Can I Take the X Component Out When Differentiating with Respect to Z?

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SUMMARY

The discussion centers on the differentiation of a force component with respect to a variable while performing a cross product. Specifically, the user is examining the expression \(\nabla \times F(x) = \frac{\partial}{\partial z} F_y\) where \(F_y = x\). The confusion arises from the inability to treat the x component as a constant during differentiation with respect to z, leading to the conclusion that the differentiation process must account for the variable dependencies involved.

PREREQUISITES
  • Understanding of vector calculus, specifically cross products and gradients.
  • Familiarity with partial differentiation and its rules.
  • Knowledge of force components in physics and their mathematical representation.
  • Basic understanding of determinants and their application in vector calculus.
NEXT STEPS
  • Study the properties of cross products in vector calculus.
  • Learn about partial derivatives and their implications in multivariable functions.
  • Explore the application of determinants in solving vector equations.
  • Investigate the relationship between force components and their respective variables in physics.
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Students and professionals in physics, mathematics, and engineering who are dealing with vector calculus and differentiation of force components.

Oblio
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I'm confused on a procedural idea...

If I'm doing the cross product of a gradient and 'the x component of a force' , so:

[tex]\nabla X F(x) = \frac{\partial}{\partial z} Fy[/tex]

and Fy = x..

even though I am differentiating with respect to z , I am solving for an x component, which means I can not take x out with the other constants... yes? (if there were some)

I guess I'm confused because I know this shouldn't = 0 afterwards, but I'm not differentiating wrt z, so I'm trying to figure out why.


Thanks a lot !

(F(z) is 0 in case your wondering why its only Fy.)
 
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Is one looking at something like this?

[tex]det \[ \left[ \begin{array}{ccc}<br /> \hat{x} & \hat{y} & \hat{z} \\\<br /> \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}} \\\<br /> F_x & F_y & F_z \end{array} \right]\][/tex]
 
yes I am...
thats how I set up the equation, but I don't know if I may take x out or not.
 

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