1.1 Single Variable Calculus versus Multivariable Calculus Rectangular Coordinate Systems... 4

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1 MATH2202 Notebook 1 Fall 2015/2016 prepared by Professor Jenny Baglivo Contents 1 MATH2202 Notebook Single Variable Calculus versus Multivariable Calculus Rectangular Coordinate Systems Vectors, Origin, Position and Displacement Vectors Vector Addition and Scalar Multiplication Vector Length, Unit Vectors, Standard Bases Dot Product (Scalar Product) of Vectors Timeout: Square Matrices, Determinants, Areas, Volumes Cross Product of Vectors in R Equations of Lines and Planes in R Scalar Functions of Two Variables, Surfaces in R Vector Functions, Curves in R 2 and R Calculus with Vector Functions, Tangents, Arc Length Parametrized Surfaces in R c Copyright by Jenny A. Baglivo. All Rights Reserved. 1

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3 1 MATH2202 Notebook 1 This notebook is concerned with introductory topics covered in Sections 9.1 through 9.6, and parts of Sections 10.1 through 10.5, of the Stewart textbook. 1.1 Single Variable Calculus versus Multivariable Calculus In single variable calculus we study functions of the form y = f(x), where the independent variable (the input) x and the dependent variable (the output) y are real numbers. In multivariable calculus, either the independent variable (the input) or the dependent variable (the output) or both are lists of real numbers. For example, Left plot: The left plot is the graph of the function with rule f(x, y) = (3x 2 + y 2 )e (x2 +y 2), where x and y are real numbers. A point (x, y, z) is on the graph if z = f(x, y) for some real numbers x and y. The graph of f has two mountains and a valley in between. 2. Right plot: The right plot is a visualization of the function with rule f(x, y) = ( y, x), where x and y are real numbers. An arrow representing the directed line segment from the origin (0, 0) to ( y, x) is translated to start at (x, y) for various choices of (x, y). The lengths of all arrows are rescaled (shortened) to give a pleasing graph. f could represent movement on the surface of a body of water, for example, where f(x, y) is the velocity (speed and direction) at position (x, y). The visualization above suggests circular movement around a central point. 3

4 1.2 Rectangular Coordinate Systems Most of our work this term will be in the rectangular coordinate system. 1. Two-Space: A point in 2-space (the plane) is represented by an ordered pair of real numbers. The set of all such pairs is denoted by R 2 = {(x, y) x, y R}, where R is our notation for the set of real numbers. 2. Three-Space: A point in 3-space can be represented by an ordered triple of real numbers. The set of all such triples is denoted by 3. n-space: More generally, is the set of ordered n-tuples. R 3 = {(x, y, z) x, y, z R}. R n = {(x 1, x 2,..., x n ) x 1, x 2,..., x n R} If P (x 1, x 2,..., x n ) R n, then x i is the i th coordinate of P. For example, the point P ( 4, 2, 3) in 3-space is shown below from two viewpoints: In each case, the coordinate axes (the solid lines) intersect at the origin O(0, 0, 0), and the sides of the rectangular box (defined by the dashed lines) lie on the three coordinate planes or on planes parallel to the coordinate planes. 4

5 The equations of the three coordinate planes are as follows: Coordinate Plane: Set Notation Equation xy-plane {(x, y, 0) x, y R} z = 0 xz-plane {(x, 0, z) x, z R} y = 0 yz-plane {(0, y, z) y, z R} x = 0 The equations of the three planes parallel to the coordinate planes in the plots on the last page are (please complete) Sheets or cylinders in three-space. Surfaces in 3-space whose equations are in two variables only are often called sheets or cylinders. A curve in the coordinate plane of the two variables is repeated infinitely often in the third coordinate. For example, the following three plots illustrate surfaces where one of the coordinates is missing from the equation of the surface: (a) z = y (b) z = x 2 (c) x 2 + 4y 2 = 25 (a) The surface z = y is an example of a tilted plane. (b) The surface z = x 2 is an example of a parabolic sheet. (c) The surface z = x 2 + 4y 2 = 25 is an example of an elliptic cylinder. In each case, the thick curve shows the graph in the appropriate coordinate plane. 5

6 Distance between points. If P (p 1, p 2,..., p n ) and Q(q 1, q 2,..., q n ) are points in R n, then the distance between P and Q can be found using the formula P Q = (q 1 p 1 ) 2 + (q 2 p 2 ) (q n p n ) 2. Exercise. Let P (2, 2, 1), Q(3, 0, 1), and R(1, 1, 0) be points in 3-space. (a) The distance between P and Q is. (b) The distance between P and R is. (c) The distance between Q and R is. (d) Based on distances, are P, Q and R collinear (that is, do the points lie on one line in 3-space)? Do they form a right triangle in 3-space? Why? Exercise. Find an equation of the set of all points equidistant from A( 1, 5, 3) and B(6, 2, 2). Describe the set. 6

7 Spheres in three-space. The sphere with radius r and center C(c 1, c 2, c 3 ) is the set of all points P (x, y, z) satisfying P C = r. After squaring both sides, the equation becomes (x c 1 ) 2 + (y c 2 ) 2 + (z c 3 ) 2 = r 2. Exercise. The surface to the right represents the set of points satisfying the equation 4x 2 + 4y 2 + 4z x 48y 12z = 27. Show that the equation represents a sphere and find its center and radius. 7

8 1.3 Vectors, Origin, Position and Displacement Vectors The term vector is used by scientists to indicate a quantity (such as displacement or velocity) that has both magnitude and direction. The notation for a vector with n coordinates is v = v 1, v 2,..., v n, where each v i is a real number. The i th component (or coordinate) of v is v i. Note: Vectors with 2 coordinates are often represented as directed line segments in the plane. Vectors with 3 coordinates are often represented as directed line segments in 3-space. It is common practice to use the notation R n for both the set of all points with n coordinates and the set of all vectors with n coordinates, and it is common practice to use the expression a vector in n-space to refer to a vector with n coordinates. Origin. The origin in n-space is the vector all of whose coordinates are zero: O = 0, 0,..., 0. Position and displacement vectors. a directed line segment in two ways: If v is not the origin, then v can be represented as 1. As the position vector drawn from the origin to the point P (v 1, v 2,..., v n ): v = OP. 2. As the displacement vector drawn from point P (p 1, p 2,..., p n ) to point Q(q 1, q 2,..., q n ), provided that v i = q i p i for each i: v = P Q. For example, let v = 4, 3. In the plot below, v is represented as the position vector OP, and displacement vectors AB and V W If point B has coordinates (2, 3), then A has coordinates. 2. If point V has coordinates (2, 4), then W has coordinates. 8

9 1.4 Vector Addition and Scalar Multiplication Let v = v 1, v 2,..., v n and w = w 1, w 2,..., w n be vectors in R n, and let c R be a scalar (a real number). Then 1. The vector sum, v + w, is the vector obtained by using componentwise addition: v + w = v 1 + w 1, v 2 + w 2,..., v n + w n. 2. The scalar multiple, cv, is the vector obtained by multiplying each component of the vector v by the constant c: cv = cv 1, cv 2,..., cv n. For example, if v = 8, 3, 2, 7 and w = 9, 2, 3, 1, then v + w = 4v = 4v + 2w = Properties of vector addition and scalar multiplication. Since the operations of addition and scalar multiplication are defined componentwise, they enjoy properties that are similar to the properties of numbers. Specifically, let u, v, w R n and c, d R. Then, 1. Commutative: v + w = w + v. 2. Associative: (u + v) + w = u + (v + w), c(dv) = (cd)v. 3. Distributive: (c + d)v = cv + dv, c(v + w) = cv + cw. 4. Identity: 1v = v, 0v = O, v + O = v. 5. Vector difference: The difference between vectors is well-defined by v w = v + ( 1)w. 9

10 Parallelogram and triangle rules. Addition in 2-space or 3-space can be visualized by using a parallelogram (left plot), and subtraction by using a triangle (right plot). + =- 1. Left plot: In the left plot, you can obtain v + w by (a) starting on the lower left and going across the bottom and then up the right side, or (b) starting on the lower left and going up the left side and across the top (since v + w = w + v). Geometrically, the vector sum lies on the diagonal of the parallelogram. 2. Right plot: In the right plot, observe that w + u = v implies that u = v w. Parallel vectors; same and opposite directions. Vectors v, w are said to be parallel when w = mv for some constant m. If m > 0, then the vectors are pointing in the same direction; if m < 0, then the vectors are pointing in the opposite direction. The plot below is a graphical way to think about scalar multiplication in 2-space and 3-space. - 2v 1.5v The vector 2v is a vector pointing in the same direction as v, and of twice the length of v. The vector 1.5v is a vector pointing in the opposite direction of v, and of 1.5 times the length of v. 10

11 Exercise. Let P (2, 1), Q(4, 2), R( 3, 4), S( 5, 2) be points in the plane, as shown to the right Write each vector in coordinate form. - - a = P Q = a + b = b = QR = a + b + c = c = RS = a + b + c + d = d = SP = 1.5 Vector Length, Unit Vectors, Standard Bases The length of the vector v = v 1, v 2,..., v n R n is defined as follows: v = v1 2 + v v2 n. For example, if v = 3, 1, 4, 2, then v =. Unit vector. A unit vector is a vector of length 1. If v O, then u = 1 v v is the unit vector in the direction of v and u = 1 v v is the unit vector in the direction opposite to v. For example, if v = 5, 12, then the unit vector in the direction of v is, and the unit vector in the opposite direction is. 11

12 Exercise. You are travelling in your boat at 20 miles per hour in the direction of 30 degrees east of north (vector a in the plot to the right). A wind of 5 miles per hour from the east (vector b) kicks up, pulling you off course and changing your speed. Find 1. the coordinates of your new velocity vector (that is, find a + b), and 2. your new speed (the length of a + b). - - Standard basis vectors. Vectors in R n can be written in terms of standard basis vectors representing unit vectors along each coordinate axis. Specifically, 1. In R 2, the standard basis is i= 1, 0 and j= 0, 1, and vector v is v = v 1, v 2 = v 1 i + v 2 j. 2. In R 3, the standard basis is i= 1, 0, 0, j= 0, 1, 0, k= 0, 0, 1, and vector v is v = v 1, v 2, v 3 = v 1 i + v 2 j + v 3 k. 3. When n > 3, the standard basis is e 1 = 1, 0,..., 0, e 2 = 0, 1, 0,..., 0,..., e n = 0, 0,..., 0, 1 and the vector v can be written as follows: v = v 1, v 2,..., v n = v 1 e 1 + v 2 e v n e n. 12

13 In each case, we write a vector as a linear combination of vectors representing different directions in space. The representation is often useful when we want to focus on change in the given directions. Continuing with the speed boat exercise from the last page, a = i + j b = i + j = a + b = i + j 1.6 Dot Product (Scalar Product) of Vectors If v, w are vectors in R n, then their dot product (or scalar product) is the number v w = v 1 w 1 + v 2 w v n w n. For example, 5, 7 2, 3 = and 5, 1, 7, 0 2, 8, 3, 4 =. Properties of the dot product. Given u, v, w R n and c R: 1. Commutative: v w = w v. 2. Distributive: u (v + w) = (u v) + (u w). 3. Scalar: (cu) v = u (cv) = c(u v). 4. Lengths: v v = v 2 0 for all v. Further, v v = 0 only when v = O. Exercise. Demonstrate that a b 2 = a 2 + b 2 2(a b). 13

14 Law of cosines; angle between vectors. Let P, Q and R be the corners of a triangle in 2-space or 3-space, and let θ be the angle at Q. Define three vectors as shown to the right in the plot below: = = θ - = Then the law of cosines says that a b 2 = a 2 + b 2 2 a b cos(θ) and θ is said to be the angle between vectors a and b. Theorem (Relationship Between Dot Product and Angles). If a and b are vectors in 2-space or 3-space and θ is the angle between a and b, then a b = a b cos(θ). Exercise. Use the law of cosines, and the result of the exercise on the last page, to demonstrate the equality a b = a b cos(θ). 14

15 Using the dot product to find angles. Since ( ) a b a b = a b cos(θ) = θ = cos 1 a b we can solve for θ if we know the coordinates of vectors a and b. Exercise. Sketch the triangle with corners P ( 3, 0), Q(0, 4) and R(6, 2), and use vector methods to find the angle at Q. Exercise. Find the angle between a = 1, 1, 1 and b = 0, 2, 1. 15

16 Cauchy-Schwarz inequality. Using the algebraic definition of dot product, the mathematicians Cauchy and Schwarz demonstrated that v w v w, for all v, w in R n. Angle between vectors. Since v w v w 1 v w v w 1 for nonzero vectors, we can formally define the angle, θ, between nonzero vectors as follows: ( ) θ = cos 1 v w v w when v and w are nonzero vectors. Based on this definition, we can say that 1. Parallel: The nonzero vectors v and w are parallel when θ = 0 or π. 2. Orthogonal: The nonzero vectors v and w are orthogonal (or perpendicular) when θ = π 2. Equivalently, v and w are orthogonal when v w = 0. Note that if v w = 0, then either v = O or w = O or v and w are orthogonal. Exercise. For what values of x are the vectors x, x, 1 and 1, x, 6 orthgonal? Exercise. For what values of x is the angle between 1, 2, 1 and 1, 0, x equal to 60 o? 16

17 Projection of b onto a. Let θ be the angle between the nonzero vectors a and b. Then the projection of b onto a, proja(b), is the thick solid vector shown in each plot below. θ θ θ < π/2: If θ < π/2, the length of the projection vector is proja(b) = b cos(θ) and the vector points in the direction a. θ > π/2: If θ > π/2, the length of the projection vector is proja(b) = b cos(π θ) and the vector points in the direction opposite a. Algebraic method for finding the projection of b onto a. Geometry, trigonometry and the relationship a b = a b cos(θ) can be used to demonstrate that ( ) a b proja(b) = a. a a Note: The projection is a scalar multiple of a, with scalar equal to the ratio (a b)/(a a). Exercise. Sketch a = 6, 2 and b = 1, 5 as position vectors. Find proja(b) and b proja(b). 17

18 Exercise. Let a and b be nonzero vectors. Use the properties of dot product to demonstrate that a and b proja(b) are orthogonal. Exercise. Let a = P Q and b = P R, where P (0, 1, 1), Q(1, 3, 0) and R(4, 6, 3) are the corners of a triangle in 3-space. Find proja(b) and b proja(b). Use these answers to find the area of triangle P QR. 18

19 1.7 Timeout: Square Matrices, Determinants, Areas, Volumes [ ] A matrix is a rectangular array of numbers. For example, A = is a matrix with rows and 3 columns. A square matrix is a matrix with the same number of rows and columns. In this section, we will use square matrices to find areas of parallelograms in 2-space (left) and volumes of parallelepipeds in 3-space (right). Parallelogram Parallelopiped Vectors along adjacent sides will be used in the computations. 2-by-2 matrices, determinants, areas of parallelograms in 2-space. Consider a parallelogram with adjacent sides given by vectors a = a 1, a 2 and b = b 1, b 2. Let A be the 2-by-2 matrix whose rows are the components of a and b, respectively: [ ] a1 a A = 2. b 1 b 2 Then 1. The determinant of A is the number: det(a) = a 1 a 2 b 1 b 2 = a 1b 2 b 1 a Analytic geometry can be used to show that the area of the parallelogram is equal to the absolute value of the determinant of the matrix A: Area of parallelogram = det(a). 19

20 Exercise. Sketch the position vectors a = 0, 3 and b = 4, 2. Find the area of the parallelogram with adjacent sides a and b using the method above. 3-by-3 matrices, determinants, volumes of parallelopipeds. Consider a parallelopiped with adjacent sides given by vectors a = a 1, a 2, a 3, b = b 1, b 2, b 3 and c = c 1, c 2, c 3. Let A be the 3-by-3 matrix whose rows are the components of the 3 vectors: a 1 a 2 a 3 A = b 1 b 2 b 3. c 1 c 2 c 3 Then 1. The determinant of A is defined recursively: a 1 a 2 a 3 det(a) = b 1 b 2 b 3 c 1 c 2 c 3 = a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c Analytic geometry can be used to show that the volume of the parallelopiped with adjacent sides a, b and c is the absolute value of the determinant of the matrix A: Volume of parallelopiped = det(a). 20

21 Exercise. Find the volume of the parallelopiped in 3-space with a = 1, 5, 3, b = 2, 4, 1 and c = 0, 2, 0 along adjacent sides. 1.8 Cross Product of Vectors in R 3 If a, b R 3, then the cross product a b is the vector obtained as follows i j k a b = a 1 a 2 a 3 b 1 b 2 b 3 = a 2 a 3 b 2 b 3 i a 1 a 3 b 1 b 3 j + a 1 a 2 b 1 b 2 k, where i = 1, 0, 0, j = 0, 1, 0 and k = 0, 0, 1 are the standard basis vectors in R 3. In this formula, the absolute values indicate computing a determinant. For example, (a) If a = 1, 3, 2 and b = 2, 0, 2, then a b equals i j k = i j k =. (b) If a = 1, 2, 3 and b = 2, 4, 6, then i j k a b = = i j k =. 21

22 The cross product a b has several interesting geometric interpretations, including: (1) If a and b lie along adjacent sides of a parallelogram of positive area and θ is the angle between the vectors, then analytic geometry and trigonometry can be used to demonstrate that a b = a b sin(θ) is the area of the parallelogram. θ (2) a b is orthogonal to both a and b. (3) the ordered triple (a, b, a b) satisfies the right hand rule. (If you sweep from a to b with your right hand, then the cross product a b will point in the direction of your thumb.) Exercise. A parallelogram has corners P (2, 1, 0), Q(1, 3, 1), R( 1, 3, 4) and S(0, 1, 3) as you go around the boundary. Use cross product methods to find the area of the parallelogram P QRS and to find the area of the triangle P QR. 22

23 Properties of the cross product. Given a, b, c R 3 and d R, 1. Anticommutative: a b = (b a). 2. Distributive: a (b + c) = (a b) + (a c) and (a + b) c = (a c) + (b c). 3. Scalars: (da) b = a (db) = d(a b). 4. Parallel: If a and b are parallel, then a b = O. Notes: Cross products of the standard basis vectors yield unit vectors lying along the three coordinate axes. In particular, i j = k j k = i k i = j The cross product is not associative, in general. That is, (a b) c a (b c), as a general rule. Scalar triple product. The scalar triple product gives an interesting relationship between dot and cross products for vectors in 3-space: a 1 a 2 a 3 (a b) c = a (b c) = b 1 b 2 b 3 c 1 c 2 c 3. The quantity on the right is the determinant of the matrix whose rows are the components of the three vectors. Note: From the work we did in the last section, we know that if the three vectors lie along adjacent sides of a parallelepiped (as illustrated on page 19), then the volume of the parallelopiped is the absolute value of one of the two scalar triple products above. That is, Volume of parallelopiped = (a b) c = a (b c). Since the volume is the absolute value of the determinant of a 3 3 matrix, there is no need to first compute a cross product and then compute a dot product. 23

24 1.9 Equations of Lines and Planes in R 3 Lines. Let L be a line in 3-space and let P (x, y, z) be an arbitrary point on the line. L can be determined uniquely once we know an initial point P 0 (x 0, y 0, z 0 ) lying on L and a direction vector v = a, b, c parallel to L. (1) Vector Equation: Let r 0 = OP 0 and r = OP. Then the vector equation for L is r = r 0 + tv, where t is a scalar. (2) Parametric Equations: By equating the coordinates in the vector equation, we get x = x 0 + ta, y = y 0 + tb, z = z 0 + tc. These are the parametric equations of the line. (3) Symmetric Equations: The symmetric equations are equations involving x, y and z only. The symmetric equations are obtained by eliminating t from the parametric equations. Exercise. Let L be the line containing the points (1, 4, 2) and ( 4, 5, 1). Find a vector equation for L and find parametric and symmetric equations for L. 24

25 Exercise. Let L be the line containing the points (1, 4, 8) and (6, 4, 2). Find a vector equation for L and find parametric and symmetric equations for L. Exercise. Let L 1 = {(1 + 2t, 3 + 2t, 2 t) t R} and L 2 = {(6 + s, 2 s, s) s R}. Do L 1 and L 2 have a point in common? If so, find it. If not, explain why. 25

26 Planes. Let Π be a plane in 3-space and let P (x, y, z) be an arbitrary point on the plane. Π can be determined uniquely once we know an initial point P 0 (x 0, y 0, z 0 ) lying on Π and a vector n = a, b, c orthogonal to Π. (n is called a normal vector to the plane.) (1) Vector Equation: Let r 0 = OP 0 and r = OP. Then the vector equation for Π is n (r r 0 ) = 0, - or simply, n P 0 P = 0. (2) Scalar equation: The scalar equation is obtained by writing the dot product algebraically: a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. (3) Linear Equation: The linear equation is obtained by simplifying the scalar equation: ax + by + cz = d. Exercise. Find a linear equation for the plane containing the point (8, 1, 2) and orthogonal to the vector 5, 3, 2. Exercise. Find a vector equation for the plane 3x + 2y z =

27 Exercise. Find a vector equation for the plane containing the points (1, 2, 3), (2, 1, 2), (2, 3, 5). 27

28 Exercise. Let L 1 = {(1 + 2t, 1 t, 1 + t) t R} and L 2 = {( 4s, 2 + 2s, 4 2s) s R}. (a) Demonstrate that L 1 and L 2 are parallel. (b) Find a linear equation for the plane containing L 1 and L 2. 28

29 Exercise. Let L = {(1 + 2t, 5 t, 7 4t) t R} and Π = {(x, y, z) R 3 5x 2y + 3z = 16}. (a) Demonstrate that the line L lies on the plane Π. (b) Find a linear equation for the plane orthogonal to Π and containing L. 29

30 Distance from a point to a plane. Let Π be the plane with vector equation n P 0 P = 0 and let Q be a point in 3-space. Then, the distance between Q and Π is given by the following formula: D = projn(p 0 Q) = n P 0Q n That is, the distance is equal to the length of the projection of P 0 Q onto n. Question: Can you explain why the expressions above actually equal the distance D? Exercise. Find the distance between the point (2, 8, 5) and the plane x 2y 2z = 1. 30

31 1.10 Scalar Functions of Two Variables, Surfaces in R 3 Scalar functions of two variables. A scalar function of two variables f is a rule that assigns to each element of a set D R 2 a unique real number: f : D R 2 R. We often write z = f(x, y). The variables x and y are the independent variables, and z is the dependent variable. In this setting, (1) D is the domain of f. (2) R is the codomain of f. (3) R = {f(x, y) (x, y) R 2 } R is the range of f. R is the set of values that the function f assumes. (4) G = {(x, y, f(x, y)) (x, y) D} R 3 is the graph of f. G can be visualized as a surface in 3-space. Exercise. In each case, find the domain and range of the function. (a) z = x + y 1 (b) z = 16 x 2 4y 2 31

32 Traces. The cross-section of a surface in 3-space determined by holding one variable constant is often called a trace. Cross-sections help with visualization. For example, consider again the part of the surface z = x + y 1 for pairs (x, y) in the domain and in the rectangular region [0, 8] [ 8, 9]. The plots below show traces on the surface in one direction only: x = 2, 4, 6 y = 5, 0, 5 z = 1.5, 2.5, 3.5 Problem. Find equations for the three traces z = 1.5, 2.5, 3.5 shown above, and sketch all three curves on the xy-plane. Be sure to label the curves. 32

33 As a second example, consider again the part of the surface z = 16 x 2 4y 2 lying on and above the xy-plane. The plots below show traces on the surface in one direction only: x = 2, 0, 2 y = 1, 0, 1 z = 4, 8, 12 Problem. Find equations for the three traces z = 4, 8, 12 shown above, and sketch all three curves on the xy-plane. Be sure to label the curves. 33

34 Other surfaces. Surfaces in 3-space are not necessarily the graphs of functions, as illustrated by the following examples: (a) 4x 2 + 9y z 2 = 36 (b) x 2 + z 2 = 1 Exercise. In each case shown above, find the formula for a typical trace of the form y = k and state the range of values of k to which your formula applies. 34

35 1.11 Vector Functions, Curves in R 2 and R 3 Vector functions with n coordinates. A vector function r is a rule that assigns to each element of an interval I R a unique vector with n coordinates. We often write r(t) = x 1 (t), x 2 (t),..., x n (t), t I. The parametric equations for the vector function are x i = x i (t), i = 1, 2,..., n space: Vector functions with 2 coordinates are often written as r(t) = x(t), y(t) = x(t)i + y(t)j, t I, and visualized by graphing the curve C = {P (x(t), y(t)) t I} R space: Vector functions with 3 coordinates are often written as r(t) = x(t), y(t), z(t) = x(t)i + y(t)j + z(t)k, t I, and visualized by graphing the curve C = {P (x(t), y(t), z(t)) t I} R 3. Circles and ellipses in the plane. Two examples of particular interest are parametric descriptions of circles and ellipses in the plane. Let t be the radian measure of an angle measured counterclockwise from the positive x-axis. Then the following are parametric equations for 1. Circle: The circle with equation (x c 1 ) 2 + (y c 2 ) 2 = a 2 : x = c 1 + a cos(t), y = c 2 + a sin(t), 0 t < 2π. 2. Ellipse: The ellipse with equation (x c 1) 2 a 2 + (y c 2) 2 b 2 = 1: x = c 1 + a cos(t), y = c 2 + b sin(t), 0 t < 2π. For example, (a) Parametric equations for the circle (x 1) 2 + y 2 = 9 are as follows: x = π π/ y = - - π/ 35

36 (b) Parametric equations for the ellipse are as follows: x = x (y+1)2 9 = 1 π π/ y = - - π/ Other parametrizations of circles and ellipses are possible. For example, consider traversing the circle (x + 1) 2 + (y 2) 2 = 16 / clockwise in one unit of time, where locations for times 0, 1/4, 1/2 and 3/4 are shown to the right. / Find parametric equations for x and y. - - / - 36

37 Vector functions with 3 coordinates. Vector functions with three coordinates are often visualized as curves lying on surfaces in 3-space. For example, the function r(t) = 2 cos(t), t, sin(t) is visualized in the plot to the right lying on the elliptic cylinder x 2 + 4z 2 = 4. The curve is an example of an elliptic helix. Question: Can you see how I derived the equation of the cylinder from the equation for r(t)? 1.12 Calculus with Vector Functions, Tangents, Arc Length Calculus with vector functions is done component by component. Let r(t) = x 1 (t), x 2 (t),..., x n (t), t I, be a vector function with n coordinates, and let a I be an element of the domain. 1. Limit: The limit of r(t) as t a is defined as follows: lim r(t) = lim x 1(t), lim x 2 (t),..., lim x n (t), t a t a t a t a provided the limits of the component functions exist. 2. Continuity: The vector function r is continuous at a if lim r(t) = r(a). t a Thus, r is continuous at a iff each component function is continuous at a. 37

38 3. Differentiable: the vector function r is differentiable at a if r(t) r(a) lim t a t a exists. Thus, r is differentiable at a iff each component function is differentiable at a. Further, if r is differentiable at a, the value of the limit is written as r (a) and r (a) = x 1(a), x 2(a),..., x n(a). Notes: (1) The vector in the numerator of the definition of r (a) lies on a secant line to the curve C. As t a, the secant line approaches a tangent line, if a tangent line exists. (2) The following theorem specifies when a tangent line exists and provides a vector equation for the line. Theorem (Tangent Lines). If each coordinate function is differentiable in a neighborhood of a and if r (a) O, then a tangent line exists at a. A vector equation for the tangent line is l(t) = r(a) + r (a)t. Note that a neighborhood of a refers to all points in an interval of the form (a δ, a + δ), where δ is a positive number. (3) When a tangent line exists, the vector r (a) points in the direction of increasing t. (4) When t is a time parameter, then r (a) is interpreted as the velocity vector at time a, and its length is interpreted as the speed at time a. Derivative function. The derivative function for a vector function r(t) is defined as follows: r (t) = lim t 0 r(t + t) r(t), if this limit exists. t When the limit exists, the value is r (t) = x 1 (t), x 2 (t),..., x n(t). 38

39 Exercise. Let r(t) = 4 cos(t), 2 sin(t), t [0, 2π]. (a) Find the derivative function r (t). (b) Find a vector equation and parametric equations for the tangent line when t = π/3. (c) Find an equation in x and y describing the underlying curve of r(t). π π/ π/ π/ 39

40 Exercise. Let r(t) = e t, e 2t, t R. (a) Find the derivative function r (t). - (b) Find a vector equation and parametric equations for the tangent line when t = 0. (c) Find an equation in x and y describing the underlying curve of r(t). 40

41 Exercise. Let r(t) = t 2 1, t 2 + 1, t + 1, t R. (a) Find the derivative function r (t). (b) Find a vector equation and parametric equations for the tangent line when t = 1. (c) Write an equation for the surface drawn in the plot using the variables x, y, z only. 41

42 Exercise. Let r(t) = 2 cos(t), 2 sin(t), 4 cos(2t), when t [0, 2π]. (a) Find the derivative function r (t). (b) Find a vector equation and parametric equations for the tangent line when t = π/6. (c) Write an equation for the surface drawn in the plot using the variables x, y, z only. 42

43 Exercise. Let r(t) = t 2, t 3, for t [ 2, 2]. (a) Sketch r(t), and indicate the direction of increasing t. (b) Find a vector equation and parametric equations for the tangent line when t = 1. (c) Explain why a tangent to the curve at t = 0 does not exist

44 Arc length. Suppose that r(t) is defined and differentiable for all t [a, b] and that r (t) O, for all t (a, b). If the curve C underlying r(t) is traversed exactly once as t increases from a to b, then it can be shown that the length of C is given by the following formula: L = b a r (t) dt. Note: When t is a time parameter, then arc length can be interpreted as total distance travelled. The total distance travelled is obtained by integrating the speed over the time interval [a, b]. Exercise. Let r(t) = t 2, t 3, t 2, t [0, 2]. (a) Find a simplified formula for r (t). (b) Find the length of the curve underlying r(t) for t [0, 2]. 44

45 Exercise. Let r(t) = 2 cos(t), t, 2 sin(t), t [0, 6π]. (a) Find a simplified formula for r (t). (b) Find the length of the curve underlying r(t) for t [0, 6π] Parametrized Surfaces in R 3 A vector function of two variables r is a rule that assigns to each element of a set D R 2 a unique vector with n coordinates. We often write r(u, v) = x 1 (u, v), x 2 (u, v),..., x n (u, v), (u, v) D. Each coordinate function, x i = x i (u, v), is a function of the two parameters u and v. Parametrized surfaces in 3-space. Vector functions with 3 coordinates are written as r(u, v) = x(u, v), y(u, v), z(u, v), (u, v) D, and visualized by graphing the surface S = {P (x(u, v), y(u, v), z(u, v)) (u, v) D}. In this case, the surface S has been parametrized using the vector function r. 45

46 Graphs of scalar functions of two variables. Let G = {(x, y, f(x, y)) (x, y) D} be the graph of the scalar function f, where D is the domain of f. Then G can be written as a parametrized surface using the vector function r(x, y) = x, y, f(x, y), (x, y) D. For example, the graph of f(x, y) = x 2 y 3xy 2, (x, y) R 2, can be parametrized with vector function r(x, y) = x, y, x 2 y 3xy 2, (x, y) R 2. Exercise. In each case, find a vector function r that can be used to parametrize the surface. (a) y = 7xz + z 2 (b) x = z 2 46

47 Circular and elliptic cylinders. Circular and elliptic cylinders cannot be viewed as the graphs of a scalar function of two variables, no matter which variables are chosen as the independent variables. However, the identity sin 2 (θ) + cos 2 (θ) = 1, for all θ, can be used to find parametric descriptions of these surfaces. Exercise. In each case, find a vector function r that can be used to parametrize the surface. (a) (y 1) 2 + (z + 1) 2 = 9 (b) x 2 + 4z 2 = 4 47

48 Parametric descriptions of planes. The plane Π containing initial point P 0 (x 0, y 0, z 0 ) and parallel to direction vectors a and b can be parametrized using the vector function r(u, v) = OP 0 + OP = OP 0 + ua + vb, (u, v) R 2, where P (x, y, z) is an arbitrary point on the plane. The direction vectors a and b must be non-zero and nonparallel. For example, the plot to the right illustrates the parametrized surface r(u, v) = v 1, u 1, u + v 1, for (u, v) R 2, where the initial point and direction vectors are as follows (please complete) Exercise. If you connect the points P (0, 5, 10), Q(2, 1, 8), R(4, 0, 7), S(2, 4, 9) in the order given, you will get a parallelogram in 3-space. Find a vector function r that can be used to parametrize the plane containing this parallelogram. 48