Lecture Notes for MATH6106. March 25, 2010
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1 Lecture Notes for MATH66 March 25, 2
2 Contents Vectors 4. Points in Space Distance between Points Scalars and Vectors Vectors and Coordinate Systems Vector Operations Revision of Vectors Quiz Topics in 3D Geometry 9 2. Planes Lines Surfaces Functions of several variables Change of Coordinates Partial Derivatives 3 3. First Order Partial Derivatives Higher Order Partial Derivatives Chain Rule Maxima and Minima A little Vector Calculus Gradient Directional Derivative Curl and Divergence Laplacian Linear Equations Introduction Row Echelon Form Soving Systems of Linear Equations Linear Independence and Gram Schmidt Linear Combination of Vectors Linear Independence of Vectors Orthogonal and orthonormal vectors
3 6.4 Gram-Schmidt Process Matrices Introduction Matrix Operations Special matrices Determinants Inverses Eigenvalues, Eigenvectors and Diagonalization Introduction Method to find eigenvalues and eigenvectors Dimension of Eigenspace Diagonalization Diagonalization of symmetric matrices Operators and Commutators 2 9. Operators Linear Operators Composing Operators Commutators Fourier Series 25. Introduction Definition of a Fourier Series Why the coefficients are found as they are Fourier Series Odd and even functions Functions with period 2L Parseval s Result
4 Vectors. Points in Space A point in two dimensional (2D) space is represented on the XY -plane using two coordinates (x, y). Example. Plot P (, 2). 3D Coordinate system The coordinate system in 3D space consists of three axes X,Y and Z. To determine the directions of these axes, we follow the right hand rule: We first define a plane in 3D space as the XY plane. Then place your right hand on the X-axis, curl your fingers in the direction of the Y -axis, and let your thumb stick out. Your thumb then points in the direction of the positive Z axis. A point in three dimensional (3D) space is represented using three coordinates (x, y, z). Example. Plot P (, 2, 3)..2 Distance between Points In 2D-space, the distance between two points P (x, y ) and Q(x 2, y 2 ) is d(p, Q) = (x 2 x ) 2 + (y 2 y ) 2.
5 Similarly in 3D-space, the distance between two points P (x, y, z ) and Q(x 2, y 2, z 2 ) is d(p, Q) = (x 2 x ) 2 + (y 2 y ) 2 + (z 2 z ) 2. Example. Find the distance between P (, 2, 3) and Q(,, ). d(p, Q) = (x 2 x ) 2 + (y 2 y ) 2 = ( ) 2 + ( 2) 2 + ( 3) 2 = 3..3 Scalars and Vectors Certain quantities in physical applications of mathematics are represented by vectors. To understand what a vector is it is perhaps better to first understand what a scalar is: Definition. A scalar quantity is defined completely by a single number, e.g. length, area, volume, mass, time etc. It has only magnitude. Definition. (Physical) A vector quantity is defined by magnitude and direction, e.g. force, velocity, acceleration. Example.. Force is reresented by a vector; the direction of the vector describes the direction in whch the force is applied, and the magnitude of the vector indicates the strength applied. 2. Velocity of a car is represented by a vector; the direction gives the direction of motion and the magnitude gives the speed of the car. Note. Two vectors with the same length and same direction are equal, we do not care about their position in space..4 Vectors and Coordinate Systems Vectors can occur in two, three or even higher dimensions. Definition. (Geometric) A vector in 2D-space can be represented by a directed line segment, with an initial point and a terminal point. We denote a vector by an over head arrow, e.g. v. Then the magnitude of v is given by the length of the directed line segment, and is denotd by v, and the direction of v is given by the direction of the line segment.
6 Example. Plot the vector P Q with initial point P (2, 3) and terminal point Q(, 4). Find the magnitude of the vector P Q. P Q = ( 2) 2 + (4 3) 2 = Example. We can also represent the vector vector OQ and OP (Use diagram.) P Q = OQ OP = (, 4) (2, 3) = ( 3, ) P Q as the difference of the Then the directed line segment from O(, ) to R( 3, ) also represents the vector P Q. Verify that the magnitude of both line segments is the same. OR = ( 3 ) 2 + ( ) 2 = The two vectors also point in the same direction. Definition. (Algebraic) A vector in 2D-space is represented by two components (x, y), indicating the change in the X and Y directions from the origin (, ). Note. When we say a point P ( 3, ) in 2D-space, we mean the point with x-coordiante 3 and y-coordinate. When we say the vector v = ( 3, ) in 2D-space, we mean the directed line segment from the point (, ) to the point ( 3, ), which has magnitude or length ( 3) = and points in the direction of the line segment. Definition. A vector in 3D-space is given by a three tuple (x, y, z), indicating the change in the X, Y and Z directions from the origin. Note. We represent a vector v = (x, y, z) in 3D-space as the directed line segment from the point (,, ) to the point (x, y, z), which has length x2 + y 2 + z 2 and points in the direction of the line segment. Example. Plot the vector v = (, 2, 3) and determine it s length.
7 .5 Vector Operations Addition and Subtraction We add and subtract vectors component-wise. Example. Let v = (,, ) and v 2 = (, 2, 3). Then v + v 2 = (, + 2, + 3) = (, 2, 4) and v v 2 = ( ( ), 2, 3) = (2, 2, 2). Scalar Multiplication If c is a real number and v = (x, y, z), then c v = (cx, cy, cz). This is known as scaling - we change the length but not the direction of v. Special vectors c v = c 2 x 2 + c 2 y 2 + c 2 z 2 = c x 2 + y 2 + z 2 = c v The zero vector: o = (,, ). Definition. A unit vector is a vector with length equal to one. i.e. v =. In 2D space, the unit vectors in the X and Y direction are special and denoted by: i = (, ) j = (, ) Similarly, in 3D space, the unit vectors in the X, Y and Z directions are denoted by: i = (,, ) j = (,, ) k = (,, )
8 Example. Using the special unit vectors i, j, k, we get a different way of representing a vector, Consider v = (3, 6, 8). Then, v = (x, y, z) = x i + y j + z k. v = (3, 6, 8) = 3(,, ) + 6(,, ) 8(,, ) = 3 i + 6 j 8 k This representation is often useful and we will use both notations in the course. Example. Find a unit vector in the direction of the vector v = (6, 8, ). Dot Product Definition. The dot product of 2 vectors a = (a, a 2, a 3 ) and b = (b, b 2, b 3 ) in 3-dimensional space is defined to be the scalar a. b = a b + a 2 b 2 + a 3 b 3. This is the algebraic definition of the dot product On the other hand, if θ is the angle between the vectors a and b, using the Law of Cosines on the triangle with sides a, b, a b we get, a b 2 = a 2 + b 2 2 a. b cos θ (a b ) 2 + (a 2 b 2 ) 2 + (a 3 b 3 ) 2 = a 2 + a a b 2 + b b a. b cos θ a b + a 2 b 2 + a 3 b 3 = a. b cos θ a. b = a. b cos θ Definition. Therefore we could equivalenty define the dot product to be a. b = a. b cos θ. This is the geometric definition of the dot product.
9 Remark. The two main applications of taking dot products; (i) to find the angle between vectors and (ii) to determine if two vectors are perpendicular. Example. Find the dot product of a = (,, 3) and b = (2,, ). Also find the angle between the two vectors. Solution: a. b =.2 + ( )( ) + 3. = 3 a = = 5 b = 4 + = 5 cos θ = a. b a b = 3/5 θ = cos ( 3 5 ) Proposition Let a and b be non-zero vectors. The vectors, a and b, are perpendicular to each other if and only if a b =. Solution: Let a and b be non zero vectors, then a and b. Let θ be the angle between the two vectors. a b = a. b cos θ = cos θ = ; Since a, b, we can divide the equation by them. θ = π/2 Therefore the angle between a and b is 9 degrees. Example. Determine if the vectors a = (,, ) and b = (, 2, ) are perpendicular to each other. a b = ()() + ( )(2) + ()() = Therefore by the above proporsition, the vectors are perpendicular to each other.
10 Cross Product Definition. Given 2 vectors a = (a, a 2, a 3 ) and b = (b, b 2, b 3 ) in 3-dimensional space the cross product a b is a vector defined as follows: a b = ( a. b sin θ) n where n is a unit vector which is perpendicular to the plane containing a and b, determined by the right-hand-rule. i.e. place your right hand on a and curl it towards b, n now points in the direction of your thumb. This is the geometric definition of the cross-product. We also have an algebraic definition of the cross-product, but before we can define it, we need to be able to compute determinants! Definition. The determinant of a 2 2 matrix is defined as follows: a b c d = ad bc. Example. Compute the determinant =.6 3( 5) = 9. Definition. The determinant of a 3 3 matrix is defined as follows: a a 2 a 3 a 2 a 22 a 23 a = a 22 a 23 a 3 a 32 a 33 a 32 a 33 a 2 a 2 a 23 a 3 a 33 + a 3 a 2 a 22 a 3 a 33 = a (a 22 a 33 a 32 a 23 ) a 2 (a 2 a 33 a 3 a 23 ) + a 3 (a 2 a 32 a 3 a 22 ) Example. Compute the determinant = = 4( 5) 2(2) + ()( 52) = 54
11 Definition. Given 2 vectors a = (a, a 2, a 3 ) and b = (b, b 2, b 3 ) in 3-dimensional space the cross product a b can also be defined as follows: a i j k b = a a 2 a 3 b b 2 b 3 where i = (,, ), j = (,, ), k = (,, ) are unit vectors along the X, Y, Z axes respectively. However while calculating the determinant they only have symbolic value. Remark. The cross product operates on 2 vectors and produces a new vector. Example. Consider the two vectors a = (2,, ) and b = (,, ) in 3- dimensional space. Find the cross-product of the two vectors and verify that it is indeed perpendicular to both a and b. Also find the magnitude of a b. Solution: a b = = i j k 2 i 2 = i j + 3 k j + 2 = (,, ) (,, ) + 3(,, ) = (,, 3) k Observe that the two vectors lie in the XY -plane and the cross product lies on the Z-axis and is indeed perpendicular to the plane containing the two vectors. But we can verify this more concretely using the dot product. Let n = a b = (,, 3). n a = (2,, ).(,, 3) = + + =. n b = (,, ).(,, 3) = + + =.
12 Since both the dot products are zero, we can conclude that a b is perpendicular to both a and b. Finally the magnitude of a b is given by, a b = = 3. Remark. The two important properties of cross products are: () The direction of the cross product a b is perpendicular to both a and b. In particular it points in the direction of your thumb if the fingers of your right hand curl from a to b. (2) The magnitude of a b is given by a. b sin θ, if θ is the angle between a and b. It is the area of the parallelogram determined by a and b..6 Revision of Vectors Example. Find a unit vector in the direction of the following vectors: (i) v = (2, 3). (ii) v 2 = (, 2, 5). (iii) v 3 = (,,, ). Solution: To get a unit vector in the direction of a given vector, all we need to do is scale the size of the vector by the inverse of its length. (i) v = (2, 3). The length of v is v = ( 3) 2 = 3. A unit vector u in the direction of v may then be obtained as follows: u = 3 (2, 3) = ( 2 3, 3 3 ). (ii) v 2 = (, 2, 5). The length of v 2 is v 2 = 2 + ( 2) = 3. A unit vector u 2 in the direction of v 2 may then be obtained as follows: u 2 = (, 2, 5) = (, 2 5, )
13 (iii) v 3 = (,,, ). The length of v 3 is v 3 = = 4 = 2. A unit vector u 3 in the direction of v 3 may then be obtained as follows: u 3 = 2 (,,, ) = ( 2, 2, 2, 2 ). Example. Consider the two vectors a = (2,, 4) and b = (3, 2, 5) in 3-dimensional space. Find the cross-product of the two vectors. Solution: a b = = i j k i j k = 3 i 22 j 7 k = ( 3, 22, 7)
14 .7 Quiz. Find d dx (tan(x2 ) cos(e x ). Solution: d dx (tan(x2 ) cos(e x ) = tan(x 2 ) d dx (cos(ex ) + cos(e x ) d dx (tan(x2 )) = tan(x 2 )( sin(e x ))e x + cos(e x )(sec 2 (x 2 ))2x 2. Find the stationary points of y = x 3 3x. Solution: Let f(x) = x 3 3x. Since f(x) is continuous, the stationary points of f(x) are the solutions of the equation f (x) =. f(x) = x 3 3x f (x) = 3x 2 3 3x 2 3 = 3(x )(x + ) = The stationary points are: {, }. 3. Find x ln xdx. Solution: u = ln(x); v = x2 2 du = dx ; dv = xdx x x ln(x)dx = udv = uv vdu = x2 ln(x) 2 = x2 ln(x) 2 = x2 ln(x) 2 x 2 2 dx x x 2 dx x2 4 + C. 4. Find ( x 2 ) /2 dx.
15 Solution: Therefore, a2 u 2 du = sin ( u a ) + c. ( x 2 ) /2 dx = sin (x) + c. 5. Find the determinant of ( ). Solution: = (2)(2) ( )( 6) = 2 6. Let u = (2, 3, 5), v = (,, ). Find (a) The unit vector in the direction of v = (,, ) is, (b) (,, ) = (,, ) = (,, ) ( ) u 5 v = 3(2, 3, 5) 5(,, ) = (6, 9, 5) (5, 5, 5) = (, 4, 2). (c) (d) u v = = u v = (2, 3, 5).(,, ) = =. i j k i 2 5 j k = (3( ) (5)()) i (2( ) (5)()) j + ((2)() (3)()) k = 8 i + 7 j k = ( 8, 7, )
16 7. Determine if u = ( 2, 2,, ) is perpendicular to v = ( 2, 3,, ). Solution: u. v = ( 2)( 2) + (2)( 3) + ()() + ( )() = 2. Hence the two vectors are not perpendicular.
17 Summary. The dot product of 2 vectors a = (a, a 2, a 3 ) and b = (b, b 2, b 3 ) in 3-dimensional space is defined to be the scalar a. b = a b + a 2 b 2 + a 3 b 3. or a. b = a. b cos θ. 2. Let a and b be non-zero vectors. The vectors, a and b, are perpendicular to each other if and only if a b =. 3. Given 2 vectors a = (a, a 2, a 3 ) and b = (b, b 2, b 3 ) in 3-dimensional space the cross product a b can be defined as follows: a i j k b = a a 2 a 3 b b 2 b 3 where i = (,, ), j = (,, ), k = (,, ) are unit vectors along the X, Y, Z axes respectively. 4. The direction of the cross product a b is perpendicular to both a and b. In particular it points in the direction of your thumb if the fingers of your right hand curl from a to b. 5. The magnitude of a b is given by a. b sin θ, if θ is the angle between a and b. It is the area of the parallelogram determined by a and b.
18 References. A complete set of notes on Pre-Calculus, Single Variable Calculus, Multivariable Calculus and Linear Algebra. Here is a link to the chapter on Vectors A collection of examples, animations and notes on Multivariable Calculus. jflores/multicalc2/webbook/chapter 3/Graphics/Chapter3 /Demo 3. Gallery of animated and graphical demonstrations of calculus and related topics, from the University of Minnesota. nykamp/m2374/readings/completelist.html. 4. Links to various resources on Calculus Engineering Mathematics, by K. A. Stroud.
19 2 Topics in 3D Geometry In two dimensional space, we can graph curves and lines. In three dimensional space, there is so much extra space that we can graph planes and surfaces in addition to lines and curves. Here we will have a very brief introduction to Geometry in three dimensions. 2. Planes Just as it is easy to write the equation of a line in 2D space, it is easy to write the equation of a plane in 3D space. The point-normal equation of a plane A vector perpendicular to a plane is said to be normal to the plane and is called a normal vector, or simply a normal. To write the equation of a plane we need a point P (x, y, z ) on the plane and a normal vector n = (a, b, c) to the plane. Let P = (x, y, z ) be a point on the plane and n be a vector perpendicular to the plane. Then a point Q(x, y, z) lies on the plane, the vector P Q lies on the plane, P Q and n are perpendicular, n P Q =, (a, b, c) (x x, y y, z z ) =, a(x x ) + b(y y ) + c(z z ) =. Definition. The point-normal equation of a plane that contains the point P (x, y, z ) and has normal vector n = (a, b, c) is a(x x ) + b(y y ) + c(z z ) =. Example. Let P be a plane determined by the points A = (, 2, 3), B = (2, 3, 4), and C = ( 2,.3). Find a vector which is normal to the plane. Find an equation of the plane.
20 Solution: We need a point on the plane and a normal to the plane. The vector AB AC = (2, 3, ) is a normal to the plane and we take A = (, 2, 3) as a point on the plane (you can choose B or C instead of A if you want). The equation on the plane in point-normal form is: or equivalently, 2(x ) 3(y 2) + (z 3) = 2x 3y + z = Observe that the coefficients of x, y and z are (2, 3, ) which is the normal to the plane. 2.2 Lines Vector equation of a line To write the vector equation of a line, we need a point P (x, y, z ) on the line and a vector v = (a, b, c) that is parallel to the line. Definition. The vector equation of a line that contains the point P (x, y, z ) and is parallel to the vector v = (a, b, c) is: or, Parametric equation of a line P + t v = r, where t is scalar. (x, y, z ) + t(a, b, c) = (x, y, z) (x + ta, y + tb, z + tc) = (x, y, z) The parametric equation of a line is derived from the vector equation of a line. Definition. The parametric equation of a line that contains the point P (x, y, z ) and is parallel to the vector v = (a, b, c) is: x = x + ta y = y + tb z = z + tc
21 Example. Let L which passes through the points P (,, ) and Q(3, 2, ). Find a vector which is parallel to the line. Find the vector-equation and parametric equation of the line. Solution: The vector P Q = (2,, ) is parallel to the line and we take the point P (,, ) on the line. The vector equation of the line: The parametric equations of the line: (,, ) + t(2,, ) = (x, y, z) x = + 2t y = + t z = Example. Find the equation of the plane which contains the point (,, 2) and is perpendicular to the line (,, ) + t(2,, ) = (x, y, z). 2.3 Surfaces The graph in 3D space of an equation in x, y and z is a surface. Often the graph is too difficut to draw, but here we sketch the graph of a few special types of equations whose graphs are easy to visualize. Cylindrical surfaces The graph in 3D space of an equation containing only one or two of the three variables x, y, z is called a cylindrical surface. Example. Plot y = x 2. Plot x 2 + y 2 = 5.
22 Quadric Surfaces The graph in 2D space of a second degree equation in x and y is an ellipse, parabola or hyperbola. In 3D space, the graph of a second degree equation in x, y and z is one of six quadric surfaces.. Ellipsoid x 2 a 2 + y2 b 2 + z2 c 2 = 2. Elliptic Cone z 2 = x2 + y2 a 2 b 2 3. Elliptic Paraboloid z = x2 + y2 a 2 b 2 4. Hyperbolic Parabolid z = y2 b 2 x2 a 2 5. Elliptic Hyperboloid of one sheet x 2 a 2 + y2 b 2 z2 c 2 = 6. Elliptic Hyperboloid of two sheets x2 y2 + z2 = a 2 b 2 c 2 Cross-sections of some quadric surfaces
23 2.4 Functions of several variables So far you have studied about functions of one variable, e.g. f(x) = x + x 2. You have learned how to graph these functions, perhaps how to determine their domain and range. You have gone much further in your quest to understand functions; you have learned how to differentiate them, then to calculate the maximum and minimum, then to integrate them. At the end of this term (in March) we will learn how to write a function as a sum of simpler functions, i.e. a Fourier Series Expansion of a function. However now we will learn something new. We will learn about functions of several variables, e.g. f(x, y) = x 2 + y 2, g(x, y) = 2xy + 7, h(x, y) = e x + 2y. You will very quickly see that although the concepts are new, the techniques are old and familiar. Graphs and Level Sets To draw the graph of f(x) = x 2, we drew the graph of the equation y = x 2. Similarly, to draw the graph of the equation f(x, y) = x 2 + y 2, we draw the graph of the equation z = x 2 + y 2. We now use the methods developed in the last lecture to draw graphs. Note however that it is difficult to graph general surfaces. Remark. The graph of a function f(x) of one variable is the graph of the equation y = f(x), a curve in 2D space. The graph of a function f(x, y) of two variables is the graph of the equation z = f(x, y), a surface in 3D space. The graph of a function f(x, y, z) is a set of points in 4D space and we cannot draw the graph.
24 One way to understand functions of two or more variables is by using level sets. Example. An example of level sets is a topographic map, which maps hills and valleys in a region by drawing curves indicating height or elevation. If h(x, y) is the height function over a region, say the Himalayas, then if we mark all the points (x, y) on the ground at which the height of the mountain is h(x, y) = 3m, we get the level set for L = 3. If we draw the level sets for different heights, e.g 2m, 3m, 4m, 5m, 6m, 7m, 8m, we get a rough topographic map for the Himalayas. Note that we draw the level sets on the ground, i.e. in the domain. Definition. We fix a number C. The level set of a function of two variables f(x, y) is the set of points (x, y) in the domain which satisfy the equation f(x, y) = C. For every real number C, we get a level set. In general, for a fixed number C and a function of several variables f, we define the level set to be the collection of points in the domain which satisfy the equation f = C. Example. Let f(x) = x 2 and say the constant c =. The level set is the set of points such that x 2 = x =, Hence the level set for c = is {, -}. Example. Find the level sets of the function f(x, y) = x 2 +y 2 for C =, 4, 9. Solution: The level sets are C = : x 2 + y 2 = ; a circle of radius C = 4 : x 2 + y 2 = 4; a circle of radius 2 C = 9 : x 2 + y 2 = 9; a circle of radius 3
25 Example. Let f(x, y, z) = x 2 + y 2 + z 2 and say the constant c =. The level set is the set of points such that x 2 + y 2 + z 2 = Hence the level set for c = is the set of points lying on the unit sphere. 2.5 Change of Coordinates Before we describe cylindrical and spherical coordinate systems, we will recall the polar coordinate system in 2D space. Polar Coordinates The polar coordinate system is equivalent to the rectangular coordinate system. It locates points using two coordinates r and θ. The coordinate r is the distance from a point to the origin, and θ is the angle used in trignometry which measures the counterclockwise rotation from the positive X-axis. Conversion from rectangular to polar coordinates: Let (x, y) be a point in the rectangular coordinate system. r = x 2 + y 2 tan θ = y x Conversion from polar to rectangular coordinates: Let (r, θ) be a point in the polar coordinate system. x = r cos θ y = r sin θ Example. Express in polar coordiantes the portion of the unit disc that lies in the first quadrant. Solution: The region may be expressed in polar coordinates as r ; θ π/2
26 Example. Express in polar coordinates the function f(x, y) = x 2 + y 2 + 2yx. Solution: We substitute x = r cos θ and y = r sin θ in f, to get f(r, θ) = r 2 cos 2 (θ) + r 2 sin 2 (θ) + 2r 2 sin(θ) cos(θ) = r 2 ( + sin(2θ)). Cylindrical Coordinates This is a three dimensional extension of plane polar coordinates. Cylindrical coordinates are given by the 3-tuple (r, θ, z), the polar coordinates of the X, Y plane and the rectangular coordinate z. Given a point (x, y, z) in 3- dimensional space, to calculate r and θ we project the point to the XY-plane (x, y, z) (x, y). Then calculate (r, θ) as in the previous section. Cylindrical to Rectangular Conversion Formulas: x = r cos θ y = r sin θ z = z Rectangular to Cylindrical Conversion Formulas: r = x 2 + y 2 θ = tan ( y x ) z = z Example.. Express in cylindrical coordinates the function f(x, y, z) = x 2 + y 2 + z 2 2z x 2 + y 2 2. Express in rectangular coordinates the equation r = sin θ
27 Spherical coordinates Spherical coordinates consist of the 3-tuple (ρ, θ, φ). These are determined as follows:. ρ = the distance from the origin to the point. 2. θ = the same angle that we saw in polar/cylindrical coordinates. 3. φ = the angle between the positive z-axis and the line from the origin to the point. Spherical to Rectangular Conversion Formulas: x = ρ sin(φ) cos(θ) y = ρ sin(φ) sin(θ) z = ρ cos φ Rectangular to Spherical Conversion Formulas: ρ = x 2 + y 2 + z 2 θ = tan ( y x ) φ = cos ( z ρ ) Example.. Express in spherical coordinates the function f(x, y, z) = x 2 + y 2 + z 2 2. Express in rectangular coordinates the equation ρ = 5
28 Summary. The point-normal equation of a plane that contains the point P (x, y, z ) and has normal vector n = (a, b, c) is a(x x ) + b(y y ) + c(z z ) =. 2. The vector equation of a line that contains the point P (x, y, z ) and is parallel to the vector v = (a, b, c) is: P + t v = r, where t is scalar. 3. The parametric equation of a line that contains the point P (x, y, z ) and is parallel to the vector v = (a, b, c) is: x = x + ta y = y + tb z = z + tc 4. We fix a number C. The level set of a function of two variables f(x, y) is the set of points (x, y) in the domain which satisfy the equation f(x, y) = C. 5. The polar coordinate system has two coordinates r and θ. The coordinate r is the distance from a point to the origin, and θ is the angle used in trignometry which measures the counterclockwise rotation from the positive X-axis. Conversion from rectangular to polar coordinates: Let (x, y) be a point in the rectangular coordinate system. r = x 2 + y 2 tan θ = y x Conversion from polar to rectangular coordinates: Let (r, θ) be a point in the polar coordinate system. x = r cos θ y = r sin θ
29 6. Cylindrical Coordinate System is a three dimensional extension of plane polar coordinates. Cylindrical coordinates are given by the 3-tuple (r, θ, z), the polar coordinates of the X, Y plane and the rectangular coordinate z. Cylindrical to Rectangular Conversion Formulas: x = r cos θ y = r sin θ z = z Rectangular to Cylindrical Conversion Formulas: r = x 2 + y 2 θ = tan ( y x ) z = z 7. Spherical coordinates consist of the 3-tuple (ρ, θ, φ). These are determined as follows:. ρ = the distance from the origin to the point. 2. θ = the same angle that we saw in polar/cylindrical coordinates. 3. φ = the angle between the positive z-axis and the line from the origin to the point. Spherical to Rectangular Conversion Formulas: x = ρ sin(φ) cos(θ) y = ρ sin(φ) sin(θ) z = ρ cos φ Rectangular to Spherical Conversion Formulas: ρ = x 2 + y 2 + z 2 θ = tan ( y x ) φ = cos ( z ρ )
30 References. A complete set of notes on Pre-Calculus, Single Variable Calculus, Multivariable Calculus and Linear Algebra. Here is a link to the chapter on Lines, Planes and Quadric Surfaces. Also read the section on cylindrical and spherical coordiantes Another gallery of animated and graphical demonstrations of calculus and related topics, from the University of Minnesota Links to various resources on Calculus.
31 3 Partial Derivatives 3. First Order Partial Derivatives A function f(x) of one variable has a first order derivative denoted by f (x) or df dx = lim f(x + h) f(x). h h It calculates the slope of the tangent line of the function f at x. A function f(x, y) of two variables has two first order partials f, f. x y Just like in the one variable case, the partial derivatives are also related to the tangent plane of the function at a point (x, y) Definition. f is defined as the derivative of f with respect to x with y x treated as a constant. f y f x = lim h f(x + h, y) f(x, y). h is defined as the derivative of f with respect to y with x treated as a constant. f y = lim k f(x, y + k) f(x, y). k Similarly for f(x, y, z) we can define three first order partial derivatives: f f, f. y z Example.. 2. f(x, y, z) = ze 2x+3y+4z 3. f(x, y) = x 2 + y 2 f(x, y) = x 2 y 3 f x = x2 y 3 x f = x2 y 3 y y x2 = y3 x = 2xy3 = 3x 2 y 2 Example. Consider the function f(x, y) = x 2 /y. Calculate the first order partial derivatives and evaluate them at the point P (2, ). x,
32 Solution: f(x, y) = x2 y f x = 2x y f x {x=2,y=} = 4 f y = x2 y 2 f y {x=2,y=} = 9 Remark. Partial derivatives are used in the same manner as the derivative of a function of one variable. The partial of f(x, y) with respect to x is the rate of change (or the slope) of f with respect to x as y stays constant. Similarly the partial of f(x, y) with respect to y is the rate of change (or the slope) of f with respect to y as x stays constant. For instance in the above example, the slope of the function at the point P (2, ) in the x direction is 4, and the slope of the function at P in the y direction is /9. Theorem Let f(x, y) be a function of two variables and let P = (x, y, z ) be a point on the graph. The equation of the tangent plane to the graph of f(x, y) at the point P is given by, z z = f x {x,y }(x x ) + f y {x,y }(y y ). We will see why this is true after we study about Directional Derivative and the Gradient. For now we accept it as a result and use it in some examples. Example. We consider the function f(x, y) = x 3 + 2xy y + 3. Compute the tangent plane at the point P = (, 2, 6) on the graph of the function. Solution: (Solution:) We first check that the point P (x, y, z ) = (, 2, 6) is on the graph of the function: f(, 2) = = = 6
33 So the point P does indeed lie on the graph of f. Then the equation of the tangent plane is given by; z z = f (x x ) + f x {x,y } y {x,y }(y y ). f x {,2} = 3x 2 + 2y {,2} = 7 f y {,2} = 2x {,2} = z 6 = 7(x ) + (y 2) = 7x 7 + y 2 z = 7x + y 3 Therefore the tangent plane to the graph of f at (, 2, 6) is 7x + y z 3 =. 3.2 Higher Order Partial Derivatives If f is a function of several variables, then we can find higher order partials in the following manner. Definition. If f(x, y) is a function of two variables, then f f and are x y also functions of two variables and their partials can be taken. Hence we can differentiate them with respect to x and y again and find, 2 f x 2, the derivative of f taken twice with respect to x, 2 f x y 2 f y x, the derivative of f with respect to y and then with respect to x,, the derivative of f with respect to x and then with respect to y, 2 f y 2, the derivative of f taken twice with respect to y. We can carry on and find, which is taking the derivative of f first with x y 2 respect to y twice, and then differentiating with respect to x, etc. In this manner we can find nth-order partial derivatives of a function. 3 f
34 Theorem 2 f and 2 f x y y x when 2 f x y and 2 f y x are called mixed partial derivatives. They are equal are continuous. Note. In this course all the fuunctions we will encounter will have equal mixed partial derivatives. Example.. Find all partials up to the second order of the function f(x, y) = x 4 y 2 x 2 y 6. Solution: Notations: f x = 4x3 y 2 2xy 6 f y = 2x 4 y 6x 2 y 5 2 f x 2 = 2x 2 y 2 2y 6 2 f y x = 8x3 y 2xy 5 2 f y 2 = 2x 4 3x 2 y 4 2 f x y = 8x 3 2xy 5 f x = f x f y = f y f xx = 2 f x 2 f yy = 2 f y 2 f xy = 2 f x y
35 3.3 Chain Rule You are familiar with the chain rule for functions of one variable: if f is a function of u, denoted by f = f(u), and u is a function of x, denoted u = u(x). Then df dx = df du du dx. Chain Rules for First-Order Partial Derivatives For a two-dimensional version, suppose z is a function of u and v, denoted and u and v are functions of x and y, z = z(u, v) u = u(x, y) and v = v(x, y) then Example. z x = z u u x + z v v y z y = z u u y + z v v y z z z u v u v u u x v v y x y x y x y. Find the first partial derivatives using chain rule. z = z(u, v) u = xy v = 2x + 3y
36 Solution: z x = z u u = y z z y = z u x + z v u + 2 z v u y + z v = x z u + 3 z v v x v y Chain Rule for Second Order Partial Derivatives To find second order partials, we can use the same techniques as first order partials, but with more care and patience! Example. Let. Find 2 z y 2. z = z(u, v) u = x 2 y v = 3x + 2y Solution: We will first find 2 z y 2. z y = z u u y + z v v y = x2 z u + 2 z v. Now differentiate again with respect to y to obtain 2 z y 2 = z (x2 y u ) + y (2 z v ) = x 2 y ( z u ) + 2 y ( z v ) Note that z is a function of u and v, and u and v are functions of x and y. Then the partial derivatives z z and are also initially functions of u v u and v and eventually functions of x and y. In other words we have the same dependence diagram as z.
37 ( z ) 2 z ). Never write 2 z as it is mathematically mean- y u y u y u Note. ingless. 2 z u 2 z u 2 z v u u v u u x v v y x y x y x y Therefore 2 z u v z v 2 z v 2 u v u u x v v y x y x y x y 2 z = x 2 ( 2 z u y 2 u 2 y + 2 z v v u y ) + 2( 2 z u u v y + 2 z v v 2 y ) = x 2 (x 2 2 z u z 2 v u ) + 2 z 2(x2 u v + z 2 2 v ) 2 The mixed partials are equal so the answer simplifies to 2. Find 2 z x y Solution: We have 2 z y = 2 z 2 x4 u + 2 4x2 z y = z u u y + z v v y 2 z u v + z 4 2 v 2. = x2 z u + 2 z v.
38 Now differentiate again with respect to x to obtain 2 z x y = z (x2 x u ) + x (2 z v ) = x 2 x ( z z ) + 2x u u + 2 x ( z = x 2 ( 2 z u u 2 x + 2 z v u v ) v z ) + 2x x u + 2( 2 z u u v x + 2 z v v 2 x ) = x 2 (2xy 2 z u z z ) + 2x 2 v u u + 2(2xy 2 z u v + z 3 2 v ) 2 = 2x 3 y 2 z u + 2 (3x2 + 4xy) 2 z z ) + 2x v u u + z 6 2 v Maxima and Minima Recall from -dimensional calculus, to find the points of maxima and minima of a function, we first find the critical points i.e where the tangent line is horizontal f (x) =. Then (i) If f (x) > the gradient is increasing and we have a local minimum. (ii) If f (x) < the gradient is decreasing and we have a local maximum. (iii) If f (x) = it is inconclusive. Critical Points of a function of 2 variables We are now interested in finding points of local maxima and minima for a function of two variables. Definition. A function f(x, y) has a relative minimum (resp. maximum) at the point (a, b) if f(x, y) f(a, b) (resp. f(x, y) f(a, b)) for all points (x, y) in some region around (a, b) Definition. A point (a, b) is a critical point of a function f(x, y) if one of the following is true (i) f x (a, b) = and f y (a, b) = (ii) f x (a, b) and/or f y (a, b) does not exist.
39 Classification of Critical Points We will need two quantities to classify the critical points of f(x, y):. f xx, the second partial derivative of f with respect to x. 2. H = f xx f yy f 2 xy the Hessian If the Hessian is zero, then the critical point is degenerate. If the Hessian is non-zero, then the critical point is non-degenerate and we can classify the points in the following manner: case(i) If H > and f xx < then the critical point is a relative maximum. case(ii) If H > and f xx > then the critical point is a relative minimum. case(iii) If H < then the critical point is a saddle point. Example. Find and classify the critical points of 2x 3 + y 3 + 2x 2 y 75y. Solution: We first find the critical points of the function. f x = 36x xy = 2x(3x + 2y) f y = 3y 2 + 2x 2 75 = 3(4x 2 + y 2 25). The critical points are the points where f x = and f y = f x = 2x(3x + 2y) = Therefore either x = or 3x + 2y =. We handle the two cases separately; case(i) x =. Then substituting this in f y we get f y = 3(y 2 25) = implies y = ±5.
40 case(ii) 3x + 2y =. Then y = 3x/2 and substituting this in f y we get, f y = 3(4x 2 + 9x2 4 25) = 3 4 (6x2 + 9x 2 ) = 3 4 (25x2 ) = 75 4 (x2 4) f y = 75 4 (x2 4) = x 2 4 = x = ±2 Thus we have found four critical points: (, 5), (, 5), (2, 3), ( 2, 3). We must now classify these points. f xx = 72x + 24y = 24(3x + y) f xy = 24x f yy = 6y Points f xx H Type (, 5) 2 36 Minimum (, 5) Maximum (2, 3) Saddle ( 2, 3) Saddle H = f xx f yy fxy 2 = (24)(3x + y)(6y) (24x) 2
41 Summary. The First Order Partial Derivative f is defined as the derivative of x f with respect to x, with y treated as a constant. Similarly we can define f y. f f 2. and are also functions of two variables and hence their partials x y can be taken. We can differentiate them with respect to x and y again and find Higher Order Partial Derivatives, 2 f x, 2 f 2 x y, 2 f y x, 2 f y, 3 f 2 x, etc Chain Rule for Partial Dericatives. Suppose z is a function of u and v, denoted z = z(u, v) and u and v are functions of x and y, u = u(x, y) and v = v(x, y), then z x = z u u x + z v v y, z y = z u u y + z v v y 4. A point (a, b) is a critical point of a function f(x, y) if one of the following is true (i) f x (a, b) = and f y (a, b) = (ii) f x (a, b) and/or f y (a, b) does not exist. 5. We will need two quantities to classify the critical points of f(x, y):. f xx, the second partial derivative of f with respect to x. 2. H = f xx f yy f 2 xy the Hessian If the Hessian is zero, then the critical point is degenerate. If the Hessian is non-zero, then the critical point is non-degenerate and we can classify the points in the following manner: case(i) If H > and f xx < then the critical point is a relative maximum. case(ii) If H > and f xx > then the critical point is a relative minimum. case(iii) If H < then the critical point is a saddle point.
42 References. Engineering Mathematics, by K. A. Stroud. 2. Engineering Mathematics, by K. A. Stroud. 3. A complete set of notes on Pre-Calculus, Single Variable Calculus, Multivariable Calculus and Linear Algebra. Here is a link to the chapter on Partial Differentiation Here is a link to the chapter on Higher Order Partial Differentiation. Here is a link to the chapter on Chain Rules. Here is a link to the chapter on Maxima and Minima A collection of examples, animations and notes on Multivariable Calculus. jflores/multicalc2/webbook/chapter 5/ 6. Links to various resources on Calculus.
43 4 A little Vector Calculus 4. Gradient Vector Function/ Vector Fields The functions of several variables we have so far studied would take a point (x, y, z) and give a real number f(x, y, z). We call these types of functions scalar-valued functions i.e. for example f(x, y, z) = x 2 + 2xyz. We are now going to talk about vector-valued functions, where we take a point (x, y, z) and the value of f(x, y, z) is a vector. i.e. for example f(x, y, z) = (y, x, z 2 ) = y i + x j + z 2 k. Definition. A vector function is a function that takes one or more variables and returns a vector. Example.. A vector function of a single variable: r(t) = (2 + t, 3 + 2t, 3t). Let us look at a few values. f(2) = (4, 7, 5), f( 2)= (,, 7), f() = (3, 5, 2), f( ) = (,, 4), f() = (2, 3, ) 2. A vector function of three variables: f(x, y, z) = (y, x, z 2 ). Let us look at a few values. f(,, ) = (,, ), f(,, )= (,, ), f(,, ) = (,, ), f(,, )= (,, ), f(, 2, ) = (2,, ), f(2,, 2) = (, 2, 4)
44 Definition. A vector field is an assignment of a vector to every point in space. Example.. In a magnetic field, we can assign a vector which describes the force and direction to every point in space. 2. Normal to a surface. For a surface, at every point on the surface, we can get a tangent plane and hence a normal vector to every point on the surface. Example. Let f(x, y, z) = x 2 + 2xyz be a scalar-valued function. We then define a vector-valued function by taking its partial derivatives. f = ( f x, f y, f z ) = (2x + 2yz, 2xz, 2xy) This kind of vector function has a special name, the gradient. Definition. Suppose that f(x, y) is a scalar-valued function of two variables. Then the gradient of f is the vector function defined as, f = ( f x, f y ) = f x i + f y j. Similarly if f(x, y, z) is a scalar-valued function of three variables. Then the gradient of f is the vector function defined as, f = ( f x, f y, f z ) = f x i + f y j + f z k. Remark. The gradient of a scalar function is a vector field which points in the direction of the greatest rate of increase of the scalar function, and whose magnitude is the greatest rate of change. Example. Consider the scalar-valued function defined by f(x, y) = x 2 + y 2. Find the gradient of f at the point x = 2, y = 5. Solution: Then gradient of f, is a vector function given by, f = ( f x, f y ) = (2x, 2y) f(2, 5) = (4, )
45 4.2 Directional Derivative For a function of 2 variables f(x, y), we have seen that the function can be used to represent the surface z = f(x, y) and recall the geometric interpretation of the partials: (i) f x (a, b)-represents the rate of change of the function f(x, y) as we vary x and hold y = b fixed. (ii) f y (a, b)-represents the rate of change of the function f(x, y) as we vary y and hold x = a fixed. We now ask, at a point P can we calculate the slope of f in an arbitrary direction? Recall the definition of the vector function f, We observe that, f = ( f x, f ). y f î = f x f ĵ = f y This enables us to calculate the directional derivative in an arbitrary direction, by taking the dot product of f with a unit vector, u, in the desired direction. Definition. The directional derivative of the function f in the direction u denoted by D u f, is defined to be, D u f = f u u Example. What is the directional derivative of f(x, y) = x 2 + xy, in the direction i + 2 j at the point (, )?
46 Solution: We first find f. f = ( f x, f y ) = (2x + y, x) f(, ) = (3, ) Let u = i + 2 j. u = = + 4 = 5. D u f(, ) = = = f u u (3, ).(, 2) 5 (3)() + ()(2) 5 = 5 5 = 5 Properties of the Gradient deduced from the formula of Directional Derivatives D u f = f u u = f u cos θ u = f cos θ. If θ =, i.e. i.e. u points in the same direction as f, then D u f is maximum. Therefore we may conclude that (i) f points in the steepest direction. (ii) The magnitude of f gives the slope in the steepest direction.
47 2. At any point P, f(p ) is perpendicular to the level set through that point. Example.. Let f(x, y) = x 2 + y 2 and let P = (, 2, 5). Then P lies on the graph of f since f(, 2) = 5. Find the slope and the direction of the steepest ascent at P on the graph of f Solution: We use the first property of the Gradient vector. The direction of the steepest ascent at P on the graph of f is the direction of the gradient vector at the point (, 2). f = ( f x, f y ) = (2x, 2y) f(, 2) = (2, 4). The slope of the steepest ascent at P on the graph of f is the magnitude of the gradient vector at the point (, 2). f(, 2) = = Find a normal vector to the graph of the equation f(x, y) = x 2 + y 2 at the point (, 2, 5). Hence write an equation for the tangent plane at the point (, 2, 5). Solution: We use the second property of the gradient vector. For a function g, g(p ) is perpendicular to the level set. So we want our surface z = x 2 + y 2 to be the level set of a function. Therefore we define a new function, Then our surface is the level set g(x, y, z) = x 2 + y 2 z. g(x, y, z) = x 2 + y 2 z = z = x 2 + y 2
48 g = ( g x, g y, g z ) = (2x, 2y, ) g(, 2, 5) = (2, 4, ) By the above property, g(p ) is perpendicular to the level set g(x, y, z) =. Therefore g(p ) is the required normal vector. Finally an equation for the tangent plane at the point (, 2, 5) on the surface is given by 4.3 Curl and Divergence 2(x ) + 4(y 2) (z 5) =. We denoted the gradient of a scalar function f(x, y, z) as f = ( f x, f y, f z ) Let us separate or isolate the operator = (,, ). We can then define x y z various physical quantities such as div, curl by specifying the action of the operator. Divergence Definition. Given a vector field v(x, y, z) = (v (x, y, z), v 2 (x, y, z), v 3 (x, y, z)), the divergence of v is a scalar function defined as the dot product of the vector operator and v, Div v = v = ( x, y, z ) (v, v 2, v 3 ) = v x + v 2 y + v 3 z Example. Compute the divergence of (x y) i + (x + y) j + z k.
49 Solution: v = ((x y), (x + y), z) = ( x, y, z ) Div v = v = ( x, y, ) ((x y), (x + y), z) z (x y) = + x = + + = 3 (x + y) y + z z Curl Definition. The curl of a vector field is a vector function defined as the cross product of the vector operator and v, i j k Curl v = v = x y z v v 2 v 3 = ( v 3 y v 2 z )i ( v 3 x v z )j + ( v 2 x v y )k Example. Compute the curl of the vector function (x y) i + (x + y) j + z k. Solution: Curl v = v = i j k x y (x y) (x + y) z = ( z y z (x + y) )i ( z z x = ( ) i ( ) j + ( ( )) k = 2 k (x y) (x + y) )j + ( z x (x y) )k y
50 4.4 Laplacian We have seen above that given a vector function, we can calculate the divergence and curl of that function. A scalar function f has a vector function f associated to it. We now look at Curl( f) and Div( f). Curl( f) = f = ( f z y f y z )i + ( f x z f z x )j + ( f y x f x y )k = (f yz f zy )i + (f zx f xz )j + (f xy f yx )k = Div( f) = f = ( x, y, z ) ( f x, f y, f z ) = 2 f x f y f z 2 Definition. The Laplacian of a scalar function f(x, y) of two variables is defined to be Div( f) and is denoted by 2 f, 2 f = 2 f x f y 2. The Laplacian of a scalar function f(x, y, z) of three variables is defined to be Div( f) and is denoted by 2 f, 2 f = 2 f x f y f z 2. Example. Compute the Laplacian of f(x, y, z) = x 2 + y 2 + z 2. Solution: 2 f = 2 f x + 2 f 2 y + 2 f 2 z 2 = 2x x + 2y y + 2z z = = 6.
51 We have the following identities for the Laplacian in different coordinate systems: Rectangular : 2 f = 2 f x + 2 f 2 y + 2 f 2 z 2 P olar : 2 f = ( f ) 2 f r + r r r r 2 θ 2 Cylindrical : 2 f = ( f ) 2 f r + r r r r 2 θ + 2 f 2 z 2 Spherical : 2 f = ( ρ 2 f ) + ρ 2 ρ ρ ρ 2 sin θ θ ( sin θ f θ ) 2 f + ρ 2 sin 2 θ φ 2 Example. Consider the same function f(x, y, z) = x 2 + y 2 + z 2. We have seen that in rectangular coordinates we get 2 f = 2 f x f y f z 2 = 6. We now calculate this in cylindrical and spherical coordinate systems, using the formulas given above.. Cylindrical Coordinates. We have x = r cos θ and y = r sin θ so Using the above formula: f(r, θ, z) = r 2 cos 2 θ + r 2 sin 2 θ + z 2 = r 2 + z 2. 2 f = r r = r ( f ) 2 f r + r r 2 θ + 2 f 2 z 2 (2z) (r2r) + + r z = r (4r) + 2 = = 6 2. Spherical Coordinates. We have x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ and ρ = x 2 + y 2 + z 2, so f(r, θ, z) = ρ 2.
52 Using the above formula: 2 f = ρ 2 ρ ( ρ 2 f ) + ρ ρ 2 sin θ = ρ 2 ρ (ρ2 2ρ) + + = ρ 2 ρ (2ρ3 ) = ρ 2 (6ρ2 ) = 6. θ ( f ) 2 f sin θ + θ ρ 2 sin 2 θ φ 2 These three different calculations all produce the same result because 2 is a derivative with a real physical meaning, and does not depend on the coordinate system being used.
53 Summary. If f(x, y) is a scalar-valued function of two variables. Then the gradient of f is the vector function defined as, f = ( f x, f y ) = f x i + f y j. 2. Important Properties about the Gradient. f points in the steepest direction. The magnitude of f gives the slope in the steepest direction. At any point P, f(p ) is perpendicular to the level set through that point. 3. The directional derivative of the function f in the direction u denoted by D u f, is defined to be, D u f = f u u 4. Given a vector field v(x, y, z) = (v (x, y, z), v 2 (x, y, z), v 3 (x, y, z)), the divergence of the vector field, v is a scalar function defined as the dot product of the vector operator and v, Div v = v = ( x, y, z ) (v, v 2, v 3 ) = v x + v 2 y + v 3 z 5. The curl of a vector field is a vector function defined as the cross product of the vector operator and v, i j k Curl v = v = x y z v v 2 v 3 = ( v 3 y v 2 z )i ( v 3 x v z )j + ( v 2 x v y )k
54 6. We have the following identities for the Laplacian in different coordinate systems: Rectangular : 2 f = 2 f x + 2 f 2 y + 2 f 2 z 2 P olar : 2 f = ( f ) 2 f r + r r r r 2 θ 2 Cylindrical : 2 f = ( f ) 2 f r + r r r r 2 θ + 2 f 2 z 2 Spherical : 2 f = ( ρ 2 f ) + ρ 2 ρ ρ ρ 2 sin θ θ ( sin θ f θ ) 2 f + ρ 2 sin 2 θ φ 2
55 References. A briliant animated example, showing that the maximum slope at a point occurs in the direction of the gradient vector. The animation shows: a surface a unit vector rotating about the point (,, ), (shown as a rotating black arrow at the base of the figure) a rotating plane parallel to the unit vector, (shown as a grey grid) the traces of the planes in the surface, (shown as a black curve on the surface) the tangent lines to the traces at (,, f (, )), (shown as a blue line) the gradient vector (shown in green at the base of the figure) 2. A complete set of notes on Pre-Calculus, Single Variable Calculus, Multivariable Calculus and Linear Algebra. Here is a link to the chapter on Directional Derivatives. Here is a link to the chapter on Curl and Divergence.
56 5 Linear Equations 5. Introduction A linear equation is one in which all the unknown variables occur with a power of one. i.e. 2x + y = 5 is a linear equation, but 2x 2 + y = 5 is not a linear equation, since x has power 2. Systems of linear equations are common in all branches of science. In school you have learned how to simultaeously solve a system of two linear equations. Example. Solve the system of two linear equations: 3x + 2x 2 = 7 x + x 2 = 6 Then x = and x 2 = 5 is a solution. Geometrically the solution, (, 5) is the point of intersection of the two lines given by the two equations. Here is an example of larger system of equations from Chemistry. Example. Under certain controled conditions, mix toulene C 7 H 8 and nitric acid HNO 3 to produce trinitrotoluene (TNT) C 7 H 5 O 6 N 3 along with water. In what proportion should those components be mixed, i.e. solve for x, y, z, w? Solution: The number of atoms of each element before the reaction must equal the number present afterward! So we get the system: xc 7 H 8 + yhno 3 zc 7 H 5 O 6 N 3 + wh 2 O 7x = 7z 8x + y = 5z + 2w y = 3z 3y = 6z + w To find the answer we need to solve the above system of linear equations. In this chapter we will learn an easy method to solve such a system.
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