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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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    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**.2012-07-03
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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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    I got to this stage in the solution. What should I do after I will find the 2 answers of u1,u2?2012-07-03
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    @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.2012-07-03
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    There are 2 sets of $(u1,u2)$. The solution will be: $\vec U = u1*\vec i + u2*\vec j$ ?2012-07-03
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    Yes, you are cutting a circle with a straight line. You'll find two intersections, which are exactly the two unit vectors that solve your problem.2012-07-03
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    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?2012-07-03
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    Also, if I going backward, and calculating the derivative in the $\vec U$ direction, it will be 6..?2012-07-03
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    (1) Find $u_1$ and $u_2$. (2) The vector $\mathbf{U}=(u_1,u_2)$ solves your problem. (3) If you are not convinced, compute the derivative along $(u_1,u_2)$ and check if it works.2012-07-03
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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