ dS , where F  18ziˆ 12 ˆj  3ykˆ and S is that part of the plane S 2x  3y  6z  12 which is located in the first octant. If S is part of the level surface G(x,y,z)=C then nˆ may be taken to be one of the two vectors nˆ   or nˆ  depending on which vector gives the preferred direction. Thus let G  2x  3y  6z nˆ    2iˆ  3 ˆj  6kˆ 7 dS   kˆ dA  7 6 On the surface S, the value F  nˆ is given by the formula ˆ Thus the flux of F outward through S is 36z 36 18 y 7 Rxy where 6z  12 2x 3y Now, along the x-axis; y=0, z=0 and we integrate from 0 to 6. From the given surface on the xy-plane, z=0 so that along the y-axis, we integrate from 0 to 12 2x 3 6 2 3 Example 5.12 A vector field F  yiˆ  2 ˆj  kˆ exists over a surface S defined by x 2  y 2  z 2  9 bounded by x  0, y  0, z  0 in the first octant. Evaluate F  d S over the indicated surface. Z nˆ S : x 2  y 2  z 2 9  0 ; X 3   p 3 Y x 2  y 2  9, z  0  2xiˆ  2 yˆj  2zkˆ
CHAPTER 5 nˆ  3 VECTOR INTEGRATION AND INTEGRAL THEOREMS dS  dA  6 2z dA for z 0 F  n    1   xy2 yz F  nˆ dS  S S xy2 yz 3  3z dA xy2zyz dA S The projection of S on the xy- plane is the region R bounded by the circle x 2  y 2  9, z  0 . Then xy2 yz S R z dxdy Converting to spherical polar coordinates; x  3sin  cos, y  3sin  sin , z  3cos 0   2 ; 0   2 So that x, y  , and we have 1 3   2 2 0 0 2  sin cos  6sin sin  3cos 9sin d d  2 0 sin 2  2   2 2 0 3 4  Example 5.13 Evaluate F  nˆ dS where F  ziˆ  xˆj 3y 2 z  kˆ S and S is the surface of the cylinder x 2  y 2  16 included in the first octant between z  0 and z  5
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Here projection is upon the xz- plane. We could also have projected on the yz-plane but not on the xy since the surface x 2  y 2  16 has no z component. 4 xz  xy 4 Hence S S xzxy 4 5 x0 z0 z0 5 Example 5.14 Z Evaluate F  nˆ dS where S is the surface of the S sphere x 2  y 2  z 2  a 2 above the xy- plane and F  xiˆ  yˆj 2zkˆ . X nˆ dS S Y x 2  y 2  a 2 , z  0 Projection of S on the xy- plane is the region R bounded by the circle x 2  y 2  a 2 , z  0 . nˆ    2 yˆj  2zkˆ 4    yˆj  zkˆ a dS   ˆj dA  a dA  a z dz a a 2 2 2 F  nˆ dS  S S x2  y 2 2z 2 a a z S x2  y 2 2z 2 z dA In spherical polar coordinates x  a sin  cos, y  a sin  sin , z  a cos and dA  a 2 sin  d d
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Z Since we have a cylinder, we convert coordinate system into the cylindrical one. 4 dV Thus we have x   cos, y   sin, z  z where 0  2, 0  3, 0 z 4 O  3 Y 3 Hence  2ˆ  2zˆ  yk dxdydz V X 2 3 4 0 0 z0 2 3 4 0 0 0 2 0 3 0 2 2 0  3 3 2 0 iˆ  2 ˆj  sin  kˆd  72iˆ  2 ˆj Example 5.17 Evaluate 2x  ydV where V is the closed region bounded by the cylinder and the planes x=0, y=0, y=z and z=0 We have 2 2 Example 5.18 Evaluate V x2  y 2 dxdydz, where V is the region bounded by z=x2+y2 and
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS For this case we use cylindrical coordinates. x   cos, y   sin, z  z x2  y 2  z  2 cos2    2 sin 2   z  2  z or    z Z Z  8 said region V 2 4 z 0 z0 z  2 ddzd Z  4 Hence  2 4 2 3 0 z0 3 2 0 z 5 4 2 0 256 15  Y 2 2 EXERCISE X P  2 2 Q1. If F  , where  is single-valued and has continuous partial derivatives, show that the work done in moving a particle from one point P1 x, y, z in this field to another point P2 x2 , y2 , z2  is independent of the path joining the two points. Q2. Show y 2 z3 cos x 4x3 zdx  2z3 y sin xdy  3y 2 z 2 sin x x4 dz is an exact differential of a function  and find  . Ans.   y 2 z3 sin x x4 z  const. Q3. Evaluate F  d r around the closed curve of figure Y 2 adjacent if F  x yiˆ  x  y ˆj . Ans. 2 3 O y  x 2 X
Q4. If F  y 2xiˆ  3x  2 y ˆj , compute the circulation of F about a circle C in the xy- plane with center at origin and radius 2, if C is traversed in the positive direction Ans. 8 Q5. a) Prove that F  y2 cos x  z3 iˆ  2 y sin x 4 ˆj  3xz2  2kˆ is a conservative force field. b) Find the scalar potential for F . 70
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS c) Find the work done in moving an object in this field from (0, 1, -1) to  2 Ans. b)   y 2 sin x  xz 3 4 y  2z  const. c) 15  4 . Q6. Evaluate F  nˆ dS for each of the following cases; S a) F  yiˆ  2xˆj zkˆ and S is the surface of the plane 2x+y=6 in the first octant cut off by the plane z=4. b) F  x  y2 iˆ 2xˆj  2 yzkˆ and S is the surface of the plane 2x+y+2z=6 in the first octant. Ans. a) 108 b) 81 Q7. Evaluate x 2  y 2 dxdy over the region in the xy- plane bounded by the plane R x 2  y 2  36 . Ans. 144 Q8. If F  2x2 3ziˆ 2xyˆj 4xkˆ Evaluate a) F dV b) F dV where V is the closed region bounded by the planes x=0, y=0, z=0 and 2x+2y+z=4 Ans. a) 8 3 b) 8 3   DIVERGENCE THEOREM (GAUSS’ THEOREM) For a closed surface S, enclosing a region V in a vector Z field F  FdV  F  nˆdS  F  d S V S S V S In general, this means that the volume integral on the left- hand side can be expressed as a surface integral on the right-hand surface. 71 X O Y
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Example 5.19 Evaluate F  nˆdS over the entire surface of the region bounded by the cylinder S x2+z2=9, x=0, y=0, z=0 and y=8, if F  6 zˆi  2x  yˆj xkˆ . Apply divergence theorem F  nˆdS  FdV S V 3 8  ˆj  3 8 3 8 8 On letting x  3 cos , dx  sind ,  2  0 0 2 9 9 cos2   sind 2 0 4 0   18 Example 5.20 Evaluate F  nˆdS over the entire surface of the region above the xy- plane bounded S by the cone z2=x2+y2 and the plane z=4, if F  4 xzˆi  xyz2 ˆj  3zkˆ . Z By divergence theorem; F  nˆdS  FdV S V z  4 x 2  y 2  z 2 Y 4 z z0 4z  2  3dydxdz X
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Then a unit normal is n    2 . xˆi  yˆj 2 F  nˆ dS3  2x2 y3 dS3 S3 S3 Now since the radius of the cylinder 2; in cylindrical polar coordinates; we have x  2 cos , y  2 sin , z  z so that dS3  2ddz 2 3 2 3 2 S3 Then the surface integral  0  36  48  84 . Hence verified. NB: Evaluation of the surface integral over S could also have been done by projection of S3 on the xz or yz coordinate planes. QUESTIONS: Q1. Verify the Gauss divergence theorem for the vector field F  xiˆ  2 ˆj  z 2kˆ taken over the region bounded by the planes z=0, z=4, x=0, y=0 and the surface x2+y2=4 in the first octant. Ans. 20 Q2. Evaluate F  nˆ dS where F  4 xzˆi y 2 ˆj  yzkˆ and S is the surface of the cube S bounded by x=0, x=1, y=0, y=1, z=0, z=1. Ans. 3 2 Q3. Evaluate r  nˆ dS , where S is a closed surface. S Ans. 3V Q4. Prove that dV   d S . V S Q5. Let S be a closed surface and let r denote the position vector of any point (x,y,z) measured from an origin O. prove that: S 3 0 if 0 lies outside S . if 0 lies inside S . 74 xˆ
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Q6. Evaluate F  d S over the hemisphere defined by x2+y2+z2=25 with z 0 , S where F  x  yˆi 2zˆj  ykˆ GREEN’S THEOREM: Let C be a simple closed curve in the xy- plane such that a line parallel to either axis cuts C in at most two points. Let M, N and and be continuous functions of (x,y) inside and on C. let R be the region inside C. R dxdy where C is traversed in the positive (counter clockwise) sense. PROOF Let the equations of the curves AEB and AFB (see adjoining figure) be y=Y1(x) and y=Y2(x) respectively. If f Y F A B we have E Y2 x b x, Y2 x yY1x dx O b X b b a a a b C Then
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS cross-cut AB connecting the exterior and interior boundaries. The region bounded by ABCDBAEFGHA is simply-connected, and so Green’s theorem is valid. Then ABCDBAEFGHA R dxdy . But the integral on the left, leaving out the integrand is equal to      Since  . Thus, if C1 is the curve AEFGHA, C2 is the curve BCDB and C is the boundary of R consisting of C1 and C2 traversed in the positive sense, then   so that R dxdy Example 5.25 Express Green’s theorem in the plane in vector notation. We have Mdx  Ndy       F  d r where F  Mˆ  Nˆ and r   yˆ so that d r  dxˆi  dyˆj Also if F  Mˆi  Nˆj then,  F  ˆi M ˆj N kˆ O  ˆ  So that   kˆ  The Green’s theorem in the plane can be written F  d r    kˆdR R where dR=dxdy A generalization of this to surface S in space having a curve C as boundary leads quite naturally to Stokes’ theorem which is treated later.
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Example 5.26 Evaluate the line integral xydx  2x ydy round the region bounded by the curve y=x2 and x=y2 by use of Green’s theorem. Points of intersection are O(0,0) and A(1,1). R Y 2 R 1 x 2 2 O X 1 31 60 In the special case when M=y and N=-x  1 and  Green’s theorem then states  Mdx  Ndy R i.e. dxdy R  ydx xdy  xdy ydx Therefore the area of the closed region A  dxdy  R 1 2 C QUESTIONS Q1. Evaluate 3x  4 ydx  2x 3 ydy where C, a circle of radius 2 with center at the origin of the xy- plane, is traversed in the positive sense. Ans. 8 Q2. Evaluate x 2 2xydx  x 2 y  3dy around the boundary of the region defined by y2=8x and x=2. a) directly b) by using Green’s theorem
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Let x=cos t, y = sin t, z  0 , 0 t 2 be parametric equations of C. Then   C o 2 2 (2 cos t sin sin t)dt   Also,  F  i j k  kˆ 2x y yz y2 z Then  s s dxdy R since nˆ  kˆds  dxdy and R is the projection of S on the xy- plane. This last integral equals 1 1 0 0 1 0 And the Stokes’ theorem is verified. Example 5.27 A surface S consists of that part of the cylinder x 2  y 2  9 between z  0 and z=4 for y≥0 and the two semicircles of radius 3 in the planes z  0 and z  4 . If  S   The surface S consists of three sections a. the curved surface of the cylinder b. the top and bottom semicircles. We therefore evaluate S   81
CHAPTER 6 CURVILINEAR COORDINATES The vectors eˆ1 , eˆ2 and eˆ3 are mutually orthogonal and form a right-handed system. Note that unlike the unit vectors iˆ, ˆj, kˆ in the Cartesian system, the unit vectors eˆ1 , eˆ2 , eˆ3 or base vectors as they are called are not fixed in directions, but change as the position of P changes. If F (r ) is a vector associated with P, then F (r )  F eˆ1  F eˆ2  Fzeˆ3  vectors eˆ1 , eˆ2 and eˆ3 . 2. Spherical Polar Coordinates Here P is a function of three variables r,  ,  . Z a) If  and  remain constant and r increases, P moves is thus a vector normal to the surface of the sphere at P and the unit X O   r eˆ1 p(r,,) eˆ2 eˆ3 Y vector eˆ , in that direction is therefore eˆ1  b) If r and  remain constant and  increases, P will move along the ‘meridian”
throughout P, i.e. is a tangent vector to this circle at P and the unit vector eˆ2  c) If r and  remain constant and  increases, P will more along the circle through P perpendicular to the z-axis. is therefore a tangent vector at P and the unit vector ˆ 3 in this direction is given by eˆ3  86
CHAPTER 6 CURVILINEAR COORDINATES A vector F (r ) associated with P can therefore be expressed as F (r )  Fr eˆ1  F eˆ2  F eˆ3 where Fr , F , F are the components of F in the directions of the base vectors eˆ1 , eˆ2 , eˆ3 . Both cylindrical and spherical polar coordinates are orthogonal. Scale Factors: 1. For cylindrical polar coordinates, the unit base vectors are eˆ1  eˆ2  ˆe3     1 h 1 h 1 h ; h  ; h  ; hz  2. For spherical polar coordinates, the unit base vectors are hr  ; h  ; h  In each case, h is called the scale factor. If U1, U2, U3 represent the coordinates in any coordinate system, we may obtain the scale factors as 2 2 dU1 dU 2 dU 3 1. Rectangular Coordinate System In rectangular coordinate system; U1 ,U 2 ,U 3   x, y, z  Then hU1  hx  1, hU2  hy  1, hU3  hz  1 2. Cylindrical Coordinate System U1 ,U 2 ,U 3   ,, z  where x   cos, y   sin, z  z  cos x Z O  r P(,, z) Y
CHAPTER 6 CURVILINEAR COORDINATES Thus eˆ  eˆ1  cos iˆ  sin  ˆj Similarly eˆ  eˆ2  sin  iˆ  cos ˆj ; eˆz  kˆ 3. Spherical Coordinate System U1,U2 ,U3   r, ,  Z P(r, , ) x  r sin  cos, y  r sin  sin , z  r cos  and hr  1, h  r, h  r sin  X x O  y r r Z Y General Curvilinear Coordinate System (U1, U2, U3) Any system of coordinates can be treated in like manner to obtain expressions for the appropriate units vectors eˆ1 , eˆ2 , eˆ3 . eˆ1  ; eˆ2  2 2 ; eˆ3  3 3 ; These unit vectors are not always at right angles to each other. If they are mutually perpendicular, the coordinate system is orthogonal. The unit vectors are orthogonal if eˆ1  eˆ2  eˆ1  eˆ3  eˆ2.  eˆ3  0.
Example 6.1 For the vectors iˆ 2 ˆj  4kˆ ; 2iˆ  3 ˆj  kˆ, 2iˆ  ˆj  kˆ, eˆ1  1 21 (iˆ 2 ˆj  4kˆ); eˆ2  1 14 (2iˆ  3  eˆ3  1 6  ˆj  kˆ) Then eˆ1  eˆ2  1 7 6 (2 6  4)  0 eˆ1.  eˆ3  eˆ2  eˆ3  1 3 14 1 2 21 2  4)  0  3  1)  0 Hence the three vectors are orthogonal. 88
CHAPTER 6 CURVILINEAR COORDINATES Example 6.2 Prove that a spherical coordinate system is orthogonal. The position vector of any point in spherical coordinates is  er  sin cos ˆ  sin sin ˆ  cos k eˆ  cos cos iˆ  cos sin ˆj sin kˆ eˆ  sin iˆ  cos ˆj eˆr  eˆ  sin cos cos2   sin cos sin2  cos sin  0 eˆr  eˆ  sin cos sin  sin cos sin  0 eˆr  eˆ  cos cos sin  cos cos sin  0 Hence orthogonal. Example 6.3   eˆ  eˆ  eˆz   cos iˆ  sin  ˆj  sin  iˆ   cos ˆj  kˆ We then solve simultaneously for iˆ, ˆj, kˆ to obtain iˆ  cos eˆ sin  eˆ , ˆj  sin  eˆ  cos eˆ , kˆ  eˆz Now   z(cos eˆ sin  eˆ ) 2 cos (sin  eˆ  cos eˆ )  2 sin  eˆz  (z cos 2 cos sin  ) eˆ (z sin   2 cos2  ) eˆ   sin  eˆz and F  z cos 2 cos sin , F  sin  2 cos2 , Fz   sin 89
CHAPTER 6 CURVILINEAR COORDINATES Arc Lengths ds and Volume Element dv Q u3 h3du3eˆ3 d r u2 h1du1e1 O h2du2e2 u1 Fig( A) a) Element of arc dS Element of arc dS from o to Q given by   2 2 2 2 2 and 2 2 2 2 2 Example 6.4 Find the square of the element of arc length in cylindrical coordinates and determine the corresponding scale factors. We have r  xiˆ  yˆj  zkˆ   cos iˆ   sin  ˆj  zkˆ     sin 2 d 2  2 sin  cos dd   2 cos2 d 2  dz 2 2 2 2 and hU1  1, hU2  , hU3  1 are the scale factors.
CHAPTER 6 CURVILINEAR COORDINATES b) Element of volume dV dV  (hU1 dU1eˆ1 )  (hU 2 dU 2eˆ2  hU3 dU 3eˆ3 )  hU1 hU2 hU3 dU1dU 2 dU 3 ) dU1dU 2dU 3  y z) U 2 U 3 ) dU1dU 2dU 3  JdU1dU 2 dU 3 where J is called the Jacobian of the transformation and is given by J  y z) U 2 U 3 ) xU1  xU2 xU3 yU1 yU2 yU3 zU1 zU2 zU3 where xU1  etc. c) Element of Area dA The area element is given by dA1  hU 2dU 2eˆ2  hU3 dU 3eˆ3 from  fig. A   hU2 hU3 eˆ2  eˆ3 dU 2 dU 3  hU2 hU3 dU 2dU 3 since eˆ2  eˆ3  eˆ1  1 similarly dA2  hU1 hU3 dU1dU 3 Example 6.5 and dA3  hU1 hU2 dU1dU 2 Evaluate V 2  y 2  z 2 )dxdydz where V is a sphere having centre at the origin and radius equal to a. We may consider the portion in the first octant and in spherical coordinates we have dxdydz  r 2 sindrdd
CHAPTER 6 Thus we have CURVILINEAR COORDINATES v   a . 5 2 2 2 5   0 0  0  0  2  0 a r0 r 2 .r 2 X Z 2 2 2 2 2 2 2  dV  dxdydz Y Grad, div and Curl in Orthogonal Curvilinear Coordinates (a) Grad V ) Let a scalar field  exist in space and let d be the change in  from a point p to Q say. If the position vector of p is r then that of Q is r  dr u3 Q Let grad   2 dU 2  3 dU 3 O iˆ u2 P u1  )U 1 eˆ1  )U2 eˆ2  )U3 eˆ3 Where )U1, U2, U3 are the components of grad  in the U1, U2, U3 directions. Also dU1  2 dU 2  3 dU 3  hU1 dU1eˆ1  hU2 dU 2eˆ2  hU3 dU 3eˆ3 Now since dU  dU 2  dU3 2    )U1 e1  )U2 e2  )U3 e3    U1 hU1 dU1  U2 hU2 dU 2  U3 hU3 dU3
CHAPTER 6 So that CURVILINEAR COORDINATES grad    1 hU1 eˆ1  1 hU2 2 eˆ2  1 hU3 3 eˆ3 and  e1 hU1 e hU2 2 e hU3 3 (b) DivF    e hU1 hU2 2 eˆ hU3 3      1 hU1  1 2 2  1 hU3 3  1 hU1 hU2 hU3   2   3 hU1 (c)    FU1 e1   FU2 e2   FU3 e3   e2 U3 U1 3 U1 U1 U1 U1 3 U1 U2 3    e3 hU1 hU2 e1 hU2 hU3 2   e1 hU2 hU3 3 e2 hU3 hU1
     eˆ1 hU2 hU3  3 eˆ2 hU3 hU1  FU3 hU3  eˆ3 hU1 hU2  2  which is written in compact form as  F  1 hU1 hU2 hU3 hU1 eˆ1 FU1 hU1 hU2 eˆ2 2 FU2 hU2 hU3 eˆ3 3 FU3 hU3 94
CHAPTER 6 CURVILINEAR COORDINATES Example 6.6 Express 2 in orthogonal curvilinear coordinates.  eˆ1 hU1  2 hU2 2  3 hU3 3 e hU1 e hU2 eˆ hU3 e hU1  2 hU2 eˆ hU3  1 hU1 hU2 hU3 2 3 hU1 Example 6.7 2 3 at the point (5, 4, 3).  1, 2  2U 2 , 3  3U 32 hU1  2, hU2  1, hU3  1 1 2 3  Example 6.8 Express the heat equation ut  2u in spherical coordinates. We have 
CHAPTER 6 CURVILINEAR COORDINATES If u is independent of  ; we have r etc QUESTIONS Q1. Determine the unit vectors in the directions of the following three vectors and test whether they form an orthogonal set 3iˆ 2 ˆj  kˆ, iˆ  2 ˆj  kˆ, 2iˆ ˆj  4kˆ. Ans. Yes, an orthogonal set  Ans. hu = 1, hv = 2V, h = 2u. Q3. A vector field F is given in cylindrical coordinates by  Determine (a)  F (b)  F Ans. (a) (2 cos  2 cos 2  1) (b) (2sin 2  sin)kˆ  Fr , F and F . Hint: You determine eˆr , eˆ , eˆ in terms of iˆ, ˆj and kˆ. Use Cramer’s rule to  Ans.   where Fr  2r sin 2  sin  cos r sin cos sin   3r sin cos cos F  2r sin cos sin  r cos2  sin  3r sin 2  cos F  sin sin 2  r cos cos Q5. Find the element of arc length on a sphere of radius a. Ans. dS 2  a 2  