Divergence measures the change in density of a fluid flowing according to a given vector field.
Log in Chiarandini Pandetta 8 years agoPosted 8 years ago. Direct link to Chiarandini Pandetta's post “Will there be exercises f...” Will there be exercises for multivariable calculus in the future? • (51 votes) Arman Rafian 7 years agoPosted 7 years ago. Direct link to Arman Rafian's post “I'm almost sure that KA's...” I'm almost sure that KA's goal is to develop practice and then mastery exercises for every subject, but since it is a work in progress, they are probably focusing on what they feel is most important at the moment. I'm guessing that the majority of users on KA are not here for more advanced math like this course, so they may be trying to focus on other subjects, or lower level math. For example this July/August they are coming out with quite a few more exercises here all the way up to BC Calculus. You can find this here: It seems in the past six months the team has been working more on statistics, history, and more tools for teachers/coaches. It must be a tough task to try to work on so many things at once, but hopefully they can get some exercises for this stuff in the next few years as well! (22 votes) Connor Doherty 8 years agoPosted 8 years ago. Direct link to Connor Doherty's post “Why not just call the neg...” Why not just call the negative of divergence the "convergence" and eliminate all confusion? You can look at a source or sink and instantly tell if the particles are converging, coming toward each other. Faster than telling if they are doing the opposite of that, and eliminates confusion (converge = high density). • (17 votes) Oleg Novikov 6 years agoPosted 6 years ago. Direct link to Oleg Novikov's post “On the contrary, it will ...” On the contrary, it will be more confusing. Example 1. Consider the inverse of what the author uses as a reminder, the function which takes the point (x, y) to the vector (-x,-y): f(x,y)=-xi-yj The resulting vector field has all vectors pointing to the origin, the divergence is negative and equals to -2. In this case the divergence is negative at any point of the field but intuitively you think about it as coverging only at the origin. Example 2. Consider the function which takes the point (x, y) to the vector (-exp(x), 0): f(x,y)=-exp(x)i The resulting vector field has all vectors pointing horisontally to the left since theres no y component and the divergence is again strictly negative at any point. Nevertheless this flow never converges anywhere (except maybe when x equals minus infinity). (12 votes) Alexander Wu 8 years agoPosted 8 years ago. Direct link to Alexander Wu's post “Technically operators are...” Technically operators are functions. When I first learned about them, functions are anything that takes in something and outputs an unambiguous something else. For example, there was f(Paul) = Mary or something like that. So functions probably can also input and output other functions. Therefore, divergence is a function that inputs and outputs functions that input and output vectors and scalars, respectively. Right......? • (2 votes) Connor Doherty 8 years agoPosted 8 years ago. Direct link to Connor Doherty's post “Not quite. Operators are ...” Not quite. Operators are more fundamental... The plus sign "+" is an operator, for example, although it is "infix", meaning the first argument/parameter goes in front of it rather than after. If it wasn't infix, x+y would look something like this: +(x,y) or more clearly, sum(x,y). Likewise with multiply(x,y) and subtract(x, y), etc. Other operators are not infix, such as derivative(x) or grad(x), written with their respective symbols. But how would you define these "functions" with only other symbols, without explaining their abstract concepts? I mean, some function f(x) could be defined as f(x)=3+x. Simple. g(y) = 4/y. and so on. But notice how I had to use those fundamental operators to define them? (+ and /) Yet those operators can't be written in such a way. What's the definition of sum(x,y)? "You just gotta add them". F(x) might be equal to 1, and derivative(x) also. But can you write out a function f( ) that always puts out exactly what derivative( ) puts out? Try it. Something that "takes something in and outputs something else" is not specific enough to define functions, nor operators, for this reason. You will see that they are not the same thing. You can describe a function in terms of operators ("this function adds three to its input"), but you can't describe those operators without more operators ("what do you mean, 'adds'? What is this 'addition' you speak of?") You could think of operators as "verbs". They're like "actions" you can take on some function/expression that keeps it a function/expression. Like, "take the derivative". That's a verb that does things to functions, not just variables and constants. Sure, you can define a function like f(x) in terms of another, such as f(x) = g(x) +3, where g(x) is x+2. But for x=7 you'd first evaluate g(7)=9 and pass it into f(7)=9+3=12, whereas you cannot do that with deriv( m(x) ) where m(x) = x^2, because passing the constant (7)^2=49 into deriv(49) gets you zero, when you should have passed the x^2 as it is, getting you deriv(x^2) = 2x, which you can then evaluate to get 14. Operators cannot be "evaluated" at a value like functions, they fundamentally change the function (x^2 --> 2x) and must always be done first (highest precedence, so before anything else in PEMDAS) so are called left-associative instead of right-associative (done from the outside in. instead of instead out). So in F(deriv(x(y(t)))+2), that derivative must be done before its insides (x,y...) and even before it can be passed to the function outside (F) that contains it. It's kind of like the difference between "pass by reference" and "pass by value" in computer science. You work directly on the input function itself ("reference") rather than it's output ("value"). (22 votes) Dinghao Luo 7 years agoPosted 7 years ago. Direct link to Dinghao Luo's post “I would suggest rememberi...” I would suggest remembering the sign-divergence stuff just by understanding that the word 'divergence' means 'the quality of going out from a single point,' and when there is a positive divergence, that means there IS divergence, whilst when there is a negative one, that means there IS a sort of 'anti-divergence.' • (5 votes) Andres 6 years agoPosted 6 years ago. Direct link to Andres's post “The "quick diagram" is so...” The "quick diagram" is so tiny that it's practically impossible to see... • (3 votes) scourge 4 years agoPosted 4 years ago. Direct link to scourge's post “why don't we use the y co...” why don't we use the y coordinate in the first partial derivative, and the x coordinate in the second partial derivative for the divergence? • (2 votes) Charles Morelli 7 months agoPosted 7 months ago. Direct link to Charles Morelli's post “Read the next article or ...” Read the next article or watch the series of videos starting with this one: (1 vote) senthilkumar 4 years agoPosted 4 years ago. Direct link to senthilkumar's post “what is the difference be...” what is the difference between divergence and gradient? are they same • (2 votes) Adit Stegosaurous 4 years agoPosted 4 years ago. Direct link to Adit Stegosaurous's post “The gradient of a functio...” The gradient of a function is a vector that consists of all its partial derivatives. For example, take the function f(x,y) = 2xy + 3x^2. The partial derivative with respect to x for this function is 2y+6x and the partial derivative with respect to y is 2x. Thus, the gradient vector is equal to <2y+6x, 2x>. Divergence, on the other hand, is described above in the article and can be thought of as the dot product between a vector of partial derivatives and the vector function that determines the vector field. Hope this helps! (1 vote) Gregory 5 years agoPosted 5 years ago. Direct link to Gregory's post “Hi, I would like to know ...” Hi, I would like to know if I have the function • (1 vote) MBCory 4 years agoPosted 4 years ago. Direct link to MBCory's post “You do not get a 2x2 matr...” You do not get a 2x2 matrix of derivatives (called the Jacobian, and referenced in this article), because you are not taking all possible derivatives when calculating divergence. You just take d/dx(vsubx) + d/dy(vsuby), where vsubx is the function that is the x-component of v and vsuby is the y-component of v. Hope this helps! (3 votes) apndolby297 5 years agoPosted 5 years ago. Direct link to apndolby297's post “Is it correct to say that...” Is it correct to say that on simple addition of the terms of gradient ,we obtain the divergence? And also why is it that the summation of these terms determines why the fluid is getting collected or distributed...? • (1 vote) MBCory 4 years agoPosted 4 years ago. Direct link to MBCory's post “To answer your first ques...” To answer your first question, it is not quite accurate to say that we are adding the terms of the gradient, because the gradient is only defined for functions that output scalar values, and divergence is defined for vector-valued functions. (2 votes) Michaelson Britt 7 years agoPosted 7 years ago. Direct link to Michaelson Britt's post “Looking at the Incompress...” Looking at the Incompressible Fluid Flow Example; it appears that both the X and Y coordinate values of the vectors change while moving from left-to-right through any point on the graph, yet neither the X or Y coordinates change while moving from bottom-to-top through any point on the graph. That seems to imply that the partial derivative of the first vector coordinate v1 with respect to x is nonzero, while the partial derivative of of the second vector coordinate v2 with respect to y remains zero. The divergence is defined as the sum of these two partial derivative scalars (is that correct?). Adding the two scalars yields a nonzero scalar everywhere on the graph, thus the divergence would be nonzero everywhere, indicating some divergence or convergence (compressibility). This must reflect an incorrect understanding of divergence, as the example is for incompressibilty. How is this incorrect? • (1 vote) Charles Morelli 7 months agoPosted 7 months ago. Direct link to Charles Morelli's post “I'm afraid i found your q...” I'm afraid i found your question confusing, but i think i understand your point now. (1 vote)Want to join the conversation?
https://khanacademy.zendesk.com/hc/en-us/articles/115007408888
https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/divergence-grant-videos/v/divergence-intuition-part-1
vf(x,y) = [xy, y^2], why it is being done just by two partial derivatives like:
[y,2y] but not it this way [[y,x],[2y,0]]
*The summation in itself doesn't appear to be something so significant in terms of first impressions..*
First of all, whilst accepting that a vector field is very helpful as a visualisation, IMHO functions should generally be considered to output into their own space, so your use of x and y 'coordinates' for the output vector confused me (you later refer to them as v1 and v2 'coordinates', i think).
I will use '1st' and '2nd' components of the output vector (and functions v1(x,y) and v2(x,y) respectively for those entities), and the 'x' and 'y' coordinates of the input to avoid confusion of terminology.
I understand your point suggests that the 1st and 2nd components (‘v1’ and ‘v2’) of the output vector both seem to change continuously as you look from left to right (i.e., as the x-input coordinate increases) across the input space implying ∂v1(x,y)/∂x ≠ 0 (also ∂v2(x,y)/∂x ≠ 0, but that is irrelevant to the answer)
[whereas: i think the 1st ('v1') component of the output vector is, in fact, constant across the whole x-y plane as i discuss shortly];
also, that both output components appear constant as only the y-input changes (∂v2(x,y)/∂y = 0; also ∂v1(x,y)/∂y = 0, but again, the latter is irrelevant to the answer).
I think the main issue here is that the unanimated vector arrows are all the same length and there is no colour-coding to indicate variation in magnitude. I don't think the arrows really have the same magnitude, but that this is a simplification for visualisation purposes.
[I thought Grant had given the function behind that animation somewhere but i can't find it now, so can’t confirm the maths, so this is what i think is going on]
1) the arrows only indicate vector direction, there is no magnitude information given (and magnitude here is NOT constant if my other suppositions are correct: i.e., vectors with a vertical component should have greater magnitude than those without);
2) there is no change to the 1st component of the output vector as the x-input changes, but the 2nd output component varies with x
(i.e.,: ∂v/∂x = (0)î + (∂v2(x,y)/∂x)ĵ - using another notation format)…
so i think if you track the dots moving from left to right in the animation and ignore the vertical oscillation, you see that they move at a constant horizontal velocity;
3) there is no change to the output vector at all as only the y-input varies
(i.e: ∂v/∂y = (0)î + (0)ĵ ).
In summary, ∂v1(x,y)/∂x = 0 and ∂v2(x,y)/∂y = 0 hence their sum is also 0, hence zero divergence.
