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The directional derivatives of $f$ at the origin in the directions of $\overrightarrow v =\langle1,-1\rangle$ and $\overrightarrow w =\langle\sqrt{3}, 1\rangle$ are $-\sqrt{2}$ and $4 + 3\sqrt{3}$ respectively. Find the maximum rate of change of $f$ at the origin.

Can someone help me answer? Or give me an idea or the steps I should do to answer this?
Thank you!

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    It wasn't given in the problem.2012-11-24

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The direction derivate along the vector $\langle a,b \rangle$ at origin is given as $\langle a,b \rangle \cdot \langle f_x,f_y \rangle_{(0,0)} = af_x(0,0) + bf_y(0,0)$ The direction derivate along the direction $\langle a,b \rangle$ at origin is given as $\dfrac{\langle a,b \rangle}{\Vert \langle a,b \rangle \Vert} \cdot \langle f_x,f_y \rangle_{(0,0)} = \dfrac{af_x(0,0) + bf_y(0,0)}{\langle a,b \rangle}$ You are given that $\dfrac{f_x(0,0) - f_y(0,0)}{\sqrt{2}} = -\sqrt{2}$ and $\dfrac{\sqrt{3} f_x(0,0) + f_y(0,0)}2 = 4 + 3\sqrt{3}$ Solving for $f_x(0,0)$ and $f_y(0,0)$ gives us $f_x(0,0) = 6; \,\,\,\, f_y(0,0) = 8$ The maximum rate of change is along $\vec{\nabla} f$ i.e. along the direction $\dfrac{\langle f_x,f_y \rangle}{\Vert \langle f_x,f_y \rangle \Vert}$. Hence, the maximum rate of change at the origin is $\langle f_x,f_y \rangle \cdot \dfrac{\langle f_x,f_y \rangle}{\Vert \langle f_x,f_y \rangle \Vert} = \Vert \langle f_x,f_y \rangle \Vert = \sqrt{f_x(0,0)^2 + f_y(0,0)^2}$ which gives us the answer as $10$.

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    Oh I see. Thanks!2012-11-25