Finding Curl(F) of a vector-valued function

  • Thread starter Thread starter JJBladester
  • Start date Start date
  • Tags Tags
    Function
Click For Summary

Homework Help Overview

The discussion revolves around finding the curl of a vector-valued function, specifically the function \(\boldsymbol{F}(x,y,z) = 3x^2\boldsymbol{i} + 7e^xy\boldsymbol{j}\). Participants are exploring the application of the curl operator and the correct interpretation of partial derivatives in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the curl using the determinant of a 3x3 matrix and question the accuracy of partial derivatives applied to the components of the vector function. There is an exploration of the correct interpretation of terms in the determinant expansion.

Discussion Status

Some participants have provided clarifications regarding the calculation process and the application of partial derivatives. There is an acknowledgment of misunderstandings in the multiplication of functions and the need to treat partial derivatives as operators. The discussion is ongoing with various interpretations being explored.

Contextual Notes

Participants are navigating through potential errors in their calculations and the implications of treating partial derivatives in the context of the determinant. There is a focus on ensuring the correct application of mathematical principles without reaching a definitive solution yet.

JJBladester
Gold Member
Messages
281
Reaction score
2

Homework Statement



Determine the curl of the vector function below.

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

Homework Equations



curl\mathbf{F}=\mathbf{\nabla}\times \mathbf{F}
=\begin{vmatrix}<br /> \mathbf{i}&amp; \mathbf{j}&amp; \mathbf{k}\\ <br /> \frac{\partial}{\partial x}&amp; \frac{\partial}{\partial y}&amp; \frac{\partial}{\partial z}\\ <br /> P(x,y,z)&amp; Q(x,y,z)&amp; R(x,y,z)<br /> \end{vmatrix}

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}.
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: 1 person
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.
 
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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
33
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K