f : R 2 R (x, y) x 2 + y 2

Transcription

1 Chapter 2 Vector Functions 2.1 Vector-Valued Functions Definitions Until now, the functions we studied took a real number as input and gave another real number as output. Hence, when defining a function, it was usually enough to simply specify a formula for it. This is no longer the case. In the future, when we define a function we will also need to specify the kind of objects it acts on (numbers, vectors,...) and the kind of output it produces (numbers, vectors,...). For example, to define a function which takes as input a pair of numbers and produces a real number as output, defined by the formula f (x, y) = x 2 + y 2, we would write: f : R 2 R (x, y) x 2 + y 2 We could also say: f : R 2 R defined by f (x, y) = x 2 + y 2. The part f : R 2 R indicates that the inputs come from R 2 (they are pairs) and the outputs are in R. With this in mind, we now define vector valued functions. Definition (vector-valued functions) A vector-valued function or simply a vector function is a function whose domain is a set of real numbers and range a set of vectors (2-D or 3-D). In other words, it is a function of the form r : R R 2 or r : R R 3. You will note that we used vector notation because the function is actually a vector. 1. In the plane, such a function will be of the form r : R R 2 defined by r (t) = f (t), g (t). 2. In space, such a function will be of the form r : R R 3 defined by r (t) = f (t), g (t), h (t). 121

2 122 CHAPTER 2. VECTOR FUNCTIONS 3. We usually use t to denote the independent variable because many applications of these functions come from physics and the variable is time. t is often called a parameter. 4. The functions f, g, h are called component functions. They are functions for which both the input and output values are real numbers. In other words, f : R R, g : R R, and h : R R. They are functions like the functions you used in diff erential and integral calculus. 5. The domain of a vector function is the set of values of t for which r (t) is defined. That is, it is the set of values of t for which all the component functions are defined. One way to find it is to find the intersection of the domains of each component function. Example Consider the vector function r (t) = sin t, cos t. Its domain it the set of values of t for which both sin t and cos t are defined. Since both functions are always defined, the domain of r (t) is R. Example Consider the vector function r (t) = t 2, ln (3 t), t. Its domain it the set of values of t for which t 2, ln (3 t) and t are defined. t 2 is always defined. ln (3 t) is defined when 3 t > 0 that is when t < 3 or t (, 3). t is defined when t [0, ). The set of values of t in which all three functions are defined is [0, 3) Limits and Continuity Limit and continuity of vector functions are defined in terms of limits and continuity of their components. Definition (limit of a vector function) If r (t) = f (t), g (t), h (t) then lim r (t) = lim f (t), lim g (t), lim h (t) t a t a t a t a Remark Since the component functions are real-valued functions of one variable, we can use all the tools we learned in calculus I to find their limits. 2. Because taking the limit of a vector function amounts to taking the limits of real-valued functions, it can be shown that the limit rules we learned in calculus I are also true for vector functions. Example Find lim t 0 r (t) for r (t) = t 2, ln (3 t), t. From the definition, we have lim t 0 r (t) = lim t 0 t2, lim ln (3 t), lim t t 0 t 0 = 0, ln 3, 0

3 2.1. VECTOR-VALUED FUNCTIONS 123 Definition (Continuity of a vector function) The vector function r (t) is continuous if and only if lim r (t) = r (a) t a Remark From the definition, we see that a vector function is continuous if and only if its component functions are continuous. Since the component functions are real-valued functions of one variable, we can use all the theorems studied in calculus I. Example Find where r (t) = t 2, ln (3 t), t is continuous. We need to find the set of values of t at which each component of r is continuous. t 2 is continuous for all values of t. ln (3 t) is continuous where it is defined that is on (, 3). t is continuous where it is defined that is on [0, ). It follows that r (t) is continuous on [0, 3) Plane and Space Curves Definition (plane and space curves) Let f, g, h be three continuous real-valued functions. 1. If r (t) = f (t), g (t) then the set on points (x, y) where { x = f (t) y = g (t) (2.1) and t varies throughout some interval I is called a plane curve. r (t) is the position vector of the point (f (t), g (t)). 2. If r (t) = f (t), g (t), h (t) then the set on points (x, y, z) where x = f (t) y = g (t) z = h (t) (2.2) and t varies throughout some interval I is called a space curve. r (t) is the position vector of the point (f (t), g (t), h (t)). 3. Equations 2.1 and 2.2 are called the parametric equations of the curve. The curve is being traced by the tip of the position vector as t varies through the interval I as illustrated in figure 2.1. t is called the parameter. For each value of t corresponds a point on the curve. Example Find the point on the curve given by r (t) = sin t, cos t, t when t = 0, t = π 2. When t = 0, the corresponding point is r (0) = (sin 0, cos 0, 0) = (0, 1, 0).

4 124 CHAPTER 2. VECTOR FUNCTIONS Figure 2.1: A space curve and its position vector at different values of the parameter When t = π 2, the corresponding point is r ( 1, 0, π ). 2 ( π 2 ) = ( sin π 2, cos π 2, π ) = 2 Example Find the value of t which corresponds to the point (0, 1, 0) for the curve given by r (t) = sin t, cos t, t. We are looking for the value of t such that (0, 1, 0) = (sin t, cos t, t), so we see sin t = 0 that we must have cos t = 1. We use the easiest equation to solve and verify t = 0 that the solution we ound works for the other equations. Here, the last equation gives us the solution. This solution also works in the other equations. So, t = 0 gives us the point (0, 1, 0). We will not focus on plotting plane or space curves by hand. However, students should have a basic knowledge of some basic curves. They are illustrated in the examples below. Students should also know how to plot plane and space curves using their favorite computer program. Example Recall the parametric equation of the line through (x 1, y 1 z 1 )

5 2.1. VECTOR-VALUED FUNCTIONS 125 with direction vector a, b, c is where t R. x = x 1 + at y = y 1 + bt z = z 1 + ct Example The parametric equations of the line through two points (x 0, y 0, z 0 ) and (x 1, y 1 z 1 ) are x = (1 t) x 0 + tx 1 y = (1 t) y 0 + ty 1 (2.3) z = (1 t) z 0 + tz 1 for t R This is easy to see. A direction vector will be x 1 x 0, y 1 y 0, z 1 z 0. Thus, the equation of the line using the previous example will be Rearranging gives the desired equation. x = x 0 + t (x 1 x 0 ) y = y 0 + t (y 1 y 0 ) z = z 0 + t (z 1 z 0 ) Example From the above example, one can see that the equation of the line segment between two points (x 0, y 0, z 0 ) and (x 1, y 1 z 1 ) are x = x 0 + t (x 1 x 0 ) y = y 0 + t (y 1 y 0 ) z = z 0 + t (z 1 z 0 ) for t [0, 1]. { x = cos (t) Example The plane curve given by the parametric equations y = sin (t) for t [0, 2π] is a circle of radius 1 centered at the origin. This can be seen by noticing that x 2 + y 2 = 1 which is the equation of a circle of radius 1 centered at the origin. { x = cos (2πt) Example The plane curve given by the parametric equations y = sin (2πt) for t [0, 1] is also circle of radius 1 centered at the origin. Remark The last two examples illustrate the fact that diff erent parametric equations can trace the same curve. If we think of t as time, then in the second case, the circle is being traced faster than in the first. Remark When looking at a curve, its shape is important. The direction in which it is being traced is also important.

6 126 CHAPTER 2. VECTOR FUNCTIONS Example { If y = f (x), then the corresponding parametric equations are x = t. This shows that it is easy to write the parametric equations of y = f (t) a curve given a function representing { the curve. For example, the parametric x = t equations of a parabola y = x 2 are y = t 2. However, this is not the only way { x = at to write y = f (x) in parametric form. We can also use where a y = f (at) { x = t is a constant. The diff erence between the two parametrization and y = f (t) { x = at is the speed at which the curve is traversed. If in addition a < 0, y = f (at) the order in which the curve is traversed will be reversed. Example Another curve we will use often is the helix. Its parametric equations are of the form x = cos t y = sin t z = t for t R. Its trace in the xy-plane is a unit circle of radius 1 centered at the origin. As t increases, the z coordinate of the point corresponding to the position vector will increase. The position vector will actually trace an upward spiral. Example Fint the points at which the curve given by r (t) = 1 t 2, 2t 1, ln 1 t intersects with the xy-plane, the xz-plane and the yz-plane. Intersection with the xy-plane: To be on the xy-plane the z-coordinate of a point must be 0, so we must have ln 1 t = 0 which means that 1 t = 1 or t = 1. The corresponding point is r (1) = (0, 1, 0). Intersection with the xz-plane: To be on the xz-plane the y-coordinate of a point must be 0, so we must have 2t 1 = 0 or t = 1. The corresponding 2 point is ( ) 1 r = (1 14 ) ( ) 3 2, 0, ln 2 =, 0, ln 2. 4 Intersection with the yz-plane: To be on the yz-plane the x-coordinate of a point must be 0, so we must have 1 t 2 = 0 or t = ±1. However, since ln 1 t is defined only when 1 > 0 we must have t > 0, so we only keep t t = 1. The corresponding point is r (1) = (0, 1, 0) Things to Know Know what a space curve is. Know simple examples of space curves.

7 2.1. VECTOR-VALUED FUNCTIONS 127 Be able to compute the limit of a space curve. Be able to tell if and where a space curve is continuous Problems 1. Find the domain of each vector function below. (a) r (t) = t, t 2, sin t (b) r (t) = ln(3 t) t 1, sin t, et (c) r (t) = ln (3 t), t 4 2. Compute the limits below. (a) lim sin t t 0 t, cos t, e t 1 (b) lim ln t, e t t 1 (c) lim t t e, e t, 1 t t 3. Find where the vector functions below are continuous. (a) r (t) = t, t 2, sin t (b) r (t) = ln(3 t) t 1, sin t, et (c) r (t) = ln (3 t), t 4 4. Using your favorite computer program, sketch the graph of the vector functions below and also plot the position vector for various values of the parameter. (a) r (t) = t, t 2, e t (b) r (t) = (cos t, sin t, e t ) (c) r (t) = cos 4t, t, sin 4t (d) r (t) = t sin t, 1 cos t (e) r (t) = sin t, cos t, sin 2t 5. Suppose that two objects are travelling along the curves given by r 1 (t) = t 2, 7t 12, t 2 and r 2 (t) = 4t 3, t 2, 5t 6. Answer the questions below. (a) Do the curves intersect? If yes, find at which point. (b) Verify the answer you obtained in part a by graphing both curves. (c) Would the two objects travelling on these curves collide? (d) Are parts a and c the same question? Explain. 6. What is the value of t that corresponds to the point (4, 2, 4) for r (t) = t 2, 7t 12, t 2.

8 128 CHAPTER 2. VECTOR FUNCTIONS Answers 1. Find the domain of each vector function below. (a) r (t) = t, t 2, sin t All components are always defined, so the domain is R. (b) r (t) = ln(3 t) t 1, sin t, et The last two components are always defined. The numerator of the first one is defined when 3 t > 0 or t < 3 and the denominator cannot be 0 so the domain is (, 1) (1, 3) (c) r (t) = ln (3 t), t 4 The first component is defined when t < 3 and the second when t > 4 so the domain is empty. 2. Compute the limits below. (a) lim t 0 sin t t, cos t, e t 1 = 1, 1, 0 (b) lim t 1 ln t, e t = 0, e (c) lim t t e, e t, 1 t t = 0, 0, 0 3. Find where the vector functions below are continuous. (a) r (t) = t, t 2, sin t All components are always continuous, so r (t) is continuous on R. (b) r (t) = ln(3 t) t 1, sin t, et The last two components are always continuous. The numerator of the first one is defined hence continuous when 3 t > 0 or t < 3 and the denominator cannot be 0 so r (t) is continuous on (, 1) (1, 3) (c) r (t) = ln (3 t), t 4 The first component is defined when t < 3 and the second when t > 4 so the domain is empty hence r (t) is continuous nowhere. 4. Using your favorite computer program, sketch the graph of the vector functions below and also plot the position vector for various values of the parameter. (a) r (t) = t, t 2, e t

9 2.1. VECTOR-VALUED FUNCTIONS z x y (b) r (t) = cos t, sin t, e t 150 z y x (c) r (t) = cos 4t, t, sin 4t

10 130 CHAPTER 2. VECTOR FUNCTIONS y0.5 2 z x 4 (d) r (t) = t sin t, 1 cos t y x (e) r (t) = sin t, cos t, sin 2t

11 2.1. VECTOR-VALUED FUNCTIONS z y x Suppose that two objects are travelling along the curves given by r 1 (t) = t 2, 7t 12, t 2 and r 2 (t) = 4t 3, t 2, 5t 6. Answer the questions below. (a) Do the curves intersect? If yes, find at which point. We begin by renaming the parameter of the second equation, call it s, t 2 = 4s 3 then set the two equal and solve. We need to solve 7t 12 = s 2. t 2 = 5s 6 From equations 1 and 3 we see that 4s 3 = 5s 6 that is s = 3. This means that t = ±3. From equation 2, we see that 7t 12 = 9 or t = 3. So, yes, the curves do intersect at the point (9, 9, 9) that is when s = t = 3. (b) Verify the answer you obtained in part a by graphing both curves. The graphs are shown below, r 1 (t) is in blue and r 2 (t) in red.

12 132 CHAPTER 2. VECTOR FUNCTIONS z y x (c) Would the two objects travelling on these curves collide? Yes they would because they reach their common point at the same time. (d) Are parts a and c the same question? Explain. They are different questions. For two curves to intersect, it is enough for their path to cross. For objects travelling on these curves to intersect, they have to reach the point of intersection at the same time, that is for the same value of the parameter. 6. What is the value of t that corresponds to the point (4, 2, 4) for r (t) = t 2, 7t 12, t 2. t = 2 is the answer.

13 Bibliography [1] Joel Hass, Maurice D. Weir, and George B. Thomas, University calculus: Early transcendentals, Pearson Addison-Wesley, [2] James Stewart, Calculus, Cengage Learning, [3] Michael Sullivan and Kathleen Miranda, Calculus: Early transcendentals, Macmillan Higher Education,