# Finding Curl(F) of a vector-valued function

1. Jun 24, 2014

### JJBladester

1. The problem statement, all variables and given/known data

Determine the curl of the vector function below.

$$\boldsymbol{F}\left ( x,y,z \right )=3x^2\boldsymbol{i}+7e^xy\boldsymbol{j}$$

2. Relevant equations

curl$\mathbf{F}=\mathbf{\nabla}\times \mathbf{F}$
$$=\begin{vmatrix} \mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ P(x,y,z)& Q(x,y,z)& R(x,y,z) \end{vmatrix}$$

3. The attempt at a solution

This problem is solved by my FE review as below. I understand how to solve for the determinant of a 3x3 matrix by rewriting the first two columns to the right of the matrix and obtaining three "+" terms and three "-" terms. I think my partials for x and y are off, perhaps.

$$\mathbf{i}\left ( \frac{\partial }{\partial y}0-\frac{\partial }{\partial z}7e^xy \right )-\mathbf{j}\left ( \frac{\partial }{\partial x}0-\frac{\partial }{\partial z}3x^2 \right )+\mathbf{k}\left ( \frac{\partial }{\partial x}7e^xy-\frac{\partial }{\partial y}3x^2 \right )$$

$$=\mathbf{i}(0-0)-\mathbf{j}(0-0)-\mathbf{k}\left ( 7e^xy-0 \right )=7e^xy\mathbf{k}$$

The expressions I calculated for $\mathbf{i} and \mathbf{j}$ match what the book has. However, my expression for $\mathbf{k}$ seems to be incorrect. Here is what I calculated for the values of the matrix:

$$\frac{\partial }{\partial x}=6x$$
$$\frac{\partial }{\partial y}=7e^x$$
$$\frac{\partial }{\partial z}=0$$
$$P(x,y,z)=3x^2$$
$$Q(x,y,z)=7e^xy$$
$$R(x,y,z)=0$$

So my expression for $\mathbf{k}$ was:

$$\left [(6x)7e^xy-\left ( 7e^x \right )3x^2 \right ]\mathbf{k}$$

I think I went wrong with my calculation of $\frac{\partial }{\partial x}$.

2. Jun 24, 2014

### lurflurf

It is unclear what exactly you have done.
My guess is that you found
$$\dfrac{\partial}{\partial x}3\, x^2=6\, x$$
when you should have found
$$\dfrac{\partial}{\partial x}7 \, e^x \, y=7 \, e^x \, y$$
also you are multiplying functions together for some reason

Look at the solution, of the six terms 5 are clearly zero so only the sixth remains.

3. Jun 24, 2014

### JJBladester

This is why I love physicsforums. You answered me and helped me understand the problem within 20 minutes of my post.

You're right, I was multiplying the functions when I should not have been. Normally when working with matrices containing only constants, you *do* just multiply elements together. So that seems not to be the case when you have partial derivatives as elements.

4. Jun 24, 2014

### SteamKing

Staff Emeritus
For the curl, it's helpful to think of the partial derivatives in the second row as operators to be applied to the functions P, Q, and R when the determinant is expanded. Expressing the curl in terms of a determinant is just a handy way to remember how to calculate its various components.

Taking the i-component of the curl, for example, expanding the determinant gives:

Ry - Qz, where the subscripts y and z indicate the variable w.r.t. which the partial derivative is taken.

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