Sentence
stringlengths 131
8.39k
| video_title
stringlengths 12
104
|
|---|---|
So the overall effect of the composition function is to just take a real number to a real number. So it's a single-variable function, so that's why we're taking this ordinary derivative, rather than a partial derivative, or gradient, or anything like that. But because it goes through a multi-dimensional space, and you have this intermediary multivariable nature to it, that's why you have a gradient and a vector-valued derivative. So with the formal argument, the first thing you might do is just write out the formal definition of a derivative. And in this case, it's a limit. Definitions of derivatives are always gonna be some kind of limit as a variable goes to zero. And here, you're loosely thinking about h as being dt.
|
More formal treatment of multivariable chain rule.mp3
|
So with the formal argument, the first thing you might do is just write out the formal definition of a derivative. And in this case, it's a limit. Definitions of derivatives are always gonna be some kind of limit as a variable goes to zero. And here, you're loosely thinking about h as being dt. And you could write delta t, but it's common to use h just because that can be used for whatever your differential quantity is. So that's on the denominator, because you're thinking of it as dt. And the top is whatever the change to this whole function is when you nudge that input by t. And what I mean by that is you'll take f of v, not of t, but of t plus h, that kind of nudged output value, and you're wondering how different that is from f of v of t, your original value, v of t. So this is what happens when you just apply the formal definition of the derivative, the ordinary derivative, to your composition function.
|
More formal treatment of multivariable chain rule.mp3
|
And here, you're loosely thinking about h as being dt. And you could write delta t, but it's common to use h just because that can be used for whatever your differential quantity is. So that's on the denominator, because you're thinking of it as dt. And the top is whatever the change to this whole function is when you nudge that input by t. And what I mean by that is you'll take f of v, not of t, but of t plus h, that kind of nudged output value, and you're wondering how different that is from f of v of t, your original value, v of t. So this is what happens when you just apply the formal definition of the derivative, the ordinary derivative, to your composition function. And now, what do you do as you're trying to reason about what this should equal? And a good place to start, actually, is to look back to the intuition that I was giving for the multivariable chain rule in the first place. You imagine nudging your input by some dt, some tiny change, and I was saying, oh, so that causes a change in the intermediary space of some kind of, you know, you could call it dv, a change in the vector.
|
More formal treatment of multivariable chain rule.mp3
|
And the top is whatever the change to this whole function is when you nudge that input by t. And what I mean by that is you'll take f of v, not of t, but of t plus h, that kind of nudged output value, and you're wondering how different that is from f of v of t, your original value, v of t. So this is what happens when you just apply the formal definition of the derivative, the ordinary derivative, to your composition function. And now, what do you do as you're trying to reason about what this should equal? And a good place to start, actually, is to look back to the intuition that I was giving for the multivariable chain rule in the first place. You imagine nudging your input by some dt, some tiny change, and I was saying, oh, so that causes a change in the intermediary space of some kind of, you know, you could call it dv, a change in the vector. And the way that you're thinking about that is you take the vector value derivative and multiply it by dt. It's the proportionality constant between the size of your nudge and the resulting vector. And loosely, you might imagine those dts crossing out as if they were fractions.
|
More formal treatment of multivariable chain rule.mp3
|
You imagine nudging your input by some dt, some tiny change, and I was saying, oh, so that causes a change in the intermediary space of some kind of, you know, you could call it dv, a change in the vector. And the way that you're thinking about that is you take the vector value derivative and multiply it by dt. It's the proportionality constant between the size of your nudge and the resulting vector. And loosely, you might imagine those dts crossing out as if they were fractions. Doesn't really matter. And then you say, what does this change, does this change by a dv cause for f? And by definition, the resulting nudge to the output space of f is the directional derivative in the direction of whatever your vector nudge is of the function f. So this is the loose intuition, and where does that carry over to formality?
|
More formal treatment of multivariable chain rule.mp3
|
And loosely, you might imagine those dts crossing out as if they were fractions. Doesn't really matter. And then you say, what does this change, does this change by a dv cause for f? And by definition, the resulting nudge to the output space of f is the directional derivative in the direction of whatever your vector nudge is of the function f. So this is the loose intuition, and where does that carry over to formality? You say, well, in this intermediary space, we had to deal with the vector value derivative of v. So it might be a good thing to just write down that definition, right? Write down the fact that the definition for the vector value derivative of v, again, it looks almost identical. All these derivative definitions really do look kind of the same, because what you're doing is you're taking the limit as h goes to zero.
|
More formal treatment of multivariable chain rule.mp3
|
And by definition, the resulting nudge to the output space of f is the directional derivative in the direction of whatever your vector nudge is of the function f. So this is the loose intuition, and where does that carry over to formality? You say, well, in this intermediary space, we had to deal with the vector value derivative of v. So it might be a good thing to just write down that definition, right? Write down the fact that the definition for the vector value derivative of v, again, it looks almost identical. All these derivative definitions really do look kind of the same, because what you're doing is you're taking the limit as h goes to zero. h we're still thinking of as being dt. So it kind of sits on the bottom. But here, you're just wondering how your vector changes.
|
More formal treatment of multivariable chain rule.mp3
|
All these derivative definitions really do look kind of the same, because what you're doing is you're taking the limit as h goes to zero. h we're still thinking of as being dt. So it kind of sits on the bottom. But here, you're just wondering how your vector changes. And the difference, even though, you know, we're kind of writing this the same way, and it looks almost identical notationally, is on the numerator here, this v of t plus h, and this v of t, these are vectors. So this is kind of a vector minus a vector. So when you take the limit, you're getting a limiting vector, something in your high-dimensional space.
|
More formal treatment of multivariable chain rule.mp3
|
But here, you're just wondering how your vector changes. And the difference, even though, you know, we're kind of writing this the same way, and it looks almost identical notationally, is on the numerator here, this v of t plus h, and this v of t, these are vectors. So this is kind of a vector minus a vector. So when you take the limit, you're getting a limiting vector, something in your high-dimensional space. It's not just a number. And now, another way to write this, one that's more helpful, more conducive to manipulation, is to say, not that it equals the limit of this value, and I'm gonna go ahead and just kind of copy this value here, kind of down here, and say the value of our derivative actually equals this, subject to some kind of error, which I'll just write as e of h, like an error function of h. And what you should be thinking is that that error function goes to zero as h goes to zero. This is just writing things so that we're able to manipulate it a little bit more easily.
|
More formal treatment of multivariable chain rule.mp3
|
So when you take the limit, you're getting a limiting vector, something in your high-dimensional space. It's not just a number. And now, another way to write this, one that's more helpful, more conducive to manipulation, is to say, not that it equals the limit of this value, and I'm gonna go ahead and just kind of copy this value here, kind of down here, and say the value of our derivative actually equals this, subject to some kind of error, which I'll just write as e of h, like an error function of h. And what you should be thinking is that that error function goes to zero as h goes to zero. This is just writing things so that we're able to manipulate it a little bit more easily. So I'll give ourselves some room here. And what you can do with this is multiply all sides by h. So this is our vector value derivative, just rewriting it, multiplied by h. And you're thinking of this h as a dt, so maybe in the back of your mind, you're kind of thinking of canceling this dt with the h. And what it equals is this top, this numerator here, which was v of t plus h, minus v of t. And in the back of your mind, you might be thinking this whole thing represents, you know, dv, a change in v, so the idea of canceling out that dt with the h really does kind of come through here. But the difference between the more hand-waving argument before of canceling those out, and what we're doing here, is now we're accounting for that error function, that, in this case, it's now multiplied by h, because everything was multiplied by h error function.
|
More formal treatment of multivariable chain rule.mp3
|
This is just writing things so that we're able to manipulate it a little bit more easily. So I'll give ourselves some room here. And what you can do with this is multiply all sides by h. So this is our vector value derivative, just rewriting it, multiplied by h. And you're thinking of this h as a dt, so maybe in the back of your mind, you're kind of thinking of canceling this dt with the h. And what it equals is this top, this numerator here, which was v of t plus h, minus v of t. And in the back of your mind, you might be thinking this whole thing represents, you know, dv, a change in v, so the idea of canceling out that dt with the h really does kind of come through here. But the difference between the more hand-waving argument before of canceling those out, and what we're doing here, is now we're accounting for that error function, that, in this case, it's now multiplied by h, because everything was multiplied by h error function. And there's actually another way that I'm gonna write this. There's a very useful convention in analysis where I'll take something like this, and instead I'll write it as little o of h. And this isn't literally a function, it's just a stand-in to say whatever this is, whatever function that represents, it satisfies the property that when we take that function and divide it by h, that will go to zero as h goes to zero, right? Which is true here, because you imagine taking this and dividing it by h, and that would be, this h cancels out, and you just have your error function is gonna go to zero.
|
More formal treatment of multivariable chain rule.mp3
|
But the difference between the more hand-waving argument before of canceling those out, and what we're doing here, is now we're accounting for that error function, that, in this case, it's now multiplied by h, because everything was multiplied by h error function. And there's actually another way that I'm gonna write this. There's a very useful convention in analysis where I'll take something like this, and instead I'll write it as little o of h. And this isn't literally a function, it's just a stand-in to say whatever this is, whatever function that represents, it satisfies the property that when we take that function and divide it by h, that will go to zero as h goes to zero, right? Which is true here, because you imagine taking this and dividing it by h, and that would be, this h cancels out, and you just have your error function is gonna go to zero. So now what I do is I use this entire expression to write this v of t plus h. And the reason I wanna do that, if we kind of scroll back up, is because we see v of t plus h showing up in the original definition we care about. So this is just a way of starting to get a grapple on that a little bit more firmly. So what I'd write, I'd say that that v of t plus h, v of t plus h, that nudged output value, is equal to the original value that I have, v of t, plus, and it's gonna be plus this derivative term, and you can kind of think that it's almost like a Taylor polynomial, where this is our first order term.
|
More formal treatment of multivariable chain rule.mp3
|
Which is true here, because you imagine taking this and dividing it by h, and that would be, this h cancels out, and you just have your error function is gonna go to zero. So now what I do is I use this entire expression to write this v of t plus h. And the reason I wanna do that, if we kind of scroll back up, is because we see v of t plus h showing up in the original definition we care about. So this is just a way of starting to get a grapple on that a little bit more firmly. So what I'd write, I'd say that that v of t plus h, v of t plus h, that nudged output value, is equal to the original value that I have, v of t, plus, and it's gonna be plus this derivative term, and you can kind of think that it's almost like a Taylor polynomial, where this is our first order term. You know, we're evaluating it at whatever that t is, but we're multiplying it by the value of that nudge, that linear term. And then the rest of the stuff is just some little o of h. And maybe you'd say, shouldn't you be subtracting off that little o of h? And it's not an actual function, it just represents anything that shrinks.
|
More formal treatment of multivariable chain rule.mp3
|
So what I'd write, I'd say that that v of t plus h, v of t plus h, that nudged output value, is equal to the original value that I have, v of t, plus, and it's gonna be plus this derivative term, and you can kind of think that it's almost like a Taylor polynomial, where this is our first order term. You know, we're evaluating it at whatever that t is, but we're multiplying it by the value of that nudge, that linear term. And then the rest of the stuff is just some little o of h. And maybe you'd say, shouldn't you be subtracting off that little o of h? And it's not an actual function, it just represents anything that shrinks. And maybe I should say it's the absolute value, like the magnitude, because in this case, this is a vector-valued quantity, you know that error is a vector. So it's the size of that vector divided by the size of h goes to zero. So this is the main tool that we're gonna end up using.
|
More formal treatment of multivariable chain rule.mp3
|
And it's not an actual function, it just represents anything that shrinks. And maybe I should say it's the absolute value, like the magnitude, because in this case, this is a vector-valued quantity, you know that error is a vector. So it's the size of that vector divided by the size of h goes to zero. So this is the main tool that we're gonna end up using. This is the way to represent v of t plus h. And now if we go back up to the original definition of the vector-valued derivative, and I'll go ahead and copy that. Go ahead and copy that guy. A little bit of debris.
|
More formal treatment of multivariable chain rule.mp3
|
So this is the main tool that we're gonna end up using. This is the way to represent v of t plus h. And now if we go back up to the original definition of the vector-valued derivative, and I'll go ahead and copy that. Go ahead and copy that guy. A little bit of debris. So copied that original definition for the ordinary derivative of the composition function. And now when I write things in according to all the manipulations that we just did, this is really, it's still a limit, as h goes to zero. But what we put on the inside here is it's f of, now instead of writing v of t plus h, I'm gonna use everything that I did up there.
|
More formal treatment of multivariable chain rule.mp3
|
A little bit of debris. So copied that original definition for the ordinary derivative of the composition function. And now when I write things in according to all the manipulations that we just did, this is really, it's still a limit, as h goes to zero. But what we put on the inside here is it's f of, now instead of writing v of t plus h, I'm gonna use everything that I did up there. It's the value of v of t plus the derivative at our point times the size of h. So again, it's kind of like a Taylor polynomial, this is your linear term. And then it's plus something that we don't care about, something that's gonna get really small as h goes small, and really small in comparison to h, more importantly. And from that, you subtract off f of v of t. Kind of running off the edge, I always keep running off the edge.
|
More formal treatment of multivariable chain rule.mp3
|
But what we put on the inside here is it's f of, now instead of writing v of t plus h, I'm gonna use everything that I did up there. It's the value of v of t plus the derivative at our point times the size of h. So again, it's kind of like a Taylor polynomial, this is your linear term. And then it's plus something that we don't care about, something that's gonna get really small as h goes small, and really small in comparison to h, more importantly. And from that, you subtract off f of v of t. Kind of running off the edge, I always keep running off the edge. And all of that is divided by h. Now the point here is when you look at this limit, because we're taking it as h goes to zero, we'll basically be able to ignore this o of h component, because as h goes to zero, this gets very, very small in comparison to h. So everything that's on the inside here is basically just the v of t plus this vector value. And this is h times some kind of vector. But if you think back, I made a video on the formal definition of the directional derivative.
|
More formal treatment of multivariable chain rule.mp3
|
And from that, you subtract off f of v of t. Kind of running off the edge, I always keep running off the edge. And all of that is divided by h. Now the point here is when you look at this limit, because we're taking it as h goes to zero, we'll basically be able to ignore this o of h component, because as h goes to zero, this gets very, very small in comparison to h. So everything that's on the inside here is basically just the v of t plus this vector value. And this is h times some kind of vector. But if you think back, I made a video on the formal definition of the directional derivative. And if you remember it, or if you kind of go back and take a look now, this is exactly the formal definition of the directional derivative. We're taking h to go to zero. The thing we're multiplying it by is a certain vector quantity.
|
More formal treatment of multivariable chain rule.mp3
|
But if you think back, I made a video on the formal definition of the directional derivative. And if you remember it, or if you kind of go back and take a look now, this is exactly the formal definition of the directional derivative. We're taking h to go to zero. The thing we're multiplying it by is a certain vector quantity. That vector is the nudge to your original value. And then we're dividing everything by h. So by definition, this entire thing is the directional derivative in the direction of the derivative of the function at t. I'm writing v prime t instead of getting the whole dv dt down there. All of that of f evaluated at where?
|
More formal treatment of multivariable chain rule.mp3
|
The thing we're multiplying it by is a certain vector quantity. That vector is the nudge to your original value. And then we're dividing everything by h. So by definition, this entire thing is the directional derivative in the direction of the derivative of the function at t. I'm writing v prime t instead of getting the whole dv dt down there. All of that of f evaluated at where? Well, the place that we're starting is just v of t. So that's v of t. And that's it, that's the answer. Because when you evaluate the directional derivative, the way that you do that, you take the gradient of f, evaluate it at whatever point you're starting at, in this case, it's the output of v of t, and you take the dot product between that and the vector value derivative, well, I mean, the dot product between that and whatever your vector is, which in this case is the vector value derivative of v. And that's the multivariable chain rule. And if you look back through the line of reasoning, it all really did match the thoughts of kind of nudging, nudging, and seeing how that nudged.
|
More formal treatment of multivariable chain rule.mp3
|
All of that of f evaluated at where? Well, the place that we're starting is just v of t. So that's v of t. And that's it, that's the answer. Because when you evaluate the directional derivative, the way that you do that, you take the gradient of f, evaluate it at whatever point you're starting at, in this case, it's the output of v of t, and you take the dot product between that and the vector value derivative, well, I mean, the dot product between that and whatever your vector is, which in this case is the vector value derivative of v. And that's the multivariable chain rule. And if you look back through the line of reasoning, it all really did match the thoughts of kind of nudging, nudging, and seeing how that nudged. Because the reason we thought to use the vector value derivative was because of that intuition. And the reason for all the manipulation that I did is just because I wanted to be able to express what a nudge to the input of v looks like, and what that looks like is the original value plus a certain vector here. This was the resulting nudge in the intermediary space.
|
More formal treatment of multivariable chain rule.mp3
|
And if you look back through the line of reasoning, it all really did match the thoughts of kind of nudging, nudging, and seeing how that nudged. Because the reason we thought to use the vector value derivative was because of that intuition. And the reason for all the manipulation that I did is just because I wanted to be able to express what a nudge to the input of v looks like, and what that looks like is the original value plus a certain vector here. This was the resulting nudge in the intermediary space. I wanted to express that in a formal way. And sure, we have this kind of O of H term that expresses something that shrinks really fast. But once you express it like that, you just end up plopping out the original, the definition of the directional derivative.
|
More formal treatment of multivariable chain rule.mp3
|
This was the resulting nudge in the intermediary space. I wanted to express that in a formal way. And sure, we have this kind of O of H term that expresses something that shrinks really fast. But once you express it like that, you just end up plopping out the original, the definition of the directional derivative. So I hope that gives kind of a satisfying reason for those of you who are a little bit more rigor-inclined for why the multivariable chain rule works. I should also maybe mention there's a more general multivariable chain rule for vector-valued functions. I'll get to that at another point when I talk about the connections between multivariable calculus and linear algebra.
|
More formal treatment of multivariable chain rule.mp3
|
I've rewritten Stokes' theorem right over here. What I want to focus on in this video is the question of orientation. Because there are two different orientations for our boundary curve, we could go in that direction like that, or we could go in the opposite direction. We could be going like that. And there are also two different orientations for this normal vector. The normal vector might pop out like that, or it could actually go in to the surface like that. So we have to make sure that our orientations are consistent.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
We could be going like that. And there are also two different orientations for this normal vector. The normal vector might pop out like that, or it could actually go in to the surface like that. So we have to make sure that our orientations are consistent. And what I want to do is give you two different ways of thinking about it, and you might think of others, but these are the ones that work for me. In order for Stokes' theorem to hold, we have to make sure that we're not actually picking the negative of one or the other orientations. And so the easiest way for me to remember it is if our normal vector, if the normal vector is, let's say it goes in that direction, and if you have some hypothetical person traversing the boundary of our surface, and their direction is pointed, and their head is pointed in the same direction as the normal vector, so this is the normal vector, so their head is pointed in the exact same direction as the normal vector, so you could say maybe their body, or really their head, so that's them, then the direction that you would have to actually traverse the boundary is the direction that would allow this person to keep the surface to their left.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
So we have to make sure that our orientations are consistent. And what I want to do is give you two different ways of thinking about it, and you might think of others, but these are the ones that work for me. In order for Stokes' theorem to hold, we have to make sure that we're not actually picking the negative of one or the other orientations. And so the easiest way for me to remember it is if our normal vector, if the normal vector is, let's say it goes in that direction, and if you have some hypothetical person traversing the boundary of our surface, and their direction is pointed, and their head is pointed in the same direction as the normal vector, so this is the normal vector, so their head is pointed in the exact same direction as the normal vector, so you could say maybe their body, or really their head, so that's them, then the direction that you would have to actually traverse the boundary is the direction that would allow this person to keep the surface to their left. So over here, he would have to go in this direction, in order to keep the surface to his left. So he would have to go, he would have to go just like that. If we oriented the surface differently, so let me redraw the surface right over here, draw a similar surface.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
And so the easiest way for me to remember it is if our normal vector, if the normal vector is, let's say it goes in that direction, and if you have some hypothetical person traversing the boundary of our surface, and their direction is pointed, and their head is pointed in the same direction as the normal vector, so this is the normal vector, so their head is pointed in the exact same direction as the normal vector, so you could say maybe their body, or really their head, so that's them, then the direction that you would have to actually traverse the boundary is the direction that would allow this person to keep the surface to their left. So over here, he would have to go in this direction, in order to keep the surface to his left. So he would have to go, he would have to go just like that. If we oriented the surface differently, so let me redraw the surface right over here, draw a similar surface. So if we had a surface, so this surface looks very similar. If this surface, this is a very similar looking surface that I'm drawing right over here, just to give an idea of some of the contours, but if we said that the normal vector for this surface, we orient it in the opposite way. So if we said that the normal vector here was actually pointing downward like that, then we would have to, in order for Stokes' Theorem to hold, we would have to traverse the boundary in a different direction.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
If we oriented the surface differently, so let me redraw the surface right over here, draw a similar surface. So if we had a surface, so this surface looks very similar. If this surface, this is a very similar looking surface that I'm drawing right over here, just to give an idea of some of the contours, but if we said that the normal vector for this surface, we orient it in the opposite way. So if we said that the normal vector here was actually pointing downward like that, then we would have to, in order for Stokes' Theorem to hold, we would have to traverse the boundary in a different direction. Because once again, if I draw my little character right over here, his head is pointed in the direction of the normal vector. He is now upside down, so let me draw him. So this is him, this is him running right over here.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
So if we said that the normal vector here was actually pointing downward like that, then we would have to, in order for Stokes' Theorem to hold, we would have to traverse the boundary in a different direction. Because once again, if I draw my little character right over here, his head is pointed in the direction of the normal vector. He is now upside down, so let me draw him. So this is him, this is him running right over here. I could draw a better job. This is him running right over here. Now, in order to keep, and from his point of view, this would kind of look like some type of a pool or a ditch of some kind.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
So this is him, this is him running right over here. I could draw a better job. This is him running right over here. Now, in order to keep, and from his point of view, this would kind of look like some type of a pool or a ditch of some kind. It would actually go down. Here it looks like a hill to him. But since he's upside down, in order for him to keep the boundary to his left, he would have to now go in, he would have to now go in the other direction.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
Now, in order to keep, and from his point of view, this would kind of look like some type of a pool or a ditch of some kind. It would actually go down. Here it looks like a hill to him. But since he's upside down, in order for him to keep the boundary to his left, he would have to now go in, he would have to now go in the other direction. So depending on the orientation of your normal vector, or which is really the orientation of your actual surface, will dictate how you need to traverse the path. Now, another way to think about it, and this idea was introduced by one of the viewers on YouTube but it's a valid way of thinking about it, is to imagine that the surface is a bottle cap. And so let me draw some type of a bottle over here.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
But since he's upside down, in order for him to keep the boundary to his left, he would have to now go in, he would have to now go in the other direction. So depending on the orientation of your normal vector, or which is really the orientation of your actual surface, will dictate how you need to traverse the path. Now, another way to think about it, and this idea was introduced by one of the viewers on YouTube but it's a valid way of thinking about it, is to imagine that the surface is a bottle cap. And so let me draw some type of a bottle over here. So I'll draw, let me draw a bottle. So this is, you can imagine some type of a glass soda bottle. So, but what we really care about is the cap of the bottle.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
And so let me draw some type of a bottle over here. So I'll draw, let me draw a bottle. So this is, you can imagine some type of a glass soda bottle. So, but what we really care about is the cap of the bottle. So, make it feel like it's glass. So there, that's our bottle. And let me draw, let me draw its cap.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
So, but what we really care about is the cap of the bottle. So, make it feel like it's glass. So there, that's our bottle. And let me draw, let me draw its cap. Let me draw the cap of the bottle because that's what we care about. We can kind of imagine that being the surface. So this is the cap of our bottle.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
And let me draw, let me draw its cap. Let me draw the cap of the bottle because that's what we care about. We can kind of imagine that being the surface. So this is the cap of our bottle. And you just need to think about, well, which way would I have to twist the cap in order to make the cap move up, in order to take the cap off? And you could think of the normal vector as the direction that the cap would move, is the direction that the cap would move. And the twisting is the direction that you would have to traverse the path.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
So this is the cap of our bottle. And you just need to think about, well, which way would I have to twist the cap in order to make the cap move up, in order to take the cap off? And you could think of the normal vector as the direction that the cap would move, is the direction that the cap would move. And the twisting is the direction that you would have to traverse the path. So you would have to twist the bottle that way. Or you can think about the other way. If you twisted the bottle the other way, then the cap would move down.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
And the twisting is the direction that you would have to traverse the path. So you would have to twist the bottle that way. Or you can think about the other way. If you twisted the bottle the other way, then the cap would move down. So the normal vector is the direction that the cap would move. And the direction that you would traverse the boundary is how you would actually twist it. So either of these are ways of thinking about it, but they're important to keep in mind, especially once the shapes start getting a little bit more convoluted and oriented in strange ways.
|
Orientation and stokes Multivariable Calculus Khan Academy.mp3
|
And we'll start with an example of a torus, or more commonly known as a donut shape. And we know what a donut looks like. Let me draw it in a suitable, well I don't have any suitable donut colors, so I'll just use green. So a donut looks something like this. So it has a hole in the center, and maybe the other side of the donut looks something like that. And we could shade it in like that, and then maybe shade it in like that. That is what a donut looks like.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So a donut looks something like this. So it has a hole in the center, and maybe the other side of the donut looks something like that. And we could shade it in like that, and then maybe shade it in like that. That is what a donut looks like. So how do we construct that using two parameters? So what we want to do is you can just visualize that a donut, if we were to draw some axes here, so that's our donut, let me draw some axes. So let's say I have the z-axis that goes straight up and down.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
That is what a donut looks like. So how do we construct that using two parameters? So what we want to do is you can just visualize that a donut, if we were to draw some axes here, so that's our donut, let me draw some axes. So let's say I have the z-axis that goes straight up and down. So the way I've drawn it here, the donut's a little at a tilt so the z-axis, I'll tilt it a little bit. So our z-axis goes straight through the center of the donut. So that right there.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So let's say I have the z-axis that goes straight up and down. So the way I've drawn it here, the donut's a little at a tilt so the z-axis, I'll tilt it a little bit. So our z-axis goes straight through the center of the donut. So that right there. This is going to be an exercise in drawing more than anything else. So that is my z-axis, and then you can imagine the z-axis goes from there, and then this coming out of here will be my x-axis, that right there is my x-axis, and then maybe my y-axis comes out like that. And the whole reason why I drew it this way is that if you imagine the cross-section of this donut, I'll draw it a little bit neater, but the cross-section of this donut in the xz-axis is going to look something like this.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So that right there. This is going to be an exercise in drawing more than anything else. So that is my z-axis, and then you can imagine the z-axis goes from there, and then this coming out of here will be my x-axis, that right there is my x-axis, and then maybe my y-axis comes out like that. And the whole reason why I drew it this way is that if you imagine the cross-section of this donut, I'll draw it a little bit neater, but the cross-section of this donut in the xz-axis is going to look something like this. If I were to just slice it in the xz-axis, it would look something like that. I'm slicing, this is the xz-axis. That would be the slice.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And the whole reason why I drew it this way is that if you imagine the cross-section of this donut, I'll draw it a little bit neater, but the cross-section of this donut in the xz-axis is going to look something like this. If I were to just slice it in the xz-axis, it would look something like that. I'm slicing, this is the xz-axis. That would be the slice. It would trace out, and we're thinking about not a full donut, just the surface of a donut. So it would trace out a circle like this. If you were to cut the donut in the positive zy-axis, it's going to trace out a circle that looks something like that right there.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
That would be the slice. It would trace out, and we're thinking about not a full donut, just the surface of a donut. So it would trace out a circle like this. If you were to cut the donut in the positive zy-axis, it's going to trace out a circle that looks something like that right there. And if you go out here, you're going to get circles. It's just a bunch of circles. So if you think about it, it's a bunch of circles rotated around the z-axis.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
If you were to cut the donut in the positive zy-axis, it's going to trace out a circle that looks something like that right there. And if you go out here, you're going to get circles. It's just a bunch of circles. So if you think about it, it's a bunch of circles rotated around the z-axis. If you think of it that way, it'll give us some good intuition for the best way to parameterize this thing. So let's do it that way. So let's start off with just the zy-axis.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So if you think about it, it's a bunch of circles rotated around the z-axis. If you think of it that way, it'll give us some good intuition for the best way to parameterize this thing. So let's do it that way. So let's start off with just the zy-axis. I'll draw it a little bit neater than I've done here. So that is the z-axis, and that is y, just like that. And let's say that the center of these circles, let's say it lies on, it can lie, when you cross the zy-axis, the center sits on the y-axis.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So let's start off with just the zy-axis. I'll draw it a little bit neater than I've done here. So that is the z-axis, and that is y, just like that. And let's say that the center of these circles, let's say it lies on, it can lie, when you cross the zy-axis, the center sits on the y-axis. I didn't draw it that neatly here, but I think you can visualize. So it sits right there on the y-axis. And let's say that it is a distance b away from the center of the torus, or from the z-axis.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And let's say that the center of these circles, let's say it lies on, it can lie, when you cross the zy-axis, the center sits on the y-axis. I didn't draw it that neatly here, but I think you can visualize. So it sits right there on the y-axis. And let's say that it is a distance b away from the center of the torus, or from the z-axis. It's a distance of b. It's always going to be a distance of b. If you imagine the top of the donut, let me draw the top of the donut.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And let's say that it is a distance b away from the center of the torus, or from the z-axis. It's a distance of b. It's always going to be a distance of b. If you imagine the top of the donut, let me draw the top of the donut. If you're looking down on a donut, so let me draw a donut right here. So if you're looking down on a donut, it just looks something like that. The z-axis is just going to be popping straight out.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
If you imagine the top of the donut, let me draw the top of the donut. If you're looking down on a donut, so let me draw a donut right here. So if you're looking down on a donut, it just looks something like that. The z-axis is just going to be popping straight out. The x-axis would come down like this, and then the y-axis would go to the right like that. So you can imagine I'm just flying above this. I'm sitting on the z-axis looking down at the donut.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
The z-axis is just going to be popping straight out. The x-axis would come down like this, and then the y-axis would go to the right like that. So you can imagine I'm just flying above this. I'm sitting on the z-axis looking down at the donut. It'll look just like this. And if you imagine the cross-section, this circle right here will look just like that. This circle right here, that top part of the circle, if you're looking down, would look just like that.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
I'm sitting on the z-axis looking down at the donut. It'll look just like this. And if you imagine the cross-section, this circle right here will look just like that. This circle right here, that top part of the circle, if you're looking down, would look just like that. And this distance, this distance b, is the distance from the z-axis to the center of each of these circles. So this distance, let me draw it in the same color, from the center to the center of these circles, that is going to be b. And it's just going to keep going to the center of the circles.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
This circle right here, that top part of the circle, if you're looking down, would look just like that. And this distance, this distance b, is the distance from the z-axis to the center of each of these circles. So this distance, let me draw it in the same color, from the center to the center of these circles, that is going to be b. And it's just going to keep going to the center of the circles. b. That's going to be b. That's going to be b.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And it's just going to keep going to the center of the circles. b. That's going to be b. That's going to be b. All of, from the center of our torus to the center of our circle that defies this torus, it's a distance of b. So this distance right here, that distance right there is b. And from b, we can imagine we have a radius of length a.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
That's going to be b. All of, from the center of our torus to the center of our circle that defies this torus, it's a distance of b. So this distance right here, that distance right there is b. And from b, we can imagine we have a radius of length a. So these circles have radius of length a. So this distance right here is a. This distance right here is a.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And from b, we can imagine we have a radius of length a. So these circles have radius of length a. So this distance right here is a. This distance right here is a. This distance right there is a. That distance right there is a. If I were to look at these circles, these circles have radius a.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
This distance right here is a. This distance right there is a. That distance right there is a. If I were to look at these circles, these circles have radius a. And what we're going to do is have two parameters. One is the angle that this radius makes with the xz-plane. So you can imagine the x-axis coming out.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
If I were to look at these circles, these circles have radius a. And what we're going to do is have two parameters. One is the angle that this radius makes with the xz-plane. So you can imagine the x-axis coming out. Let me do that in the same color. You can imagine the x-axis coming out here. So this is the xz-plane.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So you can imagine the x-axis coming out. Let me do that in the same color. You can imagine the x-axis coming out here. So this is the xz-plane. So one parameter is going to be the angle between our radius and the xz-plane. We're going to call that angle, or that parameter, s. And so as s goes between 0 and 2 pi, when 0 is just going to be, you're going to be at this point right here. And then as it goes to 2 pi, you're going to trace out a circle that looks just like that.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So this is the xz-plane. So one parameter is going to be the angle between our radius and the xz-plane. We're going to call that angle, or that parameter, s. And so as s goes between 0 and 2 pi, when 0 is just going to be, you're going to be at this point right here. And then as it goes to 2 pi, you're going to trace out a circle that looks just like that. Now, we only have one parameter. What we want to do is then spin this circle around. What I just drew is just that circle right there.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And then as it goes to 2 pi, you're going to trace out a circle that looks just like that. Now, we only have one parameter. What we want to do is then spin this circle around. What I just drew is just that circle right there. What we want to do is spin the entire circle around. So let's define another parameter. We'll call this one t. And I'll take the top view again.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
What I just drew is just that circle right there. What we want to do is spin the entire circle around. So let's define another parameter. We'll call this one t. And I'll take the top view again. This one's getting a little bit messy. Let me draw another top view. As you can see, this is all about visualization.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
We'll call this one t. And I'll take the top view again. This one's getting a little bit messy. Let me draw another top view. As you can see, this is all about visualization. So let's say this is my x-axis, that is my y-axis. And we said we started here on the zy-plane. We are b away from the z-axis.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
As you can see, this is all about visualization. So let's say this is my x-axis, that is my y-axis. And we said we started here on the zy-plane. We are b away from the z-axis. So that distance is b. In this diagram, the z-axis is just popping out at us. It's popping out of the page.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
We are b away from the z-axis. So that distance is b. In this diagram, the z-axis is just popping out at us. It's popping out of the page. We're looking straight down. It's just like the same view is right there. And what I just drew, when s is equal to 0 radians, we're going to be out here.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
It's popping out of the page. We're looking straight down. It's just like the same view is right there. And what I just drew, when s is equal to 0 radians, we're going to be out here. We're going to be exactly 1 radius further along the y-axis. Then we're going to rotate. As we rotate around, we're going to rotate and then come all the way over here.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And what I just drew, when s is equal to 0 radians, we're going to be out here. We're going to be exactly 1 radius further along the y-axis. Then we're going to rotate. As we rotate around, we're going to rotate and then come all the way over here. That's when we're right over there. And then come back down. So if you looked on the top of the circle, it's going to look like that.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
As we rotate around, we're going to rotate and then come all the way over here. That's when we're right over there. And then come back down. So if you looked on the top of the circle, it's going to look like that. Now, to make the donut, we're going to have to rotate this whole setup around the z-axis. Remember, the z-axis is popping out. It's looking straight up at us.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So if you looked on the top of the circle, it's going to look like that. Now, to make the donut, we're going to have to rotate this whole setup around the z-axis. Remember, the z-axis is popping out. It's looking straight up at us. It's coming out of your video screen. Now, to rotate it, we're going to rotate this circle around the z-axis. And to do that, we'll define a parameter that tells us how much we have rotated it.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
It's looking straight up at us. It's coming out of your video screen. Now, to rotate it, we're going to rotate this circle around the z-axis. And to do that, we'll define a parameter that tells us how much we have rotated it. So this is when we've rotated 0 radians. At some point, we're going to be over here. And we would have rotated it.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And to do that, we'll define a parameter that tells us how much we have rotated it. So this is when we've rotated 0 radians. At some point, we're going to be over here. And we would have rotated it. This is b as well. And our circle is going to be looking like this. This is maybe this point on our donut right there.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And we would have rotated it. This is b as well. And our circle is going to be looking like this. This is maybe this point on our donut right there. At that point, we would have rotated it, let's say, t radians. So this parameter of how much have we rotated it around the z-axis, how much have we gone around that way, we're going to call that t. And t is also going to vary between 0 and 2 pi. And I want to make this very clear.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
This is maybe this point on our donut right there. At that point, we would have rotated it, let's say, t radians. So this parameter of how much have we rotated it around the z-axis, how much have we gone around that way, we're going to call that t. And t is also going to vary between 0 and 2 pi. And I want to make this very clear. Let's actually draw the domain that we're mapping from to our surface so that we understand this, I guess, fully. So let me draw some, and then we'll talk about how we can actually parameterize that into a position vector-valued function. So right here, let's call that the t-axis.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And I want to make this very clear. Let's actually draw the domain that we're mapping from to our surface so that we understand this, I guess, fully. So let me draw some, and then we'll talk about how we can actually parameterize that into a position vector-valued function. So right here, let's call that the t-axis. Let's remember how much we rotated around the z-axis right there. And let's call this down here our s-axis. And I think this will help things out a good bit.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So right here, let's call that the t-axis. Let's remember how much we rotated around the z-axis right there. And let's call this down here our s-axis. And I think this will help things out a good bit. So when s is equal to 0, and we vary just t. So they're both going to vary between 0 and 2 pi. So this right here is 0. This right here is 2 pi.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And I think this will help things out a good bit. So when s is equal to 0, and we vary just t. So they're both going to vary between 0 and 2 pi. So this right here is 0. This right here is 2 pi. Let me do some things in between. This is pi. This would be pi over 2, obviously.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
This right here is 2 pi. Let me do some things in between. This is pi. This would be pi over 2, obviously. This would be 3 pi over 4. You do the same thing on the t-axis. It's going to go up to 2 pi.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
This would be pi over 2, obviously. This would be 3 pi over 4. You do the same thing on the t-axis. It's going to go up to 2 pi. Let's do that. So we're going to go up to 2 pi. I really want you to visualize this, because then the parameterization, I think, will be fairly straightforward.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
It's going to go up to 2 pi. Let's do that. So we're going to go up to 2 pi. I really want you to visualize this, because then the parameterization, I think, will be fairly straightforward. So that's 2 pi. This is pi. This is pi over 2.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
I really want you to visualize this, because then the parameterization, I think, will be fairly straightforward. So that's 2 pi. This is pi. This is pi over 2. And then this is 3 pi over 4. So let's think about what it looks like if you just hold s constant at 0, and we just vary t between 0 and 2 pi. And let me do that in magenta right here.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
This is pi over 2. And then this is 3 pi over 4. So let's think about what it looks like if you just hold s constant at 0, and we just vary t between 0 and 2 pi. And let me do that in magenta right here. So we're holding s constant, and we're just varying the parameter 2 pi. So this, if you think about it, should just form a curve in three dimensions, not a surface, because we're only varying one parameter right here. So let's think about what this is.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And let me do that in magenta right here. So we're holding s constant, and we're just varying the parameter 2 pi. So this, if you think about it, should just form a curve in three dimensions, not a surface, because we're only varying one parameter right here. So let's think about what this is. Remember, s is, let me draw my axes. So that is my x-axis. That is my y-axis.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So let's think about what this is. Remember, s is, let me draw my axes. So that is my x-axis. That is my y-axis. And then this is my, I'm getting messier and messier, that is my z-axis. Actually, let me draw it a little bit bigger than that. I think it'll help all of our visualizations.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
That is my y-axis. And then this is my, I'm getting messier and messier, that is my z-axis. Actually, let me draw it a little bit bigger than that. I think it'll help all of our visualizations. All right. So this is my x-axis. That is my y-axis.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
I think it'll help all of our visualizations. All right. So this is my x-axis. That is my y-axis. And then my z-axis goes up like that. Now remember, when s is equal to 0, that means we haven't rotated around this circle at all. That means we're out here.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
That is my y-axis. And then my z-axis goes up like that. Now remember, when s is equal to 0, that means we haven't rotated around this circle at all. That means we're out here. We're going to be b away and then a away again. We haven't rotated around this at all. We're setting s is equal to 0.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
That means we're out here. We're going to be b away and then a away again. We haven't rotated around this at all. We're setting s is equal to 0. So essentially, we're going to be b away. So this is going to be a distance of b away. And then we're going to be another a away.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
We're setting s is equal to 0. So essentially, we're going to be b away. So this is going to be a distance of b away. And then we're going to be another a away. The radius, the b is the center of the circle. And then we're going to be another a away. We're going to be right over there.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
And then we're going to be another a away. The radius, the b is the center of the circle. And then we're going to be another a away. We're going to be right over there. So this is a plus b away. And then we're going to vary t. Remember, t was how much we've gone around the z-axis. These were top views over here.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
We're going to be right over there. So this is a plus b away. And then we're going to vary t. Remember, t was how much we've gone around the z-axis. These were top views over here. So this line right here in our st domain, we can say, when we map it or when you parameterize it, it'll correspond to the curve that's essentially the outer edge of our donut. If this is the top view of the donut, it will be the outer edge of our donut, just like that. So let me draw the outer edge.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
These were top views over here. So this line right here in our st domain, we can say, when we map it or when you parameterize it, it'll correspond to the curve that's essentially the outer edge of our donut. If this is the top view of the donut, it will be the outer edge of our donut, just like that. So let me draw the outer edge. And to do that a little bit better, let me draw the axes in both the positive and the negative domain. It might make my graph a little bit easier to visualize things. Positive and negative domain.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So let me draw the outer edge. And to do that a little bit better, let me draw the axes in both the positive and the negative domain. It might make my graph a little bit easier to visualize things. Positive and negative domain. This is negative z right there. So this line in our t-s plane, I guess we could say, this magenta line, where we hold s at 0 radians and we increase t. This is t is 0. This is t is equal to 2 pi.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
Positive and negative domain. This is negative z right there. So this line in our t-s plane, I guess we could say, this magenta line, where we hold s at 0 radians and we increase t. This is t is 0. This is t is equal to 2 pi. That's t is equal to pi. This is t is equal to 3 pi over 2, all the way back to t is equal to 2 pi. This line corresponds to that line as we increase t and hold s constant at 0.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
This is t is equal to 2 pi. That's t is equal to pi. This is t is equal to 3 pi over 2, all the way back to t is equal to 2 pi. This line corresponds to that line as we increase t and hold s constant at 0. Now let's do another point. Let's say when s is at pi. Remember, when s is at pi, we've gone exactly, pi is 180 degrees, when s is at pi, we've gone exactly 180 degrees around the circle, around each of these circles.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
This line corresponds to that line as we increase t and hold s constant at 0. Now let's do another point. Let's say when s is at pi. Remember, when s is at pi, we've gone exactly, pi is 180 degrees, when s is at pi, we've gone exactly 180 degrees around the circle, around each of these circles. So we're right over there. And now let's hold it constant at pi and then rotate it around to form our donut. So we're going to form the inside of our donut.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
Remember, when s is at pi, we've gone exactly, pi is 180 degrees, when s is at pi, we've gone exactly 180 degrees around the circle, around each of these circles. So we're right over there. And now let's hold it constant at pi and then rotate it around to form our donut. So we're going to form the inside of our donut. So when s is at pi, and we're going to take t from 0. So when s is pi and t is 0, this was the center of our circle. We're going to be a below that.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
So we're going to form the inside of our donut. So when s is at pi, and we're going to take t from 0. So when s is pi and t is 0, this was the center of our circle. We're going to be a below that. We're going to be right over there. And then as we vary, as we increase t, so as we move up along holding s at pi, and we increase t, we're going to trace out the inside of our donut. That will look something like that.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
We're going to be a below that. We're going to be right over there. And then as we vary, as we increase t, so as we move up along holding s at pi, and we increase t, we're going to trace out the inside of our donut. That will look something like that. That was my best shot at drawing it. And then we could do that multiple times. When s is pi over 2, I want to do multiple different colors.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
That will look something like that. That was my best shot at drawing it. And then we could do that multiple times. When s is pi over 2, I want to do multiple different colors. When s is pi over 2, up here we've rotated exactly 90 degrees, right? Pi over 2 is 90 degrees at this point. And then if we vary t, we're essentially tracing out the top of the donut.
|
Introduction to parametrizing a surface with two parameters Multivariable Calculus Khan Academy.mp3
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.