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Find the points $(x,y)\in \mathbb R^2$ and unit vectors $\vec u$ such that the directional derivative of $f(x,y)=3x^2+y$ has the maximum value if $(x,y)$ is in the circle $x^2+y^2=1$

My attempt:

I know that the directional derivative is $D_{\vec u}f=\nabla f\cdot \vec u=6xu_1+u_2$ which does not depends on $y$ if $u_1$ is positive then $D_uf$ attains it´s maximum in $x=1$ and if $u_1$ is negative then $D_uf$ attains it´s maximum in $x=-1$

then we have two cases 1)$6u_1+u_2$ and 2)$-6u_1+u_2$ then I need to find the points $(u_1,u_2)$ such that $u_1^2+u_2^2=1$

Using Lagrange multipliers I found that $u_1=u_2={\pm {1\over \sqrt{2}}}$ hence $(x,y)=(\pm 1,0)$ and $(u_1,u_2)=(\pm {1\over \sqrt{2}},\pm{1\over \sqrt{2}})$

Is this approach correct? or Am I missing something?

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    $\begingroup$ Hint: The directional derivative has its maximum value in the direction of the gradient. Check your result: is the directional derivative at $(1,0)$ in the direction of $(u_1,u_2)$ really greater than the directional derivative in the direction of $\nabla f_{(1,0)}$? $\endgroup$ – amd Jun 2 '16 at 5:09
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From $f(x,y):=3x^2+y$ we obtain $\nabla f(x,y)=(6x,1)$, hence $|\nabla f(x,y)|=\sqrt{36x^2+1}$.

It is obvious that the points on the closed unit disk where the largest value $|\nabla f(x,y)|$ occurs are the two points $(\pm1,0)$, with $\nabla f(\pm1,0)=(\pm6,1)$ and $|\nabla f(\pm1,0)|=\sqrt{37}$.

Now at each point the direction of maximal increase of $f$ is given by the direction of $\nabla f$ at this point. It follows that the two unit vectors we are interested in are given by $${\bf u}_\pm=\left({\pm6\over\sqrt{37}}, \>{1\over\sqrt{37}}\right)\ ,$$ where ${\bf u}_+$ refers to $(1,0)$ and ${\bf u}_-$ to $(-1,0)$.

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