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.


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

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  • $\begingroup$ 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

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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