Given $ f(x,y,z) = x^2 +3xz + \dfrac{y^3}{3}$ I'm asked to find the maximum value, over all unit vectors, of $D_{u} f(1,2,-1)$.
Attempt :
$$ \nabla f = \langle f_x , f_y, f_z \rangle = \langle 2x + 3z, y^2 , 3x \rangle $$
$$D_u f = || \nabla f(1,2,-1)|| = || (-1,4,-3)|| = \sqrt{26} $$
I feel like i'm missing a step somewhere, particularly something concerning the " all unit vectors".
Thank you for help.
Best Answer
Recall that $D_u f(P) = \nabla f(P) \cdot u$, so $$ D_u f(P) = \Vert \nabla f(P) \Vert \, \Vert u \Vert \,\cos\theta $$ where $\theta$ is the angle between $\nabla f(P)$ and $u$. Remember $\Vert u \Vert =1$ and $|\cos\theta|\leq 1$. So the maximum value of $D_u f(P)$, over all unit vectors $u$, occurs when $u$ points in the same direction as $\nabla f(P)$ (i.e., $\cos\theta =1 \implies \theta = 0$). In this case, $D_u(P) = \Vert \nabla f(P) \Vert$.