Directional Derivative in the direction of steepest descent.

• March 6th 2009, 04:18 AM
Solo
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 $f(x, y, z) = x^2 + 3xy + 2z$ and asked to find the directional derivative at the point $(2, 0, -1)$ in the direction of the steepest descent.

Partial Derivatives at $(2, 0, -1)$:
$f_x = 2x + 3y = 4$
$f_y = 3x = 6$
$f_z = 2$

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

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

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

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

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

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 $\mid u \mid = \sqrt{14}$ ?

Thanks a lot.
• March 6th 2009, 05:56 AM
Opalg
That is correct up to the final step, and yes, you do then have to substitute in the point (2,0,–1).

Also, the magnitude of the rate of steepest descent is $2\sqrt{14}$, because that is what you divided $\nabla f$ by in order to get a unit vector.