VECTOR CALCULUS. B.Sc. Mathematics (V SEMESTER) CORE COURSE (2011 ADMISSION ONWARDS) UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION

Transcription

1 VECTOR CALCULUS B.Sc. Mathematics (V SEMESTER) CORE COURSE (2011 ADMISSION ONWARDS) UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION Calicut University, P.O. Malappuram, Kerala, India

2 UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION B.SC. MATHEMATICS (2011 ADMISSION ONWARDS) V SEMESTER CORE COURSE: VECTOR CALCULUS Prepared by: Sri. Nandakumar. M. Assistant Professor, NAM College, Kallikkandi, Kannur. Scrutinized by: Dr. Anil Kumar. V Head of the Dept. Dept. of Maths, University of Calicut. Layout & Settings: Computer Section, SDE Reserved Vector Calculus 2

3 CONTENTS PAGES MODULE - I 5 70 MODULE - II MODULE - III MODULE - IV SYLLABUS 223 Vector Calculus 3

4 Vector Calculus 4

5 MODULE - 1 ANALYTIC GEOMETRY IN SPACE VECTORS 1. A vector is a quantity that is determined by both its magnitude and its direction; thus it is an arrow or a directed line segment. For example force is a vector. A velocity is a vector giving the speed and direction of motion. We denote vectors by lowercase boldface letter a,b,v, etc. 2. A scalar is a quantity that is determined by its magnitude, its number of units measured on a scale. For Problem, length temperature, and voltage are scalars. 3. A vector has a tail, called its initial point, and a tip, called its terminal point. 4. The length of a vector a is the distance between its initial point and terminal point. 5. The length (or magnitude) of a vector a is also called the norm (or Euclidean norm) of a and is denoted by. 6. A vector of length 1 is called a unit vector. 7. Two vector and are equal, written, a=b, if they have the same length and the same direction. Hence a vector can be arbitrarily translated, that is, its initial point can be chosen arbitrarily. COMPONENTS OF A VECTOR We consider a Cartesian coordinate system in space, that is, a usual rectangular coordinate system with the same scale of measurement on the three mutually perpendicular coordinate axes. Then if a given vector has initial point and terminal point, then the three numbers..(1) Are called the components of the vector a with respect to that coordinated system, and we write simply In terms of components, length of a is given by.(2) Problem Find the components and length of the vector a with initial point and terminal point Vector Calculus 5

6 . The components of a are Hence. Using (2), the length of the vector is POSITION VECTOR A Cartesian coordinate system being given, the position vector r of a point is the vector with the origin as the initial point and as the terminal point. From (1), the components are given by So that Theorem I (Vectors as ordered triples of real numbers) A fixed Cartesian coordinate system being given each vector is uniquely determined by its ordered triple of corresponding components. Conversely, to each ordered triple of real numbers there corresponds precisely one vector, with corresponding to the zero vector 0, which has length 0 and no direction. VECTOR ADDITION Definition (Addition of Vectors) The sum of two vectors and is obtained by adding the corresponding components....(3) Basic Properties of Vector Addition a) (commutativity) b) (associativity) c) d) where denotes the vector having the length and the direction opposite to that of SCALAR MULTIPLICATION Definition (Scalar Multiplication by a Number) The product ca of any vector and any scalar c (real number c) is the vector obtained by multiplying each component of a by c. That is, Vector Calculus 6

7 ...(4) Geometrically, if, then with has the direction of and with has the direction opposite to. In any case, the length of is and if or (or both) Basic Properties of Scalar Multiplication (a) (b) (c) (written cka) (5) (d) Remarks (4) and (5) imply for any vector a (a) (b) Instead of we simply write Problem Given the vectors and Find and : ; ; Unit Vectors i,j,k ; A vector can also be represented as a=a 1 i+a 2 j+a 3 k (6) In this representation i,j,k are the unit vectors in the positive directions of the axes of a Cartesian coordinate system. Hence..(7) Problem The vectors and can also be written as and Inner Product Definition (Inner Product (Dot Product) of vectors) The inner product or dot product (read a dot b ) of two vectors a and b is the product of their lengths times the cosine of their angle. Vector Calculus 7

8 if if (1) The angle between a and b in measured when the vectors have their initial points coinciding. In components,...(2) Definition A vector a is called orthogonal to a vector b if Then b is also orthogonal to a and we call these vectors orthogonal vectors. 1. The zero vector is orthogonal to every vector. 2. For nonzero vectors if and only if thus Theorem 1 (Orthogonality) The inner product of two nonzero vector is zero if and only if these vectors are perpendicular. Length and Angle in terms of inner Product From (1), with we get. Hence (3) From (3) and (1) we obtain for the angle between two nonzero vectors (4) Problem Find the inner product and the lengths of and as well as the angle between these vectors Properties of Inner Products radians. and (4) gives the angle For any vectors and scalars (a) (b) (Linearity) (Symmetry) (c) if and only if (Positive definiteness) Hence dot multiplication is commutative and is distributive with respect to vector addition, in fact from the above with and we have (Distributivity) Furthermore, from (1) and we see that (Schwarz inequality) Vector Calculus 8

9 Result: Prove the following triangle inequality: Proof using (3) since using (3) and Schwarz inequality Taking square roots on both sides, we obtain Result: Prove the Parallelogram equality (parallelogram identity) Proof, using (3) Derivation of (2) from (1) We can write the given vectors a and b in components as and Since i,j and k are unit vectors, we have from (3) Since they are orthogonal (because the coordinate axes perpendicular) Orthogonality Theorem gives Hence if we substitute those representations of a and b into Symmetry, we first have a sum of nine inner products. and use Distributivity and Since six of these products are zero, we obtain (2) Vector Calculus 9

10 APPLICATIONS OF INNER PRODUCTS Work done by a force as inner product Consider a body which a constant force p acts. Let the body be given a displacement d. Then the work done by p in the displacement d is defined as that is, magnitude of the force times length of the displacement times the cosine of the angle between p and d. If, then. If p and d are orthogonal, then the work is zero. If, then which means that in the displacement one has to do work against the force. PROJECTION OF A VECTOR IN THE DIRECTION OF ANOTHER NON ZERO VECTOR Components or projection of a vector a in the direction of a vector is defined by..(5) where is the angle between a and b. Thus is the length of the orthogonal projection o a on a straight line l parallel to b, taken with the plus sign if b has the direction of b and with the minus sign if has the direction opposite to b Multiplying (5) by, we have ie (b (6) if b is a unit vector,as it is often used for fixing a direction then (6) simply gives p=a.b ( b =1) Definition An orthonormal basis I,j k associated with a Cartesian coordinate system. Then {i, j, k} form an orthonormal basis, called standard basis An orthonormal basis has the advantage that the determination of the coefficients in representation v=l 1 a+l 2 b+l 3 c (v a given vector) is very simple. This is illustrated in the following Problem. Problem When (v a given vector), show that Vector Calculus 10

11 , since and Similarly, it can be shown that and Normal Vector to a given line Two non-zero vectors and in the plane are perpendicular (or orthogonal) if i,e, if Consider a line The line though the origin and parallel to is when can also be written where and. Now implies that a is perpendicular to position vector of each points on the line. Hence a is perpendicular to the line and also to because and are parallel is called a normal vector to (and to ). is another normal vector to (and to ). Two straight lines and are perpendicular if their normal vectors are perpendicular. Since and are normal vectors of and respectively and are perpendicular if Problem Find a normal vector to the line Let given line be. Then the through the origin and parallel to is which can also be written as where and Hence by the discussion above a normal vector to the line is Problem Find the straight line through the point in the xy-plane and perpendicular to the straight line Also find the point of intersection of the lines and Suppose the required straight line be. Then is a normal vector to and is perpendicular to the normal vector of the line. That is i.e,.(8) Now, if we take and we have is a normal vector to and hence Since it passes through by substituting in the equation of we have or Hence the equation of the required line is Vector Calculus 11

12 or Now the point of intersection of of and can be obtained by solving the following systems of equations. and Solving, we obtain and Hence the point of intersection of the lines is NORMAL VECTOR TO A PLANE Let be a plane in space. It can also be writer as.(9) where and The unit vector in the direction of a is Dividing (9) by, we get..(10) where Representation (10) is called Hesse s normal form of a plane. In (10), p is the projection of r in the direction of n. Note that the projection has the same constant value for the position vector r of any point in the plane. (10) holds if and only if n is perpendicular to the plane. n is called unit normal vector to the plane (the other being n) Remark From the above discussion it follows that is the distance of the plne from the origin. Problem 6 Find a unit vector perpendicular to the plane. Also find the distance of the plane from the origin. A normal vector to the given line is Hence and the unit normal vector is given by Vector Calculus 12

13 and since we have and the plane has distance from the origin Assignments For the vectors, and find 1., 2,, Find the work done by the force p acting on a body if the body displaced from a point A to a point B along the straight segment AB. Sketch p and AB Can work be zero or negative? In what cases? Let. Find the angle between Find the angle between the straight lines and 12 Find the angle between the planes and 13 Find the angles of the triangle with vertices 14. Find the angles of the parallelogram with vertices Find the component of a in the direction of b: Vector Product Definition The vector product (cross product) of two vector and is a vector as follows: If a and b have the same direction, If a and b have the opposite direction, In any, other case, has the length..(1) Vector Calculus 13

14 where is the angle between a and b. This is the area of the parallelogram with a and b as adjacent sides The direction of is perpendicular to both a and b and such that a,b,v in this order, form a right handed triple. CROSED PRODUCT IN COMPONENTS In components, the cross product is given by.(2) Notice that in (2) Hence is the expansion of the determinant by the first row. Definition A Cartesian coordinate system is called right-handed if the corresponding unit vectors I,j,k in the positive directions of the axes form a right handed triple. The system is called left-handed if the sense of k is reversed. Problem Find the vector product of and in right-handed coordinates. or Problem With respect to a right-handed Cartesian coordinate system, let and Then Vectors Product and Standard basis vectors Since i,j,k are orthogonal (mutually perpendicular) unit vectors, the definition of vector product gives some useful formulas for simplifying vector products; in right-handed Vector Calculus 14

15 coordinates these are (3) For left-handed coordinates, replace k by k, Thus GENERAL PROPERTIES OF VECTOR PRODUCT Cross multiplication has the following properties: For every scalar l It is distributive with respect to vector addition, that is, (a) (b) Vector product is not commutative but anticommutative, that is, It is not associative, that is in general (4).(5) (6) Proof. so that the parentheses cannot be omitted. (4) follows directly from the definition. (8) The sum of the two determinants in (8) is the first component of side of (5a). For the other components in (5a) and in (5b), equality follows by the same idea. the right Now to get (6), note that, using (2**), as the interchange of rows 2 And 3 multiples the determinants by -1 again using (2**) Vector Calculus 15

16 We can confirm this geometrically if we set and then by (11) and for b, a,w to from a right handed triple, we must have For the proof of (7), note that, where as SCALAR TRIPLE PRODUCT Definition (Scalar Triple Product) The scalar product of three vectors a,b and c, denoted by, is defined as : The scalar triple product is also denoted by and is also called the box product of the three vectors Remarks The scalar triple product is a scalar quantity. Since the scalar triple product involves both the signs of cross and dot it is some times called the mixed product. Geometrical meaning of Scalar triple product The scalar triple product has a geometrical interpretation. Consider the parallelepiped with a, b and c as co terminus edges. Its height is the length of the component of a on. To be precise, we should say that this height is the magnitude of, where is the angle between a and. Now, Where the sign or depends on which is positive or negative according is acute or obtuse that is according as a, b, c is right handed or left handed. Hence the volume of the parallelepiped with co terminal edges a, b and c is, up to sign, the scalar triple product. Expression for the scalar triple product as a determinant Let and Then the scalar triple product can be easily evaluated using the following formula: Vector Calculus 16

17 Problem Compute if and Problem Find the volume of the parallelepiped whose co-terminal edges are arrows representing the vectors and Problem and Find the volume of the tetrahedron with co terminal edges representing the vectors The volume of the tetrahedron Hence, the volume of the tetrahedron is Theorem (Linear independence of three vectors) Three vectors form a linearly independent set if and only if their scalar triple product is not zero. The following is restatement of the above Theorem. Theorem Scalar triple product of three coplanar vectors is zero. Proof. Let be three coplanar vectors. Now represent a vector which is perpendicular to the plane containing b and c in which also lies the vector a and hence is perpendicular to a. Therefore Thus when three vectors are coplanar. Vector Calculus 17

18 Conversely, suppose that. That is, which shows that is perpendicular to a. But is vector perpendicular to the plane containing b and c hence a should also lie in the plane of b and c. That is a, b, c are coplanar vectors. Problem 11 Show that the vectors are linearly independent. By the Theorem, it is enough to show that the scalar triple product of the given vectors is not zero. It can be seen that the scalar triple product is not zero. Hence the given vectors are linearly independent. Problem Find the constant so that the vectors are coplanar. Three vectors a, b, c are coplanar if i.e, if or if or if Problem Prove that the points and are coplanar. Let the given points be respectively and if these four points are coplanar then the vectors are coplanar, so that their scalar triple product is zero. i.e Now sition vector of position vector of Similarly Now Hence the given points are coplanar. Equation of a Plane with three points Let and c=x 3 i+y 3 j+z 3 k be the position vectors of three points and Let us assume that the three points and do not lie in the same straight line. Hence they determine a plane. Let be the position vector of any point Vector Calculus 18

19 in the plane. Consider the vectors which all lie in the plane. That is and are coplanar vectors. Now we apply the condition for coplanar vectors. or or Problem Find the equation for the plane determine by the points and The equation of a plane with three points and is given by Hence here, the equation of the plane with the points is given by and or i.e, the equation of the plane is or Assignments In Assignments 1-9 with respect to a right-handed Cartesian coordinate system, let Find the following expressions Vector Calculus 19

20 10 What properties of cross multiplication do Assignments 1,4 and 5 illustrate? 11 A wheel is rotating about the x-axis with angular speed 3 The rotation appears clockwise if one looks from the origin in the positive x-direction. Find the velocity and the speed at the point. 12 What are the velocity and speed in Exercise 11 at the point if the wheel rotates about the y-axis and? A force p acts on a line through a point A. Find the moment vector m of p about a point are Find the area of the parallelogram if the vertices are 16 Find the area of the triangle in space if the vertices are 17 Find the plane through 18 Find the volume of the parallelepiped if the edge vectors are 19 Find the volume of the tetrahedron with the vertices 20 Are the vectors linearly independent? LINES AND PLANES IN SPACE In this chapter we show how to use scalar and vector products to write equations for lines, line segments, and planes in space. Lines and Line Segments in Space Suppose is line in space passing through a point and is parallel to a vector Then is the set of all points for which is parallel to v. That is, lies on if and only if is a scalar multiple of. Vector equation for the line through and parallel to v is given by (1) Expanding Eq. (1), we obtain Equating the corresponding components of the two sides gives three scalar equations involving the parameter t: Vector Calculus 20

21 When rearranged, these equations give us the standard parameterization of the line for the interval as follows : (2) Standard parameterization of the line through is given by above. and parallel to Problem Find parametric equations for the line through and parallel to. With equal to and equal to Eq (2) become Problem Find parametric equations for the line through and. The vector is parallel to the line, and Eg.(2) with base point give..(3) If we choose as the base point we obtain..(4) The equations in (4) serve as well as the equations in (3); they simply place you at a different point for a given value of. Line Segment Joining Two Points To parameterize a line segment joining two points, we first parameterize the line through the points. We then find the values for the end points and restrict to lie in the closed interval bounded by these values. The line equations together with this added restriction parameterize the segment. Problem Parameterize the line segment joining the points and. We begin with equations for the line through and, which obtained in Problem 2:..(5) Vector Calculus 21

22 We observe that the point Passes through at and at We add the restriction to Eq (3) to parameterize the line segment:..(6) The Distance from a Point to a Line in Space To find the distance from a point to to a line that passes through a point parallel to a vector v, we find the length of the component of normal to the line. In the notation of the figure, the length is which is Distance from a Point to a Line Through parallel to is given by...(7) Problem Find the distance from the point to the line...(8) Putting we see from the equations for that passes through and is parallel to (v is obtained by comparing (8) with (2) to get With and Eq. (7) gives Equations for planes in Space Suppose plane passes through a point and is normal (perpendicular) to the nonzero vector Then is the set of all points for which is orthogonal to That is, lies on if and only if This equation is equivalent to or Vector Calculus 22

23 i.e. Plane Through normal to is given by the following equivalent equatations: Vector equation :.(9) Component equatation :.(10) Problem Find an equatation for the plane through perpendicular to. Using Eq. (10), Problem Find the plane through We find a vector normal to the plane and use it with one of the points (it does not matter which) to write an equatation for the plane. The cross Product is normal to the plane. Note that Similiarly, Hence We substitute the components of this normal vector and the co ordinates of the point (0,0,1) into Eq. (10) to get i.e Problem Find the point where the line Intersects the plane The point Vector Calculus 23

24 (12) Lies in the plane if its coordinates satisfy the equatation of the plane; that is, if Putting the point of intersection is The Distance from a Point to a Plane ProblemFind the distance from to the plane We follow the above algorithm. Using (11) vector normal; to the given plane is given by The poins on the plane easiest to find from the plane s equation are the intercepts.if we take to be the y-intercept, then putting and in the equation of the plane or Hence is, and then The distance from to the plane is Angles Between Planes; Lines of Intersection The angle between two intersecting planes is defined to be the?(acute) angle determined by their normalvectors Vector Calculus 24

25 Problem Find the angle between the planes and Using (11), it can be seen that the vectors are normals to the given planes and respectively. The angle between them (using the definition of dot product) is Problem Find the vector parallel to the line of intersection of the planes. and The line of intersection of two planes is perpendicular to the plane s normal vector s and, and therefore parallel to.in particular is a vector parallel to the plane s line of intersection. In our case,, We note that any nonzero scalar multiple of is also a vector parallel to the line of intersection of the planes and. Problem Find parametric equations for the line in which the planes intersect. and v =14i+2j+15k as a vector parallel to the line. To find a point on the line, we can take any point common to the two planes. Substituting z=0 in the plane equations we obtain and solving for the x and y simultaneously gives x=3, y= -1. Hence one of the point common to the plane is (3,-1,10) The line is [Using eq.(2)] Assignments Find the parametric equations for the lines in Exercise 1-6 Vector Calculus 25

26 1. The line through the point parallel to the vector i+j+k 2. The line through and 3. The line through the origin parallel to the vector 2j+k 4. The line through (1,1,1) parallel to the z-axis 5. The line through perpendicular to the plane 6. The x-axis Find the parametrizations for the line segments joining the points in Assignments Draw coordinate axes and sketch each segments indicating the direction of increasing t for your parametrization. 7., 8., 9,. 10, Find equations for the planes in Assignments The plane through normal to 12 The plane through and 13 The plane through perpendicular to the line 14 Find the point of intersection of the lines and 15 Find the plane determined by the intersection of the lines: 16 Find a plane through and perpendicular to the line of intersection of the planes In Assignments 17-19, find the distance from the point to the line In Exercise 20-22, find the distance from the point to the plane Vector Calculus 26

27 22 23 Find the distance from the plane to the plane, 24 Find the angles between the following planes: Use a calculator to find the acute angles between planes in Assignments to the nearest hundredth of a radial In Exercise 27-28, find the point in whixh the line meets the given planes Find parametrizations for the lines in which the planes in Assignments 29 intersect Given two lines in space, either they are parallel, or they intersect or they are skew (imagine, for Problem the flight paths of two planes in the sky). Exercise 31 give three lines. Determine whether the lines, taken two at a time, are parallel, intersect or are skew. If they interesect, find the point of intersection. 31 CYLINDERS,SPHERE,CONE AND QUADRIC SURFACES Definition A cylinder is a surface generated by a line which is always parallel to a fixed line passes through (intersects) a given curve. The fixed line is called the axis of the cylinder and the given curve is called a guiding curve or generating curve Remark If the guiding curve is a circle, the cylinder is called a right circular cylinder. Since the generator is a straight line, it extends on either side infinitely. As such, a cylinder is an infinite surface. Vector Calculus 27

28 The degree of the equation of a cylinder depends on the degree of the equation of the guiding curve. A cylinder, whose equation is of second degree, is called a quadric cylinder. When graphing a cylinder or other surface by hand or analyzing one generated by a computer, it helps to look at the curves formed by intersecting the surface with planes parallel to the coordinate planes. These curves are called cross sections or traces. We now consider a cylinder generated by a parabola. Problem Find an equation for the cylinder made by the lines parallel to the z-axis that pass through the parabola Suppose that the point lies on the parabola in the plane. Then, for any value of z, the point will lie on the cylinder because it lies on the line through parallel to the z-axis. Conversely any point whose y- coordinate is the squre of it x-coordinate lie on the cylinder because it lies on the line through parallel to the z-axis Remark Regardless of the value of z, therefore, the points on the surface are the points whose coordinates satisfy the equation. This makes an equation for the cylinder. Because of this, we call the cylinder the cylinder. As Problem 1 or the Remark follows it suggests, any curve in the - plane defines a cylinder parallel to the z-axis whose equation is also Problem The equation defines the circular cylinder made by the lines parallel to the z-axis that pass through the circle in the xy-plane. Vector Calculus 28

29 Problem The equation defines the elliptical cylinder made by the lines parallel to the z-axis that passes through the ellipse in the In the similar way, we have the following: Any curve in the defines a cylinder parallel to the y-axis whose space equation is also. Any curve defines a cylinder parallel to the x-axis whose space equation is also. We summarize the above as follows : Problem The equation defines surface made by the lines parallel to the x- axis that passes through the hyperbola in the plane Quadric Surfaces A quadric surface is the graph in space of a second-degree equation in the and z. The most general form is Where A,B,C and so on are constants, but the equation can be simplified y translation and rotation, as in the two-dimensional case. We will study only the simpler equations. Although the definition did not require it, the cylinders considered so far in this chapter were also Problem of quadric surfaces. We now examine ellipsoids (these include spheres as a special case), paraboloids, cone, and hyperboloids. Problem The ellipsoid (1) cuts the coordinate axes at and. It lies within the rectangular box defined by the inequalities The surface is symmetric with respect to each of the coordinate planes because the variables in the defining equation are squared. The curves in which the three coordinate planes cut the surface are ellipses. They are Vector Calculus 29

30 The section cut from the surface by the plane is the ellipse..(2) Special Cases: If any two of the semi axes a, b and c are equal, the surface is an ellipsoid of revolution. If all three are equal, the surface is sphere. Problem The elliptic paraboloid..(3) is symmetric with respect to the planes and as the variables x and y in the defining equation are squared. The only intercept on the axes is the origin (0,0,0). Except for this point, the surface lies above or entirely below the xy-plane, depending on the sign of c. The sections cut by the coordinate planes are..(4) Each plane above the xy-plane cuts the surface in the ellipse Problem The circular paraboloid or paraboloid of revolution..(5) is obtained by taking by b=a in Eg. (3) for the elliptic paraboloid. The cross sections of the surface by planes perpendicular to the z-axis are circles centered on the z-axis. The cross sections by planes containing the z-axis are congruent parabolas with a common focus at the point (0,0,. Application : Shapes cut from circular paraboloids are used for antennas in radio telescopes, satellite trackers, and microwave radio links. Definition A cone is a surface generated by lines all of which pass through a fixed point (called vertex) and (i) all the lines intersect a given curve (called guiding curve) or (ii) all the lines touch a given surface Vector Calculus 30

31 or (iii) all the lines are equally inclined to a fixed line through the fixed point. The moving lines which generate a cone are known as its generators. When the moving lines satisfy condition (ii) in the definition of a cone, we term the cone as enveloping cone. Problem The elliptic cone..(6) is symmetric with respect to the three coordinate planes Fig (7). The sections cut by coordinate planes are (7) (8) The secions cut by planes above and below the xy-plane are ellipse whose enters lie on the z-axis and whose vertices lie on the lines in Eq.(7) and (8). If a=b, the cone is a right circular cone. Problem 9 The hyperboloid of one sheet.(9) is symmetric with respect to each of the three coordinate planes. The sections cut out by the coordinate planes are..(10) The plane cuts the surface in an ellipse with center on the z-axis and vertices on one of the hyperbolas in (10) If a=b, the hyperboloid is a surface of revolution Remark : The surface in Problem 9 is connected, meaning that it is possible to travel from one point on it to any other without leaving the surface. For this reason it is said to have one sheet, in contrast to the hyperboloid in the next Problem, which as two sheets. Problem The hyperboloid of two sheets.(11) Vector Calculus 31

32 is symmetric with respect to the three coordinate planes. The plane z=0 does not intersect the surface; in fact, for a horizontal plane to intersect the surface, we must have The hyperbolic sections have their vertices and foci on the. The surface is separated into two portions, one above the plane and the other below the plane. This accounts for the name, hyperboloid of two sheets. Eq.(9) and (11) have different numbers of negative terms. The number in each case is the same as the number of sheets of the hyperboloid. If we replace the I on the right side of either Eq.(9) or Eq.(11) by 0, we obtain the equation for an elliptic cone (Eq.6). The hyperboloids are asymptotic to this cone in the same way that the hyperbolas are asymptotic to the lines In the xy-plane. Problem The hyperbolic paraboloid..(12) Has symmetry with respect to the planes and. The sections in these planes are. (13).(14) In the plane x=0, the parabola opens upward from the origin. The parabola in the plane y=0 opens downward. If we cut the surfaces by a plane, the section is a hyperbola. (15) With its focal axis parallel to the y-axis and its vertices on the parabola in (13). Vector Calculus 32

33 If is negative, the focal axis is parallel to the x-axis and the vertices lie on the parabola in (14). Near the origin, the surface is shaped like a saddle. To a person travelling along the surface in the yz-plane, the origin looks like a minimum. To a person travelling along the surface in the xz-plane, the origin looks like a maximum. Such a point is called a minimax or saddle point of surface. Assignments Sketch the surfaces in Assignments CYLINDRICAL AND SPHERICAL COORDINATES Cylindrical and Spherical Coordinates This section introduces two new coordinate systems for space: the cylindrical coordinate system and the spherical coordinate system. Cylindrical coordinates Vector Calculus 33

34 simplify the equations of cylinders. Spherical coordinates simplify the equations of spheres and cones. Cylindrical Coordinates We obtain cylindrical coordinates for space by combining polar coordinates in the xyplane with the usual z-axis. This assigns to every point in space one or more coordinates triples of the form (r, θ, z), as shown in Fig 1. Definition Cylindrical coordinates represent a point P in space by ordered triples (r, θ, z) in which 1. r and θ are polar coordinates for the vertical projection of P on the xy - plane, 2. z is the rectangular vertical coordinate. The values of x, y, r, and θ in rectangular and cylindrical coordinates are related by the usual equations. Equations Relating Rectangular (x,y,z) and Cylindrical (r, θ, z) Co ordinates x = r cos θ, y = r sin θ, z = z ; r 2 = x 2 + y 2 tan θ = In cylindrical coordinates, the equation r = a describes not just a (1) circle in the xyplane but an entire cylinder about the z-axis. The z-axis is given by r = 0. The equation θ= θ 0 describes the plane that contains the z- axis and makes an angle θ 0 with the positive x-axis. And, just as in rectangular coordinates, the equation z= z 0 describes a plane perpendicular to the z-axis. Problem What points satisfy the equations r = 2, θ = These points make up the line in which the cylinder r = 2 cuts the portion of the plane θ = where r is positive. This is the line through the point (2,, 0)parallel to the z -axis. Along this line, z varies while r and θ have the constant values r = 2 and θ = Problem Sketch the surface r = 1 + cos θ The equation involves only r and θ; the coordinate variable z is missing. Therefore, the surface is a cylinder of lines that pass through the cardioid r = 1 + cos θ in the r θ-plane and lie parallel to the z-axis. The rules for sketching the cylinder are the same as always: sketch the x-, y-, and z-axes, draw a few Vector Calculus 34

35 perpendicular cross sections, connect the cross sections with parallel lines, and darken the exposed parts. Problem Find a Cartesian equation for the surface z = r 2 and identity the surface. From Eqs. (1) we have z=r 2 = x 2 + y 2. The surface is the circular paraboloid x 2 + y 2 = z Problem Find an equation for the circular cylinder 4x 2 + 4y 2 = 9 in cylindrical coordinates. The cylinder consists of the points whose distance from the z-axis is = The corresponding equation in cylindrical coordinates is r = Problems Find an equation for the cylinder x 2 + (y 3) 2 = 9 in cylindrical coordinates The equation for the cylinder in cylindrical coordinates is the same as the polar equation for the cylinder's base in the xy- plane: Spherical Coordinates Definition Spherical coordinates represent a point P in space by ordered triples (, in which 1. is the distance from P to the origin. 2. is the angle makes with the positive z-axis (0 ), 3. is the angle from cylindrical coordinates. The equation = a describes the sphere of radius a centered at the origin. The equation = 0 describes a single cone whose vertex lies at the origin and whose axis lies along the z-axis. (We broaden our interpretation to include the xy-plane as the cone ) If 0 is greater than, the cone = 0 opens downward. Equations Relating Spherical Coordinates to Cartesian and Cylindrical Coordinates r =, x = r cos =, r =, y = r sin =, (2) Problem Find a spherical coordinate equation for the sphere We use Eqs. (2) to substitute for x, y and z: Vector Calculus 35

36 1 1 Problem Find a spherical coordinate equation for the cone z = Use geometry. The cone is symmetric with respect to the z-axis and cuts the first quadrant of the yz-plane along the line z = y. The angle between the cone and the positive z-axis is therefore radians. The cone consists of the points whose spherical coordinates have equal to so its equations is Assignments In Assignments 1-26, translate the equations and inequalities from the given coordinates system (rectangular, cylindrical, spherical ) into equations and inequalities in the other two systems. Also, identify the figure being defined. 1. r = z = 0 4. z = 2 5. z = 6. z = = = 5 cos 12. = 6 cos 13. r = csc θ 14. r = 3 sec θ Vector Calculus 36

37 21. z = 4 4, z = 4 r, 0 23., 0 24., z + = Find the rectangular coordinates of the center of the sphere VECTOR- VALUED FUNCTIONS AND SPACE CURVES Space Curves In this chapter, we shall consider the equations of the form..(1) where, and are real valued functions of the scalar variable t. As t increases from its initial value to the value the point trace out some geometric object in space; it may be straight line or curve. This geometric object is called space curve or arc. Simply, the equations represent a curve in space. A space curve is the locus of the point functions of a single variable whose co ordinates are Definitions When a particle moves through space during a time interval I, then the particle s coordinates can be considered as functions defined on I. The points make up the curve in space that we call the particle s path. The equations and interval in (1) parameterize the curve. The vector Vector Calculus 37

38 from the origin to the particle s position at time is the particle s position vector. The functions and h are the component functions (components) of the position vector. We think of the particle s path as the curve traced by during the time interval Equation (1) defines as a vector function of the real variable on the interval More generally, a vector function or vector valued function on a domain set D is a rule that assigns a vector in space to each element in D. For now the domains will be intervals of real numbers. There are situations when the domains will be regions in the plane or in space. Vector functions will then be called vector fields. A detailed study on this will be done in later chapter. We refer to real valued functions as scalar functions to distinguish them from vector functions. The components of r are scalar functions of t. When we define a vector valued function by giving its component functions, we assume the vector function s domain to be the common domain of the components. Problem Consider the circle It is most convenient to use the trigonometric functions with t interpreted as the angle that varies from. Then we have In this chapter we show that curves in space constitute a major field of applications of vector calculus. To track a particle moving in space, we run a vector r from the origin to the particle and study the changes in r. Problem A straight line L through a point A with position vector direction of constant vector can be represented in the form in the (2) If b is a unit vector, its components are the direction cosines of L. In this case distance of the points of L from A. measures the Problem The vector function is defined for all real values of t. The curve traced by r is a helix (from an old Greek word for spiral ) that winds around the circular cylinder. The curve lies on the cylinder because the i and j components of r, being the x and y coordinates of the tip of r, satisfy the cylinder s equation: The curve rises as the k component increase. Each time increases by the curve completes one turn around the cylinder. The equations Parameterize the helix, the interval being understood. Vector Calculus 38

39 Limits The way we define limits of vector valued functions is similar to the way we define limits of realvalued functions. Definition Let be a vector function and a vector. We say that has limit as approaches and write If, for every number, there exists a corresponding number such that for all If, then precisely when The equation provides a practical way to calculate limits of vector functions. Problem If then and We define continuity for vector functions the same way we define continuity for scalar functions. Continuity Definition A vector function is continuous at a point in its domain if. The function is continuous if it is continuous at every point in its domain. Component Test for Continuity at a Point Since limits can be expressed in terms of components, we can test vector functions for continuity by examining their components. The vector function is continuous at if and only if the component functions are continuous at Vector Calculus 39

40 Problem The function,cost,sint and t are continuous is contitinuous because the components Consider the function is discontinuous at every integer. We note that the components and are continuous everywhere. But the function is discontinuous at every integer Hence is discontinuous at every integer. Derivatives and Motion Suppose that is the position vetor of a particles moving along a curve in space and that are differentiable functions of Then the difference between the particle s positions at time and time is In terms of components, As approaches zero, three things seem to happen simultaneously. First, approaches along the curve. Second, the secant line seems to approach a limiting position tangent to the curve at Third, the quotient approaches the following limit We are therefore led by past experience to the following definitions. Definitions The vector function is differentiable at.also, is said to be differentiable if it is differentiable at every point of its domain. At any point at which is differentiable, its derivative is the vector Vector Calculus 40

41 Problem If find Given Hence Definition The curve traced by is smooth if is continuous and never 0, i.e., if have continuous first derivatives that are not simultaneously 0. Remark The vector when different from0, is also a vector tangent to the curve. Definition The tangent line to the curve at a point is defined to be the line through the point parallel to at. Remark We require for a smooth curve to make sure the curve has a continuously turning tangent at each point. On a smooth curve there are no sharp corners of cusps. Definition A curve that is made up of a finite number of smooth curves pieced together in a continuous fashion is called piecewise smooth Definitions If r is the position vector of a particle moving along smooth curve in space, then is the particle s velocity vector, tangent to the curve. At any time the direction of is the direction of motion, the magnitude of v is the particle s speed, and the derivative when it exists, is the particle s acceleration vector. In short, 1. Velocity is the derivative of position : 2 Speed is the magnitude of velocity : Speed= 3 Acceleration is the derivative of velocity : 4 The vector is the direction of motion at time. We can express the velocity of a moving particle as the product of its speed and direction. Problem The vector Vector Calculus 41

42 gives the position of a moving body at time. Find the body s speed and acceleration when times, if any, are the body s velocity and acceleration orthogonal? At At, the body s speed and direction are Speed : Direction To find the times when and are orthogonal, we look for values of for which The only value is Problem 8 A particle moves along the curve Find the velocity and acceleration at Here the position vector of the particle at time is given by Then the velocity is given by and the acceleration a is given by When, and Problem Show that if b, c, d are constant vectors, then is a path of a point moving with constant acceleration. The velocity v is given by Vector Calculus 42

43 The acceleration a is given by since the derivative of the constant vector d is 0 The above is a constant vector, being a scalar multiple of the constant vector b. Hence the result. Problem A particle moves so that its position vector is given by where is a constant. Show that (i) the velocity of the particle is perpendicular to r (ii) the acceleration is directed towards the origin and has magnitude proportional to the distance from the origin. (iii) is a constant vector. (i) The velocity v is given by Now Hence velocity of the particles is perpendicular to r. (ii) The acceleration a is given by Thus direction of acceleration is opposite to that vector r and as such it is directed towards the origin and the magnitude is proportional to. i.e., the acceleration is directed towards the origin and has magnitude proportional to the distance from the origin (iii) a constant vector. Vector Calculus 43

44 Problem (Centripetal acceleration) suppose a particle moves along a circle C having radius R in the counter clock wise sense. Then its motion is given by the vector function. (1) The velocity vector Is a tangent at each point to the circle C and it magnitude is constant, Hence so that is the called the angular speed. The acceleration vector is with magnitude. Since are constants this implies that there is an acceleration of constant magnitude towards the origin.( due to negative sign). This acceleration is called centripetal acceleration. It results from the fact that the velocity vector is changing direction at a constant rate. The centripetal force is Where is the mass of. The opposite vector is called centrifugal force, and the two forces are in equilibrium at each instant of the motion. Differentiation Rules Because the derivatives of vector functions may be computed component by component, the rules for differentiating vector functions have the same form as the rules for differentiating scalar functions They are : Constant Function Rule : (any constant vector C) If and are differentiable vector functions of t, then Scalar Multiple Rules : (any number C) (any differentiable scalar function f(t)) Sum Rule : Difference Rule : Dot Product Rule : Vector Calculus 44

45 Cross product rule Chain Rule (Short Form) : If r is a differentiable function of t and t is a differentiable function of s, then We will prove the dot and cross product rules and Chain Rule and leaving the others as Assignments. Proof of the Dot Product Rule Suppose that and Then u'.v u.v Proof of the Cross Product Rule According to the definition of derivative, To change this fraction into an equivalent one that contains the difference quotients for the derivates of u and v, we subtract and add in the numerator. Then The last of these equalities holds because the limit of the cross product of two vector functions is the cross product of their limits if the latter exist. As approaches zero, approaches because v, being differentiable at is continuous at The two fractions approach the values of and at In short Vector Calculus 45

46 Proof the Chain Rule Suppose that is a differentiable vector function of and that t is a differentiable scalar functions of some other variable. Then are differentiable functions of and the Chain Rule for differentiable real valued functions gives Problem Show that Let then and Also or Constant Vectors A vector changes if either its magnitude changes or its direction changes or both direction and magnitude change. Theorem A The necessary and sufficient condition for the vector function to be constant is that the zero vector, Necessary Part If is constant, then. Hence Vector Calculus 46

47 Sufficiency Part Conversely, suppose that Let Then Hence using the assumption we have Equating the coefficients, we get and this implies that and are constants, (in other words this means that and are independent of. Therefore is a constant vector. Theorem B The necessary and sufficient condition for the vector function to have constant magnitude is that Let F be a vector function of the scalar variable Suppose to have constant magnitude, say F, so that Then Therefore, or or or Conversely, suppose that. Then or or or and this implies that is a constant or is a constant. i.e. the vector function have constant magnitude. Theorem C The necessary and sufficient condition for the vector function to have constant direction is that Let be a vector function of the scalar variable and n be a unit vector in the direction of If be the magnitude of then Vector Calculus 47

48 (since by theorem B).(1) Necessary Part Suppose that F has constant direction. Then n is a constant vector. Therefore,. Hence by (1) above we have Sufficiency Part Conversely, suppose that. Then from (1), we get or.(2) Since n is of constant length, by the last theorem, we have (3) From (2) and (3), we obtain. Hence n is a constant vector. Therefore the direction of F is constant. Vector Functions of a Constant Length When we track a particle moving on a sphere centered at the origin the position vector has a constant length equal to the radius of the sphere. The velocity vector, tangent to the path of motion, is tangent to the sphere and hence perpendicular to r, This is always the case for a differentiable vector function of constant length (as seen in Theorem above): The vector and its first derivative are orthogonal. With the length constant, the change in the function is a change in direction only, and direct changes take place at the right angles. If is differentiable vector functions of of constant length, then..(3) For the proof of (3) see Theorem above. Problem Show that Vector Calculus 48

49 has constant length and is orthogonal to its derivative. Integrals of Vector Functions A differentiable vector function is an anti derivative of a vector function on an interval if at each point of If is an anti derivative of on it can be shown, working one component at a time, that every anti derivative of on has the form of for some constant vector. The set of all anti derivates of on is the indefinite integral on Definition The indefinite integral or with respect to is the set of all anti derivatives of, denoted by If is any anti derivative of then The usual arithmetic rules for indefinite integrals apply. Problem (4) (5) with As in the integration of scalar functions, it is recommended that you skip the steps in (4) and (5) and go directly to the final form. Find an anti derivative for each component and add a constant vector at the end. Problem 15 Evaluate where A is a vector function in the variable We know that Hence Vector Calculus 49

50 where is an arbitrary constant vector. Problem Prove that We know that Differentiating with respect to on both sides, we get On integration, we get Definite integrals Definite integrals of vector functions are defined in terms components. Definition If the component of are integral over then so is r, and the definite integral of from to is The usual arithmetic rules for definite integrals apply. Problem If find Vector Calculus 50