MATH 2730: Multivariable Calculus. Fall 2018 C. Caruvana

Transcription

1 MATH 273: Multivariable Calculus Fall 218 C. Caruvana

2 Contents 1 Vectors Vectors in the Plane Vector Addition Vector Components Vectors in Space A Note on Lines The Dot Product Projections Work The Cross Product Geometric Properties of the Cross Product Algebraic Properties of the Cross Product Torque Lines and Curves in Space Lines More General Curves Motion in Space Length of Curves The Polar Case Parameterizing by Arc Length Curvature Principal Unit Normal Vector Torsion Functions of Several Variables Surfaces in Space Functions of Two Variables and Their Graphs Level Curves Functions of Three Variables Level Surfaces Limits and Continuity Limits and Continuity in Three Variables Partial Derivatives A Preview of Differentiability Partial Derivatives in Three Variables The Chain Rule More Variables Implicit Differentiation Directional Derivatives and the Gradient Directional Derivatives and the Gradient in Three Variables Tangent Planes Linear Approximation Differentials Differentials with Three Variables Extreme Values for Functions of Two Variables Lagrange Multipliers ii

3 3 Integration in Multiple Variables Double Integrals Over Rectangular Regions Over Other Regions Using Polar Coordinates Triple Integrals Using Rectangular Coordinates Using Cylindrical Coordinates Using Spherical Coordinates Change of Variables Vector Fields Line Integrals Conservative Vector Fields iii

4 1 Vectors 1.1 Vectors in the Plane Definition 1. A vector (in the plane) is something which has magnitude and direction. Let s elaborate on the differences between planar vectors and points. Below, we plot some points P, Q, and R in the plane. P Q R Notice that these are just locations. Without any particular point of reference, magnitude and direction are not coherent properties of these points. Now, we will plot some vectors v, w, and x in the plane. v w x Now, the magnitude of each vector corresponds to the length of the vector and the direction is that direction to which the arrow points. Notation. For a vector v, we will let v denote its magnitude. One may also notice that, inspired by the two previous plots, vectors can be determined by pairs of points in the plane. Namely, if P and Q are points in the plane, then we can define the vector P Q to be the one starting at point P, ending at point Q, and pointing in the direction of Q. Graphically, 1

5 Q P Q P With the same points P and Q, compare P Q to QP : Q QP P We will say that two vectors v and w are equal, written v w, provided that v and w have the same magnitude and point in the same direction. The key point in this equality is that location doesn t matter. To illustrate this, in the following plot, both v and w are equal as vectors: v w Definition 2. The zero vector, which we will denote by, is the unique vector which has no magnitude and no direction. 2

6 In the context of points, for any point P in the plane, the zero vector can be thought of as P P. Definition 3. By a scalar, we mean any a R. That is, a scalar is a real number. The scalars act on vectors in a very natural way. Definition 4. We define scalar multiplication between scalars and vectors in the following way: Given a scalar a and a vector v, we define the vector a v by cases: if a >, then a v is the vector with magnitude a v and the same direction as v; if a, then a v is the zero vector; if a <, then a v is the vector with magnitude a v and the opposite direction of v. Definition 5. a v w. We say that two vectors v and w are parallel if there exists a scalar a so that For example, the three vectors plotted below are parallel Vector Addition Vector addition is best seen geometrically. For example, suppose we wish to add the following two vectors together. v w 3

7 Since the location of the vectors doesn t matter, we will show two geometric ways to construct the addition: w w v + w v v + w v v w Figure 1: Triangle Rule Figure 2: Parallelogram Rule Just as we can add vectors, we can make sense of subtraction. w w v v v Before we move on to developing a calculus of vectors, let s combine scalar multiplication and vector addition: v v w 2 v w 3 w w 3 w 2 v 4

8 1.1.2 Vector Components Since the position of vectors in the plane is immaterial for our calculus, we can standardize vectors by situating them in the Cartesian plane with their base at the origin. Then each vector will be uniquely determined by the location of its head. Let s first sketch some vectors along with the vectors in standard position. v v w u u w Figure 3: Standard Positions In Figure 3, we see that the head of v in standard position is located at the point (1, 2), the head of w in standard position is located at the point (2, 1), and the head of u in standard position is located at the point ( 3, 1). To keep the distinction between points and vectors clear, we introduce special notation for position vectors. Definition 6. We define the position vector v x, y to be the unique vector with its base at (, ) and its head at the point (x, y ). Moreover, we call the number x the x-component of v and the number y the y-component of v. So, referring back to Figure 3, the vector v is equal to the position vector 1, 2, the vector w is equal to the position vector 2, 1, and the vector u is equal to the position vector 3, 1. What one should immediately notice is that to vectors v 1 x 1, y 1 and v 2 x 2, y 2 are equal if and only if x 1 x 2 and y 1 y 2. With these standardized position vectors, it may be difficult for one to keep the difference between points and vectors present. But they truly are categorically different. For example, a point is a particular location in the Cartesian plane but a vector, as we have so set forth, is really the collection of all planar vectors which are equal to the standard position vector. In Figure 4, some particular planar vectors which are all equal are plotted. The black vector is the position vector. 5

9 Figure 4: Some equal planar vectors Recall, from Definition 1, that vectors have magnitude. We can compute the magnitude of any vector by calculating the magnitude of the corresponding position vector. Namely, given a position vector v x, y, the magnitude of v is given by v x 2 + y2. This is clearly an application of the Pythagorean Theorem: y x, y x More generally, if you have a vector in the plane with its base at the point (x, y ) and its head at the point (x 1, y 1 ), the corresponding position vector is x 1 x, y 1 y. Notice that x 1 x, y 1 y (x 1 x ) 2 + (y 1 y ) 2. Position vectors also give us an elegant way of doing scalar multiplication and vector addition. In particular, if a and b are scalars and v v 1, v 2 and w w 2, w 2 are position vectors, a v + b w av 1 + bw 1, av 2 + bw 2. 6

10 Definition 7. A unit vector is any vector v so that v 1. We also distinguish the coordinate unit vectors î 1, and ĵ, 1. For any non-zero vector v, notice that 1 v v is a unit vector with the same direction as v. Also, an important property of î and ĵ is that, the position vector x, y xî + yĵ. Let s summarize the algebraic properties of vectors. For all scalars a and b and vectors u, v, and w: + u u + u Additive identity u + v v + u Commutativity of vector addition ( u + v) + w u + ( v + w) Associativity of vector addition u + ( u) Additive inverse a( u + v) a u + a v Distributive property of scalars (a + b) u a u + b u Distributive property of vectors u Multiplication by zero scalar a Multiplication by zero vector 1 u u Multiplicative identity a(b u) (ab) u Associativity of scalar multiplication These properties allow us to solve algebraic expressions. Example 1. Solve for u. Solution. Notice that 3 u + 4 v 12 w 3 u + 4 v 12 w 3 u + 4 v 4 v 12 w 4 v 3 u 12 w 4 v u 1 (12 w 4 v) 3 u 4 w 4 3 v. If we know the magnitude of a vector and the angle it makes with the positive x-axis, we can find its components in the following way: 7

11 v v sin(θ) θ v cos(θ) That is, for the vector v pictured above, the corresponding position vector is v cos(θ), v sin(θ). Example 2. A 4-lb engine is suspended from two chains that form a 6 angle with a horizontal ceiling. How much weight does each chain support? Solution. Let s start with a picture: 6 6 v 1 v 2 Observe that w 4-lb v 1 v 1 cos(6 ), v 1 sin(6 ) and v 2 v 2 cos(6 ), v 2 sin(6 ) and that To be at a state of equilibrium, we need w, 4. v 1 + v 2 + w, v 1 + v 2 w ( v 2 v 1 ) cos(6 ), ( v 1 + v 2 ) sin(6 ), 4. Notice that ( v 2 v 1 ) cos(6 ) v 1 v 2. It follows that 2 v 1 sin(6 ) 4 v 1 v So each chain must be able to support at least lbs. 8

12 Example 3. Three forces are applied to an object as shown below. F 2 1-lb F 1 6-lb F 3 15-lb Find the magnitude and direction of the sum of the forces. Solution. First note that F 1 6 cos(3 ), 6 sin(3 ) F 2 1 cos(45 ), 1 sin(45 ) F 3 15 cos(6 ), 15 sin(6 ) Then, compute 6 cos(3 ) 1 cos(45 ) 15 cos(6 ) That is, 93.75; 6 sin(3 ) + 1 sin(45 ) 15 sin(6 ) F 1 + F 2 + F , To get an approximation of the angle this vector makes with the x-axis, observe that ( ) arctan Graphically,

13 1.2 Vectors in Space We will now extend our investigations to three dimensions. The world around us seems to have properties like length, width, and height. For example, an ant on a lily pad can be located using only two dimensions (relative to the lily pad) but the position of a frog in the trees looking for a quick lunch is best modeled using three dimensions. There is some location relative to the ground component and an altitude component. Here is the standard xyz-coordinate system: z y A point in the xyz-coordinate system is an ordered triple: x (a, b, c) where a, b, c R Then a is the x-coordinate, b is the y-coordinate, and c is the z-coordinate. We call the point (,, ) the origin. This is similar to the affairs of the plane. Remark 1. The xyz-coordinate system will also be colloquially referred to as space. To plot a point in the xyz-coordinate system using its ordered triple of components (a, b, c), one graphs by traveling a units along the x-axis, b units along the y-axis, and c units along the z-axis. An example of a plotted point (a, b, c) follows: z (a, b, c) y x Before we get into vectors, let s discuss a conventional orientation for the z-axis. The convention gives us an orientations according to the right hand rule. We will give three descriptions of this in words: 1

14 1. If one curls the fingers of the right hand so that they curl from the positive x-axis to the positive y-axis, then the thumb points in the direction of the positive z-axis. 2. If one puts the right hand with the fingers pointing in the direction of the positive x-axis (perhaps with the palm at the origin), then two cases are possible. Either, one must curl one s fingers in the counter-clockwise direction to approach the positive y-axis or one must curl one s fingers in the clockwise direction to approach the positive y-axis. In either case, the direction in which the thumb points is still the direction of the positive z-axis. 3. If one orients the xy-plane as usual (with the positive x-axis pointing to the right and the positive y-axis pointing to the sky), then the positive z-axis is that axis which points from the plane to the viewer. Just as we had in two dimensions, vectors (things with magnitude and direction) exist in three dimensions and we will say, just as before, that two vectors (in space) are equal provided they have the same magnitude and direction. Hence, points in space are categorically different than vectors in space. Recall that we can express curves in the plane with equations. For example, we can think of the equation x 2 + y 2 1 as representing the curve {(x, y) : x 2 + y 2 1}, the unit circle. In a similar way, we can express some collections of points in space using algebraic expressions. In particular, we can identify the xy-plane with {(x, y, z) : z } which can be concisely expressed with just the equation z. Then the equation y expresses the xz-plane and the equation x expresses the yz-plane. Let s take a look: z z z y y y x x x xy-plane xz-plane yz-plane Similarly, we can look at other planes like z 3: 11

15 z (,, 3) (,, ) y x We will give a rigorous definition for what we mean by a plane later on when we discuss planes more in depth. But for now, let s offer the notion of parallel planes. Definition 8. Example 4. Two planes are said to be parallel if they have no points in common. The following are pairs of parallel planes: The planes x and x 2 are parallel. The planes x 4 and x π are parallel. The planes y 12 and y 2 are parallel. The following are pairs of planes which are not parallel: The planes x 1 and y 2 are not parallel. The point (1, 2, 3) is in both planes. The planes x 2 and z π are not parallel. The point ( 2, 1, π ) is in both planes. For points in the plane, the distance formula is analogous to the familiar distance formula in two dimensions. Namely, given two points P 1 (x 1, y 1, z 1 ) and P 2 (x 2, y 2, z 2 ), the distance between them is given by (x2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2. It turns out that the line connecting P 1 and P 2 can be described by the set {(1 t)p 1 + tp 2 : t [, 1]} {((1 t)x 1 + tx 2, (1 t)y 1 + ty 2, (1 t)z 1 + tz 2 ) : t [, 1]}. In particular, if t 1, we get the midpoint 2 ( x1 + x 2, y 1 + y 2, z ) 1 + z on the line segment joining P 1 and P 2. One may also notice that the description given above is particularly a directed line starting at P 1 and ending at P 2 if we give it the natural order of. With the distance formula in hand, we can now define spheres and balls in space. Particularly, given a point (a, b, c), then the 12

16 sphere centered at (a, b, c) with radius r > is described by the equation and the (x a) 2 + (y b) 2 + (z c) 2 r 2 ball centered at (a, b, c) with radius r > is described by the equation In words, (x a) 2 + (y b) 2 + (z c) 2 r 2. the sphere centered at (a, b, c) with radius r > is the set of all points which have distance r from the point (a, b, c) and the ball centered at (a, b, c) with radius r > is the set of all points which have distance at most r from the point (a, b, c). Definition 9. Just as in the planar case, a position vector is a vector with its base at the origin. Then, such a vector is determined uniquely by the location of its head. So we will let a, b, c be the position vector which has its base at the origin (,, ) and the head at the point (a, b, c). We will say that a is the x-component, b is the y-component, and c is the z-component. Graphically, z (a, b, c) v y x where v a, b, c. Fortunately, the operations of scalar multiplication and vector addition generalize easily to the three-dimensional case. Namely, given two vectors v v 1, v 2, v 3 and w w 1, w 2, w 3 in space and scalars a and b, a v + b w av 1 + bw 1, av 2 + bw 2, av 3 + bw 3. Moreover, all of the same algebraic properties hold in the space case as in the planar case. Definition 1. Also as in the two-dimensional case, we will say that two vectors v and w in space are parallel if there exists a scalar a so that a v w. To find the magnitude of any vector in space we use the distance formula. If a vector v is a position vector a, b, c, then v a 2 + b 2 + c 2. 13

17 Definition 11. A unit vector is any vector with a magnitude of 1. Now, since we are in three dimensions, we distinguish the three following coordinate unit vectors: î 1,, ĵ, 1, ˆk,, 1 Exercise 1. Give a geometric description of the set of points satisfying x 2 + y 2 + z 2 4x + 9y 11. Exercise 2. radius? Exercise 3. If P (1, 3, 5) and Q( 5, 1, 7) lie on the diameter of a sphere, what is the center and Find two unit vectors parallel to P Q where P (1,, 1) and Q(3, 2, ) A Note on Lines Recall that any line in the plane can be written using an equation of the form or y mx + b x a in the case that we wish to describe a vertical line. If we view these as curves in the plane, we have in the former case and {(x, y) : y mx + b} {(x, mx + b) : x R} {(a, y) : y R} in the latter. Now, let s take a different view. What about the line joining the two points (x 1, y 1 ) and (x 2, y 2 )? We can draw this easily: (x 2, y 2 ) (x 1, y 1 ) It turns out that this line will be the set of points { x (1 t)x1 + tx 2 But why? y (1 t)y 1 + ty 2 14

18 Definition 12. We say that three points P, Q, and R are co-linear if there exists a line l so that each point P, Q, and R lies on l. Let s examine three points which are not co-linear and some lines connecting them: R R P P Q Q Figure 5 For points P and Q, let d(p, Q) denote the distance between P and Q. What Figure 5 suggests is the following inequality, known as the triangle inequality: For any three points P, Q, and R, In Figure 5, we in fact see that d(p, R) d(p, Q) + d(q, R). d(p, R) < d(p, Q) + d(q, R). One can easily verify that if P, Q, and R are co-linear, equality holds; i.e. d(p, R) d(p, Q) + d(q, R). Now, supposing that we have three points P, Q, and R so that d(p, R) d(p, Q) + d(q, R), perhaps by resorting to trigonometric methods, one could verify that P, Q, and R are indeed co-linear. Let P (x 1, y 1 ), R (x 2, y 2 ), and Q (x t, y t ) where x t (1 t)x 1 +tx 2 and y t (1 t)y 1 +ty 2 for some fixed t [, 1]. Then notice that d(p, Q) (x 1 x t ) 2 + (y 1 y t ) 2 (x 1 ((1 t)x 1 + tx 2 )) 2 + (y 1 ((1 t)y 1 + ty 2 )) 2 (x 1 x 1 + tx 1 tx 2 ) 2 + (y 1 y 1 + ty 1 ty 2 ) 2 t 2 (x 1 x 2 ) 2 + t 2 (y 1 y 2 ) 2 t (x 1 x 2 ) 2 + (y 1 y 2 ) 2 15

19 and that d(q, R) (x 2 x t ) 2 + (y 2 y t ) 2 (x 2 ((1 t)x 1 + tx 2 )) 2 + (y 2 ((1 t)y 1 + ty 2 )) 2 (x 2 x 1 + tx 1 tx 2 ) 2 + (y 2 y 1 + ty 1 ty 2 ) 2 [(x 1 x 2 )(t 1)] 2 + [(y 1 y 2 )(t 1)] 2 (x 1 x 2 ) 2 (t 1) 2 + (y 1 y 2 ) 2 (t 1) 2 (t 1) 2 [(x 1 x 2 ) 2 + (y 1 y 2 ) 2 ] t 1 (x 1 x 2 ) 2 + (y 1 y 2 ) 2 which, along with the facts that t t and t 1 1 t since t 1, provides us with the fact d(p, Q) + d(q, R) t d(p, R) + (1 t) d(p, R) d(p, R). So we see that P, Q, and R are co-linear. When we put things in terms of vectors, we can say things more simply. Fact. Three points P, Q, and R in space are co-linear if and only if the vectors P Q and P R are parallel. 1.3 The Dot Product Let s start with some motivation. Sometimes it s useful to determine whether or not two vectors are perpendicular to each other; i.e., the angle between them is a 9 angle. Consider the following picture: v w v w w If it were the case that the angle between v and w were 9, then the Pythagorean Theorem would provide us with the fact that v 2 + w 2 v w 2. Even further, v 2 + w 2 v w 2 v v v w w w 2 3 (v 1 w 1 ) 2 + (v 2 w 2 ) 2 + (v 3 w 3 ) 2 which, after expanding the right-hand side, provides us with v 1 w 1 + v 2 w 2 + v 3 w 3. 16

20 Definition 13. For two vectors v and w, we define the dot product by the rule v w v 1 w 1 + v 2 w 2 + v 3 w 3. The dot product between two vectors gives a scalar as a result. For this reason, it is sometimes called the scalar product. Remark 2. For two-dimensional vectors v and w, v w v 1 w 1 + v 2 w 2. The following properties of the dot product, which can be easily verified from the definition, are recorded for convenient reference. For all scalars a and b and vectors u, v, and w: v v v 2 v v v w w v a v b w (ab)( v w) u ( v + w) u v + u w ( v + w) u v u + w u Self-dot is square norm Dot with zero vector Commutativity of the dot Scalar distribution Left-distributivity Right-distributivity Exercise 4. What s wrong with the expression u v w? Remark 3. When working with only real numbers, we are accustomed to operations on pairs of objects to produce an object of the same kind. It is worth reiterating that the dot product between two vectors produces a scalar, something which is categorically different than a vector. With that said, do not make the mistake of writing because that s not true. v w v 1 w 1, v 2 w 2, v 3 w 3 Example 5. In general, even if u, it s not true that u v u w v w. Solution. Notice that î ĵ and î ˆk. Hence, î ĵ î ˆk, î, and ĵ ˆk. Exercise 5. Given that a 1, b 3, a b 2, find ( a b) (2 a + b). Definition 14. For two vectors v and w, we say that v and w are orthogonal, denoted v w, provided that v w. Theorem 15. For any two vectors v and w, let θ be the angle between them. Then v w v w cos(θ). 17

21 Hence, the notion of being orthogonal corresponds to being perpendicular, as motivated earlier. The advantage to having the dot product as defined here is that computation is seen to be more straight forwards only given vectors in standard position. For example, if one wishes to determine whether or not the vectors v : 2î + ĵ + ˆk and w : î + ĵ 3ˆk are orthogonal, one need not try to compute the angle between them. In fact, (2î + ĵ + ˆk) (î + ĵ 3ˆk) which shows us that v w. On the other hand, if we do wish to compute the angle between two vectors, the dot product can help us do that. Example 6. Find the angle between the two vectors v 1, 1, 2 and w 1, 2,. Solution. Notice that Theorem 15 provides the fact that so we can write v w v w cos(θ) cos(θ) v w v w Using computational technology, we see that ( ) θ cos Projections Definition 16. by For any non-zero vector v, we define the projection of another vector u onto v proj v u ( ) u v v. v v As one can readily recognize, proj v u is a vector which is parallel to v. The following picture demonstrates the geometric idea of projections: u u v proj v u v 18

22 In fact, consider the following identity: proj v u ( u v v v ) v ( ) u v v v v. It is immediately clear from this equality that the vector proj v u has magnitude u v v. Also, for the coordinate unit vectors and any other vector a, b, c, projî a, b, c a,,, projĵ a, b, c, b,, projˆk a, b, c,, c. If we know the angle θ between two vectors u and v and both of their magnitudes, we can compute the projection via Theorem 15 as follows: ( ) u v v proj v u v v ( ) u v cos(θ) v v v v u cos(θ) v. Example 7. Solution. Notice that Also, Find the projection from 8,, 2 onto 1, 3, 3. 1, 3, ,, 2 1, 3, Therefore, proj 1,3, 3 8,, , 3, , 42 19, Exercise 6. Suppose that u and v are parallel vectors. Show that, for any vector w, proj u w proj v w Work Given a constant force vector F and a displacement vector d, the work done is given by W F d. Sometimes we measure force in Newtons: 19

23 N kg m s 2 The units for work will usually be in joules: J kg m2 s 2 Example 8. Find the work done by a force F 2, 4, 1 (in Newtons) moving an object from (,, ) to (2, 4, 6) (in meters). Solution. Notice that the displacement vector is 2, 4, 6. Then, we calculate the work to be So the work is 26 J. 2, 4, 1 2, 4, Example 9. Find the components of the vertical force F, 1 in the directions parallel to and normal to the inclined plane which makes an angle of θ arctan(4/5) with the x-axis. Solution. Let s start with a picture of the situation: To find a vector with the same direction as the parallel, we must calculate 5 cos(θ), sin(θ), θ Notice that 4, 5 5, so we see that the normal direction is given by the vector 4, Let P 5, and N 4, Since both vectors P and N are unit vectors, observe that ( ) proj P F F P P ( ) 4 P ,

24 and that ( ) proj N F F N N ( ) 5 N , Lastly, we record that proj P F + proj N F 2 41, , 25 41, The Cross Product Given two non-zero vectors (which are not parallel to each other), there is a unique plane containing both of them. It will be useful to us to be able to identify a vector which is orthogonal to both of the given vectors. More precisely, suppose v v 1, v 2, v 3 and w w 1, w 2, w 3 are non-zero vectors and that p p 1, p 2, p 3 is orthogonal to v and w. Then we have that and v 1 p 1 + v 2 p 2 + v 3 p 3 (1) w 1 p 1 + w 2 p 2 + w 3 p 3 (2) If it were the case that v 3 w 3, we could let p,, 1 and check that v p and w p, establishing that p is orthogonal to both v and w. So, proceed under the assumption that v 3 or w 3. Notice that and that which provides v 1 w 3 p 1 + v 2 w 3 p 2 + v 3 w 3 p 3 v 3 w 1 p 1 + v 3 w 2 p 2 + v 3 w 3 p 3 (v 1 w 3 v 3 w 1 )p 1 + (v 2 w 3 v 3 w 2 )p 2. (3) Notice that (3) looks like Ap 1 + Bp 2 which has the obvious solution p 1 B, p 2 A. That is, p 1 v 2 w 3 v 3 w 2, p 2 v 3 w 1 v 1 w 3. Putting these values into (1) gives us v 1 (v 2 w 3 v 3 w 2 ) + v 2 (v 3 w 1 v 1 w 3 ) + v 3 p 3 v 1 v 2 w 3 v 1 v 3 w 2 + v 2 v 3 w 1 v 1 v 2 w 3 + v 3 p 3 v 2 v 3 w 1 v 1 v 3 w 2 + v 3 p 3 v 3 (v 2 w 1 v 1 w 2 + p 3 ) 21

25 and putting the into (2) gives us w 1 (v 2 w 3 v 3 w 2 ) + w 2 (v 3 w 1 v 1 w 3 ) + w 3 p 3 v 2 w 1 w 3 v 3 w 1 w 2 + v 3 w 1 w 2 v 1 w 2 w 3 + w 3 p 3 v 2 w 1 w 3 v 1 w 2 w 3 + w 3 p 3 w 3 (v 2 w 1 v 1 w 2 + p 3 ). Since we were assuming that either v 3 or w 3, we see that Therefore, the vector is orthogonal to both v and w. Definition 17. to be p 3 v 1 w 2 v 2 w 1. v 2 w 3 v 3 w 2, v 3 w 1 v 1 w 3, v 1 w 2 v 2 w 1 For two vectors v v 1, v 2, v 3 and w w 1, w 2, w 3, we define the cross product v w v 2 w 3 v 3 w 2, v 3 w 1 v 1 w 3, v 1 w 2 v 2 w 1. Remark 4. Unlike the dot product, the cross product between two vectors produces a vector. For this reason it is sometimes called the vector product. To facilitate these studies, we must introduce matrices and determinants. Definition 18. For a 2 2 matrix ( a b c d consisting of real numbers, the determinant is defined to be a b c d ad bc. ) For a 3 3 matrix a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 consisting of real numbers, the determinant is defined to be a 1 a 2 a 3 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 2. Then we can write the cross product in terms of determinants: v 1, v 2, v 3 w 1, w 2, w 3 v 2 v 3 w 2 w 3 î v 1 v 3 w 1 w 3 ĵ + v 1 v 2 w 1 w 2 ˆk or, loosely: The key to all of this is that v 1, v 2, v 3 w 1, w 2, w 3 22 î ĵ ˆk v 1 v 2 v 3 w 1 w 2 w 3.

26 Theorem 19. The vector v w is orthogonal to both v and w. Proof. Compute ( v w) v and ( v w) w. Example 1. Compute Solution. Check that î ĵ, ĵ ˆk, ˆk î. î ĵ ĵ ˆk ˆk î î ĵ ˆk 1 1 î ĵ ˆk 1 1 î ĵ ˆk 1 1 Example 1 inspires the following diagram: î ĵ + 1ˆk ˆk, 1î ĵ + ˆk î, î ( 1)ĵ + ˆk ĵ. î ˆk ĵ Figure 6 The directionality of Figure 6 will be elaborated on after we establish some more properties of the cross product Geometric Properties of the Cross Product Now that we have the cross product between two vectors defined and know that it is orthogonal to the given vectors, we must understand its properties of magnitude and direction. The orientation of v w follows the right-hand rule. That is, using the right hand, then pointing in the direction of v with the index finger and pointing in the direction of w with the middle finger, the direction of v w is the orthogonal direction indicated by the thumb. To address the magnitude, we have Theorem 2. If θ ( θ π) is the angle between v and w, v w v w sin(θ). Corollary 21. Two non-zero vectors v and w are parallel if and only if v w. Actually, Theorem 2 has a completely geometric interpretation. Given vectors v and w, we can consider the parallelogram determined by them: 23

27 w v v sin(θ) v θ w What is made clear from this is that the area of the parallelogram determined by v and w is v w sin(θ) v w Algebraic Properties of the Cross Product Here, we record some algebraic properties of the cross product. For all scalars a and b and vectors u, v, and w: v v Cross with zero vector v v Self-cross is zero v w w v Anti-commutativity (a v) (b w) (ab)( v w) Scalar distribution u ( v + w) u v + u w Left-distributivity ( v + w) u v u + w u Right-distributivity u ( v w) ( u v) w u ( v w) ( u w) v ( u v) w Recall from Example 1 that î ĵ ˆk, ĵ ˆk î, ˆk î ĵ. Using the property of anti-commutativity of the cross product, ĵ î ˆk, ˆk ĵ î, î ˆk ĵ. Hence, we see that, referring back to Figure 6, that the reverse direction gives us a way to remember any cross product between coordinate unit vectors. Notice that we didn t offer any associativity property for the cross product. In fact, associativity fails in general for the cross product. Exercise 7. Show that î (î ĵ) (î î) ĵ. The cross product also fails to satisfy the cancellation property discussed in Example 5. Exercise 8. Show that, in general, it is not true that, given u, u v u w v w. 24

28 1.4.3 Torque Consider a force vector F acting on a rigid body represented by a vector r. Then the rotational vector representing the twisting effect is captured by the vector For example, consider the following diagram: τ r F. r F p F Figure 7: Torque If we can imagine the vector r in Figure 7 being rotated about point p by the force vector F, we see that the direction of the cross product r F is into the page. That is, the twisting motion, if point p were representing a bolt and r a wrench, is tightening the bolt into the page, figuratively speaking. This should suggest the proverbial righty tighty, lefty loosey. To elaborate even more, consider the following diagram, similar to Figure 7: F p r F Figure 8 Exercise 9. In Figure 8, what is the direction of r F? Example 11. Consider the following diagram: r θ F p Assuming F is fixed, find θ ( θ π) so that the magnitude of the torque r F is maximized. Solution. By Theorem 2, we have that r F r F sin(θ). 25

29 Notice that sin(θ) 1 for each θ π. We also know that sin(π/2) 1, so sin(θ) sin(π/2) r F r F sin(θ) r F sin(π/2) for all θ π. Therefore, θ π/2 maximizes the magnitude of the torque. 1.5 Lines and Curves in Space If one were to model an object s trajectory through space, one could parameterize based on time and position. In particular, we can use functions f : I R 3 where I R is a non-degenerate interval. Usually, when modeling motion, I [, ), or so. It will be quite useful to write such functions parametrically; that is, f(t) (x(t), y(t), z(t)) or x x(t) y y(t) z z(t) where x(t), y(t), and z(t) are real-valued functions. To begin with our investigation of trajectories in space, we can start with a simple kind of trajectory: linear trajectories Lines We will start with some motivation in the plane before we discuss lines in space in any detail. Recall that any line in the plane is uniquely determined by a point and the slope. We can actually identify the slope with a collection of vectors in the following way. The slope of a line l can be computed by looking at rise over run, or y x y 2 y 1 x 2 x 1 where (x 1, y 1 ) and (x 2, y 2 ) are distinct points on the line l. Now, consider the vector v x, y. This vector carries a direction which is parallel to the line l. Finally, the line l is determined by a point (x, y ) on l and the direction given by v. We capture this with a picture: v (x, y ) v l Since we haven t defined a way to add points to vectors, we can identify the point (x, y ) with the vector x, y and see that the collection of vectors x, y + t v 26

30 give us a description of the points on the line. In parametric equations, we describe l with { x x + ( x)t y y + ( y)t Let s check that this parametric form translates directly to our usual equational descriptions y mx + b. Observe that which provides us with In point-slope, x x + t x t x x x y ( ) x x y + y x y + y x (x x ). y y y x (x x ). Now, in space, we have a few more directions to move in so our line s direction can t be captured by a real number in the same way the slope is used in the planar case. Using our previous considerations as inspiration, though, we can determine lines in space by using a point and a vector which indicates the direction. In particular, given a point p (x, y, z ) in space and a vector v a, b, c, by identifying p with the vector p+ x, y, z, we can express the line passing through point p with direction indicated by v with Parametrically, p + t v. x x + at y y + bt z z + ct Definition 22. Two lines in space are parallel if the vectors describing their directions are parallel. Two lines in space intersect if they share a point. Two intersecting lines in space are said to be perpendicular if the vectors describing their directions are orthogonal. If two lines in space neither intersect nor are parallel, we say that they are skew lines. Example 12. intersect. Find the point for which the two lines x 1 + t l 1 y 6 + 2t z 2 + t and l 2 Solution. What we want to do is solve the system 1 + t 1 2t t t t t 2 27 x 2t y 4 + t z 1 + 2t

31 which is equivalent to 1 + t 1 2t t 1 t t 1 2t 2 The first and third are the same so let s make the substitution and solve t 1 2t t 1 t (2t 2 1) t 2 3t 2 12 which gives t 2 4. Moreover, t 1 7. Finally, the point of intersection is (8, 8, 9). Example 13. are skew lines. The two lines x 1 + t l 1 y 2 + 3t z 4 t x 2t and l 2 y 3 + t z 3 + 4t Solution. First, let s check that the corresponding vectors describing direction are not parallel. For l 1, two points on the line are (1, 2, 4) and (2, 1, 3). Hence, a vector describing the direction of l 1 is 1, 3, 1. For l 2, we find two points (, 3, 3) and (2, 4, 1) on the line. This gives the vector 2, 1, 4 describing its direction. To leave nothing to the imagination, suppose we have a constant k so that 1, 3, 1 k 2, 1, 4 2k, k, 4k. Since two position vectors are equal if and only if their coordinates are equal, we see that such a k would have to be 2 (because of the x-components) but 3 2 (from the y-components). So no such k exists. That is, l 1 and l 2 are not parallel. The only thing left to verify is that l 1 and l 2 don t intersect. If they were to intersect, we would have values t 1 and t 2 so that 1 + t 1 2t t 1 t 2 7 t 1 4t 2 Using t 1 2t 2 1 from the first equation and substituting in the second, we obtain 5 + 3t 1 t (2t 2 1) t t 2 t 2 8/5 t 1 11/5 28

32 Alas, from the third equation, 7 t 1 4t That is, the system has no solutions so l 1 and l 2 don t intersect. Conclusively, l 1 and l 2 are skew lines More General Curves Consider any function f : I R 3 where I R is a non-degenerate interval. Then the curve {f(t) : t I} in space has a natural orientation inherited from the order on R. For example, notice that f(t) (cos(t), sin(t), ) and g(t) (cos(t), sin(t), ) for t [, 2π] determine the same curve with opposition orientations. Recall that the distance formula between two points (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ) is (x1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2. We can go ahead and define the operations a(x 1, y 1, z 1 ) + b(x 2, y 2, z 2 ) (ax 1 + bx 2, ay 1 + by 2, az 1 + bz 2 ) between scalars and points in space. Then we will denote the distance between two points (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ) by (x 1, y 1, z 1 ) (x 2, y 2, z 2 ) (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2. Definition 23. Let f : I R 3 where I R is a non-degenerate interval. Then we say that, for a I and P R 3, lim f(t) P t a provided that lim f(t) P. t a Proposition 24. Let f : I R 3, where I R is a non-degenerate interval, be f(t) (x(t), y(t), z(t)), P (p 1, p 2, p 3 ), and a I. Then if and only if the following hold: lim f(t) P t a lim x(t) p 1, t a lim y(t) p 2, t a lim z(t) p 3. t a 29

33 Example 14. Let f : [, ) R 3 be defined by the rule Evaluate f(t) (cos(3t), sin(t) + 1, t). lim f(t). t π Solution. By Proposition 24, we need to evaluate the coordinate-wise limits. That is, ( ) lim f(t) lim cos(3t), lim(sin(t) + 1), lim t t π t π t π t π ( 1, 1, π). We define continuity for curves in the same way we defined continuity for real-valued functions. Definition 25. Let f : I R 3 where I R is a non-degenerate interval. We say that f is continuous at t a I provided that lim f(t) f(a). t a From Proposition 24, we see that a curve in space is continuous if and only if each of its coordinate functions is continuous. 1.6 Motion in Space First, we make a note about derivatives of curves. From Proposition 24, we can imagine that the derivative of a curve at a point can be computed by taking the derivative of each coordinate. Since we re talking about motion, given a position function f(t) (x(t), y(t), z(t)), the derivative should correspond to instantaneous velocity. With this in mind, there is a directionality inherent in the motion. Hence, for the position function f(t) and a point t a in its domain, the instantaneous velocity at t a is described by the vector x (a), y (a), z (a). In all honesty, this is not that different from the one-dimensional analog. Given a one-dimensional motion, the derivative not only has a magnitude, but a sign, as well. The sign of the derivative can be thought of as being the direction, though there are only two directions (positive/negative). Then the absolute value of the derivative represents the magnitude, as usual. With this sort of thing in mind, although the position function really is concerned with position (not a vector quantity), we can get a lot of theoretical mileage by making the following identification: Definition 26. f(t) id x(t), y(t), z(t). A vector-valued function is any function which outputs vectors. Now, we collect some useful properties about differentiation of vector-valued functions. 3

34 For all differentiable vector-valued functions F and G, vectors v, and differentiable scalar-valued functions f: d dt v d dt ( F + G) d dtf + d dtg d dt (f(t) G(t)) f (t) G(t) ( + f(t) d ) dtg (t) d ( dtg(f(t)) d ) dtg (f(t))f (t) d dt ( F G) ( d ) dtf G + F ( d ) dtg d dt ( F G) ( d ) dtf G + F ( d ) dtg Constant Rule Sum Rule Scalar Product Rule Chain Rule Dot Product Rule Cross Product Rule Returning to the discussion on motion in space, suppose we are given a vector-valued position function p(t). Then the velocity function is also a vector-valued function and is v(t) d p(t). We dt define the speed to be v(t), the magnitude of the velocity. We also define the acceleration to be a(t) d d2 v(t) dt dt 2 p(t). Example 15. Find the velocity and speed of at t π. f(t) cos(t), sin(t), t Solution. First, observe that the velocity at t π is given by At t π, we have Finally, the speed at t π is, 1, 1 2. f (t) sin(t), cos(t), 1. f (π), 1, 1. If we were want to graph the position function and the particular velocity vector anchored at the point on the graph, notice that the line passing through the point (x, y, z ) in the direction of the velocity a, b, c is expressed by x, y, z + t a, b, c. Let s explore this a bit in the Calculus I context; that of real-valued functions. Recall that the graph of a differentiable function f : I R, where I R is a non-degenerate interval, is the collection of points {(x, f(x)) : x I}. Then, for x I, we find the tangent line as follows: f (x ) is the slope and (x, f(x )) is the point so we use point-slope to get y f(x ) f (x )(x x ). Then check that the parametric equations { x x + t y f(x ) + tf (x ) 31

35 which can be expressed as is the same line. x, f(x ) + t 1, f (x ) Exercise 1. Consider the graph of y x 2, expressed as a vector-valued function x, x 2. Find the instantaneous velocity at x 2 and compare it to the tangent line to y x 2 at x 2. Exercise 11. t π/4. Exercise 12. t π/4. Find the instantaneous velocity of the vector-valued function cos(t), sin(t) at Find the instantaneous velocity of the vector-valued function cos(5t), sin(5t) at Proposition 27. Suppose p(t) is a differentiable vector-valued function so that p(t) is constant. Then the position vector at t t is orthogonal to the velocity vector at t t. Newton s Second Law of Motion says that the sum of all forces acting on an object is equal to the mass of the object multiplied by the acceleration. This is usually written as m a F where m is the scalar corresponding to the mass, a is the vector corresponding to acceleration, and F is the net force acting on the object. From this, and the given gravitational constant for Earth g 9.87 m sec 2, we can model projectile motion. Suppose we are given an initial position of (x, y, z ) and an initial velocity of u, v, w. Assuming that gravity is the only force acting, we have that m,, g F or that To find the velocity, we integrat: ˆ v(t) a(t),, g. a(t) dt,, gt + const where const represents a vector. Then, since we were given the initial velocity of u, v, w, we see that v(t),, gt + u, v, w u, v, w gt. Integrating once again, we get the position: ˆ p(t) v(t) dt u t, v t, w t gt2 + const. 2 Since we were given the initial position (x, y, z ), we conclude with the position function p(t) x + u t, y + v t, z + w t gt

36 1.7 Length of Curves To continue our investigation of curves, we will now discuss how to find the length of a given curve. The idea is that we can approximate the length of any smooth curve by adding up the lengths of straight line segments as in the following pictures. Theorem 28. Let f(t), g(t), h(t) be a parameterized curve where f, g, and h are differentiable with continuous derivatives. Then, the length of the curve between t a and t b is ˆ b Proof. Take a partition of the interval [a, b]: a (f (t)) 2 + (g (t)) 2 + (h (t)) 2 dt. a t < t 1 < t 2 < < t n 1 < t n b. Then, consider the line segment from (f(t k ), g(t k ), h(t k )) to (f(t k+1 ), g(t k+1 ), h(t k+1 )). The length l k of this line segment can be determined using the distance formula: l k (f(t k+1 ) f(t k )) 2 + (g(t k+1 ) g(t k )) 2 + (h(t k+1 ) h(t k )) 2. By the Mean Value Theorem, we can find q k, r k, s k [t k, t k+1 ] with f (q k ) f(t k+1) f(t k ) t k+1 t k, g (r k ) g(t k+1) g(t k ) t k+1 t k, h (s k ) h(t k+1) h(t k ) t k+1 t k. Then, letting t k t k+1 t k, we see that f (q k ) t k f(t k+1 ) f(t k ), g (r k ) t k g(t k+1 ) g(t k ), h (s k ) t k h(t k+1 ) h(t k ). It follows that l k (f(t k+1 ) f(t k )) 2 + (g(t k+1 ) g(t k )) 2 + (h(t k+1 ) h(t k )) 2 (f (q k ) t k ) 2 + (g (r k ) t k ) 2 + (h (s k ) t k ) 2 t k (f (q k )) 2 + (g (r k )) 2 + (h (s k )) 2. So the sum of the lengths of these line segments n 1 l k k approximates the length of the curve. The bigger the partition (equivalently, the smaller the t k ), the better the approximation. Since each of f, g, and h have continuous derivatives, as t k, we must have that q k t k, r k t k, s k t k. 33

37 Finally, the length of the curve is lim n k n 1 t k (f (q k )) 2 + (g (r k )) 2 + (h (s k )) 2 ˆ b a (f (t)) 2 + (g (t)) 2 + (h (t)) 2 dt. Definition 29. With this, we then define the arc length for a curve parametrized by f(t), g(t), h(t) from t a and t b to be ˆ b a (f (t)) 2 + (g (t)) 2 + (h (t)) 2 dt. Remark 5. Let u(t) f(t), g(t), h(t) be a curve with coordinate functions that have continuous derivatives. Since u (t) d dt u(t) f (t), g (t), h (t), we see that the arc length of u(t) from t a to t b is This should be familiar. Example 16. length. ˆ b a ˆ (f (t)) 2 + (g (t)) 2 + (h (t)) 2 dt u (t) dt. Verify that the circumference of a circle with radius r > is 2πr using the arc Solution. Without loss of generality, we can center our circle at the origin. In parametric form, the circle is described by { x r cos(θ) for θ 2π. Now, dx dθ which gives us the arc length: as desired. ˆ 2π r sin(θ), dy y r sin(θ) ( r sin(θ)) 2 + (r cos(θ)) 2 dθ ˆ 2π ˆ 2π 2πr, dθ r cos(θ) r 2 sin 2 (θ) + r 2 cos 2 (θ) dθ Example 17. A popped balloon climbs the atmosphere according to the following parameterized vector-valued function r dθ 12 cos(t), 12 sin(t), 5t, t. Find how far the balloon has traveled after 15 minutes. 34

38 Solution. The distance is given by ˆ 15 ( 12 sin(t)) 2 + (12 cos(t)) 2 + (5) 2 dt ˆ 15 ˆ sin 2 (t) cos 2 (t) + 25 dt 169 dt The Polar Case What about two-dimensional curves given in polar coordinates r f(θ) for a θ b? We can work with these by converting to rectangular parametrically: { x r cos(θ) y r sin(θ) By the two-dimensional analog of Theorem 28, we will want to compute ˆ b (dx ) 2 ( ) dy 2 + dθ. dθ dθ a Note that and that It follows that dx dθ dy dθ r sin(θ) + dr dθ cos(θ) r cos(θ) + dr dθ sin(θ). ( dx dθ ) 2 + ( ) dy 2 r 2 sin 2 (θ) + r 2 cos 2 (θ) + dθ r 2 + ( ) dr 2. dθ Therefore, the desired arc length is given by ˆ b Parameterizing by Arc Length a (f(θ)) 2 + (f (θ)) 2 dθ. ( ) dr 2 sin 2 (θ) + dθ ( ) dr 2 cos 2 (θ) dθ Given a smooth curve u(t) f(t), g(t), h(t) for t a, we can define a function which calculates the arc length: len(t) ˆ t a u (ξ) dξ. We will say that the curve u(t) is parameterized by arc length provided that len(t) t a. 35

39 Differentiating with respect to t gives 1 d dt len(t) d [ˆ t ] u (ξ) dξ dt a u (t). This means that, whenever a curve is parameterized by arc length, the vector corresponding to the velocity is always a unit vector. More generally, notice that d dt len(t) u (t). (4) This will be useful in the next section. Example 18. Determine whether or not cos(θ 2 ), sin(θ 2 ), θ π. is parameterized by arc length. If not, re-parameterize it so that it is. Solution. First, the arc length function is ˆ t ( 2θ sin(θ 2 )) 2 + (2θ cos(θ 2 )) 2 dθ ˆ t ˆ t t 2. 4θ 2 (sin 2 (θ 2 ) + cos 2 (θ 2 )) dθ 2θ dθ So this parametrization is not according to the arc length. Since we are interested only in t, we can take square roots to obtain a parametrization which does correspond to the arc length parametrization: cos(t), sin(t), t π. Theorem 3. mapping For any smooth parameterized curve u(t) x(t), y(t), z(t), a t b, the t len(t), [a, b] [, L], where L is the length of the whole curve, is continuous and strictly increasing. Moreover, there exists an inverse function which is continuous. Conclusively, for any smooth parameterized curve, we can give a parametrization by arc length. In fact, all real-valued functions which are continuous and strictly increasing have inverse functions though, in practice, writing such an inverse can be quite difficult. Exercise 13. Check that the function x x on the interval [1/e, 2] is continuous and strictly increasing. So x x has an inverse there but try to write it using known functions. Exercise 14. Consider the parameterized curve γ(t) 2 cos(t), 3 sin(3t), 7t. 36

40 By Theorem 3, γ can be parameterized by arc length but try to find such a parameter using known functions. 1.8 Curvature The idea of curvature is to capture how much a curve is changing direction. This suggests that curvature should be a magnitude, hence, a scalar quantity. Let s discover how we can make this mathematically precise. Given a smooth vector-valued function u(t) x(t), y(t), z(t), recall that the direction at a point t is given by the vector u (t) x (t), y (t), z (t). Since we just want to capture the direction, we can normalize the magnitudes by considering only the unit vector u (t) u (t). That is, by normalizing the magnitudes, we effectively ignore the magnitude by only capturing information about the direction. The problem with an arbitrary parameter t though is that such a parameter may not give a parameterization by arc length. The fact of the matter is, to really capture the change in direction, we must be accounting for the change in direction relative to the change in arc length. With this said, let s appeal to Theorem 3 to choose a parameterization s by arc length. Then we define the unit tangent vector at s according to x(s), y(s), z(s) to be T(s) u (s) u (s). Then we can compute the change in direction on this segment of the trajectory by looking at T(s + s) T(s) s where s is the change in arc length. As afore mentioned, we wish to capture the magnitude of the change in direction. With that said, we define the curvature of u(s) at s to be κ(s) dt ds lim s T(s + s) T(s) s Also as mentioned after Theorem 3, coming up with parameterizations by arc length can be difficult. Even though this is the case, we can actually devise a more practical way to compute curvature using any parameterization t. Recall that (4) says that, for the parameterization s by arc length and any other parameterization t, Then, by the Chain Rule, we compute Taking magnitudes, we have that which finally gives us that ds dt u (t). dt dt dt ds ds dt dt ds u (t). dt dt κ(s) u (t) κ(s) dt/dt u (t). 37.