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Math Help - Show that a differentiable function f decreases most rapidly....

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Undefdisfigure View Post
    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 f(\bold{x}): \mathbb{R}^n \mapsto \mathbb{R} is differenciable at \bold{x}_0 then the directional derivative D_{\bold{u}}f(\bold{x_0}) in the direction \bold{u} is given by \nabla f(\bold{x}_0)\cdot \bold{u}. But |\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.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    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."
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