Maximizing Function Increase: Understanding Directional Derivatives

Master J
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I have a function of 2 variables. I know it increase most rapidly in the direction of the gradient, but how about in wht direction is it not increasing?

I am thinking that the gradient (dot product)(direction in which it is not increasing) = 0

Any hints?
 
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Well, since you mentioned the word "directional derivative" anyway: you could check for which \vec v
(\vec\nabla f(x, y)) \cdot \vec v < 0
?
 
Yes, it is true that \vector{\nabla f}\cdot \vector v is the directional derivative in the directional derivative in the direction of \vec{v} (for \vec{v} of length 1). And that tells you the derivative is 0 perpendicular to the gradient.

(CompuChip, surely you didn't mean "<"?)
 
Err, no comment? :)
 
That is what is was thinking, since of course cos(pi/2) = 0. So the vector that is at a right angle to the gradient is in the direction of zero increase. But how do I go about finding this vector?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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