Vectors and the Geometry of Space

Chapter 12 Vectors and the Geometr of Space Comments. What does multivariable mean in the name Multivariable Calculus? It means we stud functions that involve more than one variable in either the input or the output: f : R! R, e.g. f() = cos(), Calc I and II, Chapters 1 11 f : R! R n, e.g. f(t) = (cos(t), sin(t)), Chapter 13. f : R n! R, e.g. f(, ) = (cos )(sin ), Chapters 14 15. f : R n! R m, e.g. f(, ) =( 2, /( 2 + 2 )), Chapter 16. Comments. Before we start studing functions involving more than one variable, we learn how to describe and picture 3-dimensional space. This is similar to when ou first learned about (, )-coordinates, what the meant, how to plot them, etc., before ou ever learned about functions. 12.1 3-D Coordinate Sstems Comments. Here s a recap of part of what ou ve learned about the main mathematical space ou ve studied so far, the 2-dimensional plane. 1. We sometimes gave the 2-dimensional plane a name, R 2,where R representsthe real numbers, and 2 represents the fact that it s 2-dimensional. 2. We gave each point in the plane an (, )-coordinate. This assumes that we pick two perpendicular aes, and that we agree to call the horiontal one the -ais and the vertical one the -ais. In other words, once we pick our aes, we can sa this R 2 = {(, ) 2 R, 2 R}. 3. We divided the plane into four quadrants: 3

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 4 QII QI QIII QIV 4. Given a function f : R! R, the graph G is defined as all the points (, ) that satisf = f(). Note that G is in R 2 and in general G is a curve. 5. We learned another coordinate sstem for the plane b giving each point an (r, )- coordinate. Comments. Now we etend these ideas two 3-dimensional space. Comments. 1. 3-dimensional space is named R 3. 2. We give each point in R 3 coordinates as follows. Pick the, and aes so that the are perpendicular

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 5 and satisf the right hand rule: If ou rotate the fingers of our right hand from the positive -ais to the positive -ais, our thumb will be pointing up along the positive -ais. Eample 1. Draw the point (1, 2, 3) in R 3. Solution: The picture below accuratel shows the point, but note that it s impossible to tell eactl where it is: For instance, ou can t tell b looking at that picture if the -value is negative or not. For our picture to be meaningful we need to add information. Di erent people might add di erent kinds of information, or the same information but pictured di erentl: in R 3 there are not universal was of drawing things! Here s one wa:

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 6 (1, 2, 3) 3 1 2 3. Note: unlike in R 2 where everone makes the -ais horiontal and the -ais vertical, people have more than one wa of drawing the (,, )-aes. Sometimes the draw as shown above, and sometimes the draw it like this: Notice that this picture still satisfies the right hand rule. 4. We divide three dimensional space into eight octants: Octants Planes dividing octants (From Wikipedia)

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 7 In R 2 the quadrants are separated b the aes, but here the octants are separated b planes. In R 2 the -ais is where equals 0 and similarl the -ais is where equals 0. In R 3 the planes that separate the octants are defined b = 0 or =0 or = 0. We name these planes for the remaining variables: (, )-plane where =0 (, )-plane where =0 (, )-plane where =0 Notice that these names introduce some ambiguit. In all our 2-dimensional pictures we call the background the (, )-plane, but now inside of 3-dimensions we also have something that we call the (, )-plane. Basicall, we think of the (, )- plane inside of R 3 as a cop of the (, )-plane that is all of R 2. If we absolutel have to tell the di erence between the (, )-plane in R 3 and the usual (, )- plane that is everthing in calculus I and II, then we can call the plane in R 3 the (,, 0)-plane. 5. Given a function f : R 2! R, the graph G is defined as all the points (,, ) that satisf = f(, ). Note that G is in R 3 and that in general G is a surface. 6. We ma eventuall learn two other coordinate sstems for 3-dimensional space: Clindrical: Spherical: each point gets coordinates (r,, ) where is height and (r, ) are as usual. each point gets coordinates (r,, ') where' is the angle between the positive ais and the point and (r, ) are as usual. Eample 2. What kinds of shapes are defined b each of the following? (a) = 2, in R 2. What about the same equations in R 3?

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 8 (b) = 2 and = 1inR 2. What about the same equations in R 3? Solution: (a) In R 2 we have a horiontal line: ( 2, 2) (1, 2) (2, 2) =2 In R 3 we have all points of the form (, 2,). This is 2-dimensional ( and are two variables). It is in fact a plane going through = 2, and parallel to the (, )-plane. (1.5, 2, 2) ( 0.5, 2, 2) (0, 2, 0) ( 0.5, 2, 1) (1.5, 2, 1) (b) In R 2 the onl point that satisfies both equations is In R 3 an point of the form (, ) =(2, 1). (,, ) =(2, 1,) satisfies both equations. In other words, the and coordinates are fied, but could be anthing. This gives us a line

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 9 (2, 1, 3) (2, 1, 1) (2, 1, 0) (2, 1, 2) It s reall hard to see that this line is in the correct place. Here are two was to see this: rotate the picture so we can see it from a di erent angle, or add a grid to the (, )-plane so we can see where the line intersects (2, 1, 3) (2, 1, 3) (2, 1, 1) (2, 1, 0) (2, 1, 1) (2, 1, 0) (2, 1, 2) (2, 1, 2) Eample 3. Using our intuition, or calculator, or computer, graph the following: = 2 in R 3. Solution: In the (, )-plane, the graph of = 2 should look like the usual graph of = 2 in the (, )-plane: with this: Here is most of the code I used for this graph. (In this eample, and most others, I ve left out some details, like how I labeled the aes, and printed the result to a file, etc.) = linspace(0,0); = linspace(-5,5); =.^2; plot3(,,);

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 10 But this isn t the actual graph we want, this is just a start, showing onl and in two dimensions, but ignoring the third dimension for. To include we can take an value on the curve we have so far, and etend it with an value. For instance, = 0, = 0, is on the graph because it satisfies = 2. But then (0, 0, 1) should be on the graph, since it also satisfies = 2. Similarl, since = 1, = 1, is on the graph, and therefore (1, 1, 1) should also be. Since = 2, = 4 is on there, so should be (2, 4, 1). If ou do this at ever point, then ou should get a cop of the parabola along the plane = 1. Here s what it looks like with four copies : = linspace(0,0); = linspace(-5,5); =.^2; hold on; plot3(,,); plot3( +0.5,,); plot3(+1,,); plot3(-0.5,,); hold off; This is where we ended on Monda, Januar 14 This is reall hard to see, and so we usuall have the computer color in the result: [,]=meshgrid( linspace(-2,2,5), linspace(-5,5)); =.^2; surf(,,); Now ou can see that it s hard to tell where the aes are. For this reason, people usuall don t draw the aes, instead the draw a bo around the picture that shows the range of values:

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 11 25 20 15 10 5-2 0-5 0 5 2 0 So what s the moral? It is possible to figure out what this graph looks like, but it takes a lot more work than in the 2D case. You, the student, will need to practice seeing these pictures and understanding what the mean, although ou will not create these graphs b hand ver often. You will probabl never be as much of an epert with 3D graphs as ou are with 2D graphs. We will rel more on computers than before, but more importantl, we will be forced to rel more on the rules of algebra and calculus than on convincing pictures. Eample 4. Using our intuition, or calculator, or computer, graph the following: =sin() in R 3. Solution: We all know what =sin() looks like in R 2, and we can start with this: = linspace(0,0); = linspace(-5,5); = sin(); plot3(,,);

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 12 But that might not look ver familiar. The problem is that the (, )-plane is pictured with coming out of the page/screen, towards us. Mabe it will look more familiar with and drawn more like the wa we usuall picture them: But, even though this probabl although this helps us figure out where the sine curve is and what it looks like, we ll stick to the usual 3D orientation for graphing things. In an case, this is onl a small part of the graph we want. All the and -values that we see on this graph should also appear on points with = 1. In other words, we should have a cop of this graph at a height of 1. Shown below is the original curve with two copies: = linspace(0,0); = linspace(-5,5); = sin(); plot3(,,); hold on; plot3(,, +0.1); plot3(,,-0.6); hold off; This is reall hard to see, and so we usuall have the computer color in the result:

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 13 1 0.5 0-0.5-10 -1 0-1 -0.5 0 0.5 1 10 [,]=meshgrid( linspace(-2*pi,2*pi), linspace(-1,1,10)); = sin(); surf(,,); So what s the moral? It is possible to figure out what this graph looks like, but it s hard. There is no single wa to view the results: ou can rotate the aes into di erent positions, ou can color the graph in di erent was, ou can move the aes outside of the picture to make a bo. All of these variations make the problem of graphing much more comple than in the 2D situation. You, the student, will need to practice seeing these pictures and understanding what the mean. You will probabl never be as much of an epert with 3D graphs as ou are with 2D graphs. We will rel more on computers than before, but more importantl, we will be forced to rel more on the rules of algebra and calculus than on convincing pictures. Rule (Distance Formula). in R 2 :A( 1, 1 ), B( 2, 2 ) AB = p ( 1 2 ) 2 +( 1 2 ) 2 in R 3 :A( 1, 1, 1 ), B( 2, 2, 2 ) AB = p ( 1 2 ) 2 +( 1 2 ) 2 +( 1 2 ) 2 Eample 5. Find the distance between A(2, 0, 1) and B(3, 1, 4). Solution: Rule (Equation of a Sphere). AB = p (2 3) 2 +(0 1) 2 +( 1 4) 2 =3 p 3 5.196. in R 2 :center (0, 0), radius r 2 + 2 = r 2

CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 14 in R 2 :center (h, k), radius r ( h) 2 +( k) 2 = r 2 in R 3 :center (0, 0, 0), radius r 2 + 2 + 2 = r 2 in R 3 :center (h, k, l), radius r ( h) 2 +( k) 2 +( l) 2 = r 2 Eample 6. Complete the square to find the center and radius of the following sphere, and then sketch the result: Solution: 2 4 + 2 +2 + 2 6 = 50 2 4 + 2 +2 + 2 6 = 50 2 4 +4+ 2 +2 +1+ 2 6 + 9 = 50 + 4 + 1 + 9 ( 2) 2 +( + 1) 2 +( 3) 2 =8 2 center = (2, 1, 3),r =8 Here s a sketch of the result. All we can reall get right is that it s a sphere, and we get it sort of in the right place: (2, 1, 3) This is where we ended on Wednesda, Januar 16