Given the function f(x; y) = 2x^2 + y + 3y^2 -1 and the unit vector u = find all points(x,y) such that the directional derivative in direction of u is 1.

Question

Given the function f(x; y) = 2x^2 + y + 3y^2 -1 and the unit vector u =<1/sqrt(5),2/sqrt(5)>
find all points(x,y) such that the directional derivative in direction of u is 1.

anonymous · Answer

simplifying,
$$
\nabla_\vec{u}f(x,y)={\partial\over\partial x}f(x,y)\times{1\over\sqrt{5}}+{\partial\over\partial y}f(x,y)\times{2\over\sqrt{5}}

$$

anonymous · Answer

$$
\nabla_\vec{u}f(x,y)={1\over\sqrt{5}}\left[(4x)+2(1+6y)\right]
$$

anonymous · Answer

yes that is right. the problem is if i expand this equation and eqaute it to 1, i get a function of two variables x and y which is (4x+12y+2)^2=5. I cant get the other equation for me to solve simultaneously.

anonymous · Answer

you'd get a "locus" of points...

anonymous · Answer

probably a circle or an ellipse

anonymous · Answer

ok, I guess you are right there thanks.