In examining the i component we have considered vector v= -yi+xj and we examined the change of x component of v due to change in y value as x co-ordinate of v is dependent on y only. And we have considered a general vector v=v1 i+v2 j And we have still considered change of v1 with respect to y only and not with respect to x. In general vector the x co-ordinate of v (i.e v1 )may also depend on x values,so why didn't we consider the change of v1 with respect to x.

Really good question. The change of v1 with respect to x won't influence the "rotational" component that we are trying to measure. Think about the simple case v1(x, y) = x, and think about how this influences the rotation in the vector field.

does anyone have any clues as to why that formula is twice the angular velocity?

Good question. We have counted the rotation twice when we added the horizontal and vertical components: d(v2)/dx - d(v1)/dy. Therefore, the actual angular velocity will be the average of the two components, or half of the 2d curl value. This is explained in more detail in the 3D curl article.

can anyone explain to me why the divergence of the curl of vector field always equals zero and what does it mean?

Recall that curl F = ∇ x F = <dh/dy - dg/dz, df/dz - dh/dx, dg/dx - df/dy> and that divergence F = ∇ * F. If we apply the divergence to the curl F we get ∇ * (∇ x F) = <d/dx, d/dy, d/dz> * <dh/dy - dg/dz, df/dz - dh/dx, dg/dx - df/dy> = dh/(dx dy) - dg/(dx dz) + df/(dy dz) - dh/(dx dy) + dg/(dx dz) - df/(dy dz) = 0 This can be interpreted as the spread of the rotation of vectors around a point is 0.

Thanks for this article. I have found it very helpful in my study of curl in Calculus! I have a question about the answer that is Pi/2 + 1. What does this number mean physically? I understand that result of the formula of curl gives us the amount of how much a particle rotates in the vector field. So, does Pi/2 + 1 mean, it rotates at 2.57 ... (what?)

Radians. Remember, angles aren't "real" in the sense that they get physical units. But perhaps you prefer degrees?

(x,y)=[−y x]=−yi^+xj^ why? why are x and y flipped? is it because curl is rotation, hence x and y flipped?

v(x, y) = [-y x] is defining a function v that maps (x, y) to (-y, x). It's the same notation as f(x, y) = (-y, x), only the name of the function is now v. I hope this helps.

Main content

Course: Multivariable calculus > Unit 2

Lesson 11: Divergence and curl (articles)

Curl warmup, fluid rotation in two dimensions

Google Classroom

Curl measures the rotation in a fluid flowing along a vector field. Formally, curl only applies to three dimensions, but here we cover the concept in two dimensions to warmup.

Background

Note: Throughout this article I will use the following convention:

$\hat{i}$ ‍ represents the unit vector in the $x$ ‍-direction.
$\hat{j}$ ‍ represents the unit vector in the $y$ ‍-direction.

What we're building to

Curl measures the "rotation" in a vector field.
In two dimensions, if a vector field is given by a function $\vec{v} (x, y) = v_{1} (x, y) \hat{i} + v_{2} (x, y) \hat{j}$ ‍, this rotation is given by the formula
$2d-curl \vec{v} = \frac{\partial v_{2}}{\partial x} - \frac{\partial v_{1}}{\partial y}$ ‍

Rotation in fluid flow

Have yourself a nice swirly vector field:

This particular vector field is defined with the following function:

\begin{aligned} \vec{v} (x, y) & = [\begin{array}{c} y^{3} - 9 y \\ x^{3} - 9 x \end{array}] \\ = (y^{3} - 9 y) \hat{i} + (x^{3} - 9 x) \hat{j} \end{aligned}

Now I want you to imagine that this vector field describes a fluid flow, perhaps in a chaotic part of a river. The following video shows a simulation of what this might look like. A sample of fluid particles, shown as blue dots, will flow along the vector field. This means that at any given moment, each dot moves along the arrow it is closest to. Focus in particular on what happens in the four circled regions.

Khan Academy video wrapper

See video transcript

Amidst all the chaos, you might notice that the fluid is rotating within the circled regions. In the left and right circles, the rotation is counterclockwise, and in the top and bottom circles, the rotation is clockwise.

Key Question: If we are given a function $\vec{v} (x, y)$ ‍ that defines a vector field, along with some specific point in space, $(x_{0}, y_{0})$ ‍, how much does a fluid flowing along the vector field rotate at the point $(x_{0}, y_{0})$ ‍?

The vector calculus operation curl answer this question by turning this idea of fluid rotation into a formula. It is an operator which takes in a function defining a vector field and spits out a function that describes the fluid rotation given by that vector field at each point.

Technically, the curl operation only applies to three dimensions. You can see what that means and how it is computed in the next article, but in this article, we warm up by describing fluid rotation in two dimensions with a formula.

Capturing two-dimensional rotation with a formula

One of the simplest examples of a vector field which describes a rotating fluid is

\begin{array}{r} \vec{v} (x, y) = [\begin{array}{c} - y \\ x \end{array}] = - y \hat{i} + x \hat{j} . \end{array}

Here's what it looks like.

Animated, all the fluid particles just go in circles.

Khan Academy video wrapper

See video transcript

In some sense, this is the most perfect example of counterclockwise rotation, and you can understand the general formula for rotation in a two-dimensional vector field just by understanding why the function

\vec{v} (x, y) = - y \hat{i} + x \hat{j}

gives counterclockwise rotation.

The $\hat{i}$ ‍-component

First, let's understand why the

- y \hat{i}

component suggests counterclockwise rotation. Imagine a small twig sitting in our fluid, oriented parallel to the

y

-axis. More specifically, let's say one end is at the origin

(0, 0)

, and the other is at the point

(0, 2)

. What does the

- y \hat{i}

component of the vector field imply for the fluid velocity at points on this twig?

This means the velocity at the top of the twig is

- 2 \hat{i}

, a leftward vector, while the velocity at the bottom of the twig is

0

For the twig, this means the important factor for counterclockwise rotation is that vectors point more to the left as we move up the vector field. Said with a few more symbols, the important point here is that the

\hat{i}

-component of a vector attached to a point

(x_{0}, y_{0})

decreases as

y_{0}

increases.

Said with even more symbols,

$\frac{\partial}{\partial y} (- y) = - 1 < 0$ ‍

Let's generalize this idea a bit.

Question: Consider a more general vector field.

$\vec{v} (x, y) = v_{1} (x, y) \hat{i} + v_{2} (x, y) \hat{j}$ ‍

The components

v_{1}

and

v_{2}

are any scalar-valued functions. If you place a small twig at some point

(x_{0}, y_{0})

, oriented parallel to the

y

-axis, how can you tell if the twig will rotate just by looking at

v_{1}, v_{2}

and

(x_{0}, y_{0})

Answer: Look at the rate of change of $v_{1}$ ‍ as $y$ ‍ varies near the point of interest, $(x_{0}, y_{0})$ ‍:
$\frac{\partial v_{1}}{\partial y} (x_{0}, y_{0}) \leftarrow Suggests counterclockwise rotation if negative$ ‍
If this is negative, it indicates that vectors point more to the left as $y_{0}$ ‍ increases, so rotation would be counterclockwise. If it is positive, vectors point more to the right as $y_{0}$ ‍ increases, indicating a clockwise rotation.

The $\hat{j}$ ‍-component

Next, let's see why the

x \hat{j}

component of the original vector field suggests counterclockwise rotation as well. This time, imagine a twig which is parallel to the

x

-axis. Specifically, put one end of the twig at the origin

(0, 0)

, and put the other at the point

(2, 0)

The vector attached to the origin is

0

, but the vector attached to the other end at

(2, 0)

2 \hat{j}

, an upward vector. Therefore, the fluid pushes the right end of the stick upwards, and the left end experiences no force, so there will be a counterclockwise rotation.

For this second twig, the vertical component of vectors increases as we move right, suggesting counterclockwise rotation. That is to say, the

y

component of a vector attached to a point

(x_{0}, y_{0})

increases as

x_{0}

increases.

In the case of a more general vector field function,

$\vec{v} (x, y) = v_{1} (x, y) \hat{i} + v_{2} (x, y) \hat{j}$ ‍

we can measure this effect near a point

(x_{0}, y_{0})

by looking at the change in

v_{2}

x

changes.

$\frac{\partial v_{2}}{\partial x} \leftarrow Suggests counterclockwise rotation if positive$ ‍

Combining both components

Putting these two components together, the rotation of a fluid flowing along a vector field

\vec{v}

near a point

(x_{0}, y_{0})

can be measured using the following quantity:

$\frac{\partial v_{2}}{\partial x} (x_{0}, y_{0}) - \frac{\partial v_{1}}{\partial y} (x_{0}, y_{0})$ ‍

When you evaluate this, a positive number will indicate a general tendency to rotate counterclockwise around

(x_{0}, y_{0})

, a negative quantity indicates the opposite, clockwise rotation. If it equals

0

, there is no rotation in the fluid around

(x_{0}, y_{0})

. If you are curious about the specifics, this formula gives precisely twice the angular velocity of the fluid near

(x_{0}, y_{0})

Some authors will call this the "two-dimensional curl" of

\vec{v}

. This isn't standard, but let's write this formula as if "2d-curl" was an operator.

$2d-curl \vec{v} = \frac{\partial v_{2}}{\partial x} - \frac{\partial v_{1}}{\partial y}$ ‍

Example: Analyzing rotation in a 2d vector field using curl

Problem: Consider the vector field defined by the function

\begin{array}{r} \vec{v} (x, y) = [\begin{array}{c} \cos (x + y) \\ \sin (x y) \end{array}] \end{array}

Determine whether a fluid flowing according to this vector field has clockwise or counterclockwise rotation at the point

\begin{aligned} p & = (0, \frac{π}{2}) \end{aligned}

Step 1: Compute the

2d-curl

of this function.

$2d-curl \vec{v} =$ ‍

We apply the 2d-curl formula we just found to this function. This will give us a new, scalar-valued function that indicates the rotation near each point. Then, we'll plug in the point $(0, π / 2)$ ‍ to see whether the rotation there is positive (counterclockwise), zero, or negative (clockwise).
$\begin{aligned} 2d-curl \vec{v} & = \frac{\partial v_{2}}{\partial x} - \frac{\partial v_{1}}{\partial y} \\ = \frac{\partial}{\partial x} (\sin (x y)) - \frac{\partial}{\partial y} (\cos (x + y)) \\ = \cos (x y) y - (- \sin (x + y)) \\ = \cos (x y) y + \sin (x + y) \end{aligned}$ ‍

Step 2: Plug in the point

(0, π / 2)

$2d-curl \vec{v} (0, π / 2) =$ ‍

Step 3: Interpret. How does the fluid tend to rotate near this point?

Choose 1 answer:
(Choice A)
Clockwise
(Choice B)
Counterclockwise
(Choice C)
No rotation

Let's watch a sample of particles in this fluid flow:

Khan Academy video wrapper

See video transcript

The point towards the top where all the particles congregate corresponds with

p = (0, \frac{π}{2})

. Particles rotate counterclockwise in this region, which should be consistent with your

2d-curl

calculations.

Summary

Curl measures the "rotation" in a vector field.
In two dimensions, if a vector field is given by a function $\vec{v} (x, y) = v_{1} (x, y) \hat{i} + v_{2} (x, y) \hat{j}$ ‍, this rotation is given by the formula
$2d-curl \vec{v} = \frac{\partial v_{2}}{\partial x} - \frac{\partial v_{1}}{\partial y}$ ‍

On to the third dimension!

The true curl operation, covered in the next article, extends this idea and this formula to three dimensions.

Want to join the conversation?

Sort by:

Gavaskar
Posted 8 years ago. Direct link to Gavaskar's post “In examining the i compon...”
In examining the i component we have considered vector v= -yi+xj and we examined the change of x component of v due to change in y value as x co-ordinate of v is dependent on y only.
And we have considered a general vector v=v1 i+v2 j
And we have still considered change of v1 with respect to y only and not with respect to x.
In general vector the x co-ordinate of v (i.e v1 )may also depend on x values,so why didn't we consider the change of v1 with respect to x.
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Grant
  Posted 8 years ago. Direct link to Grant's post “Really good question. Th...”
  Really good question. The change of v1 with respect to x won't influence the "rotational" component that we are trying to measure. Think about the simple case v1(x, y) = x, and think about how this influences the rotation in the vector field.
  Comment on Grant's post “Really good question. Th...”
  (13 votes)
bibomc
Posted 7 years ago. Direct link to bibomc's post “does anyone have any clue...”
does anyone have any clues as to why that formula is twice the angular velocity?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Evgenii Neumerzhitckii
  Posted 7 years ago. Direct link to Evgenii Neumerzhitckii's post “Good question. We have co...”
  Good question. We have counted the rotation twice when we added the horizontal and vertical components: d(v2)/dx - d(v1)/dy. Therefore, the actual angular velocity will be the average of the two components, or half of the 2d curl value. This is explained in more detail in the 3D curl article.
  Comment on Evgenii Neumerzhitckii's post “Good question. We have co...”
  (6 votes)
mohamednomear1
Posted 4 years ago. Direct link to mohamednomear1's post “can anyone explain to me ...”
can anyone explain to me why the divergence of the curl of vector field always equals zero and what does it mean?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- KLaudano
  Posted 3 years ago. Direct link to KLaudano's post “Recall that curl F = ∇ x ...”
  Recall that curl F = ∇ x F = <dh/dy - dg/dz, df/dz - dh/dx, dg/dx - df/dy> and that divergence F = ∇ * F. If we apply the divergence to the curl F we get
  
  ∇ * (∇ x F)
  = <d/dx, d/dy, d/dz> * <dh/dy - dg/dz, df/dz - dh/dx, dg/dx - df/dy>
  
  = dh/(dx dy) - dg/(dx dz) + df/(dy dz) - dh/(dx dy) + dg/(dx dz) - df/(dy dz)
  
  = 0
  
  This can be interpreted as the spread of the rotation of vectors around a point is 0.
  Button navigates to signup page
  (3 votes)
Kevin Riady
Posted 7 years ago. Direct link to Kevin Riady's post “Thanks for this article. ...”
Thanks for this article. I have found it very helpful in my study of curl in Calculus! I have a question about the answer that is Pi/2 + 1. What does this number mean physically? I understand that result of the formula of curl gives us the amount of how much a particle rotates in the vector field. So, does Pi/2 + 1 mean, it rotates at 2.57 ... (what?)
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Connor Doherty
  Posted 7 years ago. Direct link to Connor Doherty's post “Radians. Remember, angle...”
  Radians.
  Remember, angles aren't "real" in the sense that they get physical units. But perhaps you prefer degrees?
  Button navigates to signup page
  (3 votes)
a.somjp
Posted 4 years ago. Direct link to a.somjp's post “(x,y)=[−y x]=−yi^+xj^ w...”
(x,y)=[−y x]=−yi^+xj^ why?
why are x and y flipped? is it because curl is rotation, hence x and y flipped?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- lakern
  Posted 4 years ago. Direct link to lakern's post “v(x, y) = [-y x] is defin...”
  v(x, y) = [-y x] is defining a function v that maps (x, y) to (-y, x). It's the same notation as f(x, y) = (-y, x), only the name of the function is now v. I hope this helps.
  Button navigates to signup page
  (2 votes)
Sartaj Khan
Posted 4 years ago. Direct link to Sartaj Khan's post “Say we have a vector fiel...”
Say we have a vector field such that at (1,2) the vector representing the vector field is 0i+0j. Wouldn't the curl at that point(1,2)be zero? why or why not? Since the net rotational motion at that point of a fluid flow is zero
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(1 vote)
Answer
- Hunner Singh
  Posted 4 years ago. Direct link to Hunner Singh's post “Curl is not a property of...”
  Curl is not a property of the fluid flow at a point but the property of average fluid flow "around" that point think of it as you are in the eye of a cyclone. No air is pushing you but still it is curling around you. So at the point you are asking curl might not be zero.
  Button navigates to signup page
  (1 vote)

Multivariable calculus

Course: Multivariable calculus > Unit 2

Background

What we're building to

Rotation in fluid flow

Capturing two-dimensional rotation with a formula

The i^‍ -component

The j^‍ -component

Combining both components

Example: Analyzing rotation in a 2d vector field using curl

Summary

On to the third dimension!

Want to join the conversation?

The $\hat{i}$ ‍-component

The $\hat{j}$ ‍-component