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Find the maximum and minimum values of the directional derivative Duf at (1/2, 1) as u varies for the function f(x,y)= x3 -xy2-4x2+3x+x2y

What im not sure about is the phrase as u varies. I understand the formula for directional derivatives is Duf=∇f⋅u but don't understand what is meant by u varies.

My thoughts on how to do this question is letting the direction be in respect to u in (ux, uy) and I calculate the differential terms for x and y

I end up with (ux)(3x2-y2-8x+3+2xy) + (uy)(-2xy+x2)

Is this the right way of approaching this question? Also if I were to calculate the direction for u to give a maximum and minimum, how would I do so?

DDDDOO
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1 Answers1

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We know that $\nabla \vec{f}\cdot \vec{u} =\Vert f\Vert \cos \theta$, where $\theta$ is the angle between $\vec{f}$ and $\vec{u}$. So it is maximized if $\vec{f}$ and $\vec{u}$ are in the same direction while minimized if they are in opposite directions. So what I will do is evaluate $\vec{f}(1/2,1)$ and normalize it. Its normalization gives the max while the negative of the normalization minimizes the directional derivative.

June Mar Fajardo
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  • Not quite. $\pm\nabla \vec f$, whether normalized or not, give the directions of the max. and min. directional derivatives, but not the values of those derivatives. – amd Feb 23 '20 at 00:28