Mathematics 1EM/1ES/1FM/1FS Notes, weeks 18-23

Transcription

1 2 MATRICES Mathematics EM/ES/FM/FS Notes, weeks 8-2 Carl Dettmann, version May 2, 22 2 Matrices 2 Basic concepts See: AJ Sadler, DWS Thorning, Understanding Pure Mathematics, pp 59ff In mathematics, a matrix (plural: matrices is a rectangular table of numbers the table Sold tickets Mon Wed Fri Sat Sun Adults Children Free tickets 2 can also be summarised by the following matrix: A For example, the data of This matrix has three rows and five columns and is therefore said to be a 5 matrix, or a matrix of order 5 Depending on how we interpret the elements of a matrix, it might be useful to be able to add, subtract or multiply matrices etc Multiplication of a matrix by a scalar s: a a n s a m a mn sa sa n sa m sa mn Addition and subtraction of two matrices: One matrix can be added to (subtracted from another one only if they have the same number of rows and columns a a n b b n a ± b a n ± b n ± a m a mn b m b mn a m ± b m a mn ± b mn Addition of matrices is commutative A + B B + A, associative A + (B + C (A + B + C, and has distributive laws with multiplication by a scalar c(a + B ca + cb, (c + da ca + da Multiplication of two matrices: Matrices can only be multiplied if the number of columns in the first matrix equals the number of rows in the second matrix Matrix multiplication is not commutative (ie AB BA, except in special cases It is however associative: (ABC A(BC and satisfies expected properties in conjunction with addition, and with multiplication by a scalar: A(B + C AB + AC, (A + BC AC + BC, c(ab (cab A(cB a a 2 a n b b 2 b r a b + + a n b n a b r + + a n b nr a 2 a 22 a 2n b 2 b 22 b 2r a 2 b + + a 2n b n a 2 b r + + a 2n b nr a m a m2 a mn b n b n2 b nr a m b + + a mn b n a m b r + + a mn b nr The transpose of a matrix A is written A T and is found by interchanging the rows and columns Page University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

2 2 Basic concepts 2 MATRICES Examples: ( Let A ( 6, B 2 ( A + B ( 6 2 2B A 2 2 ( ( AB ( 6 4 BA 2 ( 6 + 2, C ( ( ( 6 2 ( 2 2 ( 5 2 ( 9 ( ( 5 5 As AB BA, we see that matrix multiplication is indeed not necessarily commutative ( 2 5 C T 2 The determinant of a 2 2 matrix is obtained as follows: ( a b If A, then c d det A ad bc Other ways in order to notate the determinant are: A or a c Geometrically the absolute value of the ( determinant of a( 2 2 matrix A, det A, gives us the area of the a b parallelogram based on the vectors v and v c 2 d b d The determinant of a matrix can be obtained as follows: a a 2 a b b 2 b c c 2 c a (b 2 c b c 2 a 2 (b c b x + a (b c 2 b 2 c Geometrically the absolute value ofthe determinant of a matrix is the volume of the parallelepiped based a b c on the vectors v a 2, v 2 b 2 and v c 2 a b The determinant also gives a compact method of defining the cross product: c Page 2 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

3 2 Basic concepts 2 MATRICES i j k b b 2 b c c 2 c Determinants also have the following properties: (b 2c b c 2 i (b c b c j + (b c 2 b 2 c k v 2 v AB A B A T A so the determinant is like the magnitude of a matrix An n n matrix I of the form is called identity matrix (unit matrix It is like the number one, satisfying AI IA A for any n n matrix A The inverse of a matrix A is written A and is such that AA A A I For a 2 ( 2 matrix the inverse matrix can be determined very easily: a b If A then c d ( A d b ad bc c a ie to determine the inverse of a matrix A, interchange the elements of the leading diagonal of A, change the signs of the elements of the other diagonal and divide by the determinant of the matrix NB: If the determinant of a matrix is zero, the inverse cannot be determined because n n matrix with determinant zero is called a singular matrix as it has no inverse Examples: is meaningless Any Evaluate Thus the area A of the parallelogram based on the vectors ( 5 and ( is: A 5 square units Page University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

4 22 Solving systems of linear equations using matrices 2 MATRICES 2 Evaluate 2 2 ( ( ( 2 2 ( ( 4 ( 4 9 Thus the volume V of the parallelepiped based on the vectors V 9 cubic units ( Let A 2 4 (a Calculate A ( A 4 ( (b Show that AA I and A A I ( ( AA ( A 4 A 2 2 ( 2 4 ( ( 2, and ( ( Solving systems of linear equations using matrices ( ( would be: I I A system of linear equations is a set of equations with n equations and n unknowns (x, x 2,, x n, and is of the following form: a x + a 2 x a n x n b a 2 x + a 22 x a 2n x n b 2 a n x + a n2 x a nn x n b n In matrix form the system of equations above can be written as: a a 2 a n x a 2 a 22 a 2n x 2 a n a n2 a nn x n a a 2 a n x b a 2 a 22 a 2n Let A, x x 2 and b b 2 a n a n2 a nn Then a simpler way to rewrite the system of linear equations is: Ax b x n b n b b 2 b n Page 4 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

5 22 Solving systems of linear equations using matrices 2 MATRICES Hence, whenever we can evaluate A, we can use this inverse matrix in order to solve these simultaneous linear equations, because: Examples: Ax b A Ax A b Ix A b x A b Solve the following simultaneous equations by matrix methods (Compare with the first exercise in chapter 72 I: 2x + y 7 II: x 2y 7 First we must express the equations in matrix form: ( ( ( 2 x 7 2 y 7 ( 2 Then we premultiply both sides by the inverse of : 2 ( ( ( 2 2 x 2 ( y 2 ( 2 ( ( x ( 2 y 7 7 ( ( x y ( 2 2 ( 7 7 Thus x and y 2 Solve the equations I: 2x y + z 5 II: x y + 2z 2 III: 2x + y + 4z One way of solving this system of linear equations is to eliminate the x s in two equations and then use these two equations to eliminate y: I: 2x y + z 5 II*I-2II: 5y z III*I-III: 2y z 8 I: 2x y + z 5 II*: 5y z III**2II*+5III*: 2z 42 Row III** gives: z 2 Using this value in row II* gives: 5y ( 2, ie, y Finally, using these values in row I gives: 2x ( + ( 2 5, ie, x Page 5 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

6 22 Solving systems of linear equations using matrices 2 MATRICES The same ideas are used when we solve this exercise in matrix form We express the equations in matrix form: 2 2 x 5 y z We solve this matrix equation by reducing the matrix to a triangular matrix 2 x 5 5 y 2 z x y z 5 42 Row : 2z 42, ie z 2 Row 2: 5y ( 2, ie y Row : 2x ( + ( 2 5, ie x The geometrical interpretation of the solution x, y and z 2 is that the three planes defined by the three equations have a common point, (,, 2 In this course, we will only consider systems of equations that have a unique solution If you are interested in special cases, consult: AJ Sadler, DWS Thorning, Understanding Pure Mathematics, pp 442ff The operations, called elementary row operations that we can do are just those that we could do with equations: Exchange two rows 2 Multiply all elements of a row by a non-zero number Add two rows In these calculations, we distinguish between matrices that have zeros below the diagonal, upper triangular matrices and those with zeros both above and below the diagonal, diagonal matrices To solve the linear equations, we reduced the matrix to upper triangular form, then did back substitution We could also reduce the matrix completely to diagonal form, then divide by the relevant entries to get the complete solution Each of the above elementary row operations is equivalent to multiplying on the left by an elementary matrix, for example the matrices (, (, ( applied to a 2 2 matrix (a interchange the rows, (b multiply the first row by, and (c add the first row to the second row, respectively Just as in the case of 2 2 matrices (see p 5, one could also solve the previous system of equations by determining the inverse of the corresponding matrix There are many ways to evaluate the inverse A of a matrix A The easiest one is probably to row reduce the matrix to diagonal form, then divide each row by the amount required to leave the unit matrix, to the matrix (A I until we obtain a matrix of the form (I B Each row operation is equivalent to multiplying on the left by an elementary matrix, so the combination of the row operations that converted A to I, must correspond to A and also convert I to A Thus the matrix B Page 6 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

7 22 Solving systems of linear equations using matrices 2 MATRICES obtained by this process is the inverse of A, ie A B Examples: Find the inverse A of the matrix A This is then (A I: Step is to apply the elimination method such that we get in the first column in rows 2 and : ˆ We leave the first row like it is ˆ We get the new second row by adding the first row to the second row ˆ We get the new third row by adding 2 times the first row to the third row Step 2 is to apply the elimination method such that we get in the second column in rows and : ˆ We leave the second row like it is ˆ We get the new first row by subtracting the second row from the first row ˆ We get the new third row by adding 5 times the second row to the third row Step is to apply the elimination method such that we get in the third column in rows and 2: ˆ We leave the third row like it is ˆ We get the new first row by adding times the third row to 25 times the first row ˆ We get the new second row by subtracting the third row from 5 times the second row Step 4 is to divide the rows by numbers such that we get s in the main diagonal As always, the main diagonal starts in the left upper corner of the matrix, and in this case it ends with the entry in the third row and third column Page 7 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

8 22 Solving systems of linear equations using matrices 2 MATRICES Thus: 2 Find the inverse A of the matrix A This is (A I: A Step is to apply the elimination method such that we get in the first column in rows 2 and : ˆ We leave the first row like it is ˆ We get the new second row by adding 2 times the first row to the second row ˆ We get the new third row by adding times the first row to the third row Before we can do step 2, we have to solve the problem that we have in the second column in row 2 It is allowed to exchange rows (but not columns!, so we exchange rows 2 and Step 2 is to apply the elimination method such that we get in the second column in rows and : ˆ We leave the second row like it is ˆ We get the new first row by adding the second row to 2 times the first row ˆ Here we can leave the third row like it is, as we already have in the second column in row Step is to apply the elimination method such that we get in the third column in rows and 2: ˆ We leave the third row like it is ˆ We get the new first row by adding the third row to times the first row ˆ We get the new second row by adding the third row to times the second row Page 8 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

9 2 Transformations in the plane 2 MATRICES Step 4 is to divide the rows by numbers such that we get s in the main diagonal Thus: A To summarise: We can use row reduction of (A I to compute the inverse of a matrix A In order to solve systems of equations Ax b, we can either Row reduce (A b to upper triangular form, write the result as equations, and back-substitute 2 Row reduce (A b completely, turning A into I and b into the solution Find A by the formula (2 2 or by row reduction (, then the solution is x A b The first method is probably fastest However the others are useful if we need to solve the same set of equations with different b, or we need to find A anyway 2 Transformations in the plane See also: AJ Sadler, DWS Thorning, Understanding Pure Mathematics, pp 6ff Matrices cannot only be used in order to solve systems of linear equations, they can also be used to perform transformations A transformation of a plane takes any point A in the plane and maps it( onto one and only one image A, x also lying in the plane We say that the point A(x, y with position vector has an image A y (x, y with ( x position vector y under the transformation The basic transformations in the plane are: enlargement (dilatation, reflection, rotation, shear and translation In this course, we will only study linear transformations, ie pair of equations of the form I: x ax + by II: y cx + dy transformations that can be expressed by a or, writing this in matrix form ( ( ( x a b x y c d y Thus, every linear transformation of the plane has an associated 2 2 matrix ( r Under a translation by the vector, the transformation equations are s I: x x + r II: y y + s Page 9 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

10 2 Transformations in the plane 2 MATRICES Thus a translation is not a linear transformation Here are two basic facts about linear transformations and their matrices: ( ( ˆ The associated 2 2 matrix can be determined by examining the images of and ( ( ( ( ( a b a b If and, then the transformation matrix is c d c d ˆ Under any linear transformation, the origin (, maps onto itself Examples: Find the matrices corresponding to each of the following linear transformations and consider the effect of each transformation on the triangle ABC, with A(,, B(, and C(, 2: (a a rotation of 9 anticlockwise about the origin (b a reflection in the x-axis (c a reflection in the line y x (d an enlargement by the scale factor and with the centre at (, (e a stretch ( 2 parallel to the y-axis, x-axis fixed (f a shear with x-axis fixed and (, (, (a ( ( ( ( Under this transformation and ( Thus the required matrix is We get the coordinates of the position vectors of the points A, B and C by evaluating the following multiplications: ( ( ( ˆ OA ( ( ( ˆ OB ( ( ( 2 ˆ OC 2 (b ( ( ( ( Under this transformation and ( Thus the required matrix is We get the coordinates of the position vectors of the points A, B and C by evaluating the following multiplications: Page University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

11 2 Transformations in the plane 2 MATRICES ( ˆ OA ( ˆ OB ˆ OC ( ( ( ( 2 ( ( ( 2 (c (d ( ( ( ( Under this transformation and ( Thus the required matrix is We get the coordinates of the position vectors of the points A, B and C by evaluating the following multiplications: ( ( ( ˆ OA ( ( ( ˆ OB ( ( ( 2 ˆ OC 2 ( ( ( ( Under this transformation and ( Thus the required matrix is Page University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

12 2 Transformations in the plane 2 MATRICES (e (f We get the coordinates of the position vectors of the points A, B and C by evaluating the following multiplications: ( ( ( ˆ OA ( ( ( 9 ˆ OB ( ( ( ˆ OC 2 6 ( ( ( ( Under this transformation and ( 2 Thus the required matrix is 2 We get the coordinates of the position vectors of the points A, B and C by evaluating the following multiplications: ( ( ( ˆ OA 2 2 ( ( ( ˆ OB 2 2 ( ( ( ˆ OC ( ( ( ( Under this transformation and ( Thus the required matrix is We get the coordinates of the position vectors of the points A, B and C by evaluating the following multiplications: ( ( ( 2 ˆ OA ( ( ( 4 ˆ OB Page 2 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

13 2 Transformations in the plane 2 MATRICES ( ˆ OC ( ( 2 2 ( 2 2 A linear transformation T has matrix Find (a the image of the point (2, under T, (b the coordinates of the point having an image of (7, 2 under T If (x, y is the image of (x, y then ( ( x 2 y ( x y (a In this case (x, y (2, Hence: ( x y ( x y ( 2 ( 5 ( 2 Thus (, 5 is the image of the point (2, under T (b In this case (x, y (7, 2 Hence: ( ( ( 7 2 x 2 y ( 2 Premultiplying both sides of this equation by the inverse of : ( ( ( ( 7 x 2 2 y ( ( 9 x y Hence, x and y Thus (, is the point having an image of (7, 2 under T Page University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

14 2 Transformations in the plane 2 MATRICES Find the 2 2 matrix that will transform the point (, 2 to (, and the point (, to (, ( a b Let the matrix be c d Then ( ( ( ( ( ( a b a b and c d 2 c d We can rewrite this into the following four equations: *I: a + 2b **II: c + 2d III: a + b IV: c + d I+III: b b *I: a + 2 a II+IV: d 6 d 2 Thus the required matrix is: ( 2 **II: c c The following facts regarding transformation matrices should also be noted: Let T be the matrix that corresponds to a certain linear transformation, let S be the image of a shape S under this transformation, and let A S and A S be the areas of the shapes S and S respectively Then Page 4 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

15 24 Linear transformations in -D space 2 MATRICES ˆ the inverse matrix T maps S back onto S ˆ det T gives the scale factor for any change of the area A S under the transformation, ie A S det T A S If det T, ie if T is a singular matrix, then T maps all points in the plane onto a line ˆ If the matrix P transforms (x, y to (x, y and the matrix Q transforms (x, y to (x, y, then the matrix given by the product QP will transform (x, y directly to (x, y ( 2 Exercise: Prove that the transformation matrix maps all points of the x-y plane onto a straight line 2 4 and find the equation of that line We evaluate the determinant of the transformation matrix: Thus, all points in the plane will be mapped onto a line We consider: ( ( ( ( ( x 2 x x + 2y y 2 4 y 2x + 4y Hence, all image points lie on the line y 2x 24 Linear transformations in -D space x + 2y 2(x + 2y See also: AJ Sadler, DWS Thorning, Understanding Pure Mathematics, pp 4ff In the previous chapter we saw that a linear transformation in 2-dimensional space has an associated 2 2 matrix Similarly any linear transformation in -dimensional space has an associated matrix If under this transformation the points with position vectors, and are transformed to the points with Page 5 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

16 24 Linear transformations in -D space 2 MATRICES position vectors a b c, d e f and g h i respectively, then the associated matrix is As in 2-D space, the origin (,, is mapped onto itself by a linear transformation in -D space: a d g b e h c f i Examples: Find the matrix representing the following linear transformations in -D space: reflection in the x-y plane (ie the plane z, 2 9 rotation about the x-axis such that the positive y-axis maps onto the positive z-axis, reflection in the y + x plane, 4 stretch parallel to the x-axis, scale factor 2, with y-z plane fixed, 5 orthogonal projection onto the x-z plane a d g b e h c f i 2 Under this transformation Thus the required matrix is, and Under this transformation Thus the required matrix is, and Page 6 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

17 24 Linear transformations in -D space 2 MATRICES Under this transformation Thus the required matrix is, and 4 Under this transformation 2, 2 Thus the required matrix is and 5 This projection maps any point A(a, b, c onto the point A (a,, c Thus, and and the required matrix is The following facts regarding transformation matrices should be noted (compare with the properties of 2 2 transformation matrices, p 52: Let T be the matrix that corresponds to a certain linear transformation, let S be the image of a solid S under this transformation, and let V S and V S be the volumes of the solids S and S respectively Then ˆ the inverse matrix T maps S back onto S ˆ det T gives the scale factor for any change of the volume V S under the transformation, ie V S det T V S If det T, ie if T is a singular matrix, then T maps all points in -D space either onto a line or onto a plane (as in example 5 above ˆ If the matrix P transforms (x, y, z to (x, y, z and the matrix Q transforms (x, y, z to (x, y, z, then the matrix given by the product QP will transform (x, y, z directly to (x, y, z Examples: Show that the matrix M cartesian equations of this line maps all points in -D space onto a line and find the Page 7 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

18 24 Linear transformations in -D space 2 MATRICES det M ( ( ( 6 ( ( 2 ( ( ( Thus M maps all points in -D space either onto a line or onto a plane In order to find out whether the image of the transformation is a line or a plane, we consider: x x x + y + z y y 2x 2y 6z z z 4x + 4y + 2z We see that: x 2 y and x 4 z Thus this transformation maps all points onto the line with cartesian equations x 2 y 4 z Another way to solve this exercise can be found in: AJ Sadler, DWS Thorning, Understanding Pure Mathematics, p Show that the matrix M 7 maps all points in -D space onto a plane and find the 2 cartesian equation of this plane det M + ( ( ( 2 ( 2 ( 7 ( 9 + ( ( 2 Thus M maps all points in -D space either onto a line or onto a plane In order to find out whether the image of the transformation is a line or a plane, we consider again: x y 2 7 x x 2y y x + y + 7z z 2 z 2x y + z As the formulas for x, y and z are not multiples of each other, we know that M maps all points in -D space onto a plane In order to find the equation of this plane we need the coordinates of the images of three points that do not all lie on the same straight line Just by looking at the matrix, we know already that under this transformation the points A(,,, B(,, and C(,, are mapped onto A (,, 2, B ( 2,, and C (, 7, So, we have to find the plane which is determined by A, B and C However, to make the calculation easier, it is better to replace one of these images by the image of the origin O which is always O (,, under a linear transformation So, let s replace C by O We have: O A 2 and O B Thus the parametric vector equation of the required plane is: r + λ + µ Page 8 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

19 25 Eigenvectors and eigenvalues 2 MATRICES **I: x λ 2µ II: y λ + µ *III: z 2λ µ II+III: y + z 5λ λ 5 y + 5 z *III: µ 2 5 y z z µ 2 5 y + 5 z **I: x 5 y z 4 5 y 2 5 z x 5 y z Thus 5x + y 7z is the cartesian equation of the required plane Another way to solve this exercise can be found in: Pure Mathematics, p 44 AJ Sadler, DWS Thorning, Understanding 25 Eigenvectors and eigenvalues In this section, we are interested in straight lines passing through the origin that are mapped onto themselves In 2-D space, a straight line through the origin can be described by the following parametric vector equation: ( ( ( x v L : + s y v 2 And similar in -D space: L : x y z + s Hence, in general a straight line through the origin can be described by: L : r sv Let A be the matrix of any linear transformation How can we find a straight line L passing through the origin that is mapped onto itself under this linear transformation? In other words: How can we find the corresponding vector v of any such line? Let P be a point on such a special straight line L where P is distinct from the origin: then the position vector p of any such point P can be expressed as p s v where s is a constant number Since P is lying on L, its image P must lie on L as well Therefore, it must be possible to find another constant number s 2 such that the position vector p of P can be expressed as: p s 2 v v v 2 v Page 9 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

20 25 Eigenvectors and eigenvalues 2 MATRICES Thus we get: Let λ s2 s, then: Ap p A (s v s 2 v s Av s 2 v Av s 2 s v Av λv Av λv (A λiv Our goal is thus to find those vectors v which satisfy this last equation If the inverse of the matrix (A λi existed, then the (only solution of the equation would be the zero vector, but then both v and p would have to be and L could not be a straight line That s why we will only get a proper solution if (A λi is not invertible, ie, if det(a λi The values for λ that solve this equation are called eigenvalues and the corresponding vectors v which then can be evaluated and which determine the straight lines through the origin that are mapped onto themselves are called eigenvectors Exercise: Find the equations of any lines ( passing through the origin that are mapped onto themselves by the transformation defined by the matrix 4 2 As a first step, we have to find the eigenvalues of this matrix: 4 λ 2 λ Thus the eigenvalues are λ 5 and λ Now, we want to find the vectors v (4 λ(2 λ 8 6λ + λ 2 ( v v 2 λ 2 6λ + 5 λ 6 ± λ 5 or λ that solve for either eigenvalue the following equation: ( 4 λ 2 λ ( v v 2 For λ 5 we get: ( ( 4 5 v 2 5 v 2 ( ( v v 2 Page 2 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

21 FUNCTIONS OF TWO VARIABLES I: v + v 2 II: v v 2 As one equation ( is a multiple of the other one, it is not possible to find a unique solution for v and v 2 All v vectors v with v v v 2 are eigenvectors for this transformation and can be used in order to describe 2 the first straight line ( L that passes through the origin and is mapped onto itself If we choose v, we can eg write L as: ( L : r s For λ we get: ( ( 4 v 2 v 2 ( ( v v 2 I: v + v 2 II: v + v 2 As again one equation ( is a multiple of the other one, it is not possible to find a unique solution for v and v 2 v All vectors v with v v 2 v are also eigenvectors for this transformation and can be used in order 2 to describe the second ( straight line L 2 that passes through the origin and is mapped onto itself If we choose v, we can thus write L 2 as: ( L 2 : r t Examples like this one can be solved in a similar way in -D space, but we won t cover that in this lecture Functions of two variables Surfaces and contour lines So far we have dealt with functions of a single variable But, in the real world, physical quantities often depend on two or more variables In this section, we discuss functions of two variables A function f of two variables is a rule that assigns to each ordered pair (x, y of real numbers in a set D a unique real number z, written as z f(x, y The set D is the domain of f and the set of all elements f(x, y is called the range of f The variables x and y are independent variables and z is the dependent variable Like before, it is possible to visualise the behaviour of a function by considering its graph We know that the graph of a function f of one variable is a curve C with equation y f(x The graph of a function f of two variables is a surface S with equation z f(x, y As it is often not easy to draw D-graphs by hand, we will always use Maple in this course in order to visualise the behaviour of a function of two variables Examples: Page 2 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

22 Surfaces and contour lines FUNCTIONS OF TWO VARIABLES Plot the graph of the function f(x, y 6 x 2y Maple: plotd(6 x 2y,x 55,y 55,axesboxed; The function in this example is a special case of a function f(x, y ax + by + c which is called a linear function The graph of such a function has the equation z ax + by + c, or ax + by z + c, so it is a plane 2 Plot the graph of the function g(x, y 9 x 2 y 2 Maple: plotd(sqrt(9 x 2 y 2,x,y,axesnormal,grid[,]; Plot the graph of the function h(x, y 4x 2 + y 2 Maple: plotd(4x 2 + y 2,x,y,axesnormal; Once Maple has created a surface image, it is possible to explore various options for axes, shading, contours and so on by clicking on the right mouse-button By holding down a left click on the image it is possible to rotate it in any direction Page 22 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

23 Surfaces and contour lines FUNCTIONS OF TWO VARIABLES Another method for visualising functions, borrowed from mapmakers, is a contour map on which points of constant elevation are joined to form contour curves, also called contour lines or level curves The level curves of a function f of two variables are the curves with equations f(x, y c, where c is a constant (in the range of f One common example of level curves occurs in topographic maps of mountainous regions The level curves are curves of constant elevation above sea level If you walk along one of these contour lines you neither ascend nor descend Examples: Sketch the level curves of the function z f(x, y 6 x 2y for the values c 6,, 6, 2 The level curves are 6 x 2y c or y 2 x (c 6 2 This is a family of straight lines with slope 2 The four particular level curves with c 6,, 6, 2 are: y 2 x + 6 y 2 x + y 2 x y 2 x Page 2 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

24 2 Partial derivatives FUNCTIONS OF TWO VARIABLES The level curves are equally spaced parallel lines because the graph of f is, as we have seen before, a plane 2 Given the function z g(x, y 9 x 2 y 2, plot and label the contour lines where g(x, y takes the values c, c 2, c 2 and c 4 In general for any constant c, the level curves are 9 x2 y 2 c or x 2 + y 2 9 c 2 This is a family of concentric circles with center (, and radius 9 c 2 The four particular contour lines where c, c 2, c 2 and c 4 are: x 2 + y 2 9 x 2 + y 2 8 x 2 + y 2 5 x 2 + y 2 2 Partial derivatives 2 First partial derivatives Let f(x, y be a function of two variables Now keep y constant, and consider the limit: f(x + h, y f(x, y lim h h If this limit exists, its value is the derivative of f with respect to x at (x, y; it s called the (first partial derivative of f with respect to x Page 24 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

25 2 Partial derivatives FUNCTIONS OF TWO VARIABLES If z f(x, y, we can use either of these notations for the partial derivative with respect to x: f x (x, y, f x, f x, z f(x, y or x x Similarly, we can keep x constant and consider: f(x, y + h f(x, y lim h h If this limit exists, it s called the (first partial derivative of f with respect to y Similar notations, like those above, can be used Rule for finding partial derivatives of z f(x, y: ˆ To find f x, regard y as a constant and differentiate f(x, y with respect to x ˆ To find f y, regard x as a constant and differentiate f(x, y with respect to y Examples: Find the partial derivatives z x (a z x 2y 4 (b z 5x ln y + cos x and z y for z x z y 2 If f(x, y x + x 2 y 2y 2, find f x (2, and f y (2, At first, we have to evaluate f x (x, y and f y (x, y: z x z y 8y 5 ln y sin x 5x y 5x y f x (x, y x 2 + 2xy f y (x, y x 2 y 2 4y Now, we can evaluate both partial derivatives at the point (2, : Geometric interpretation: f x (2, f y (2, Suppose the graph of z f(x, y is the surface shown below Consider the partial derivative of f with respect to x at a point (x, y Holding the y-value constant and varying x, we trace out a curve that is the intersection of the surface with the vertical plane y y The partial derivative f x (x, y measures the change in z per unit increase in x along this curve That is, f x (x, y is just the slope of the curve at (x, y The geometrical interpretation of f y (x, y is analogous Similarly, contour plots can also be related to partial derivatives For example, a vertical contour at a point means that the function has zero y derivative at that point If the contours correspond to evenly spaced values of the function (eg equally spaced c, then closely spaced contours in a region are associated with large values of one or both of the partial derivatives, or in the case of a topographic map, that the terrain is very steep Page 25 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

26 Maxima and minima FUNCTIONS OF TWO VARIABLES 22 Second partial derivatives If f is function of two variables, then its partial derivatives f x and f y are also functions of two variables, so we can consider their partial derivatives (f x x, (f x y, (f y x, and (f y y, which are called the second partial derivatives of f If z f(x, y, we use the following notations: (f x x f xx ( f x x (f x y f xy ( f y x (f y x f yx ( f x y (f y y f yy ( f y y Exercise: Find the second partial derivatives of From before we already know that: 2 f x 2 2 z x 2 2 f y x 2 f x y 2 f y 2 2 z y 2 f(x, y x + x 2 y 2y 2 2 z y x 2 z x y f x (x, y x 2 + 2xy and f y (x, y x 2 y 2 4y Therefore: f xx x (x2 + 2xy 6x + 2y f xy y (x2 + 2xy 6xy 2 f yx x (x2 y 2 4y 6xy 2 f yy y (x2 y 2 4y 6x 2 y 4 Notice that f xy f yx This is not just a coincidence It turns out that the mixed partial derivatives f xy and f yx are equal for most functions that one meets in practice We won t discuss that in detail in this lecture Maxima and minima We have already studied how it is possible to determine maximum and minimum points of functions of a single variable We will see in this chapter that it works in a similar way for functions with two variables A point (a, b is called a critical point (or stationary point of f if f x (a, b and f y (a, b, or if one of these partial derivatives does not exist At a critical point, a function can have a local maximum or a local minimum or neither Page 26 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

27 Maxima and minima FUNCTIONS OF TWO VARIABLES By checking the second partial derivatives, it is (most of the time possible to decide whether a critical point is a local maximum or local minimum or neither Second Derivatives Test: Suppose the second partial derivatives of f are continuous at (a, b, and suppose that f x (a, b and f y (a, b Let D D(a, b f xx (a, bf yy (a, b [f xy (a, b] 2 If D > and f xx (a, b >, then f attains a local minimum at (a, b 2 If D > and f xx (a, b <, then f attains a local maximum at (a, b If D <, then f has a saddle point at (a, b 4 If D, then no conclusion can be drawn (any of these behaviours is possible To remember the formula for D it s helpful to write it as a determinant: D f xx f xy f xxf yy (f xy 2 f yx f yy Exercise: Find at which points the function f(x, y x 4 + y 4 4xy + has local maximum and local minimum values, and which points are saddle points We first locate the critical points: f x 4x 4y f y 4y 4x Setting these partial derivatives equal to, we obtain the equations I: x y II: y x To solve these equations we substitute y x from the first equation into the second one This gives: x 9 x x(x 8 x(x 4 (x 4 + x(x 2 (x 2 + (x 4 + So there are three real roots: x,, By substituting these values into the first equation we get the corresponding y-values Thus the three critical points are (,, (,, and (, Next we calculate the second partial derivatives and D(x, y: f xx 2x 2 f xy 4 f yy 2y 2 D(x, y f xx f yy (f xy 2 44x 2 y 2 6 Since D(, 6 <, according to the Second Derivatives Test, the origin is a saddle point Since D(, 28 > and f xx (, 2 >, there is a local minimum at (,, which is f(, Similarly, we have D(, 28 > and f xx (, 2 >, so there is also a local minimum at (,, which is also f(, In order to visualise these results, we can plot the function using Maple: Page 27 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

28 4 DIFFERENTIAL EQUATIONS 4 Differential equations See also: AJ Sadler, DWS Thorning, Understanding Pure Mathematics, pp 59ff When we solve an algebraic (eg linear or quadratic equation involving a variable x, a solution is a real or complex number If we substitute a solution for x, the equation is satisfied Some equations have no solutions, some have one and some have more For example, the solutions of x 2 4 are 2 and 2 Substituting either of these solutions gives the equation 4 4 A differential equation is an equation involving a function y f(x that contains derivatives A solution of a differential equation is a function that when substituted, leads to an equation satisfied for all x For example, y 2y/x is a differential equation If we substitute y x 2 we get 2x 2x, which is an equation valid for all x So this is a solution of the differential equation y x 2 gives the equation 6x 6x, so this is another solution We can see that y cx 2 for any constant c is a solution; in fact it turns out that these are the only solutions of this equation The order of a differential equation is the order of the highest derivative that appears in the equation Examples: ˆ x y ln x sin y is a first order differential equation ˆ (y 4 + y + 2y x + 5x 7 is a second order differential equation Of course we could use variables other than x and y to write down differential equations Often we want to describe how quantities change with time, so we would use t in place of x 4 First order differential equations Without realising it, we have already studied how to solve first order differential equations of the form y f(x Solving for y is easy We just need to integrate both sides: y f(xdx This gives the family of solutions, also called general solution, y F (x + c for the differential equation, where c is an arbitrary constant of integration Page 28 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

29 4 First order differential equations 4 DIFFERENTIAL EQUATIONS If an initial condition is given along with the differential equation, that is, a constraint of the form y y when x x, then this information can be used to determine the particular value of c In this way, one particular solution can be selected from the family of solutions the one that satisfies both the differential equation and the initial condition Examples: Find the general solution of the differential equation y 2x y (2x dx 2 x2 2 x + c x2 x + c If we were to draw the graphs of y x 2 x + c for various values of c, we would obtain a family of curves, each of which has the property that y 2x 2 Solve the differential equation y + 4 x 2 given that when x, y 2 First, we find the general solution: y + 4 x 2 y x 2 4 y (x 2 4dx But when x, y 2, thus: y x 4x + c y x 4x + c c c 5 Hence y x 4x + 5 is the required solution In this course we will only study one other type of first order differential equation, equations that can be expressed in the form y f(x g(y Differential equations of this type can be solved by separating the variables In this case it is helpful to use the notation dy dx for y So we can rewrite the equation above as: dy dx f(x g(y Page 29 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

30 4 First order differential equations 4 DIFFERENTIAL EQUATIONS Now we can put the x s with the dx and the y s with the dy and integrate both sides: g(ydy f(xdx Note that dy dx is merely a useful notation that helps us to remember how differential equations of the type from above can be solved it is NOT an actual fraction Examples: Solve the equation dy dx x(x 2 e y We separate the variables and integrate both sides: e y dy e y dy (x(x 2dx (x 2 2xdx e y x 2 x2 2 + c e y x x 2 + c (Why do we know that e y dy e y? Because we already know that if f(y e y then f (y e y As a last step we solve for y: y ln( x x 2 + c 2 Solve the differential equation y 2 sin y dy dx 2 sin y sin y dy 2 dx cos y 2x + c cos y 2x c y cos ( 2x c Note: As c is an arbitrary constant which can take either positive or negative values, we lose no generality by writing this solution as y cos (c 2x Solve the equation y y 2 subject to y( Again, we can rearrange the equation to separable form: dy dx y2 y 2 dy dx y x + c y x + c Page University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

31 4 First order differential equations 4 DIFFERENTIAL EQUATIONS Now, use the initial condition to find the constant c: + c c y x x Having studied these two types of differential equations, we are now able to solve real-world questions like in the following exercise Exercise: An animal has mass grams at birth It reaches maturity in months The rate of growth m (t is given by 2(t 2, where m grams is the mass of the animal t months after birth ˆ Find the mass of the animal when fully grown ˆ How long does it take for the animal to reach a mass of 79 grams? At first we find a formula for the mass m(t of the animal after t months This leads us to the general solution m (t 2(t 2 m (t 2(t 2 6t + 9 m (t 2t 2 72t + 8 m(t (2t 2 72t + 8dt m(t 2 t 72 t t + c m(t 4t 6t 2 + 8t + c However, only one equation from this family of solutions is right for this problem It is given that, when t (at birth, m So c must satisfy the equation giving c The mass of the animal after t months is therefore c, m(t 4t 6t 2 + 8t + The mass when fully grown is found by putting t in this formula, giving So the mass of the animal at maturity is grams m( In order to find how long it takes for the animal to reach a mass of 79 grams, we have to solve the following equation: 4t 6t 2 + 8t t 6t 2 + 8t 76 t 9t t 9 (t (t 2 8t + 9 We get only one real solution, t Hence, after month the animal reaches a mass of 79 grams Page University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

32 42 Second order differential equations 4 DIFFERENTIAL EQUATIONS 42 Second order differential equations In this course, we will consider two standard forms of second order differential equations Differential equations of the form y f(x can be solved by integrating with respect to x, twice The first integration gives y and the second integration gives y Exercise: Solve the differential equation d2 y dx 6x 2, given that when x, dy 2 dx 2 and y dy dx (6x 2dx dy dx 6 x2 2 2x + c dy dx x2 2x + c We know that, when x, dy dx 2, thus c must satisfy the equation giving c Thus: Integrating again gives c dy dx x2 2x + y x 2 x2 2 + x + c 2 x x 2 + x + c 2 But when x, y, thus c 2 must satisfy the equation giving c 2 2 Thus the required solution is y x x 2 + x c 2 The second type of second order differential equations we discuss in this course is of the form ay + by + cy f(x If the right-hand side of this equation is equal to zero, that is, if the equation is of the form ay + by + cy then it s said to be homogeneous The general solution of such a homogeneous differential equation can be determined by taking the following steps: By substituting y e px, y pe px and y p 2 e px into the differential equation, we get which leads us to the following quadratic equation: ap 2 e px + bpe px + ce px ap 2 + bp + c Page 2 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

33 42 Second order differential equations 4 DIFFERENTIAL EQUATIONS ˆ If this quadratic equation has real distinct roots p and p 2, the general solution is y Ae px + Be p2x ˆ If this quadratic equation has a repeated root p, the general solution is y (A + Bxe px ˆ If the roots are not real, ie, if the roots are complex conjugates p α + βi and p 2 α βi, then the general solution is y e αx (A cos βx + B sin βx For f(x, we first obtain the general solution of the differential equation with f(x This is called the complementary function, and then we add a particular solution, ie Examples of particular solutions are: ˆ f(x 2x + : Choose Cx + D general solution complementary function + particular solution ˆ f(x x 2 : Choose Cx 2 + Dx + E since the derivatives in the equation will lead to the lower degree terms ˆ f(x e 2x : Choose Ce 2x ˆ f(x sin x: Choose C sin x + D cos x again due to derivatives in the equation ˆ f(x e px ie something from the complementary function: The same function won t work; it turns out we can multiply by x: Choose (Cx + De px In each case, subsitute into the original equation and determine the unknown constants C, D etc Finally, if there are initial conditions, use them to find the constants A and B in the general solution (not the complementary function Examples: Find the general solution to the differential equation y 4y + y By substituting y e px, y pe px and y p 2 e px into the differential equation, we get which leads us to the following quadratic equation: p 2 e px 4pe px + e px p 2 4p + p 4 ± p 4 ± 2 2 p and p 2 As this quadratic equation has real distinct roots, the general solution to the differential equation is y Ae x + Be x Page University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only

34 42 Second order differential equations 4 DIFFERENTIAL EQUATIONS 2 Find the general solution to the differential equation 4 d2 y dx dy + y x + 4 dx and the solution satisfying y 2 and dy dx 4 at x To find the complementary function we solve 4y + 4y + y Substituting y e px, y pe px and y p 2 e px into the differential equation will lead us to the following quadratic equation: 4p 2 + 4p + p 4 ± ie p 2 Thus the complementary function is y (A + Bxe 2 x is a repeated root To find a particular solution we try a solution of the form of the RHS of the differential equation, ie y Cx + D The corresponding derivatives are: y C and y (Note: If the RHS is a quadratic polynomial, we have to try y Cx 2 + Dx + E, and so on Substitution in the differential equation gives: C + (Cx + D x + 4 Cx + (4C + D x + 4 Now we can find C and D by comparing the coefficients of the LHS and of the RHS: I: C II: 4C + D D 4 D 8 Thus a particular solution is y x 8 and the general solution to the differential equation is: y (A + Bxe 2 x + x 8 Finally, we can also find a solution satisfying the initial conditions Differentiating the general solution gives dy dx Be 2 x 2 (Ax + Be 2 x + Substituting x and y 2 gives A Then substituting x and dy dx 4 gives B A/2 + 4, so B hence B 6 So the final solution is y ( + 6xe 2 x + x 8 Page 4 University of Bristol 22 This material is copyright of the University unless explicitly stated otherwise It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only