How to prove statement regarding directional derivatives and gradients

Question

The question asked me to prove that

$$\|\nabla f\|^2 = (D_{u}f)^2 + (D_{v}f)^2$$

whenever vectors $u$ and $v$ are perpendicular.

How can I prove this?

Answer 1

Hint: $$\Vert \nabla f \Vert^2=(\nabla f\boldsymbol{\cdot}\underline{u})^2+(\nabla f \boldsymbol{\cdot}\underline{v})^2$$ Let $\theta$ and $\phi$ be the angles between $\nabla f$ and $\underline{u},\underline{v}$ respectively. Then $$(\nabla f \boldsymbol{\cdot}\underline{u})^2=\Vert \nabla f \Vert^2\Vert\underline{u}\Vert^2 \cos^2(\theta)$$ and $$(\nabla f \boldsymbol{\cdot}\underline{v})^2=\Vert \nabla f \Vert^2\Vert\underline{v}\Vert^2 \cos^2(\phi)$$ Perhaps you can take it from here?

EDIT: It's only true if the two vectors are orthonormal, not orthogonal. I.e $\Vert \underline{u} \Vert= \Vert \underline{v} \Vert =1$. Continuing on, $$1=\Vert \underline{u}\Vert^2 \cos^2(\theta) + \Vert \underline{v} \Vert^2 \cos^2(\phi)$$ Since $\underline{u}$ and $\underline{v}$ are orthogonal, $\phi = \theta \pm \pi/2$. WLOG, I'll assume $\phi = \theta + \pi/2$. But, as $\cos^2(x+\pi/2)=1-\cos^2(x)=\sin^2(x)$ (check this), $$1=\Vert \underline{u}\Vert^2 \cos^2(\theta)+ \Vert \underline{v} \Vert^2 \sin^2(\theta)$$ Which is obviously true as long as $\Vert \underline{u} \Vert= \Vert \underline{v} \Vert =1.$