Curl Of A Vector Function

Last updated: September 5, 2021 4:45 pm

2 Min Read

Disclosure: This website may contain affiliate links, which means I may earn a commission if you click on the link and make a purchase. I only recommend products or services that I personally use and believe will add value to my readers. Your support is appreciated!

Suppose I have a vector $\textbf{A}$ such as $\textbf{A} = a_x \textbf{i} + a_y \textbf{i} + a_z\textbf{i}$ .

Contents

EXAMPLE 1
SOLUTION
EXAMPLE 2
SOLUTION

STEP 1
STEP 2

Now the curl of this vector $\textbf{A}$ will be

\[ \text{curl}~\textbf{A} = \nabla \times \textbf{A} = \left|\begin{array}{ccc} \textbf{i} & \textbf{j} &\textbf{k}\\ \cfrac{\partial}{\partial x} & \cfrac{\partial}{\partial y} & \cfrac{\partial}{\partial z}\\ a_x & a_y & a_z \end{array}\right|.\]

So if I evaluate the determinant, I’ll get the curl of this vector.

If interested, you can read more about the other posts in vector analysis like directional derivative, the gradient of a scalar field, unit normal vector, unit tangent vector and so on. Also, pretty soon I’ll write about the divergence of any vector.

Now I’ll give some examples.

EXAMPLE 1

According to Stroud and Booth (2011)*, “Show that curl $(-y\textbf{i}+x\textbf{j})$ is a constant vector.”

SOLUTION

Now here the given vector is $(-y\textbf{i}+x\textbf{j})$ .

First of all, I’ll give it a name, say $\textbf{A}$ .

So I can say $\textbf{A} = -y\textbf{i}+x\textbf{j}$ . And this means $\textbf{A} = -y\textbf{i}+x\textbf{j}+0.\textbf{k}$ .

Now, according to the formula for the curl of a vector, curl of the vector $\textbf{A}$ will be

\[ \text{curl}~\textbf{A} = \nabla \times \textbf{A} = \left|\begin{array}{ccc} \textbf{i} & \textbf{j} &\textbf{k}\\ \cfrac{\partial}{\partial x} & \cfrac{\partial}{\partial y} & \cfrac{\partial}{\partial z}\\ -y & x &0 \end{array}\right|.\]

Next, I’ll evaluate this determinant to get the curl $\textbf{A}$ as

\[\text{curl}~\textbf{A} = \textbf{i} \left[\cfrac{\partial}{\partial y}(0) - \cfrac{\partial}{\partial z} (x)\right] - \textbf{j}\left[\cfrac{\partial}{\partial x}(0)-\cfrac{\partial}{\partial z}(-y)\right]+\textbf{k}\left[\cfrac{\partial}{\partial x}(x)-\cfrac{\partial}{\partial y}(-y)\right].\]

So this gives

\[\text{curl}~\textbf{A} = \textbf{i} \left[0 - 0\right] - \textbf{j}\left[0-0\right]+\textbf{k}\left[1-(-1)\right].\]

Now I’ll simplify it to get

\[\text{curl}~\textbf{A} = 2 \textbf{k}.\]

And $2 \textbf{k}$ is obviously a constant vector.

Hence I can prove that the curl of the vector $(-y\textbf{i}+x\textbf{j})$ is a constant vector.

So this is the answer to this example.

Next, I’ll give another example.

EXAMPLE 2

According to Stroud and Booth (2011)*, “If $\textbf{A} = 2xz^2\textbf{i} - xz\textbf{j} +(y+z)\textbf{k}$ , find curl curl $\textbf{A}$ .”

SOLUTION

Now in this example, the given vector is $\textbf{A} = 2xz^2\textbf{i} - xz\textbf{j} +(y+z)\textbf{k}$ .

First of all, I’ll find out the curl of the vector $\textbf{A}$ .

STEP 1

So the curl of the vector $\textbf{A}$ will be

\[ \text{curl}~\textbf{A} = \nabla \times \textbf{A} = \left|\begin{array}{ccc} \textbf{i} & \textbf{j} &\textbf{k}\\ \cfrac{\partial}{\partial x} & \cfrac{\partial}{\partial y} & \cfrac{\partial}{\partial z}\\ 2xz^2 & -xz &(y+z) \end{array}\right|.\]

Next, I’ll evaluate this determinant to get the curl $\textbf{A}$ as

\begin{eqnarray*}\text{curl}~\textbf{A} &=& \textbf{i} \left[\cfrac{\partial}{\partial y}(y+z) - \cfrac{\partial}{\partial z} (-xz)\right] - \textbf{j}\left[\cfrac{\partial}{\partial x}(y+z)-\cfrac{\partial}{\partial z}(2xz^2)\right]\\&&+\textbf{k}\left[\cfrac{\partial}{\partial x}(-xz)-\cfrac{\partial}{\partial y}(2xz^2)\right].\end{eqnarray*}

So this gives

\[\text{curl}~\textbf{A} = \textbf{i} \left[1+x\right] - \textbf{j}\left[0-4xz\right]+\textbf{k}\left[-z-0\right].\]

Now I’ll simplify it to get

\[\text{curl}~\textbf{A} =(1+x)\textbf{i} + 4xz \textbf{j} - z \textbf{k}.\]

Next I’ll get the curl of curl $\textbf{A}$ , that is, curl of the vector $(1+x)\textbf{i} + 4xz \textbf{j} - z \textbf{k}$ .

STEP 2

Thus curl curl $\textbf{A}$ will be will be

\[ \text{curl curl}~\textbf{A} = \nabla \times \textbf{A} = \left|\begin{array}{ccc} \textbf{i} & \textbf{j} &\textbf{k}\\ \cfrac{\partial}{\partial x} & \cfrac{\partial}{\partial y} & \cfrac{\partial}{\partial z}\\ (1+x) & 4xz &-z \end{array}\right|.\]

Next, I’ll evaluate this determinant to get the curl curl $\textbf{A}$ as

\begin{eqnarray*}\text{curl curl}~\textbf{A} &=& \textbf{i} \left[\cfrac{\partial}{\partial y}(-z) - \cfrac{\partial}{\partial z} (4xz)\right] - \textbf{j}\left[\cfrac{\partial}{\partial x}(-z)-\cfrac{\partial}{\partial z}(1+x)\right]\\&&+\textbf{k}\left[\cfrac{\partial}{\partial x}(4xz)-\cfrac{\partial}{\partial y}(1+x)\right].\end{eqnarray*}

So this gives

\[\text{curl curl}~\textbf{A} = \textbf{i} \left[0-4x\right] - \textbf{j}\left[0-0\right]+\textbf{k}\left[4z-0\right].\]

Now I’ll simplify it to get

\[\text{curl curl}~\textbf{A} =-4x\textbf{i} + 4z \textbf{k}.\]

Hence I can conclude that curl curl $\textbf{A} = -4x\textbf{i} + 4z \textbf{k}$ is the answer to this example.

Share This Article

Follow:

Prabhu TL is an author, digital entrepreneur, and creator of high-value educational content across technology, business, and personal development. With years of experience building apps, websites, and digital products used by millions, he focuses on simplifying complex topics into practical, actionable insights. Through his writing, Dilip helps readers make smarter decisions in a fast-changing digital world—without hype or fluff.

Previous Article Divergence Of A Vector Function

Next Article Algebra of Matrices

Leave a review Leave a review

Leave a Review Cancel reply