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��7 Vectors 7.1 Vectors in Two Dimensions (2-D) A vector in 2-D is defined by its length and the angle it makes with a reference axis (usually the x-axis). This vector is represented graphically by an arrow. The tail of the arrow is called the initial point of the vector and the tip of the arrow is the terminal point. Two vectors are equal when both their length and angle with a reference axis are equal. 7.1.1 Addition r r r The sum of two vectors u + v = w is a vector constructed graphically as fol- lows. At the tip of the first vector, draw a vector equal to the second vector, such that its tail coincides with the tip of the first vector. The resultant vector has as its tail that of the first vector, and as its tip, the tip of the just-drawn second vector (the Parallelogram Rule) (see Figure 7.1). The negative of a vector is that vector whose tip and tail have been exchanged from those of the vector. This leads to the conclusion that the dif- ference of two vectors is the other diagonal in the parallelogram (Figure 7.2). 7.1.2 Multiplication of a Vector by a Real Number r If we multiply a vector v by a real number k, the result is a vector whose r r length is k times the length of v , and whose direction is that of v if k is pos- itive, and opposite if k is negative. 7.1.3 Cartesian Representation It is most convenient for a vector to be described by its projections on the x-axis and on the y-axis, respectively; these are denoted by (v1, v2) or (vx, vy). In this representation: 0-8493-????-?/00/$0.00+$.50 � 2000 by CRC Press LLC � 2001 by CRC Press LLC FIGURE 7.1 Sum of two vectors. FIGURE 7.2 Difference of two vectors. r u = (u1, u2) = (u1)�1 + (u2)�2 (7.1) where �1 and �2 are the unit vectors (length is 1) parallel to the x-axis and y-axis, respectively. In terms of this representation, we can write the zero vec- tor, the sum of two vectors, and the multiplication of a vector by a real num- ber as follows: � 2001 by CRC Press LLC r 0 = (0, 0) = 0�1 + 0�2 (7.2) r r r u + v = w = (u1 + v1, u2 + v2) = (u1 + v1)�1 + (u2 + v2)�2 (7.3) r ku = (ku1, ku2) = (ku1)�1 + (ku2)�2 (7.4) Preparatory Exercise Pb. 7.1 Using the above definitions and properties, prove the following identities: r r r r u + v = v + u r r r r r r (u + v) + w = u + (v + w) r r r r r u + 0 = 0 + u = u r r r u + (-u) = 0 r r k(lu) = (kl)u r r r r k(u + v) = ku + kv r r r (k + l)u = ku + lu The norm of a vector is the length of this vector. Using the Pythagorean the- orem, its square is: r 2 2 2 u = u1 + u2 (7.5) r and therefore the unit vector in the u direction, denoted by �u, is given by: 1 �u = (u1, u2) (7.6) 2 2 u1 + u2 All of the above can be generalized to 3-D, or for that matter to n-dimensions. For example: 1 �u = (u1, u2 ,& , un) (7.7) 2 2 2 u1 + u2 + & un � 2001 by CRC Press LLC 7.1.4 MATLAB Representation of the Above Results MATLAB distinguishes between two kinds of vectors: the column vector and the row vector. As long as the components of the vectors are all real, the dif- ference between the two is in the structure of the array. In the column vector case, the array representation is vertical and in the row vector case, the array representation is horizontal. This distinction is made for the purpose of including in a consistent structure the formulation of the dot product and the definition of matrix multiplication. Example 7.1 Type and execute the following commands, while interpreting the output at each step: V=[1 3 5 7] W=[1;3;5;7] V' U=3*V Z=U+V Y=V+W %you cannot add a row vector and a column %vector You would have observed that: 1. The difference in the representation of the column and row vectors is in the manner they are separated inside the square brackets. 2. The single quotation mark following a vector with real components changes that vector from being a column vector to a row vector, and vice versa. 3. Multiplying a vector by a scalar simply multiplies each component of this vector by this scalar. 4. You can add two vectors of the same kind and the components would be adding by pairs. 5. You cannot add two vectors of different kinds; the computer will give you an error message alerting you that you are adding two quantities of different dimensions. The MATLAB command for obtaining the norm of a vector is norm. Using this notation, it is a simple matter to define the unit vector in the same direc- tion as a given vector. Example 7.2 Find the length of the vector and the unit vector u = [1 5 3 2] and the unit vector parallel to it. � 2001 by CRC Press LLC u=[1 5 3 2] lengthu=norm(u) %length of vector u unitu=u/(norm(u)) %unit vector parallel to u lengthunitu=norm(unitu) %verify length of unit vector FIGURE 7.3 The geometry of the generalized Pythagorean theorem. 7.2 Dot (or Scalar) Product r r If the angle between the vectors u and v is �, then the dot product of the two vectors is: r r r r u �" v = u v cos(�) (7.8) The dot product can also be expressed as a function of the vectors compo- nents. Referring to Figure 7.3, we know from trigonometry the relation relat- ing the length of one side of a triangle with the length of the other two sides and the cosine of the angle between the other two sides. This relation is the generalized Pythagorean theorem. Referring to Figure 7.3, this gives: r r r r 2 2 2 PQ = u + v - 2 u v cos(�) (7.9) but since: � 2001 by CRC Press LLC r r r PQ = v - u (7.10) r r r r r r 1 2 2 2 �! u v cos(�) = ( u + v - v - u ) (7.11) 2 and the dot product can be written as: r r 1 2 2 2 2 u �" v = (u1 + u2 + v1 + v2 - (v1 - u1)2 - (v2 - u2)2 = u1v1 + u2v2 (7.12) 2 In an n-dimensional space, the above expression is generalized to: r r u �" v = u1v1 + u2v2 + & + unvn (7.13) and the norm square of the vector can be written as the dot product of the vector with itself; that is, r r r2 2 2 2 u = u �" u = u1 + u2 + & + un (7.14) Example 7.3 r Parallelism and orthogonality of two vectors in a plane. Let the vectors u r r r and v be given by: u = 3�1 + 4�2 and v = a�1 + 7�2. What is the value of a if the vectors are parallel, and if the vectors are orthogonal? Solution: Case 1: If the vectors are parallel, this means that they make the same angle with the x-axis. The tangent of this angle is equal to the ratio of the vector x-component to its y-component. This means that: a 3 = �! a = 21/ 4 7 4 Case 2: If the vectors are orthogonal, this means that the angle between them is 90�, and their dot product will be zero because the cosine for that angle is zero. This implies that: 3a + 28 = 0 �! a = -28 / 3 Example 7.4 Find the unit vector in 2-D that is perpendicular to the line ax + by + c = 0. � 2001 by CRC Press LLC Solution: Choose two arbitrary points on this line. Denote their coordinates by (x1, y1) and (x2, y2); being on the line, they satisfy the equation of the line: ax1 + by1 + c = 0 ax2 + by2 + c = 0 Substracting the first equation from the second equation, we obtain: a(x2 - x1) + b(y2 - y1) = 0 which means that (a, b) �" (x2 - x1, y2 - y1), and the unit vector perpendicular to the line is: �� a b ��" = , �� a2 + b2 a2 + b2 Example 7.5 Find the angle that the lines 3x + 2y + 2 = 0 and 2x y + 1 = 0 and make together. Solution: The angle between two lines is equal to the angle between their nor- mal unit vectors. The unit vectors normal to each of the lines are, respectively: �� 3 2 �� 2 -1 �� n1 = , and n2 = , �� 13 13 5 5 Having the two orthogonal unit vectors, it is a simple matter to compute the angle between them: 4 � � cos(�) = n1 �" n2 = �! � = 1.0517 radians 65 7.2.1 MATLAB Representation of the Dot Product The dot product is written as the product of a row vector by a column vector of the same length. Example 7.6 Find the dot product of the vectors: � 2001 by CRC Press LLC u = [1 5 3 7] and v = [2 4 6 8] Solution: Type and execute each of the following commands, while interpret- ing each output: u=[1 5 3 7] v=[2 4 6 8] u*v' v'*u u*v %you cannot multiply two rows u'*v u*u' (norm(u))^2 As observed from the above results, in MATLAB, the dot product can be obtained only by the multiplication of a row on the left and a column of the same length on the right. If the order of a row and column are exchanged, we obtain a two-dimensional array structure (i.e., a matrix, the subject of Chap- ter 8). On the other hand, if we multiply two rows, MATLAB gives an error message about the non-matching of dimensions. Observe further, as pointed out previously, the relation between the length of a vector and its dot product with itself. In-Class Exercises Pb. 7.2 Generalize the analytical technique, as previously used in Example 7.4 for finding the normal to a line in 2-D, to find the unit vector in 3-D that is perpendicular to the plane: ax + by + cz + d = 0 (Hint: A vector is perpendicular to a plane if it is perpendicular to two non- collinear vectors in that plane.) Pb. 7.3 Find, in 2-D, the distance of the point P(x0, y0) from the line ax + by + c = 0. (Hint: Remember the geometric definition of the dot product.) Pb. 7.4 Prove the following identities: r r r r r r r r r r r r r r r u �" v = v �" u, u �"(v + w) = u �" v + u �" w, k �"(u �" v) = (ku)�" v � 2001 by CRC Press LLC 7.3 Components, Direction Cosines, and Projections 7.3.1 Components The components of a vector are the values of each element in the defining r n-tuplet representation. For example, consider the vector u = [1 5 3 7] in real 4-D. We say that its first, second, third, and fourth components are 1, 5, 3, and 7, respectively. (We are maintaining, in this section, the arrow nota- tion for the vectors, irrespective of the dimension of the space.) The simplest basis of a n-dimensional vector space is the collection of n unit vectors, each having only one of their components that is non-zero and such that the location of this non-zero element is different for each of these basis vectors. This basis is not unique. For example, in 4-D space, the canonical four-unit orthonormal basis vec- tors are given, respectively, by: �1 = [1 0 0 0] (7.15) �2 = [0 1 0 0] (7.16) �3 = [0 0 1 0] (7.17) �4 = [0 0 0 1] (7.18) r and the vector u can be written as a linear combination of the basis vectors: r u = u1�1 + u2�2 + u3�3 + u4�4 (7.19) The basis vectors are chosen to be orthonormal, which means that in addi- tion to requiring each one of them to have unit length, they are also orthogonal two by two to each other. These properties of the basis vectors leads us to the following important result: the mth component of a vector is obtained by tak- ing the dot product of the vector with the corresponding unit vector, that is, r um = �m �" u (7.20) 7.3.2 Direction Cosines The direction cosines are defined by: � 2001 by CRC Press LLC r um �m �" u cos(� ) = = (7.21) r r m u u In 2-D or 3-D, these quantities have the geometrical interpretation of being r the cosine of the angles that the vector u makes with the x, y, and z axes. 7.3.3 Projections r r The projection of a vector u over a vector a is a vector whose magnitude is r r the dot product of the vector u with the unit vector in the direction of a, denoted by �a, and whose orientation is in the direction of �a: r r r r r r r r u �" a a u �" a r proja(u) = (u �" �a)�a = = a (7.22) r r r 2 a a a r r The component of u that is perpendicular to a is obtained by subtracting r r r from u the projection vector of u over a. MATLAB Example r r Assume that we have the vector u = �1 + 5�2 + 3�3 + 7�4 and the vector a = 2�1 + 3�2 + �3 + 4�4. We desire to obtain the components of each vector, the projec- r r r r tion of u over a, and the component of u orthogonal to a. Type, execute, and interpret at each step, each of the following commands using the above definitions: u=[1 5 3 7] a=[2 3 1 4] u(1) a(2) prjuovera=((u*a')/(norm(a)^2))*a orthoutoa=u-prjuovera prjuovera*orthoutoa' The last command should give you an answer that is zero, up to machine r r r round-up errors because the projection of u over a and the component of u r orthogonal to a are perpendicular. 7.4 The Dirac Notation and Some General Theorems* Thus far, we have established some key practical results in real finite dimen- sional vector spaces; namely: � 2001 by CRC Press LLC 1. A vector can be decomposed into a linear combination of the basis vectors. 2. The dot product of two vectors can be written as the multiplication of a row vector by a column vector, each of whose elements are the components of the respective vectors. 3. The norm of a vector, a non-negative quantity, is the square root of the dot product of the vector with itself. 4. The unit vector parallel to a specific vector is that vector divided by its norm. 5. The projection of a vector on another can be deduced from the dot product of the two vectors. To facilitate the statement of these results in a notation that will be suitable for infinite-dimensional vector spaces (which is very briefly introduced in Section 7.7), Dirac in his elegant formulation of quantum mechanics intro- duced a simple notation that we now present. The Dirac notation represents the row vector by what he called the bra- vector and the column vector by what he called the ket-vector, such that when a dot product is obtained by joining the two vectors, the result will be the scalar bra-ket quantity. Specifically: r Column vector u �! u (7.23) r Row vector v �! v (7.24) r r Dot product v �" u �! v u (7.25) The orthonormality of the basis vectors is written as: mn =�mn (7.26) , where the basis vectors are referred to by their indices, and where �m,n is the Kroenecker delta, equal to 1 when its indices are equal, and zero otherwise. The norm of a vector, a non-negative quantity, is given by: 2 (norm of u )2 = u = u u (7.27) The Decomposition rule is written as: u = n (7.28) n "c n where the components are obtained by multiplying Eq. (7.28) on the left by m . Using Eq. (7.26), we deduce: � 2001 by CRC Press LLC mu = mn = �m,n = cm (7.29) n n "c "c n n Next, using the Dirac notation, we present the proofs of two key theorems of vector algebra: the Cauchy-Schwartz inequality and the triangle inequality. 7.4.1 Cauchy-Schwartz Inequality Let u and v be any non-zero vectors; then: 2 u v d" u u v v (7.30) PROOF Let � = �1, (�2 = 1); then �� = 1 if u v e" 0 �� u v = � u v such that (7.31) �� =-1 if u v d" 0 Now, consider the ket �u + tv ; its norm is always non-negative. Computing this norm square, we obtain: �u + tv �u + tv = �2 u u + �t u v + t� v u + t2 v v = u u + 2�t u v + t2 v v (7.32) = u u + 2t u v + t2 v v The RHS of this quantity is a positive quadratic polynomial in t, and can be written in the standard form: at2 + bt + c e" 0 (7.33) The non-negativity of this quadratic polynomial means that it can have at most one real root. This means that the descriminant must satisfy the inequality: b2 4ac d" 0 (7.34) Replacing a, b, c by their values from Eq. (7.32), we obtain: 2 4 u v - 4 u u v v d" 0 (7.35) 2 �! u v d" u u v v (7.36) � 2001 by CRC Press LLC which is the desired result. Note that the equality holds if and only if the two vectors are linearly dependent (i.e., one vector is equal to a scalar multiplied by the other vector). Example 7.7 Show that for any three non-zero numbers, u1, u2, and u3, the following ine- quality always holds: 9 d" u1 + u2 + u3 �� + + (7.37) ()�� 1 1 1 �� u1 u2 u3 �� PROOF Choose the vectors v and w such that: 1/2 1 1 v = u1 , u2/2 , u3/2 (7.38) 1/2 1/2 1/2 �� 1 1 1 w = , , (7.39) �� u1 u2 u3 �� then: v w = 3 (7.40) v v = (u1 + u2 + u3) (7.41) �� 1 1 1 w w = + + (7.42) �� u1 u2 u3 �� Applying the Cauchy-Schwartz inequality in Eq. (7.36) establishes the desired result. The above inequality can be trivially generalized to n-ele- ments, which leads to the following important result for the equivalent resis- tance for resistors all in series or all in parallel. Application The equivalent resistance of n-resistors all in series and the equivalent resis- tance of the same n-resistors all in parallel obey the relation: R series n2 d" (7.43) R parallel � 2001 by CRC Press LLC PROOF The proof is straightforward. Using Eq. (7.37) and recalling Ohm s law for n resistors {R1, R2, & , Rn}, the equivalent resistances for this combina- tion, when all resistors are in series or are all in parallel, are given respec- tively by: Rseries = R1 + R2 + & + Rn (7.44) and 1 1 1 1 = + + & + (7.45) R R1 R2 Rn parallel Question: Can you derive a similar theorem for capacitors all in series and all in parallel? (Remember that the equivalent capacitance law is different for capacitors than for resistors.) 7.4.2 Triangle Inequality This is, as the name implies, a generalization of a theorem from Euclidean geometry in 2-D that states that the length of one side of a triangle is smaller or equal to the sum of the the other two sides. Its generalization is u + v d" u + v (7.46) PROOF Using the relation between the norm and the dot product, we have: 2 u + v = u + v u + v = u v + 2 u v + v v (7.47) 22 2 2 = u + 2 u v + v d" u + 2 u v + v Using the Cauchy-Schwartz inequality for the dot product appearing in the previous inequality, we deduce that: 2 2 2 2 u + v d" u + 2 u v + v = u + v (7.48) ( ) which establishes the theorem. Homework Problems Pb. 7.5 Using the Dirac notation, generalize to n-dimensions the 2-D geom- etry Parallelogram theorem, which states that: The sum of the squares of the diag- onals of a parallelogram is equal to twice the sum of the squares of the side; or that: � 2001 by CRC Press LLC r r r r r r 22 2 2 u + v + u - v = 2 u + 2 v Pb. 7.6 Referring to the inequality of Eq. (7.43), which relates the equivalent resistances of n-resistors in series and in parallel, under what conditions does the equality hold? 7.5 Cross Product and Scalar Triple Product* In this section and in Sections 7.6 and 7.7, we restrict our discussions to vectors in a 3-D space, and use the more familiar conventional vector notation. 7.5.1 Cross Product r r DEFINITION If two vectors are given by u = (u1, u2 , u3) and v = (v1, v2 , v3) r r then their cross product, denoted by u � v, is a vector given by: r r u � v = (u2v3 - u3v2 , u3v1 - u1v3 , u1v2 - u2v1) (7.49) By simple substitution, we can infer the following properties for the cross product as summarized in the preparatory exercises below. Preparatory Exercises Pb. 7.7 Show, using the above definition for the cross product, that: r r r r r r r r r r a. u �"(u � v) = v �"(u � v) = 0 �! u � v is orthogonal to both u and v r r r r r r 2 2 2 b. u � v = u v - (u �" v)2 Called the Lagrange Identity r r r r c. u � v = -(v � u) Noncommutativity r r r r r r r d. u � (v + w) = u � v + u � w Distributive property r r r r r r e. k(u � v) = (ku) � v = u � (kv) r r r f. u � 0 = 0 r rr g. u � u = 0 Pb. 7.8 Verify the following relations for the basis unit vectors: � 2001 by CRC Press LLC �1 � �2 = �3 ; �2 � �3 = �1; �3 � �1 = �2 Pb. 7.9 Ask your instructor to show you how the Right Hand rule is used to determine the direction of a vector equal to the cross product of two other vectors. 7.5.2 Geometric Interpretation of the Cross Product As noted in Pb. 7.7a, the cross product is a vector that is perpendicular to its two constituents. This determines the resultant vector s direction. To deter- r mine its magnitude, consider the Lagrange Identity. If the angle between u r and v is �, then: r r r r r r 2 2 2 2 2 u � v = u v - u v cos2(�) (7.50) and r r r r u � v = u v sin(�) (7.51) that is, the magnitude of the cross product of two vectors is the area of the parallelogram formed by these vectors. 7.5.3 Scalar Triple Product r r r r r DEFINITION If u, v, and w are vectors in 3-D, then u �" v � w is called the (r ) r r r scalar triple product of u, v, and w. PROPERTY r r r r r r r r r u �"(v � w) = v �"(w � u) = w �"(u � v) (7.52) This property can be trivially proven by writing out the components expan- sions of the three quantities. 7.5.3.1 Geometric Interpretation of the Scalar Triple Product r r r If the vectors u, v, and w original points are brought to the same origin, these three vectors define a parallelepiped. The absolute value of the scalar triple product can then be interpreted as the volume of this parallelepiped. We have r r r r shown earlier that v � w is a vector that is perpendicular to both v and w, � 2001 by CRC Press LLC and whose magnitude is the area of the base parallelogram. From the defini- r tion of the scalar product, dotting this vector with u will give a scalar that is the product of the area of the parallelepiped base multiplied by the parallel- epiped height, whose magnitude is exactly the volume of the parallelepiped. The circular permutation property of Eq. (7.52) then has a very simple geo- metric interpretation: in computing the volume of a parallelepiped, it does not matter which surface we call base. MATLAB Representation r r The cross product of the vectors u = (u1, u2 , u3) and v = (v1, v2 , v3) is found using the cross(u,v) command. r r r The triple scalar product of the vectors u, v, and w is found through the det([u;v;w]) command. Make sure that the vectors defined as arguments of these functions are defined as 3-D vectors, so that the commands work and the results make sense. Example 7.8 r r r Given the vectors u = (2, 1, 0), v = (0, 3, 0), w = (1, 2, 3), find the cross prod- uct of the separate pairs of these vectors, and the volume of the parallelepi- ped formed by the three vectors. Solution: Type, execute, and interpret at each step, each of the following com- mands, using the above definitions: u=[2 1 0] v=[0 3 0] w=[1 2 3] ucrossv=cross(u,v) ucrossw=cross(u,w) vcrossw=cross(v,w) paralvol=abs(det([u;v;w])) paralvol2=abs(cross(u,v)*w') Question: Verify that the last command is an alternate way of writing the vol- ume of the parallelepiped expression. In-Class Exercises Pb. 7.10 Compute the shortest distance from New York to London. (Hint: (1) A great circle is the shortest path between two points on a sphere; (2) the angle between the radial unit vectors passing through each of the cities can be obtained from their respective latitude and longitude.) � 2001 by CRC Press LLC Pb. 7.11 Find two unit vectors that are orthogonal to both vectors given by: r r a = (2, -1, 2) and b = (1, 2, -3) Pb. 7.12 Find the area of the triangle with vertices at the points: A(0, -1,1), B(3,1, 0), and C(-2, 0, 2) Pb. 7.13 Find the volume of the parallelepiped formed by the three vectors: r rr u = (1, 2, 0), v = (0, 3, 0), w = (1, 2, 3) Pb. 7.14 Determine the equation of a plane that passes through the point (1, 1, 1) and is normal to the vector (2, 1, 2). Pb. 7.15 Find the angle of intersection of the planes: x + y - z = 0 and x - 3y + z - 1 = 0 Pb. 7.16 Find the distance between the point (3, 1, 2) and the plane z = 2x 3y. Pb. 7.17 Find the equation of the line that contains the point (3, 2, 1) and is perpendicular to the plane x + 2y 2z = 2. Write the parametric equation for this line. Pb. 7.18 Find the point of intersection of the plane 2x 3y + z = 6 and the line x - 1 y + 1 - 2 z = = 3 1 2 Pb. 7.19 Show that the points (1, 5), (3, 11), and (5, 17) are collinear. r r r Pb. 7.20 Show that the three vectors u, v, and w are coplanar: r rr u = (2, 3, 5); v = (2, 8,1); w = (8, 22,12) Pb. 7.21 Find the unit vector normal to the plane determined by the points (0, 0, 1), (0, 1, 0), and (1, 0, 0). Homework Problem Pb. 7.22 Determine the tetrahedron with the largest surface area whose ver- tices P0, P1, P2, and P3 are on the unit sphere x2 + y2 + z2 = 1. � 2001 by CRC Press LLC (Hints: (1) Designate the point P0 as north pole and confine P1 to the zero meridian. With this choice, the coordinates of the vertices are given by: P0 = (�0 = � / 2, �0 = 0) P1 = (�1, �1 = 0) P2 = (�2 , �2) P3 = (�3 , �3) (2) From symmetry, the optimal tetrahedron will have a base (P1, P2, P3) that is an equilateral triangle in a plane parallel to the equatorial plane. The lati- tude of (P1, P2, P3) is �, while their longitudes are (0, 2�/3, 2�/3), respec- tively. (3) The area of the tetrahedron is the sum of the areas of the four triangles (012), (023), (031), (123), where we are indicating each point by its subscript. (4) Express the area as function of �. Find the value of � that maxi- mizes this quantity.) 7.6 Vector Valued Functions As you may recall, in Chapter 1 we described curves in 2-D and 3-D by para- metric equations. Essentially, we gave each of the coordinates as a function of a parameter. In effect, we generated a vector valued function because the position of the point describing the curve can be written as: r Rt) = x(t)�1 + y(t)�2 + z(t)�3 (7.53) ( r If the parameter t was chosen to be time, then the tip of the vector Rt) would ( be the position of a point on that curve as a function of time. In mechanics, r finding Rt) is ultimately the goal of any problem in the dynamics of a point ( particle. In many problems of electrical engineering design of tubes and other microwave engineering devices, we need to determine the position of elec- trons whose motion we control by a variety of electrical and magnetic fields geometries. The following are the kinematics variables of the problem. The dynamics form the subject of mechanics courses. To help visualize the shape of a curve generated by the tip of the position r vector Rt), we introduce the tangent vector and the normal vector to the ( curve and the curvature of the curve. The velocity vector field associated with the above position vector is defined through: � 2001 by CRC Press LLC r dR(t) dx(t) dy(t) dz(t) = �1 + �2 + �3 (7.54) dt dt dt dt and the unit vector tangent to the curve is given by: r dR(t) dt � T(t) = r (7.55) dR(t) dt This is, of course, the unit vector that is always in the direction of the velocity of the particle. r LEMMA r dV(t) If a vector valued function V(t) has a constant value, then its derivative is dt orthogonal to it. PROOF The proof of this lemma is straightforward. If the length of the vector r r is constant, then its dot product with itself is a constant; that is, V(t)�"V(t) = C. r r dV(t) Differentiating both sides of this equation gives �"V(t) = 0, and the dt orthogonality between the two vectors is thus established. � The tangential unit vector T(t) is, by definition, constructed to have unit length. We construct the norm to the curve by taking the unit vector in the direction of the time-derivative of the tangential vector; that is, � dT(t) dt � Nt) = (7.56) ( � dT(t) dt The curvature of the curve is � dT(t) dt � = r (7.57) dR(t) dt � 2001 by CRC Press LLC Example 7.9 Find the tangent, normal, and curvature of the trajectory of a particle moving in uniform circular motion of radius a and with angular frequency �. Solution: The parametric equation of motion is r Rt) = a cos(�t)�1 + asin(�t)�2 ( The velocity vector is r dR(t) =-a� sin(�t)�1 + a� cos(�t)�2 dt and its magnitude is a�. The tangent vector is therefore: � T(t) = - sin(�t)�1 + cos(�t)�2 The normal vector is � Nt) =- cos(�t)�1 - sin(�t)�2 ( The radius of curvature is � dT(t) dt -� cos(�t)�1 - � sin(�t)�2 1 �(t) = = = = constant r -a� sin(�t)�1 + a� cos(�t)�2 a dR(t) dt Homework Problems Pb. 7.23 Show that in 2-D the radius of curvature can be written as: x2 y2 2 - y2 x2 2 � = ((x2 )2 + (y2 )2)3/2 where the prime refers to the first derivative with respect to time, and the double-prime refers to the second derivative with respect to time. � 2001 by CRC Press LLC Pb. 7.24 Using the parametric equations for an ellipse given in Example 1.13, find the curvature of the ellipse as function of t. a. At what points is the curvature a minimum, and at what points is it a maximum? b. What does the velocity do at the points of minimum and maximum curvature? c. On what dates of the year does the planet Earth pass through these points on its trajectory around the sun? 7.7 Line Integral As you may have already learned in an elementary physics course: if a force r r F is applied to a particle that moves by an infinitesimal distance "l , then the infinitesimal work done by the force on the particle is the scalar product of the force by the displacement; that is, r r "W = F �" "l (7.58) Now, to calculate the work done when the particle moves along a curve C, located in a plane, we need to define the concept of a line integral. Suppose that the curve is described parametrically [i.e., x(t) and y(t) are given]. Furthermore, suppose that the vector field representing the force is given by: r F = P(x, y)�x + Qx, y)�y (7.59) ( The displacement element is given by: "l = "x�x + "y�y (7.60) The infinitesimal element of work, which is the dot product of the above two quantities, can then be written as: "W = P"x + Q"y (7.61) This expression can be simplified if the curve is written in parametric form. Assuming the parameter is t, then "W can be written as a function of the sin- gle parameter t: � 2001 by CRC Press LLC dx dy dx dy �� "W = P(t) "t + Q(t) "t = P(t) + Qt) "t (7.62) ( �� dt dt dt dt and the total work can be written as an integral over the single variable t: t1 dx dy �� W = P(t) + Qt) dt (7.63) ( �� +" �� dt dt t0 Homework Problems Pb. 7.25 How much work is done in moving the particle from the point r (0, 0) to the point (3, 9) in the presence of the force F along the following two different paths? a. The parabola y = x2. b. The line y = 3x. The force is given by: r F = xy�x + (x2 + y2)�y r Pb. 7.26 Let F = y�x + x�y. Calculate the work moving from (0, 0) to (1, 1) along each of the following curves: a. The straight line y = x. b. The parabola y = x2. c. The curve C described by the parametric equations: x(t) = t3/2 and y(t) = t5 A vector field such as the present one, whose line integral is independent of the path chosen between fixed initial and final points, is said to be conser- vative. In your vector calculus course, you will establish the necessary and sufficient conditions for a vector field to be conservative. The importance of conservative fields lies in the ability of their derivation from a scalar poten- tial. More about this topic will be discussed in electromagnetic courses. 7.8 Infinite Dimensional Vector Spaces* This chapter section introduces some preliminary ideas on infinite-dimen- sional vector spaces. We assume that the components of this vector space are � 2001 by CRC Press LLC complex numbers rather than real numbers, as we have restricted ourselves thus far. Using these ideas, we discuss, in a very preliminary fashion, Fourier series and Legendre polynomials. We use the Dirac notation to stress the commonalties that unite the finite- and infinite-dimensional vector spaces. We, at this level, sacrifice the mathe- matical rigor for the sake of simplicity, and even commit a few sins in our treatment of limits. A more formal and rigorous treatment of this subject can be found in many books on functional analysis, to which we refer the inter- ested reader for further details. A Hilbert space is much the same type of mathematical object as the vector spaces that you have been introduced to in the preceding sections of this chapter. Its elements are functions, instead of n-dimensional vectors. It is infi- nite-dimensional because the function has a value, say a component, at each point in space, and space is continuous with an infinite number of points. The Hilbert space has the following properties: 1. The space is linear under the two conditions that: a. If a is a constant and � is any element in the space, then a � is also an element of the space; and b. If a and b are constants, and � and � are elements belonging to the space, then a � + b � is also an element of the space. 2. There is an inner (dot) product for any two elements in the space. The definition adopted here for this inner product for functions defined in the interval tmin d" t d" tmax is: tmax � � = �(t)�(t)dt (7.64) +" tmin 3. Any element of the space has a norm ( length ) that is positive and related to the inner product as follows: tmax 2 � = � � = �(t)�(t)dt (7.65) +" tmin Note that the requirement for the positivity of a norm is that which necessitated the complex conjugation in the definition of the bra-vector. 4. The Hilbert space is complete; or loosely speaking, the Hilbert space contains all its limit points. This condition is too technical and will not be further discussed here. In this Hilbert space, we define similar concepts to those in finite-dimen- sional vector spaces: � 2001 by CRC Press LLC " Orthogonality. Two vectors are orthogonal if: tmax � � = �(t)�(t)dt = 0 (7.66) +" tmin " Basis vectors. Any function in Hilbert space can be expanded in a linear combination of the basis vectors {un}, such that: � = un (7.67) n "c n and such that the elements of the basis vectors obey the orthonor- mality relations: um un =�m,n (7.68) " Decomposition rule. To find the cn s, we follow the same procedure adopted for finite-dimensional vector spaces; that is, take the inner product of the expansion in Eq. (7.67) with the bra um . We obtain, using the orthonormality relations [Eq. (7.68)], the following: um � = um un = �mn = cm (7.69) n n , "c"c n n Said differently, cm is the projection of the ket � onto the bra um . " The norm as a function of the components. The norm of a vector can be expressed as a function of its components. Using Eqs. (7.67) and (7.68), we obtain: 2 2 � = � � = cn (7.70) n n ""c cm un um = ""c cm�n,m = " n m n m n Said differently, the norm square of a vector is equal to the sum of the magnitude square of the components. Application 1: The Fourier Series The theory of Fourier series, as covered in your calculus course, states that a function that is periodic, with period equal to 1, in some normalized units can be expanded as a linear combination of the sequence {exp(j2�nt)}, where n is an integer that goes from minus infinity to plus infinity. The purpose here is to recast the familiar Fourier series results within the language and notations of the above formalism. � 2001 by CRC Press LLC Basis: un = exp(j2�nt) and un = exp(- j2�nt) (7.71) Orthonormality of the basis vectors: 1/2 �� 1 if m = n um un = exp(- j2�mt)exp(j2�nt)dt = (7.72) �� +" -1/2 ��0 if m `" n �� Decomposition rule: " " � = un = exp(j2�nt) (7.73) n n "c "c n=-" n=-" where 1/2 cn = un � = exp(- j2�nt)�(t)dt (7.74) +" -1/2 Parseval s identity: " 1/2 1/2 2 2 2 � = � � = �(t)�(t)dt = �(t) dt = cn (7.75) " +"+" -1/2 -1/2 n=-" Example 7.9 Derive the analytic expression for the potential difference across the capacitor in the RLC circuit of Figure 4.5 if the temporal profile of the source potential is a periodic function, of period 1, in some normalized units. Solution: 1. Because the potential is periodic with period 1, it can be expanded using Eq. (7.73) in a Fourier series with basis functions {ej2�nt}: �� n Vs(t) = Re�� ej2�nt �� (7.76) �� s "V �� n �� where |sn is the phasor associated with the frequency mode (2�n). (Note that n in the expressions for the phasors is a superscript and not a power.) � 2001 by CRC Press LLC 2. We find |cn, the capacitor response phasor associated with the |sn excita- tion. This can be found by noting that the voltage across the capacitor is equal to the capacitor impedance multiplied by the current phasor, giving: � ZnVsn n |cn = Zn ( =c (7.77) c n n Zn + ZR + ZL c where from the results of Section 6.8, particularly Eqs. (6.83) through (6.85), we have: 1 Zn = (7.78) c j2�nC n ZL = j2�nL (7.79) n ZR = R (7.80) 3. Finally, we use the linearity of the ODE system and write the solution as the linear superposition of the solutions corresponding to the response to each of the basis functions; that is, � �� ZnVsn c j2�nt Vc(t) = Re�� (7.81) �� n n n "Z + ZR + ZL e �� c �� n �� leading to the expression: �� |sn Vc(t) = Re�� (7.82) �� "1 - (2�n)2 LC + j(2�n)RC ej2�nt �� n �� Homework Problem Pb. 7.27 Consider the RLC circuit. Assuming the same notation as in Section 6.5.3, but now assume that the source potential is given by: Vs = V0 cos6(�t) a. Find analytically the potential difference across the capacitance. (Hint: Write the power of the trigonometric function as function of the different multiples of the angle.) � 2001 by CRC Press LLC b. Find numerically the steady-state solution to this problem using the techniques of Chapter 4, and assume for some normalized units the following values for the parameters: LC = 1, RC = 1, � = 2� c. Compare your numerical results with the analytical results. Application 2: The Legendre Polynomials We propose to show that the Legendre polynomials are an orthonormal basis for all functions of compact support over the interval 1 d" x d" 1. Thus far, we have encountered the Legendre polynomials twice before. They were defined through their recursion relations in Pb. 2.25, and in Section 4.7.1 through their defining ODE. In this application, we define the Legendre polynomials through their generating function; show how their definitions through their recursion relation, or through their ODE, can be deduced from their defini- tion through their generating function; and show that they constitute an orthonormal basis for functions defined on the interval 1 d" x d" 1. 1. The generating function for the Legendre polynomials is given by the simple form: " 1 Gx,t) = = (x)tl (7.83) ( l "P 1 - 2xt + t2 l=0 2. The lowest orders of Pl(x) can be obtained from the small t-expan- sion of G(x, t); therefore, expanding Eq. (7.83) to first order in t gives: 1 + xt + �(t2 ) = P0(x) + tP1(x) + �(t2 ) (7.84) from which, we can deduce that: P0(x) = 1 (7.85) P1(x) = x (7.86) 3. By inspection, it is straightforward to verify by substitution that the generating function satisfies the equation: "G (1 - 2xt + t2) + (t - x)G = 0 (7.87) "t � 2001 by CRC Press LLC Because power series can be differentiated term by term, Eq. (7.87) gives: " " (1 - 2xt + t2 ) (x)tl-1 + (t - x) (x)tl = 0 (7.88) l l "lP "P l=0 l=0 Since this equation should hold true for all values of t, this means that all coefficients of any power of t should be zero; therefore: (l + 1)Pl(x) - 2lxPl(x) + (l - 1)Pl-1(x) + Pl-1(x) - xPl(x) = 0 (7.89) or collecting terms, this can be written as: (l + 1)Pl(x) - (2l + 1)xPl(x) + lPl-1(x) = 0 (7.90) This is the recursion relation of Pb. 2.25. 4. By substitution in the explicit expression of the generating function, we can also verify that: "G (1 - 2xt + t2) - tG = 0 (7.91) "x which leads to: " " l (1 - 2xt + t2)- = (7.92) l "dP (x) "P (x)tl+1 0 dx l=0 l=0 Again, looking at the coefficients of the same power of t permits us to obtain another recursion relation: dPl+1(x) dPl(x) dPl-1(x) - 2x + - Pl(x) = 0 (7.93) dx dx dx dPl-1(x) Differentiating Eq. (7.90), we first eliminate and then dx dPl(x) from the resulting equation, and use Eq. (7.93) to obtain dx two new recursion relations: � 2001 by CRC Press LLC dPl+1(x) dPl(x) - x = (l + 1)Pl(x) (7.94) dx dx and dPl(x) dPl-1(x) x -= lPl(x) (7.95) dx dx Adding Eqs. (7.94) and (7.95), we obtain the more symmetric formula: dPl+1(x) dPl-1(x) -= (2l + 1)Pl(x) (7.96) dx dx Replacing l by l 1 in Eq. (7.94) and eliminating Pl2 (x) from Eq. -1 (7.95), we find that: dPl(x) (1 - x2) = lPl-1(x) - lxPl(x) (7.97) dx Differentiating Eq. (7.97) and using Eq. (7.95), we obtain: dPl(x) d �� + l(l + 1)Pl(x) = 0 (7.98a) ��(1 - x2 ) dx �� dx �� which can be written in the equivalent form: d2Pl(x) dPl(x) (1 - x2 ) - 2x + l(l + 1)Pl(x) = 0 (7.98b) dx2 dx which is the ODE for the Legendre polynomial, as previously pointed out in Section 4.7.1. 5. Next, we want to show that if l `" m, we have the orthogonality between any two elements (with different indices) of the basis; that is 1 Pl(x)P (x)dx = 0 (7.99) m +" -1 To show this relation, we multiply Eq. (7.98) on the left by Pm(x) and integrate to obtain: � 2001 by CRC Press LLC 1 �� dPl(x) �� d �� P (x) - x2) + l(l + 1)Pl(x) dx = 0 (7.100) �� m �� +" -1 �� dx �� dx ��(1 �� Integrating the first term by parts, we obtain: 1 dP (x) dPl(x) �� 2 m ��(x - 1) dx dx + l(l + 1)P (x)Pl(x)��dx = 0 (7.101) m +" -1 �� Similarly, we can write the ODE for Pm(x), and multiply on the left by Pl(x); this results in the equation: 1 dPl(x) dP (x) �� m (x2 ( �� - 1) + mm + 1)Pl(x)P (x) dx = 0 (7.102) �� m +" -1 �� dx dx �� Now, subtracting Eq. (7.102) from Eq. (7.101), we obtain: 1 [mm + 1) - l(l + 1)] Pl(x)P (x)dx = 0 (7.103) ( m +" -1 But because l `" m, this can only be satisfied if the integral is zero, which is the result that we are after. 6. Finally, we compute the normalization of the basis functions; that is, compute: 1 Pl(x)Pl(x)dx = Nl2 (7.104) +" -1 From Eq. (7.90), we can write: Pl(x) - (2l - 1)xPl-1(x) + (l - 1)Pl-2(x) = 0 (7.105) If we multiply this equation by (2l + 1)Pl(x) and subtract from it Eq. (7.90), which we multiplied by (2l + 1)Pl 1(x), we obtain: l(2l + 1)Pl2(x) + (2l - 1)(l - 1)Pl-1(x)Pl-2(x) (7.106) - (l + 1)(2l - 1)Pl-1(x)Pl+1(x) - l(2l - 1)Pl2 (x) = 0 -1 Now integrate over the interval [ 1, 1] and using Eq. (7.103), we obtain, for l = 2, 3, & : � 2001 by CRC Press LLC 1 1 (2l - 1) Pl2(x)dx = Pl2 (x)dx (7.107) -1 +"+" (2l + 1) -1 -1 Repeated applications of this formula and the use of Eq. (7.86) yields: 1 1 3 2 Pl2(x)dx = P12(x)dx = (7.108) +"+" (2l + 1) -1 (2l + 1) -1 Direct calculations show that this is also valid for l = 0 and l = 1. Therefore, the orthonormal basis functions are given by: 1 ul = l + Pl(x) (7.109) 2 The general theorem that summarizes the decomposition of a function into the Legendre polynomials basis states: THEOREM If the real function f(x) defined over the interval [ 1, 1] is piecewise smooth and if the 1 2 integral f (x)dx <", then the series: +" -1 " f(x) = Pl(x) (7.110) l "c l=0 where 1 1 ��l �� cl = + f(x)Pl(x)dx (7.111) �� +" 2�� -1 converges to f(x) at every continuity point of the function. The proof of this theorem is not given here. Example 7.10 Find the decomposition into Legendre polynomials of the following function: �� 0 for - 1 d" x d" a f(x) = (7.112) �� 1 for a < x d" 1 �� 2001 by CRC Press LLC Solution: The conditions for the above theorem are satisfied, and 1 1 ��l �� cl = + Pl(x)dx (7.113) �� +" 2�� a From Eq. (7.96), and noting that Pl(1) = 1, we find that: 1 c0 = (1 - a) (7.114) 2 and 1 cl =- [Pl+1(a) - Pl-1(a)] (7.115) 2 We show in Figure 7.4 the sum of the truncated decomposition for Example 7.10 for different values of lmax. FIGURE 7.4 The plot of the truncated Legendre polynomials expansion of the discontinuous function given by Eq. (7.112), for a = 0.25. Top panel: lmax = 4. Middle panel: lmax = 8. Bottom panel: lmax = 16. � 2001 by CRC Press LLC 7.9 MATLAB Commands Review ' Transposition (i.e., for vectors with real components, this changes a row into a column). norm Computes the Euclidean length of a vector. cross Calculates the cross product of two 3-D vectors. det Determinant; used here to compute the triple scalar product. � 2001 by CRC Press LLC

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