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Show, in general, that $\vert \nabla f \vert^2 =(D_uf)^2 + (D_vf)^2$ where $u$ and $v$ are perpendicular.

I have tried taking the dot product of the right side, but to no avail.

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closed as off-topic by Did, The Phenotype, Ethan Bolker, user284331, A. Goodier Mar 8 '18 at 18:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, The Phenotype, Ethan Bolker, user284331, A. Goodier
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I edited your post to $\LaTeX$ify it. Cheers! $\endgroup$ – Robert Lewis Mar 2 '18 at 6:12
  • $\begingroup$ Robert: Do you know how to answer this proof? $\endgroup$ – MathWannaBe Mar 2 '18 at 6:16
  • $\begingroup$ Are you working on functions defined on $\mathbb{R}^2$? $\endgroup$ – Frank Lu Mar 2 '18 at 6:30
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If we are considering 3-D, there are infinite $\vec{v}$ perpendicular to $\vec{u}$. So the statement "$\vec{u}$ and $\vec{v}$ are perpendicular" is ambiguous.

By considering 2-D:

$$\mid \vec{\bigtriangledown} f \mid^2=(\bigtriangledown f) ^2 (1)=(\bigtriangledown f) ^2 \cos^2 \theta + (\bigtriangledown f )^2 \sin^2 \theta=(\bigtriangledown f \cos \theta)^2+(\bigtriangledown f \cos (90-\theta))^2$$

$$=(\vec{\bigtriangledown} f. \hat{u} )^2 + ( \vec{\bigtriangledown} f. \hat{v} )^2= ( D_uf )^2+( D_vf )^2$$

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