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So, it'll be that constant, which is the value of the partial derivative of f with respect to x, evaluated at the point of interest. And that's just a constant. All right, so that's there. This next term has no x's in it, so that's just gonna go to zero. This term is interesting because it's got an x in it. So, when we... | Quadratic approximation formula, part 2.mp3 |
This next term has no x's in it, so that's just gonna go to zero. This term is interesting because it's got an x in it. So, when we take its derivative with respect to x, that two comes down. So, this will be two times a, whatever the constant a is, multiplied by x minus x naught. That's what the derivative of this com... | Quadratic approximation formula, part 2.mp3 |
So, this will be two times a, whatever the constant a is, multiplied by x minus x naught. That's what the derivative of this component is with respect to x. Then this over here, this also has an x, but it's just showing up basically as a linear term. And when we treat y as a constant, since we're taking the partial der... | Quadratic approximation formula, part 2.mp3 |
And when we treat y as a constant, since we're taking the partial derivative with respect to x, what that ends up being is b, multiplied by that, what looks like a constant, as far as x is concerned, y minus y naught. And then the last term doesn't have any x's in it. So, that is the first partial derivative with respe... | Quadratic approximation formula, part 2.mp3 |
And now we do it again. Now we take the partial derivative with respect to x, and I'll, hmm, maybe I should actually clear up even more of this guy. And now when we take the partial derivative of this expression with respect to x, f sub x of x naught, y naught, that's just a constant, so that goes to zero. Two times a ... | Quadratic approximation formula, part 2.mp3 |
Two times a times x, that's gonna, we take the derivative with respect to x, and we're just gonna get two times a. And this last term doesn't have any x's in it, so that also goes to zero. So, conveniently, when we take the second partial derivative of q with respect to x, we just get a constant. It's this constant, tw... | Quadratic approximation formula, part 2.mp3 |
It's this constant, two a. And since we want it to be the case, we want that this entire thing is equal to, well, what do we want? We want it to be the second partial derivative of f, you know, both times with respect to x. So, here I'm gonna use the subscript notation. Over here I'm using the kind of Leibniz notation.... | Quadratic approximation formula, part 2.mp3 |
So, here I'm gonna use the subscript notation. Over here I'm using the kind of Leibniz notation. But here, just second partial derivative with respect to x. We want it to match whatever that looks like when we evaluate it at the point of interest. So, what we could do to make that happen, to make sure that two a is equ... | Quadratic approximation formula, part 2.mp3 |
We want it to match whatever that looks like when we evaluate it at the point of interest. So, what we could do to make that happen, to make sure that two a is equal to this guy, is we set a equal to 1 1 2 of that second partial derivative evaluated at the point of interest. Okay, so this is something we kind of tuck a... | Quadratic approximation formula, part 2.mp3 |
We remember this, this is, we have solved for one of the constants. So, now let's start thinking about another one of them. Like, let's actually don't have to scroll off, because let's say we just want to take the mixed partial derivative here, where if instead of taking it with respect to x twice, we wanted to, let's ... | Quadratic approximation formula, part 2.mp3 |
We wanted to first do it with respect to x, and then do it with respect to y. Then we can just kind of edit what we have over here, and we say we already took it with respect to x. So, now as our second go, as our second go, we're gonna be taking it with respect to y. So, in that case, instead of getting two a, let's k... | Quadratic approximation formula, part 2.mp3 |
So, in that case, instead of getting two a, let's kind of figure out what it is that we get. When we take the derivative of this whole guy with respect to y, well, this looks like a constant. This here also looks like a constant, since we're doing it with respect to y, and no y's show up. And the partial derivative of ... | Quadratic approximation formula, part 2.mp3 |
And the partial derivative of this just ends up being b. So, again, we just get a constant. This time it's b, not, you know, two, well, previously it was two a, but now it's just b. And this time we want it to equal the mixed partial derivative. So, instead of saying f sub xx, I'm gonna say fxy, which basically says yo... | Quadratic approximation formula, part 2.mp3 |
And this time we want it to equal the mixed partial derivative. So, instead of saying f sub xx, I'm gonna say fxy, which basically says you take the partial derivative first with respect to x, and then with respect to y. We want this guy to equal the value of that mixed partial derivative evaluated at that point. So, t... | Quadratic approximation formula, part 2.mp3 |
So, that gives us another fact. That means we can just basically set b equal to that. And this is another fact, another constant that we can record. And now for c, now for c, when we're trying to figure out what that should be, the reasoning is almost identical. It's pretty much symmetric. We did everything that we did... | Quadratic approximation formula, part 2.mp3 |
And now for c, now for c, when we're trying to figure out what that should be, the reasoning is almost identical. It's pretty much symmetric. We did everything that we did for the case x, and instead we do it for taking the partial derivative with respect to y twice in a row. And I encourage you to do that for yourself... | Quadratic approximation formula, part 2.mp3 |
And I encourage you to do that for yourself. It'll definitely solidify everything that we're doing here, because it can seem kind of like a lot in a lot of computations, but you're gonna get basically the same conclusion you did for the constant a. It's gonna be the case that you have the constant c is equal to 1 1⁄2 o... | Quadratic approximation formula, part 2.mp3 |
So, you're differentiating with respect to y twice, evaluated at the point of interest. So, this is gonna be kind of the third fact. And the way that you get to that conclusion, again, it's gonna be almost identical to the way that we found this one for x. Now, when you plug in these values for a, b, and c, and these a... | Quadratic approximation formula, part 2.mp3 |
Now, when you plug in these values for a, b, and c, and these are constants, even though there's, you know, we've written them as formulas, they are constants. When you plug those in to this full formula, you're going to get the quadratic approximation. It'll have six separate terms, one that corresponds to the constan... | Quadratic approximation formula, part 2.mp3 |
And if you wanna dig into more details and kind of go through an example or two on this, I do have an article on quadratic approximations, and hopefully you can kind of step through and do some of the computations yourself as you go. But in all of this, even though there's a lot of formulas going on, it can be pretty n... | Quadratic approximation formula, part 2.mp3 |
I think this is really the hard thing to do here. But the way we can parameterize the surface of a torus, which is the surface of this donut, is to say, hey, let's take a point and let's rotate it around a circle. It could be any circle. I picked a circle in the zy-plane. And how far it's gone around that circle, we'll... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
I picked a circle in the zy-plane. And how far it's gone around that circle, we'll parameterize that by s. And s can go between 0 all the way to 2 pi. And then we're going to rotate this circle around itself. Or I guess, even a better way to say it, we're going to rotate the circle around the z-axis. And the center of ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
Or I guess, even a better way to say it, we're going to rotate the circle around the z-axis. And the center of the circles are always going to keep a distance b away. And so these were top views right there. And then we defined our second parameter, t, which tells us how far the entire circle has rotated around the z-a... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And then we defined our second parameter, t, which tells us how far the entire circle has rotated around the z-axis. And those are our two parameter definitions. And then here we tried to visualize what happens. This is kind of the domain that our parameterization is going to be defined on. s goes between 0 and 2 pi. S... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
This is kind of the domain that our parameterization is going to be defined on. s goes between 0 and 2 pi. So when t is 0, we haven't rotated out of the zy-plane. s is at 0, goes all the way to 2 pi over there. Then when t goes to 2 pi, we've kind of moved our circle. We've moved it along. We've rotated around the z-ax... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
s is at 0, goes all the way to 2 pi over there. Then when t goes to 2 pi, we've kind of moved our circle. We've moved it along. We've rotated around the z-axis a bit. And then this line in our st domain corresponds to that circle in three dimensions, or in our xyz space. Now, given that, hopefully we visualize it prett... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
We've rotated around the z-axis a bit. And then this line in our st domain corresponds to that circle in three dimensions, or in our xyz space. Now, given that, hopefully we visualize it pretty well. Let's think about actually how to define a position vector valued function that is essentially this parameterization. So... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
Let's think about actually how to define a position vector valued function that is essentially this parameterization. So let's first do the z, because that's pretty straightforward. So let's look at this view right here. What's our z going to be as a function? So our x, our y's, and our z's should all be a function of ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
What's our z going to be as a function? So our x, our y's, and our z's should all be a function of s and t. That's what it's all about. Any position in space should be a function of picking a particular t and a particular s. And we saw that over here. This point right here, let me actually do that with a couple of poin... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
This point right here, let me actually do that with a couple of points, this point right there that corresponds to that point right there. Let me pick another one. This point right here corresponds to this point right over there. I could do a few more. Let me pick this point right here. So s is still 0. That's going to... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
I could do a few more. Let me pick this point right here. So s is still 0. That's going to be this outer edge way out over there. I'll pick one more just to define this square. This point right over here where we haven't rotated t at all, but we've gone a quarter way around the circle is that point right there. So for ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
That's going to be this outer edge way out over there. I'll pick one more just to define this square. This point right over here where we haven't rotated t at all, but we've gone a quarter way around the circle is that point right there. So for any s and t, we're mapping it to a point in xyz space. So our z's, our x's,... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So for any s and t, we're mapping it to a point in xyz space. So our z's, our x's, and our y's should all be a function of s and t. So the first one to think about is just the z. And I think this will be pretty straightforward. So z as a function of s and t is going to equal what? Well, if you take any circle, remember... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So z as a function of s and t is going to equal what? Well, if you take any circle, remember s is how much the angle between our radius and the xy plane. So I could even draw it over here. Let me do it in another color. So I'm running out of colors. So let's say that this is a radius right there. That angle, we said th... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
Let me do it in another color. So I'm running out of colors. So let's say that this is a radius right there. That angle, we said that is s. So if I were to draw that circle out just like that, we can do a little bit of trigonometry. This is the angle is s. We know the radius is a, the radius of our circle. We define th... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
That angle, we said that is s. So if I were to draw that circle out just like that, we can do a little bit of trigonometry. This is the angle is s. We know the radius is a, the radius of our circle. We define that. So z is just going to be the distance above the xy plane. It's going to be this distance right there. And... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So z is just going to be the distance above the xy plane. It's going to be this distance right there. And that's straightforward trigonometry. That's going to be a. I mean, we could do Sohcahtoa and all of that. You might want to review the videos. But the sine, you could view it this way. So if this is z right there, ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
That's going to be a. I mean, we could do Sohcahtoa and all of that. You might want to review the videos. But the sine, you could view it this way. So if this is z right there, you could say that the sine of s, Sohcahtoa is the opposite over the hypotenuse, is equal to z over a. Multiply both sides by a. You have a sin... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So if this is z right there, you could say that the sine of s, Sohcahtoa is the opposite over the hypotenuse, is equal to z over a. Multiply both sides by a. You have a sine of s is equal to z. That tells us how much above the xy plane we are. And that's just some simple trigonometry. So z of s and t, it's only going t... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
That tells us how much above the xy plane we are. And that's just some simple trigonometry. So z of s and t, it's only going to be a function of s. It's going to be a times the sine of s. Not too bad. Now let's see if we can figure out what x and y are going to be. Remember, z doesn't matter. It doesn't matter how much... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
Now let's see if we can figure out what x and y are going to be. Remember, z doesn't matter. It doesn't matter how much we've rotated around the z-axis. What matters is how much we've rotated around the circle. If s is 0, we're just going to be in the xy plane. z is going to be 0. If s is pi over 2 or up here, then we'... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
What matters is how much we've rotated around the circle. If s is 0, we're just going to be in the xy plane. z is going to be 0. If s is pi over 2 or up here, then we're going to be traveling around the top of the donut. And we're going to be exactly a above the xy plane, or z is going to be equal to a. Hopefully that ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
If s is pi over 2 or up here, then we're going to be traveling around the top of the donut. And we're going to be exactly a above the xy plane, or z is going to be equal to a. Hopefully that makes reasonable sense to you. Now let's think about what happens as we rotate around. Remember, these two are top views. We are ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
Remember, these two are top views. We are looking down. We're looking down on this donut. So the center of each of these circles is b away from the origin, or from the z-axis. It's always b away. So our x and y coordinates, so if we go to the center of the circle here, we're going to be b away. And then this is going t... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So the center of each of these circles is b away from the origin, or from the z-axis. It's always b away. So our x and y coordinates, so if we go to the center of the circle here, we're going to be b away. And then this is going to be b away right over there. So let's think about where we are in the xy plane, or how fa... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And then this is going to be b away right over there. So let's think about where we are in the xy plane, or how far the part of our, I guess you could imagine, if you were to project our point into the xy plane, how far is that going to be from our origin? Well, it's always going to be, remember, let's go back to this ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
This might be the most instructive. This is just one particular circle on the zy plane, but could be any of them. If this is the z-axis over here, that is the z-axis. This distance right here is always going to be b. We know that for sure. And so what is this distance going to be? We're at b to the center, and then we'... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
This distance right here is always going to be b. We know that for sure. And so what is this distance going to be? We're at b to the center, and then we're going to have some angle s. And so depending on that angle s, this distance onto, I guess, the xy plane, if we're sitting on the xy plane, how far are we from the z... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
We're at b to the center, and then we're going to have some angle s. And so depending on that angle s, this distance onto, I guess, the xy plane, if we're sitting on the xy plane, how far are we from the z-axis, or the projection onto the xy plane, or the y position? I'm saying it as many ways as possible. I think you'... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
If z is a sine of theta, this distance right here, this little shorter distance right here, that's going to be a cosine of theta. s is that angle right there. This distance right here is going to be a cosine of s. So if we talk about just straight distance from the origin along the xy plane, our distance is always goin... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
We're going to be at that point right there. So if you look at the top views over here, no matter where we are, that is b. Let's say we've rotated a little bit. That distance right here, if you look along the xy plane, that is always going to be b plus a cosine of s. That's what that distance is to any given point, dep... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
That distance right here, if you look along the xy plane, that is always going to be b plus a cosine of s. That's what that distance is to any given point, depending on our s's and t's. Now, as we rotate around, if we're at a point here, let's say we're at a point there, and that point, we already said, is b plus a cos... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
We're sitting on the z-axis looking straight down on the xy plane right now. We're looking down on the donut. So what are your x and y's going to be? Well, you draw another right triangle right here. This angle right here is t. This distance right here is going to be this times the sine of our angle. So this right here... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
Well, you draw another right triangle right here. This angle right here is t. This distance right here is going to be this times the sine of our angle. So this right here, which is essentially our x, this is going to be our x coordinate. x as a function of s and t is going to be equal to the sine of t, t is our angle r... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
x as a function of s and t is going to be equal to the sine of t, t is our angle right there, times this radius, times b plus a cosine of s. Because remember, how far we are depends on how much around the circle we are. When we're over here, we're much further away. Here we're exactly b away if you're looking only on t... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And then over here, we're b minus a away if we're on the xy plane. So that's x as a function of s and t. And actually, the way I defined it right here, our positive x-axis would actually go in this direction. So this is x positive, this is x in the negative direction. I could have flipped the sines, but hopefully this ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
I could have flipped the sines, but hopefully this actually makes sense, that that would be the positive x, this is the negative x. It depends on whether you're using a right-handed or left-handed coordinate system. But hopefully that makes sense. We're just saying, OK, what is this distance right here? That is b plus ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
We're just saying, OK, what is this distance right here? That is b plus a cosine of s. We got that from this right here when we're taking a view just kind of a cut of the torus. That's how far we are in kind of the xy direction at any point, or kind of radially out without thinking about the height. And then if you wan... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And then if you want the x coordinate, you multiply it times the cosine, or sorry, you multiply it times the sine of t, the way I've set it up here. And the y coordinate is going to be this right here, the way we've set up this triangle. So y as a function of s and t is going to be equal to the cosine of t times this r... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
I mean, if you say that this is our y coordinate right here, you just do SOH CAH TOA, cosine of t, CAH is equal to adjacent, which is y, right? This is the angle right here, over the hypotenuse, over b plus a cosine of s. Multiply both sides of the equation times this, and you get y of s of t is equal to cosine of t ti... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And we're done with our parameterization. We could leave it just like this, but if we want to represent it as a position vector valued function, we can define it like this. Find a nice color, maybe pink. So let's say our position vector function, vector valued function, is r. It's going to be a function of two paramete... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So let's say our position vector function, vector valued function, is r. It's going to be a function of two parameters, s and t. And it's going to be equal to its x value. Let me do that in the same color. So it's going to be, I'll do this part first, b plus a cosine of s times sine of t. And that's going to go in the ... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And in this case, remember, the way I defined it, the positive x direction is going to be here. So the i unit vector will look like that. i will go in that direction, the way I've defined things. And then plus our y value is going to be b plus a cosine of s times cosine of t in the y unit vector direction. Remember, th... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And then plus our y value is going to be b plus a cosine of s times cosine of t in the y unit vector direction. Remember, the j unit vector will just go just like that. That's our j unit vector. And then finally, we'll throw in the z, which was actually the most straightforward, plus a sine of s times the k unit vector... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And then finally, we'll throw in the z, which was actually the most straightforward, plus a sine of s times the k unit vector, which is a unit vector in the z direction. So times the k unit vector. And so you give me now any s and t within this domain right here, and you put it into this position vector valued function... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So if you pick, let's just make sure we understand what we're doing. If you pick that point right there, where s and t are both equal to pi over 2, and you might even want to go through the exercise, take pi over 2 in all of these. Actually, let's do it. So in that case, so when r of pi over 2, what do we get? It's goi... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So in that case, so when r of pi over 2, what do we get? It's going to be b plus a times cosine of pi over 2. Cosine of pi over 2 is 0, right? Cosine of 90 degrees. So it's going to be b, this whole thing is going to be 0, times sine of pi over 2. Sine of pi over 2 is just 1. So it's going to be b times i plus, once ag... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
Cosine of 90 degrees. So it's going to be b, this whole thing is going to be 0, times sine of pi over 2. Sine of pi over 2 is just 1. So it's going to be b times i plus, once again, cosine of pi over 2 is 0. So this term right here is going to be b. And then cosine of pi over 2 is 0. So it's going to be 0j. | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So it's going to be b times i plus, once again, cosine of pi over 2 is 0. So this term right here is going to be b. And then cosine of pi over 2 is 0. So it's going to be 0j. So it's going to be plus 0j. And then finally, pi over 2, well, there's no t here. Sine of pi over 2 is 1. | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
So it's going to be 0j. So it's going to be plus 0j. And then finally, pi over 2, well, there's no t here. Sine of pi over 2 is 1. So plus a times k. So there's actually no j direction. So this is going to be equal to b times i plus a times k. So the point that it specifies, according to this parameterization, or the v... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
Sine of pi over 2 is 1. So plus a times k. So there's actually no j direction. So this is going to be equal to b times i plus a times k. So the point that it specifies, according to this parameterization, or the vector it specifies, is b times i plus a times k. So b times i will get us right out there. And then a times... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And then a times k will get us right over there. So the position vector being specified is right over there, just as we predicted. That dot, that point right there, corresponds to that point, just like that. And of course, I picked kind of points that was easy to calculate, but this whole, when you take every s and t i... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
And of course, I picked kind of points that was easy to calculate, but this whole, when you take every s and t in this domain right here, you're going to transform it to this surface. And this is the transformation right here. And of course, we have to specify that s is between, we could write it multiple ways. We coul... | Determining a position vector-valued function for a parametrization of two parameters Khan Academy.mp3 |
How do you describe rotation in three dimensions? So for example, I have here a globe, and it's rotating in some way, and there's a certain direction that it's rotating, and a speed with which it's rotating, and the question is, how could you give me some numerical information that perfectly describes that rotation? So... | Describing rotation in 3d with a vector.mp3 |
Well, before talking about that, let's remind ourselves of how we talk about two-dimensional rotation. So I have here a little pie creature, and I set him to start rotating about, and the way that we can describe this, we pretty much need to just give a rate to it, and you might give that rate as a number of rotations,... | Describing rotation in 3d with a vector.mp3 |
But that's a little bit ambiguous, if you just say, hey, this little pie creature is rotating at 0.2 rotations per second, someone could say, well, is it clockwise or counterclockwise? So there's some ambiguity, and the convention that people have adopted is to say, well, if I give you a positive number, if the number ... | Describing rotation in 3d with a vector.mp3 |
And with this, it's very nice, because given a single number, just one number, and it could be positive or negative, you can perfectly describe two-dimensional rotation. And there's a minor nuance here, usually in physics and math, we don't actually use rotations per unit second, but instead you describe things in term... | Describing rotation in 3d with a vector.mp3 |
And just as a quick reminder of what that means, if you imagine some kind of circle, and it could be any circle, the size doesn't really matter, and if you draw the radius to that, and then ask the question, how far along the circumference would I have to go, such that the arc length, that sort of sub-portion of the ci... | Describing rotation in 3d with a vector.mp3 |
So this, you know, it would be whatever the number you have there, times two pi. And the specific numbers aren't too important. The main upshot here is that with a single number, positive or negative, you can perfectly describe two-dimensional rotation. But if we look over here at the 3D case, there's actually more inf... | Describing rotation in 3d with a vector.mp3 |
But if we look over here at the 3D case, there's actually more information than just one number that we're gonna need to know. First of all, you want to know the axis around which it's rotating. So the line that you can draw, such that all rotation happens around that line. And then you want to describe the actual rate... | Describing rotation in 3d with a vector.mp3 |
And then you want to describe the actual rate at which it's going. You know, is it slow rotation or is it fast? So you need to know a direction along with a magnitude. And you might say to yourself, hey, direction, magnitude, sounds like we could use a vector. And in fact, that's what we do. You use some kind of vector... | Describing rotation in 3d with a vector.mp3 |
And you might say to yourself, hey, direction, magnitude, sounds like we could use a vector. And in fact, that's what we do. You use some kind of vector whose length is gonna correspond to the rate at which it's rotating, typically in radians per second. It's called the angular velocity. And then the direction describe... | Describing rotation in 3d with a vector.mp3 |
It's called the angular velocity. And then the direction describes the axis of rotation itself. But similar to how in two dimensions, there was an ambiguity between clockwise and counterclockwise, if this was the only convention we had, it would be ambiguous whether you should use this vector or if you should use one p... | Describing rotation in 3d with a vector.mp3 |
And the way I've chosen to draw these guys, by the way, it doesn't matter where they are. Remember, a vector, it just has a magnitude and a direction and you can put it anywhere in space. I figured it was natural enough to just kind of put them around the poles just so that you could see them on the axis of rotation it... | Describing rotation in 3d with a vector.mp3 |
So the question is, what vector do you use? Do you use the one pointing in this direction or do you use this green one pointing in the opposite direction? And for this, we have a convention known as the right-hand rule. So I'll go ahead and bring in a picture here to illustrate the right-hand rule. What you imagine doi... | Describing rotation in 3d with a vector.mp3 |
So I'll go ahead and bring in a picture here to illustrate the right-hand rule. What you imagine doing is taking the fingers of your right hand and curling them around in the direction of rotation. And what I mean by that is the tips of your fingers will be pointing kind of the direction that the surface of the sphere ... | Describing rotation in 3d with a vector.mp3 |
Then when you stick out your thumb, that's the direction, that is the choice of vector which should describe that rotation. So in the specific example we have here, when you stick out your right thumb, that corresponds to the white vector, not the green one. But if you did things the other way around, if, oh, move this... | Describing rotation in 3d with a vector.mp3 |
So let's see, get him to stay in place. If you move things the other way around such that the rotation were going kind of in the opposite direction, then when you imagine curling the fingers of your right hand around that direction, your thumb is gonna point according to the green vector. But with the original rotation... | Describing rotation in 3d with a vector.mp3 |
The white vector is the one to go with. And this is actually pretty cool, right? Because you're packing a lot of information into that vector. It tells you what the axis is, it tells you the speed of rotation via its magnitude, and then the choice of which direction along that axis tells you whether the globe is going ... | Describing rotation in 3d with a vector.mp3 |
It tells you what the axis is, it tells you the speed of rotation via its magnitude, and then the choice of which direction along that axis tells you whether the globe is going one way or if it's going the other. So with just three numbers, the three-dimensional coordinates of this vector, you can perfectly describe an... | Describing rotation in 3d with a vector.mp3 |
And when I say that, I mean that let's say I were to take this line integral along the path c of f dot dr. And let's say my path looks like this. That's my x and y-axis, y and x. And let's say my path looks something like this. I start there and I go over there to point c, just like, or to my end point. The curve here ... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
I start there and I go over there to point c, just like, or to my end point. The curve here is c. And so I would evaluate this line integral, this vector field, along this path. This would be a path independent vector field, or we call that a conservative vector field. If this thing is equal to the same integral over a... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
If this thing is equal to the same integral over a different path that has the same end points. So let's call this c1. So this is c1 and this is c2. This vector field is conservative if I start at the same point but I take a different path. Let's say I go something like that. If I take a different path, I start at the ... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
This vector field is conservative if I start at the same point but I take a different path. Let's say I go something like that. If I take a different path, I start at the same point, but I still get the same value. These integrals, what this is telling me is that all it cares about to evaluate these integrals is my sta... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
These integrals, what this is telling me is that all it cares about to evaluate these integrals is my starting point and my ending point. It doesn't care what I do in between. It doesn't care how I get from my starting point to my end point. These two integrals have the same start point and same end point. So irregardl... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
These two integrals have the same start point and same end point. So irregardless of their actual path, they're going to be the same. That's what it means for f to be a conservative field or what it means for this integral to be path independent. So before I prove or I show you the conditions, let's build up our tool k... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
So before I prove or I show you the conditions, let's build up our tool kit a little bit. And so you may or may not have already seen the multivariable chain rule. And I'm not going to prove it in this video, but I think it'll be pretty intuitive for you. So maybe it doesn't need to have a proof, or I'll prove it event... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
So maybe it doesn't need to have a proof, or I'll prove it eventually, but I really just want to give you the intuition. And all that says is that if I have some function, let's say I have f of x and y, but x and y are then functions of, let's say, a third variable t. So f of x of t and y of t. That the derivative of f... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
So you can take a regular derivative. So times how fast x changes with respect to t. This is a standard derivative, this is partial derivative, because at that level we're dealing with two variables. And we're not done. Plus how fast f changes with respect to y, how fast the partial of f with respect to y, times the de... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
Plus how fast f changes with respect to y, how fast the partial of f with respect to y, times the derivative of y with respect to t. So dy dt. I'm not going to prove it, but I think it makes pretty good intuition. This is saying, as I move a little bit dt, how much of a df do I get? Or how fast does f change with respe... | Path independence for line integrals Multivariable Calculus Khan Academy.mp3 |
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