# Maximal directional derivative with 3 coordinates

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!

• 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. Aug 19, 2020 at 12:07

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 $$A=\sqrt {(12)^{2}+(14)^{2}+(12)^{2}}$$.