# Show that a differentiable function f decreases most rapidly....

• Nov 8th 2007, 02:21 PM
Undefdisfigure
Show that a differentiable function f decreases most rapidly....
at x in the direction opposite to the gradient vector, that is, in the direction of -Vf(x).

I made up a function of two variables f(x, y) = ln(x^2 -y^3/2) and then calculated the gradient which came to 2x/(x^2 - y^3/2) and -3/2y^2/(x^2 - y^3/2).

I then put in a value Vf (2, 1) and got the vector < 8/7 , -3/7>. I made the vector negative and got , <-8/7 , 3/7>. How do I show that a differentiable function f like the one above (or another one) decreases most rapidly at x in the direction opposite to the gradient vector?

At this point I am stuck.
• Nov 8th 2007, 02:36 PM
ThePerfectHacker
Quote:

Originally Posted by Undefdisfigure
How do I show that a differentiable function f like the one above (or another one) decreases most rapidly at x in the direction opposite to the gradient vector?

You do not need to show, it is a theorem.

If $\displaystyle f(\bold{x}): \mathbb{R}^n \mapsto \mathbb{R}$ is differenciable at $\displaystyle \bold{x}_0$ then the directional derivative $\displaystyle D_{\bold{u}}f(\bold{x_0})$ in the direction $\displaystyle \bold{u}$ is given by $\displaystyle \nabla f(\bold{x}_0)\cdot \bold{u}$. But $\displaystyle |\nabla f(\bold{x}_0) \cdot \bold{u}| \leq |\nabla f(\bold{x}_0)| |\bold{u}| = |\nabla f(\bold{x}_0)|$ by Cauchy-Shwartz inequality.
• Nov 8th 2007, 03:45 PM
Undefdisfigure
Quote:

Originally Posted by ThePerfectHacker
You do not need to show, it is a theorem.
.

I do need to show because the question I asked is exactly how the question is worded in the book I'm using "Multivariable Calculus by James Stewart 6th Edition."