In 5:16 you hinted that you won't consider derivative of a function taking a vector as argument. But in 9:01 we are doing just that, isn't it? Ok, the "function" in 9:01 , the "limit", is almost like the identity function, but you probably know how Engineers like to tickle Mathematicians (for all the times Mathematicians tickle us) . :-)
He is not considering the derivative of that function. He is using distance to be able to take limits. As far as I know, in order to be able to take limits, you need atleast a topological space. A topological space with a distance function is a metric space. He is just making a metric space with the vectors (normed vector space) so that it is possible to take limits, that's all.
@@akashpremrajan9285 A vector is more than its length. And its derivative cannot be only the derivative of a scalar, its length or other scalar. As example, a vector with a constant length does not necessarily have as derivate a zero-length vector, as it is the case for a vector giving the position about a point in a circular motion (on a circle, each positional vector is of a constant length equals to the radius).
A vector is more than it's length, sure. But to know whether two vectors are "close" or not, all you need is the length of the difference between them. Do you disagree?!?! If you disagree, please enlighten mathematicians how you want to measure the closeness of vectors.!!
Did you watch the same video, I watched? The derivative of a vector valued function is a vector only. When did the final answer become a scalar? We used the scalar function distance just to measure the closeness of vectors, to be able to take limits, that's all.
Such a cool little fact that allows for some quick back of the napkin work on predicting say velocity at a point in a video or somwthing like that. It is also nice for a position vector function that doesnt have a closed form seen in an analysis class.
Even though differentiation of functions having vectors as arguments is weird to think about you can still make an intuitive approach to it geometrically and see where the errors first occur.
Hi Prof.! A few of your recent videos in this series, including this one, do not yet appear in your 'complete playlist'. Hopefully it's not too hard to add them to it.
Is there a good way to understand continuity around a vector of a vector valued function? It's easy to see how limits work with scalar values with continuity of points in the x,y plane but hard to picture for resultant vectors from a change in t, "around" another vector. Also seems to beg the question how we would know if there is a uniform instantaneous change for a vector of a vector valued function as the distance between it and any other vector from an increase in t, approaches zero.
If the curve is continuous then, the function is continuous. If the curve is smooth, then the function is smooth. This is the object that actually gives us the intuition behind the concept of continuity.
Starting from a scalar field, you can take a difference, sure, but you must specify the direction (you are not limited only to a single direction, with a + and a - side). A possible definition of the gradient is to get the direction where that difference is at its maximum (a gradient is a vector, after all), isn't it? So, about using gradient in that way, we first should be able to find the difference in ANY direction (at a given point), then, to find the MAX on this set. And have a recipe which works for any system of coordinates assuming that there is a system of coordinates)?
@@snnwstt Gradients are discussed in detail later in this series. But to answer briefly, you can't apply the derivative to a function of a vector argument since h would be a vector and you can't divide by vectors. However, a function can, of ccourse, be of several scalar arguments and that's how the gradient is constructed in high school calculus books.
Or you can define a derivative as [ f(x+1) - f( x-1 ) ]/[ (x+1) -- (x-1)] And no limit . the concept is the same but it's symmetric and uses the smallest number of the whole numbers. You don't need calculus.
If you think about derivative as a velocity(or speed) and measure time x in hours you actually define your current velocity as the distance f you traveled during the last 2 hours divided by two hours. So if you were driving from work for two hours exactly and now you are sitting in the kitchen your velocity NOW equals the velocity of your car. No need for cars.😮
In 5:16 you hinted that you won't consider derivative of a function taking a vector as argument. But in 9:01 we are doing just that, isn't it?
Ok, the "function" in 9:01 , the "limit", is almost like the identity function, but you probably know how Engineers like to tickle Mathematicians (for all the times Mathematicians tickle us) . :-)
I personally have nothing but admiration for engineers and wish I was one
He is not considering the derivative of that function. He is using distance to be able to take limits. As far as I know, in order to be able to take limits, you need atleast a topological space. A topological space with a distance function is a metric space. He is just making a metric space with the vectors (normed vector space) so that it is possible to take limits, that's all.
@@akashpremrajan9285 A vector is more than its length. And its derivative cannot be only the derivative of a scalar, its length or other scalar.
As example, a vector with a constant length does not necessarily have as derivate a zero-length vector,
as it is the case for a vector giving the position about a point in a circular motion
(on a circle, each positional vector is of a constant length equals to the radius).
A vector is more than it's length, sure. But to know whether two vectors are "close" or not, all you need is the length of the difference between them. Do you disagree?!?! If you disagree, please enlighten mathematicians how you want to measure the closeness of vectors.!!
Did you watch the same video, I watched? The derivative of a vector valued function is a vector only. When did the final answer become a scalar? We used the scalar function distance just to measure the closeness of vectors, to be able to take limits, that's all.
Elegant, indeed, both the idea and the explanation.
Thank you!
Such a cool little fact that allows for some quick back of the napkin work on predicting say velocity at a point in a video or somwthing like that. It is also nice for a position vector function that doesnt have a closed form seen in an analysis class.
Love these videos! A real inspiration. ❤🎉😊
So glad!Thank you, I appreciate it!
I cannot wait to teach this to someone!!
Thank you for this very rewarding comment!
Even though differentiation of functions having vectors as arguments is weird to think about you can still make an intuitive approach to it geometrically and see where the errors first occur.
Hi Prof.! A few of your recent videos in this series, including this one, do not yet appear in your 'complete playlist'. Hopefully it's not too hard to add them to it.
Thank you for pointing it out! I'll fix it ASAP
Is there a good way to understand continuity around a vector of a vector valued function? It's easy to see how limits work with scalar values with continuity of points in the x,y plane but hard to picture for resultant vectors from a change in t, "around" another vector. Also seems to beg the question how we would know if there is a uniform instantaneous change for a vector of a vector valued function as the distance between it and any other vector from an increase in t, approaches zero.
If the curve is continuous then, the function is continuous. If the curve is smooth, then the function is smooth. This is the object that actually gives us the intuition behind the concept of continuity.
nice, looking for continuation
Coming on Wed.
f(u), the scalar-valued function of a vector, is exactly what a scalar field is. What's the problem with taking its derivative, i.e. gradient?
Starting from a scalar field, you can take a difference, sure, but you must specify the direction (you are not limited only to a single direction, with a + and a - side).
A possible definition of the gradient is to get the direction where that difference is at its maximum (a gradient is a vector, after all), isn't it? So, about using gradient in that way, we first should be able to find the difference in ANY direction (at a given point), then, to find the MAX on this set. And have a recipe which works for any system of coordinates assuming that there is a system of coordinates)?
@@snnwstt Gradients are discussed in detail later in this series.
But to answer briefly, you can't apply the derivative to a function of a vector argument since h would be a vector and you can't divide by vectors.
However, a function can, of ccourse, be of several scalar arguments and that's how the gradient is constructed in high school calculus books.
nice music
Or you can define a derivative as [ f(x+1) - f( x-1 ) ]/[ (x+1) -- (x-1)] And no limit . the concept is the same but it's symmetric and uses the smallest number of the whole numbers. You don't need calculus.
If you think about derivative as a velocity(or speed) and measure time x in hours you actually define your current velocity as the distance f you traveled during the last 2 hours divided by two hours. So if you were driving from work for two hours exactly and now you are sitting in the kitchen your velocity NOW equals the velocity of your car. No need for cars.😮
I mean (x+1)-(x-1)=2
Similar idea to calculus but would end up with more complicated expressions. For example, (x³)' would be 3x²+1 instead of 3x².
Actually, what he is trying to say is the very obvious idea behind just about any differentiation. I don't see any specialty here.
Thank you for your comment!
@dQuasi2 what you just learned in this video.
Promised to explain, but didn't, asking to watch the next video. Click bait, dislike.
So it wasn't what you thought - exactly what the thumbnail said.
You indeed tricked me once but lost a potential subscriber. Was it worth it?
Oh yes!