Lec 2 | MIT 18.02 Multivariable Calculus, Fall 2007

the following content is provided under a Creative Commons license your support will help MIT OpenCourseWare continue to offer high quality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT opencourseware at ocw.mit.edu thank you so okay let's continue with vectors and operations on them so remember yesterday we saw the topic yesterday was dot product and a member of the definition of dot product well the dot product of two vectors is obtained by multiplying the first component with the first component the second with the second and so on and summing these and you get a scalar and the geometric interpretation of that is that you can also take the length of a take the length of B multiply them and multiply that by the cosine of the angle between the two vectors so we've seen several applications of that so one application is to find lengths and angles so for example where you can use this relation to give you that the cosine of the angle between two vectors is the dot product divided by the product of the lengths another application that we have is to detect whether two vectors are perpendicular so to decide if two vectors are perpendicular to each other all we have to do is compute the dot product and see if we get zero okay and well one third application that we don't have time to discuss yesterday that I will mention very quickly is to find components of so let's say find the component of a vector a along a direction u so some unit vector so let me explain let's say that I have some direction for example the horizontal axis from this blackboard but it could be any direction in space and to describe this direction maybe I have a unit vector along this axis let's say that I have any other vector a and I want to find out what is the component of a along U so that means what is the length of this projection of a to the given direction okay so this thing here is the component of a along you well how do we find that well we know that here we have a right angle so this component is just length a times cosine of the angle between a and u but now that means actually we can compute it very easily because well I could you know that's the same as length a times length U times cosine theta because U is a unit vector it's a unit vector that means this is equal to one and so that's the same as the dot product between a and u ok so that's very easy and of course the most obvious cases of that is say for example we want us to find the component along I hat the first axis you know the unit vector along the x axis then you do the dot product with I hat which is 1 0 0 what you get is the first component and that's indeed the X component of the vector similarly say you want the Z component you do the dot product with K that gives you the last component of your vector but the same works with a unit vector in any direction ok so what's an application of that well for example in physics maybe you have seen situations where you have a pendulum that swings so you have maybe some mass at the end of the string and that maths wings back and forth on the circle and to analyze this mechanically you want to use of course Newton's laws of mechanics and you want to use well you know forces and so on but I claim that components of vectors are useful here to understand what happens geometrically so whatever force is exerted on this pendulum well there's its weight which usually points downward and there's the tension of the string and these two forces together are what explains how this pendulum is going to move back and forth now you could try to understand the equations of motion using XY coordinates or maybe XZ or whatever you want to call them 6a X Y but really what causes the pendulum to swing back and forth and also to somehow stay at constant distance are phenomenal relative to this circular trajectory so for example maybe instead of taking components along the x and y axes we want to look at two other unit vectors so we can look at a vector let's call it T that's tangent to the trajectory so I can you hate it if it's not very readable so T is tangent to the trajectory and on the other hand we can introduce another vector let's call that N and that one is normal or perpendicular to the trajectory and so now if you think about it you can look at the components of the weight along the tangent direction and along the normal direction and so the component of F along the tangent direction is what causes acceleration in the direction along the trajectory it's what causes the pendulum to swing back and forth and the component along n on the other hand so that's the part of the weight that tends to pull a mess away from this point is what's going to be responsible for the tension of the string it's why the string is towed and not actually slack and you know with things moving all over the place so that one is responsible for the tension of a string and now of course if you want to compute things well maybe you will call this angle theta and when you will express things explicitly using sines and cosines and you will solve for the equations of motion okay that would be a very interesting physics problem but to save time we're not going to do it I'm sure you've seen that in 801 or similar classes and so to find these components we would just do dot products okay any questions No okay so let's move on to our next topic so here we have found things about length angles and stuff like that one important concept that we haven't understood yet in terms of vectors is area so let's say that you know we want to find the area of this Pentagon well how do we compute that using vectors can we do it using vectors yes we can and that's going to be the goal but how so the first thing we should do is probably simplify the problem you know we don't actually need to bother with Pentagon's all we need to know our triangles because for example you can cut that into three triangles and then sum the areas of the triangles okay so perhaps easier what's the area of a triangle so let's start with you know a triangle in the plane and well then we need two vectors to describe it say a and B here how do we find the area of a triangle well we all know base times height / - so what's the base would be height well so the area of this triangle is going to be one-half of the base is going to be the length of a and the height well if you call theta this angle then this is length B sine theta now that looks a lot like the formula we had there except for one little catch this is a sign instead of a cosign how do we deal with that well what we could do is first find the cosine of the angle we know how to find the cosine of the angle using that product then solve for sine using sine square plus cosine square equals 1 and then plug that back into here well that works but it's kind of a very complicated way of doing it so there's an easier way and that's going to be determinant but let me explain how we get to that maybe still doing elementary geometry and dot products first so let's see what we can do is instead of finding the sine of theta well we're not good at finding signs of angles but we are very good now at finding cosines of angles so maybe we can find another angle whose cosine is the same as the sine of theta well you probably heard about complementary angles and how you know if I take my vector a let my vector B here I have an angle theta well let's say that I rotate my vector a by 90 degrees to get a new vector a prime okay so a prime is just a rotated by 90 degrees then the angle between these two guys let's say theta prime well beta prime is 90 degrees of PI over 2 radians minus theta so in particular cosine of theta prime is equal to sine of theta so in particular that means that length a length B sine theta which is what we would need to know in order to find the area of this triangle is equal to well a and a prime have the same length so let me replace that by length of a prime I'm not changing anything length B cosine theta prime and now we have something that's much easier of course because that's just a prime dot B okay so that looks like a very good plan there's only one small thing which is we don't know yet how to find this a prime well I claim it's not very hard let's see why don't you guys do do the hablok so let's say that I have a plane vector a with two components a 1 a 2 and I want to rotate it counterclockwise by 90 degrees so it looks like maybe we should change some signs somewhere we should do something with the components can you come up with an idea of what it might be okay so I see a lot of people answering way I see some other answers but the majority vote seems to be number three - A two and A one I think I agree so let's see let's say that we have this vector a with components a 1 so a 1 is here and a 2 so a 2 is here ok let's rotate this box by 90 degrees counterclockwise so this rectangular box ends up there it's the same box just flipped on its side so this length here becomes a 1 and this length here becomes a 2 and that means our new vector a prime is going to be well the first component looks like an A - but it's pointing to the left when a 2 is positive so it relates minus a 2 and the y component is going to be the same as this guy so it's going to be a 1 if you wanted instead to rotate clockwise then you would do the opposite you would do a 2 minus a 1 okay is that reasonably clear for everyone okay so well let's continue a calculation where a prime we've decided is minus a 2 from a 1 so minus a 2 A 1 dot product with let's call B 1 and B 2 the components of D then that will be well minus a 2 B 1 plus a 1 B 2 let me write that the other way around a 1 B 2 minus a 2 B 1 and that's a quantity that you may already know another name of determinants of vectors a and B which we write symbolically using this notation so we put a and B next to each other inside a 2x2 table and we put these vertical bars and that means the determinant of these numbers it means this guy times this guy - this guy time-space guy so that's called the determinant and geometrically what it measures is the area well not of a triangle because we didn't divide by two but of a parallelogram formed by a and B so it measures it out here of the parallelogram with sides a and B and of course if you want with high angle then you will just divide by two if the triangle is half a parallelogram there's one small catch the area usually is something that's going to be positive this guy here has no reason to be positive or negative because in fact well if you compute things you will see that whether it's positive or negative depends on whether a and B are clockwise or counterclockwise from each other I mean the issue that we have well when we say the area is 1/2 length a length B sine theta that was assuming that theta is positive that it's sine is positive otherwise if theta is negative maybe we need to take the absolute value of this so just to be not hopeful I will say the determinant is either plus or minus the area okay any questions about this yes oh sorry that's not a dot product that's a usual multiplication okay that's a length a times length B times sine theta and so that's equal to the area of a parallelogram so let me explain that again so if I have two vectors a and B so I can form a parallelogram with them I can form a triangle okay and so the area of the parallelogram is equal to length a length B sine theta is equal to the determinant of a and B while the area of the triangle is one-half of that and again to be truthful I should say these things can be positive or negative so depending on whether you count the angle positively or negatively you will get either the area of - the area the area is actually the absolute value of this quantities okay is that clear okay yes so if you want to compute the area you will just take the absolute value of the determinant okay so now I will be well I should say the area of the parallelogram so that it's completely clear sorry you have a snow question explain again so with the question how the determinant equals the area of a parallelogram okay so the area of a parallelogram it's going to be the base times the height right so the base let's take this guy to be the base so the length of a base will be length of a and the height will be obtained by taking B but only looking at the vertical part so that will be length of B times the sine of theta okay so that's how I got that the area of a parallelogram is length a length B sine theta and when I did this manipulation and this track of rotating to find a nice formula yes so you're asking ahead of what I'm going to do in a few minutes so you're asking about many feet of a cross B so we're going to learn about cross product in a few minutes and the answer is yes but cross product is for vectors in space here I was simplifying things by doing things just in the plane okay so just bear with me for five more minutes and we'll do things in space yes that's correct the way you compute this in practice is you just do this okay that's how you compute the determinant yes what about so for dimensions we are going to do now okay more questions should we move to okay let's move to space so so that's two things we can do in space and you know you can find you can look for the volume of solids or you can look for the area of surfaces so let me start with easy over to let me start with volumes of solids and we'll go back to area I promise so I claim that there's also a notion of determinants in space and that's going to tell us how to find volumes so let's say that we have three vectors a B and C and then the definition of the determinant is going to be so the notation for that in terms of the components is the same as above where we put the components of a the components of B and the components of C inside the article bars okay and of course I have to give a meaning to this so this will be a number and what is that number well the definition I will take is that this is a1 times the determinant of what I get by looking in this lower right corner so the 2 by 2 determinant b2 b3 c2 c3 then I will subtract a 2 times the determinant of b1 b3 c1 c3 and then I will add a 3 times the determinant b1 b2 c1 c2 and each of these guys means again you take b2 times c3 minus c2 times b3 and this times that minus this times that and so on so in fact there's a total of six terms in here okay and maybe some of you have already seen a different formula for theta 3 determinants where you directly have the six terms it's the same it's the same definition so how to remember the structure of this formula well it's called an this is called an expansion according to the first hole so we're going to take the entries in the first hole a 1 a 2 a 3 and for each of them we get a term namely we multiply it by a 2 by 2 determinant that we get by deleting the first pole and the column where we are okay so here the coefficient next to a 1 is when we delete this column and this row and we are left with B 2 B 3 C 2 C 3 next the next one we take a 2 we delete the robots in it and the column that it's in and we are left with B 1 B 3 C 1 C 3 and similarly with a 3 we take what remains that's B 1 B 2 C 1 C 2 finally last but not least there's a minus sign here for the second game ok so well it looks like a weird formula and I mean it is a little bit weird but it's a formula that you should learn because it's really really useful for a lot of things so I should say if this looks very artificial to you and you like to know more there's more in the notes so read the notes they will tell you a bit more about what this means where it comes from and so on if you want to know a lot more then someday you should take 1806 linear algebra where you will learn a lot more about determinants in n-dimensional space with n vectors and there's a generalization of this in arbitrary dimensions in this class we'll only deal with two or three dimensions yes so why is why is the negative there well that's a very good question it has to be there so that this will actually equal well what I'm going to say right now is that this will give us the volume of the box with sides a B C and the formula just doesn't work if you don't put the negative there's a more fundamental reason which has to do with orientation of space and the fact that if you switch to coordinates in space then basically you change what's called the handedness of the coordinates if you look at your right hand and your left hand they are not actually the same they're mirror images and if you swap two coordinate axes that's what you get that's the fundamental reason for the - but again I mean we don't need to think too much about that all we need in this class is the formula so why do we care about this formula it's because of a theorem that says that geometrically the determinant of the three vectors a B C is again plus or minus this determinant could be positive or negative seed as minuses and there's all sorts of stuff plus or minus the volume of the parallelepiped but just a fancy name for a box with parallel downsides in case you wonder with sides a B and C okay so you take the five actors a B and C and you form a box whose sides are all parallelograms and when it's volume is going to be the determinant okay that questions oh I'm sorry I can't quite hear you yes we are going to see how to do it geometrically without a determinant but then you'll see that you actually need a determinant to compute it no matter what so we are going to go back to this and see another formula for volume but you'll see that when I am cheating I mean the important way and then somehow computationally the only way to compute it is ready to use a determinant yes that's correct in general I mean actually I should say if you look at the 2 by 2 determinant see you can also explain it in terms of this expansion you take a 1 and multiply by this 1 by 1 determinant b2 then you take a 2 you multiply it by this 1 by 1 day time and b1 but you put a minus sign and in general indeed when you expand you would start putting plus minus plus minus alternatingly more about that in 1806 yes there's a way to do it based on other holes as well but then you have to be very careful with the sign factors so I will refer you to the notes for that ok if you did it with I mean you could also do it with a column by the way just I mean be careful about the sign rules so given how how little will use the dominance in this class I mean will use them in a way that's fundamental but we won't compute much so let's say this is going to be enough force for now ok so after determinants now I can tell you about cross product and cross product is going to be the answer to your question about Aria okay so okay so let me move on to cross product so post product is something that you can apply to two vectors in space and they gotta mean whether in three-dimensional space okay this is something that's specific to three dimensions so the definition a plus B so it's important to really do your multiplication symbol well so that you don't mistake it with a dot product well that's going to be a vector okay so that's another reason not to confuse it with dot product dot product gives you a number cross product gives you a vector so they are very completely different operations they are both called product because well somehow one couldn't come up with a better name but they are completely different operations okay so what do we do to do the cross product of a and B well we do something very strange just after I've told you that a determinant is something where we put numbers and we get a number I'm going to violate my own rule I'm going to put together a determinant in which well the last two holes are the components of the vectors a and B with the first or strangely consists of the unit vectors ijk what does that mean well that certainly is not a determinant in the usual sense you cannot you know if you try to put that into your calculator it will tell you there's an error I don't know how to put vectors in there I want numbers what it means is it's a symbolic notation that helps you remember what the formula is so the actual formula is well you use this definition and you know if you use this definition you see that it's I hat times sum well ok so let me write it as determinant of a2 a3 b2 b3 times I hat minus determinant a1 a3 b1 b3 J hat plus a1 a2 b1 b2 k hat okay and so that's the actual definition in a way that makes complete sense but to remember this formula without too much trouble it's much easier to think of it to think about it in these terms here ok so that's a definition and it gives you a vector now as usual with definitions the question is what is it good for what is the geometric meaning of this very strange relation why do we bother to do that so here's what it does geometrically so remember a vector has two different things it has a length and it has a direction okay let's start with the length the length of the cross product is the area of the parallelogram in space formed by the vectors a and B okay so now if you have a diagram in space you can find our higher just by doing this calculation when you go to coordinates of the point you do this calculation and then you take the length so the length you know you take this squared plus red square plus red squared square root it looks like a very complicated formula but it works and actually it's the simplest way to do it okay this time we don't actually need to put plus or minus because the length of a vector is always positive we don't have to worry about that and what's even more magical is that not only is the length remarkable but the direction is also remarkable so the direction of a cross B is perpendicular to the plane of a parallelogram so you know our two vectors a and B together of a determiner plane and what I'm telling you is that the vector a cross B will Point will stick straight out of that plane perpendicular to it so in fact I have to be more precise there's two ways that you can be perpendicular perpendicular to this plane right you can be perpendicular pointing up or pointing down how do I decide which well that's something called the right hand rule okay so what does the right hand will say well there's various versions right hand of all depending in which country you learn about it in France given the culture you even learn about it in terms of a crow and a wine bottle but I think and I just use the usual version here so you take your right hand if you're left-handed remember to take your right hand not the left one the other the other right okay and then place your hand to point in the direction of a so let's say my right hand is going in that direction now curl your fingers so that they point towards B so here that would be kind of into the blackboard okay don't snap any bones if it doesn't quite work then rotate your arms so that you can actually physically do it then you know get your thumb to stick straight out well here my thumb is going to go up and that tells me that a cos B will go up okay so let me write that down while you experiment with it again try not to enjoy yourselves so first your right hand points parallel to vector a then fingers point in the direction of B then your thumb when you stick it out is going to point in the direction of a cross B okay let's do a quick example let's make quick example here let's take i cross-checked so I see some of you well most of you are going in the right direction if you have it pointing in the wrong direction it might mean that you're using your left hand for example okay so example I claim that I cross J equals K okay let's see I point to others J points to our height so I guess this is your height height I think it's kind of hard to and then your thumb is going to point up okay so that tells us it's roughly pointing up and of course the length should be one because if you take the unit square in the XY plane is area is one and the direction should indeed be vertical because it should be perpendicular to the XY plane so looks like I cross J will be K well let's check with a definition you know ijk what is I I is 1 0 0 J is 0 1 0 well so the coefficient of I will be 0 times 0 minus 0 times 1 that's 0 coefficient of J will be 1 times 0 minus 0 times 0 that's 0 minus 0 J doesn't matter a bit and the coefficient of K will be 1 times 1 that's 1 minus 0 times 0 1 K ok so we do get I plus J equals K both ways in this case it's easier to do geometrically if I give you more complicated vectors probably you will actually want to do the calculation okay any questions yes so the coefficient of K remember I take so I delete the first row and the last column so I get this two-by-two determinant again that two-by-two determinant is 1 times 1 minus 0 times 0 so that gives me 1 ok that's what you do with 2 by 2 determinants similarly for viewers would be others turn out to be 0 mark questions yes let me repeat how I got the 1 in front of K so remember the definition of a determinant I expand according to the entries in the first hole when I get to K what I do is I delete the first oh and I delete the last column the column that contains K okay so I delete these guys and these guys I'm left with this two-by-two determinant now a 2 by 2 determinant you multiply according to this downward diagonal and then minus this times that okay so 1 times 1 let me say here I got 1 K because that's 1 times 1 minus 0 times 0 equals 1 so and that's really hard to read and it will be easier that way okay yes okay so let's try if I do the same for I I think I will also get zero okay so let's do the same for I so I think I I delete the first row I delete the first column I get this two-by-two determinant here and I get 0 times 0 that's 0 minus 0 times 1 that's here the trick question 0 times 1 is 0 as well so that's 0 minus 0 is 0 ok so I hope in Monday you should get more practice in recitation about how to compute determinants so hopefully it will become very easy for you all to compute these things I know the first time it's kind of a shock because there's a lot of numbers and a lot of things to do see not good okay so okay so let me return to the question that you asked a bit earlier about how do I find actually volume if I don't want to know about determinants well so let's have another look I'd prefer you okay so let's say that I have three vectors let me put them this way a B and C and let's try to see how else I could think about the volume of this box okay so probably you know that the volume of a parallelepiped is the area of a base times the height okay so sorry the volume is the area of the base times the height how do we do that in practice well what's the area of a base so the base is a parallelogram in space with sides B and C how do we find the area of a parallelogram in space well we just discovered that we can do it by taking the cross product so the area of the base well we take the cross product of B and C that's not quite it because this is a vector we would like a number where we take its length okay that's pretty good what about the height well the height is going to be the component of a in the direction that's perpendicular to the base so let's take a direction that's perpendicular to the base let's call that n a unit vector in that direction then we can get the height by taking a dot n that's what we saw at the beginning of class that a dot n will tell me how much a goes in the direction of N are you still with me okay let's keep going so let's think about this vector and how do I get it well I can get it by actually using cross product as well because I said the direction perpendicular of the two vectors I can get by taking the cross product and looking at that direction so this is still because c.length and this one is so I claim I should do an aside here and can be obtained by taking B cross C well that points in the right direction but it's not a unit vector how do I get a unit vector I divide by the length thanks so I take B cross C they divide by length because C well now I can probably simplify between these two guys and so what I will get oops to be the last thing today what I get out of this is that my volume equals a dot product with the vector B cross C and of course I have to be careful in which other I do it if I do it the other way around a dot B I get a number I cannot close that so I mean I really have to do the cross-product first I get a new vector then i dot product okay so the fact is that the determinant of ABC is equal to this so-called triple product well that looks good geometrically let's try to check whether it makes sense with the formulas just one small thing so we solve the determinant is a one times determinant it will be free si tu savais minus a2 times something plus a freight and something I let you fill in the numbers okay that's this guy what about this guy well dot product we take the first component of a that's a 1 we multiplied by the first component of B cross C what's the first column of B cross C well is this determinant b2 b3 c2 c3 right if you look at if you put B and C instead of a and B into there you'll get a 2 I component is this guy - well plus a 2 times the second component which is minus some determinant plus a 3 times the third component which is again a determinant and you can check to get exactly the same expression so everything is fine there's no contradiction in math just yet ok so well on Tuesday we'll continue with this and we'll start going into matrices equations of planes and so on meanwhile have a good weekend and please stop working on your problem sets so that you can ask lots of questions to your TAS on Monday

Channel: MIT OpenCourseWare

Views: 422,323

Rating: 4.9225297 out of 5

Keywords: calculus, vector-algebra, determinants, matrix, matrices, vector-valued-function, space-motion, scalar-function, partial-differentiation, gradient, optimization-techniques, double-integrals, line-integrals, exact-differential, conservative-fields, Green's-theorem, triple-integrals, surface-integrals, applications

Id: 9FLItlbBUPY

Length: 52min 51sec (3171 seconds)

Published: Fri Jan 16 2009