Hello I have the following problem #5

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In order to solve it I know that $L_f(x_o,y_o)(h_1,h_2) = f(x_o,y_o)$ + Gradient$f(x_o,y_o)(h_1,h_2)$

My issue is that I am unable to find the gradient. I have tried to come up with a system of equations by stating that the directional derivative in the direction of $u$ is equal to the dot product of the gradient times the vector.

DirectDerivativeU(point) = Gradient$(x_o, y_o)*(\frac{2^{1/2}}{2},\frac{2^{1/2}}{2})$

2 = $(f_x,f_y)*(\frac{2^{1/2}}{2},\frac{2^{1/2}}{2})$

2 = $\frac{2^{1/2}}{2}f_x + \frac{2^{1/2}}{2}f_y$.

Unfortunately I am only able to come up with one equation to solve the system and find my gradient. How can I find another equation to solve this or can I please get a completely alternative method to solve this. Thank you


You are given the directional derivative in the exact direction you need it, that is, from the point $(3,-1)$ towards the point where you need to approximate $f$. So you don't need the gradient to find the directional derivative in the direction of $\vec u$, because you are given the value of that directional derivative.

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    $\begingroup$ I apologize if this may seem trivial but I'm confused at this point. Would you mind solving the problem for me. $\endgroup$ – Josue Mar 9 '20 at 1:15
  • $\begingroup$ I prefer not to solve problems that might be homework problems, but I'll add that you do not need to find the gradient. You write that $L_f(x_o,y_o)(h_1,h_2)= \vec{\nabla} f(x_0,y_o)(h_1,h_2)$, and so you seem to want to find $L_f(x_o,y_o)(h_1,h_2)$. But you are given this in the question, where it is called ${\partial f\over \partial{\vec u}}(3,-1)$ and equals $2$. So you don't have to compute $\vec{\nabla} f(x_0,y_o)(h_1,h_2)$ to find it. $\endgroup$ – Steve Kass Mar 9 '20 at 1:53

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