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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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    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 $a\cdot b =\|a\|\|b\|\cos\theta$?2012-07-03

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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 $u_1^2+u_2^2=1$.

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    Probably one of your doubts comes from my notation. As a mathematician, I stopped writing vector with $\mathbf{i}$ and $\mathbf{j}$ many years ago :-) Yes, $(u_1,u_2)=u_1 \mathbf{i}+u_2 \mathbf{j}$.2012-07-03