Directional Derivative in the direction of steepest descent.

Hi

I just want to check that I am solving this correctly.

You are given a function $\displaystyle f(x, y, z) = x^2 + 3xy + 2z$ and asked to find the directional derivative at the point $\displaystyle (2, 0, -1)$ in the **direction of the steepest descent**.

Partial Derivatives at $\displaystyle (2, 0, -1)$:

$\displaystyle f_x = 2x + 3y = 4 $

$\displaystyle f_y = 3x = 6 $

$\displaystyle f_z = 2 $

Now this gives you the direction of the steepest descent : $\displaystyle -\nabla f=(-4, -6, -2)$.

So you then find the unit vector: $\displaystyle {u} = (\frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} )$

And then you dot product the above with with the partial derivatives giving:

$\displaystyle D_u f = u \bullet (2x + 3y, 3x, 2) = \frac{-13x}{\sqrt{14}} - \frac{6y}{\sqrt{14}} - \frac{2}{\sqrt{14}}$

Is that the directional derivative? Or do I then have to substitute in the point $\displaystyle (2, 0, -1)$ ?

I would appreciate if someone could verify the above for me, or correct it (more likely!).

Also, is the magnitude of the rate of steepest descent $\displaystyle \mid u \mid = \sqrt{14} $ ?

Thanks a lot.