Minimizing a directional derivative

[Pengwuino]

I know that you determine maximum value of a directional derivative at a point by finding...

$$|\nabla f(a_0 ,b_0 )|$$

But how do you find the minimum value?

I'm also kinda wondering exactly what a gradient is. It seems like if you have the gradient equation and a point... all you are getting is a single vector and a single rate of change... doesn't seem all that useful.

 

[HallsofIvy]

If going in the direction $\theta$ makes the derivative a maximum, what happens if you go in the exact opposite direction?

 

[Pengwuino]

... its the opposite and parallel vector isn't it (as in <1,1> is to <-1,-1>) hahaha... oh man... i wonder why my professor made a big fuss about it on our homework assignment then...

 

[HallsofIvy]

By the way- it also follows that the directional derivative is 0 in the direction perpendicular to the gradient. That's useful property: the gradient is always perpendicular to level curves.

 

[Pengwuino]

Wow that's helpful since I now have to figure out 2 directions where the rate of change is 0 at a certain point. Problem is, I don't remember how to figure out what vector that would be... I have...

$$f(p,q) = qe^{ - p} + pe^{ - q}$$

at (0,0)

I got….

$$\nabla f(p,q) = \frac{{\partial f}}{{\partial p}}i + \frac{{\partial f}}{{\partial q}}j \\$$
$$\frac{{\partial f}}{{\partial p}} = e^{ - q} - e^{ - p} q \\$$
$$\frac{{\partial f}}{{\partial q}} = e^{ - p} - e^{ - q} p \\$$
$$\nabla f(p,q) = (e^{ - q} - e^{ - p} q)i + (e^{ - p} - e^{ - q} p)j \\$$
$$\nabla f(0,0) = < 1,1 > \\$$
$$|\nabla f(0,0)| = \sqrt 2 \\$$

I assume this all means that at (0,0), the maximum slope is a vector of <1,1> with a rate of increase of $$\sqrt 2$$

YES, STUPID LATEX, TAKE THAT! How do you get it to automatically go to a new line without having to tex and /tex after every single line?

So... is my assumption correct or am i missing the point of gradients all together?