Vectors and Vector Arithmetic

Transcription

1 Vectors and Vector Arithmetic Introduction and Goals: The purpose of this lab is to become familiar with the syntax of Maple commands for manipulating and graphing vectors. It will introduce you to basic vector defining and manipulation using Maple. By the end of the lab we hope that you will be comfortable defining, manipulating and graphing vectors. We also hope that you will obtain the expertise to use these concepts to solve more involved problems that use vector arithmetic in their solution. Before You Start: Before you start this lab you should be familiar with the basic concepts of vectors and vector arithmetic, such as addition, subtraction, scalar multiplication, dot product and cross product. Furthermore, you should know how to calculate work and torque. Textbook Correspondence: Stewart 5 th Edition: Stewart 5 th Edition Early Transcendentals: Thomas Calculus 10 th Edition Early Transcendentals: 9.1, 9.2, 10.1, Johnston & Mathews Calculus: Maple Commands and Packages Used: Packages: linalg and plots. Commands: vector, evalm, arrow, crossprod, dotprod. History & Biographies: Maple Commands Defining a Vector: To define a vector in Maple we first need to load the linear algebra package. Maple treats vectors as 1 n matrices, much like we do. Execute the following command, > with(linalg): Warning, the protected names norm and trace have been redefined and unprotected

2 You will get a warning that the functions norm and trace have been redefined. Don t worry about it, everything is fine. As with just about everything in Maple there are many different ways to do the same thing. If you find other series of commands that accomplish the same task and you find them easier to use, feel free to use them. As always, we will use the commands that are, in our opinion, the easiest and most straightforward to accomplish the tasks needed. To define a vector v as 1,2, 3 we use the command > v:=[1, 2, 3]; v := [ 1, 2, 3 ] Note that the Maple output is in matrix form. Again, Maple is viewing a vector as an 1 n matrix. We can define vectors of any dimension. For example, the commands > v:=[1, 2, 3, 4, 5, 6]; v := [ 1, 2, 3, 4, 5, 6 ] > v:=[1, 2]; > v:=[x, y, z, w]; v := [ 1, 2 ] v := [ x, y, z, w ] define v as a 6, 2 and 4 dimensional vector respectively. Another way to define a vector is as a column vector, that is, an n 1 matrix. To do this we simply put the vector components inside angled brackets. For example, the following commands define two vectors > t:=<5,-3,2>; > s:=<1,0,-4>; t := s := Arithmetic: Define two vectors in your worksheet, v as 1,2, 3 and w as 1,3, 5 > v:=[1, 2, 3];

3 v := [ 1, 2, 3 ] > w:=[-1, 3, 5]; w := [ -1, 3, 5 ] Vector arithmetic is as easy as numeric arithmetic in Maple, simply input the vector expression. > v+w; > v-w; > 2*v+5*w; [ 0, 5, 8 ] [ 2, -1, -2] [-3, 19, 31 ] To do dot and cross products we use the two commands dotprod and crossprod. Execute the following commands, > dotprod(v,w); > crossprod(v,w); 20 [ 1, -8, 5 ] We can use column vectors in exactly the same way as row vectors. For example, execute the commands > s+t; > 3*s-2*t; The dot and cross products are done in the same way as well. Notice that the output of the cross product is in row vector form and not column vector form like the input vectors. > dotprod(s,t); -3 > crossprod(s,t);

4 [-12, -22, -3] Furthermore, we can use both forms together in arithmetic expressions, with a little alteration. For example, if we wish to add v and s together the following command produces, > v+s; [ 1, 2, 3 ] which is not what we wanted, although it is correct. If you have had or are taking a course in Linear Algebra you may be surprised with the output. In Linear Algebra you find that you can only add two matrices if they are the same size. Since v is a 1 3 matrix and s is a 3 1 matrix under the strict rules of matrix addition Maple should produce an error here. Instead Maple simply output the question because it knows that the matrix addition is undefined but it is not sure what calculation the user had in mind. Its answer is in essence asking the user to be a bit more specific. To get Maple to see v and s as vectors of the same size and perform the calculation we need to use the command evalm which is a general matrix evaluation command. Execute the following command, > evalm(v+s); [ 2, 2, -1] In this command, Maple converted s to a 1 3 matrix and then did the addition. In general, Maple tries to adjust the form of its inputs to complete the desired command and if it can t there are sometimes extra commands that will do the conversions. The commands below show that the dot and cross product functions also adjust the inputs to complete the calculation. > dotprod(v,t); > crossprod(v,s); 5 [-8, 7, -2] > dotprod(crossprod(v,s),t); -65 In many cases the column vector notation is easier to work with in Maple but since it is customary to write a vector as a row vector, at least in Calculus, we will stick with the row vector notation.

5 When calculating the cross product of two vectors by hand you usually place the vectors in a 3 3 matrix as the second and third rows with an i, j, k row at the top and then take the determinant of the matrix. Although there is a cross product command you can find the cross product using this technique. First define the matrix using the syntax below > A:=matrix([[i,j,k],[1,2,3],[-1,3,5]]); i j k A := and then calculate the determinant using the det command. > det(a); i 8 j + 5 k The major difference between these methods is that the result here is not a vector, at least Maple does not see this as a vector. Maple is treating i, j and k as variables so the output expression is really a polynomial of three variables. All in all the above method is good to see but the crossprod command is better to use for most applications. The length of a vector is obtained using the norm command. For example, to find the length of the vector v we use the following command, > norm(v,2); 14 Note the 2 as the second argument. There are many different norms in mathematics and you will see a few of them in your undergraduate career. The norm function in Maple is designed to calculate several of the norms we encounter in mathematics, so we must tell it which norm we want to find the length of a vector. Since the norm we need is the 2- norm (the Pythagorean theorem calculation) we use a 2 as the second argument. In the exercises we will use these commands to do some standard but more involved calculation such as determining the angle between two vectors, projections of a vector onto another, torque and the scalar triple product. Graphing Vectors: As with all graphics commands in Maple, the vector plotting commands have many options and styles. We will look at some of the basic options and styles for vectors here but we will not look at an exhaustive list of them. For more information on all of the vector plotting options your version of Maple has please refer to the Maple help system or one of the printed manuals. Also, most of the plotting commands needed for vector and multivariable Calculus are not loaded into Maple on startup so we need to load another package. The plots package will contain all of the extended graphics routines for

6 vector and multivariable Calculus that will be used in these labs. To load the plots package, like any other package, execute the following command. > with(plots): Warning, the name changecoords has been redefined Plotting vectors in Maple is done with the arrow command. For example the following command > arrow(w, axes=boxed); produces the figure to the right. The axes=boxed option simply places the boxed axes around the vector. If we omit this option and simply use the command arrow(w); we would get the vector arrow but not the box around it. We can change the size and style of the vectors to better suit our visual needs. For example, in the last figure the vector is a bit fat. We tend to think of vectors, graphically, as directed line segments and hence we usually draw them as thin lines and not large cylinders. We can change the width with the width option added to the arrow command. For example, execute the following command. > arrow(w, width=0.1, axes=boxed); This produces the image to the right. All it did was take the default width of the vector and made it a tenth of its size. To draw a vector as a directed line segment, with no width you might try the command, > arrow(w, width=0, axes=boxed); Unfortunately, this does not work the way we want it to. In fact, all we get from this command is the box. To draw vectors as line segments and not cylinders we use the shape option with either arrow or harpoon as the value. For example, execute the following commands. > arrow(w, shape=arrow, axes=boxed); > arrow(w, shape=harpoon, axes=boxed);

7 These commands produce the following images respectively. The main difference between the two is that the arrow option puts a double-sided arrow on the vector and the harpoon option puts a single-sided arrow on the end. If you have trouble seeing the arrows add the color option to the arrow plotting command. For example, > arrow(w, shape=arrow, color=black, axes=boxed); When displaying many vectors at once it is usually better to have the vectors as thin as possible. For example, the command > arrow({seq(<sin(i), sin(i^2), sin(i^3)>, i=1..500)}, shape=arrow, scaling=constrained, axes=framed); produces the image to the right. We will look at two more shape option values, double_arrow and cylindrical_arrow. The cylindrical_arrow option is the default that produces the cylinder type vectors we saw above and the double_arrow option produces a flat arrow with some width to it. For example, execute the following command.

8 > arrow(w, shape=double_arrow, axes=boxed); This produces the following images. When you move the display around you can see that the arrow is in fact flat. As mentioned above, the cylindrical_arrow option is the default. So executing the command, > arrow(w, shape=cylindrical_arrow, axes=boxed); produces the same image as we had with the command arrow(w, axes=boxed); Selecting the 1:1 button gives the following image. Moving this image around makes it clear that the vector is graphically produced using a cylinder and a cone.

9 To plot several vectors together you use syntax similar to the syntax used to plot several functions on the same set of axes. Combine the set of vectors you wish to plot together into a list delimited by curly brackets. For example, > arrow({w, v}, shape=arrow, color=black, axes=boxed); produces the image to the right. We can do the same thing by first creating a list of vectors and then plotting them. For example, the sequence of commands > VL:= {v,w}; VL := { v, w } > arrow(vl, shape=arrow, color=black, axes=boxed); produces the same figure as above. Let s use this to visualize the cross product of two vectors. From your reading you know that the cross product of two vectors produces a vector that is orthogonal to both vectors in the cross product. If we take the cross product of the vectors v and w and set the result to the vector s we can visualize the cross product as follows. > s:=crossprod(v,w); s := [ 1, -8, 5 ] > VL:= {v,w,s}; VL := { w, s, v } > arrow(vl, shape=arrow, color=black, axes=boxed);

10 Make sure that you select 1:1 to get the image to the right. Another feature of the arrow command is that we can draw a vector starting at any point we wish. For example, the command > arrow(v,w, shape=arrow, axes=boxed, color=black); produces the image below. Note that this is the vector w moved so that its beginning point is at the position of the vector v, as a point. We can place a set of vectors as the second argument to draw the entire set beginning at the same starting point. For example, > t:=[5,-3,2]; t := [ 5, -3, 2 ] > arrow(v,{w,t}, shape=arrow, axes=boxed, color=black); produces the image to the right. If we place a set of vectors as the first argument Maple will graph the second set of vectors at each of the beginning points defined in the first set of vectors. For example, > arrow({v,w},{w,t}, shape=arrow, axes=boxed, color=black); produces,

11 Another option in the vector plotting command arrow is difference. If we include the option difference in the list of arguments the arrow command will plot a vector from the first vector, thinking of it as a point, to the second vector, also thinking of it as a point. For example, the command > arrow(v,w, difference, shape=arrow, axes=boxed, color=black); produces the image to the right. Exercises: 1. The angle between two vectors. a. Write the mathematical formula for finding the angle between two vectors, v and w. b. Convert the mathematical formula into Maple syntax. c. Define the function AngBet(v,w) that takes two vectors v and w as input and returns the numerical angle, in radians, between the two vectors v and w. This may be the first time that you have defined a function of two variables. Defining functions of several variables is just as easy as defining functions of a single 2 variable. For example, to define the function f ( x) = x in Maple you simply input > f:=x->x^2; f := x x 2 Recall that the := is to define f as a function and the -> defines the rule that takes x to x 2. For functions of more than one variable the syntax is nearly identical, the

12 only difference is that we need to put more than one variable between the := and the ->. This is done by placing all of the variables inside parentheses separated 2 3 by commas. For example to define the function f ( x, y) = x y we use the command, > f:=(x,y)->x^2-y^3; f := ( x, y ) x 2 y 3 So the beginning of the AngBet command you will write will look like > AngBet:=(v,w)-> d. Define the function AngBetDeg(v,w) that takes two vectors v and w as input and returns the numerical angle, in degrees, between the two vectors v and w. e. Use these commands to find the angle between each of the following sets of vectors. i. 1,4, 2 and 3,5, 7. ii. 3,7, 9 and 2,1, 6. iii. 2, 5 and 5, 2. iv. 1,2,3,4, 5 and 4,6,2,7, 15. v. a, b, c and x, y, z. 2. The projection of one vector onto another. a. Write the mathematical formula for finding the projection of the vector v onto the vector w. b. Convert the mathematical formula into Maple syntax. c. Define the function VecProj(v,w) that takes two vectors v and w as input and returns the projection of the vector v onto the vector w. d. Use this command to find the projections of the first vectors onto the second for each of the following pairs of vectors. i. 1,4, 2 and 3,5, 7. ii. 3,7, 9 and 2,1, 6. iii. 2, 5 and 5, 2. e. For each pair above, plot the two vectors given along with the calculated projection. 3. The scalar triple product of three vectors. a. Write the mathematical formula for finding the scalar triple product of the vectors u, v and w. b. Convert the mathematical formula into Maple syntax. c. Define the function STP(u,v,w) that takes three vectors u, v and w as input and returns the scalar triple product of the vectors.

13 d. Use this command to find the scalar triple product of 1,4, 2, 3,5, 7 and 2,1,6. e. Using the arrow function and the display function create an image of the parallelepiped formed by the vectors 1,4, 2, 3,5, 7 and 2,1, 6. f. What is the volume of this parallelepiped? 4. Torque. a. Write the mathematical formula for finding the torque given a force vector F and a position vector r. b. Convert the mathematical formula into Maple syntax. c. Define the function Torque(F,r) that takes two vectors F and r as input and returns the torque of the system. d. Consider the following fulcrum system below. W is a large weight weighing 1500 Kilograms, the length AB is 2 meters, the length BC is 38 cm and the length BD is 20 cm. A force F is applied at A straight down. i. What is the torque at B needed to begin to lift the weight? ii. What assumptions were made in the above calculation? iii. What is the magnitude of the force F needed to begin to move the weight? iv. Assuming that the force F is always pointing down, find a function for the amount of force needed to lift the weight in terms of the angle θ which is the angle between BC and BD. v. Graph the function you derived above for θ between the at rest angle (shown in the picture) and 90 o. vi. From the graph, interpret the amount of effort you would be exerting to lift the weight 20 cm. vii. How would the calculations change if we were to remove some of the assumptions listed in the second part? That is, for each assumption you made describe the necessary alterations to the calculations if the assumption were removed. You do not need to redo the calculations. e. Consider the following fulcrum system below. W is a large weight weighing 1500 Kilograms, the length AB is 2 meters, the length BC is 38 cm and the length BD is 20 cm. A force F is applied at A perpendicular to AC.

14 i. What is the torque at B needed to begin to lift the weight? ii. What assumptions were made in the above calculation? iii. What is the magnitude of the force F needed to begin to move the weight? iv. Assuming that the force F is always perpendicular to AC, find a function for the amount of force needed to lift the weight in terms of the angle θ which is the angle between BC and BD. v. Graph the function you derived above for θ between the at rest angle (shown in the picture) and 90 o. vi. From the graph, interpret the amount of effort you would be exerting to lift the weight 20 cm. vii. How would the calculations change if we were to remove some of the assumptions listed in the second part? That is, for each assumption you made describe the necessary alterations to the calculations if the assumption were removed. You do not need to redo the calculations. f. Consider the following fulcrum system below. W is a large weight weighing 1500 Kilograms, the length AB is 2 meters, the length BC is 38 cm, the length BD is 20 cm and the length DE is 2.5 meters. A rope is tied to A and goes around a pulley at E so that the force F applied at A is always in the direction of AE. i. What is the torque at B needed to begin to lift the weight? ii. What assumptions were made in the above calculation? iii. What is the magnitude of the force F needed to begin to move the weight?

15 iv. Assuming that the force F is always in the direction of AE, find a function for the amount of force needed to lift the weight in terms of the angle θ which is the angle between BC and BD. v. Graph the function you derived above for θ between the at rest angle (shown in the picture) and 90 o. vi. From the graph, interpret the amount of effort you would be exerting to lift the weight 20 cm. vii. How would the calculations change if we were to remove some of the assumptions listed in the second part? That is, for each assumption you made describe the necessary alterations to the calculations if the assumption were removed. You do not need to redo the calculations. g. Compare and contrast the three force graphs you obtained above. From them, select the method you would use to lift the weight. What were the reasons for your choice, be as explicit as possible?