Find the directional derivative of
$f(x, y, z) = xyz$
in the direction normal to the surface
$yx^2 + xy^2 + yz^2 = 3$
at
$(1, 1, 2).$
ANSWER: $12 \sqrt{2/37}$
I am having trouble finding the vector direction to use. Since we are given the equation of a plane (I think) the normal direction should just be the coefficients of the x-y-z variables $(1)yx^2 + (1)xy^2 + (1)yz^2 = 3 $ which is (1,1,1).
So I did dot product of the gradient f(1,1,2) * (1,1,1)/$\sqrt{3}$ (to normalize vector direction) which comes out to $5\sqrt{3}$ which is wrong
Best Answer
The direction normal to the surface is given by the gradient of $$g(x, y, z) = y x^2 + x y^2 + y z^2$$ at $(1,1,2)$. You need to normalize the gradient to get a unit normal vector.
EDIT: there is a problem in the question because $(1, 1, 2)$ does not belong to the surface as $g(1,1,2) = 6$.