Function derivative in the vector direction

Let there be a function $z=x^3+xy+y^2$. Is there a vector $\vec U$ that in the point $(1,1)$ the function derivative in the vector direction is equal to 6?

Now I know that the $|\nabla F(1,1)|$ is $\sqrt{4^2+5^2} = \sqrt{41} = 6.4 > 6$. So there is a vector $\vec U$ that equal to 6.

How can I find it?

Thank!

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Hmm, it's hard to make a sense out of it... What is $\vec gradF(1,1)$ and why is it $\sqrt{4^2+5^2}$? Do you mean $\nabla F(1,1)$? Plenty of questions on your question, so please be more specific. Thank s. – draks ... Jul 3 '12 at 13:42
First of all, you mean $\|\nabla F\|$. Second, there will be two such vectors. Are you familiar with the formulas $D_v F=\nabla F\cdot v$ and $a\cdot b =\|a\|\|b\|\cos\theta$? – anon Jul 3 '12 at 13:43

For every unit vector $\mathbf{u}=(u_1,u_2)$, you know that $$\frac{\partial z}{\partial \mathbf{u}} = \nabla z \cdot \mathbf{u} = \frac{\partial z}{\partial x}u_1 + \frac{\partial z}{\partial y}u_2.$$ You can now evaluate all partial derivatives at $(1,1)$ and solve the equation $$\frac{\partial z}{\partial x}u_1 + \frac{\partial z}{\partial y}u_2=6.$$ Since $$\frac{\partial z}{\partial x} = 2x+2y$$ and $$\frac{\partial z}{\partial y} = 2x+3y^2$$ you must solve $$4u_1 +5 u_2 = 6$$ under the constraint $u_1^2+u_2^2=1$.
@Lag: Your question asks to find the vector $(u_1,u_2)$; unless there is more to your question than you let on, finding $u_1,u_2$ is the end of the issue. – anon Jul 3 '12 at 13:59
There are 2 sets of $(u1,u2)$. The solution will be: $\vec U = u1*\vec i + u2*\vec j$ ? – Gal Jul 3 '12 at 14:05
I am sorry that i'm asking this again, but this is impotent for me to understand. Is the direction vector that I need to find ($\vec U$), is $\vec U = u1*\vec i + u2*\vec j$, for every set of(u1,u2) that I found? – Gal Jul 3 '12 at 16:09