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Given $$f(x,y,z) = 2x^3y -3y^2z$$

In which directions the directional derivative at the point $a = (1,2,-1)$ is maximal (in absolute value) what is the value of the directional derivative there?

My try:

I first calculated the partial derivatives (the gradient)

$f_x = 6yx^2$
$f_y = 2x^3 -6zy$
$f_z = -3y^2$

and so $\nabla f(x,y,z) = (6yx^2, 2x^3 -6zy, -3y^2)$

Now, I need to find a direction in which the directional derivative is maximal in abs value, so I defined:
$\vec{n} = (\cos(\alpha), \cos(\beta), \cos(\gamma))$

$\frac{\partial f}{\partial \vec{n}}(1,2,-1) = \nabla f(1,2,-1) \cdot \vec{n}$

$\nabla f(1,2,-1) = (12, 14, -12)$

Now, I get a function with 3 variables! ($\alpha, \beta, \gamma$) and I am stuck as I don't know how to look for maximal points.. :

$$g(\alpha, \beta, \gamma) = 12 \cos(\alpha) +14 \cos(\beta) - 12 \cos(\gamma)$$

I don't know how to continue from here.. I would appreciate your kind help, thanks!

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  • $\begingroup$ The directional derivative at some point is always maximal in the direction of the gradient at that point . It's a rather elementary result and not that hard to prove it. $\endgroup$ – DonAntonio Aug 19 '20 at 12:07
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The directional derivative in the direction of a unit vector $(u,v,w)$ is $12 u+14v-12w$. By Cauchy -Schwarz inequality this is bounded by $\sqrt {(12)^{2}+(14)^{2}+(12)^{2}}$ and this is value is attained when $(u,v,w)=\frac 1 A (12,14,-12)$ where $A=\sqrt {(12)^{2}+(14)^{2}+(12)^{2}}$.

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