The directional derivative of $f(x,y)$ at $(1,2)$ in the direction of $\vec a =\vec i + \vec j$ is $2\sqrt{2}$.

We also know that the directional derivative of $f(x,y)$ at $(1,2)$ in the direction of $\vec b = -2\vec j$ is $-3$.

What is the derivative of $f$ at $(1,2)$ in the direction of $\vec u=-\vec i -2\vec j$

What I did:

We know at $(1,2)$ : $\frac{\langle \nabla f,\vec a\rangle}{|\vec a|}=\frac{\langle \nabla f,\vec i +\vec j\rangle}{|\vec i+\vec j|}=\frac{\langle \nabla f,\vec i\rangle}{|\vec i +\vec j|}+\frac{\langle \nabla f,\vec j\rangle}{|\vec i +\vec j|}=2\sqrt{2}$

We also know that at (1,2): $\frac{\langle \nabla f,\vec b\rangle}{|\vec b|}=\frac{\langle \nabla f,-2\vec j\rangle}{|-2\vec j|}=\frac{-2}{2}\frac{\langle \nabla f,\vec j\rangle }{|j|}=-\frac{\langle \nabla f,\vec j\rangle }{|j|}=-3$

So overall we have:

$\frac{\langle \nabla f,\vec i\rangle}{|\vec i +\vec j|}+\frac{\langle \nabla f,\vec j\rangle}{|\vec i +\vec j|}=2\sqrt{2}$ and $\frac{\langle \nabla f,\vec j\rangle }{|j|}=3$

And from this, without any knowledge of what $\vec i$ or $\vec j$ are, we are supposed to find $\frac{\langle \nabla f,-\vec i -2\vec j\rangle}{|-\vec i -2\vec j|}$

I dont Immediately see how can this be done. $\vec i$ and $\vec j$ could be anything, the results we know don't help us solve what we want.

I'd appreciate any insight.

  • $\begingroup$ I'm confused by your use of the term "directional derivative." I think the most common definition of the directional derivative of $f$ along the vector $w$ is $\langle\nabla f,w\rangle$ and not $\langle\nabla f,(w/|w|)\rangle$. $\endgroup$ – Josh Burby May 6 '15 at 22:38
  • $\begingroup$ $w$ has to be normalized. I think that's the most common definition $\endgroup$ – Oria Gruber May 6 '15 at 22:44
  • $\begingroup$ I guess what's really important is what definition of directional derivative your question is assuming. If the relevant definition actually does require normalization, then your question cannot be answered without additional information (for instance that $i$ and $j$ are orthonormal unit vectors), as noted by Dr. MV below. If the definition does not involve normalization (which is certainly common, although perhaps not the most common), then the answer is Dr. MV's times $\sqrt{5}$. $\endgroup$ – Josh Burby May 6 '15 at 23:09

Just write $\vec u=-\vec a+\frac12 \vec b$. Then,

$\nabla f \cdot \vec u=\nabla f \cdot (-\vec a+\frac12 \vec b)=-(\nabla f \cdot \vec a)+\frac12 (\nabla f \cdot \vec b)=-(2\sqrt{2})(\sqrt{2})+\frac12 (-3)(2)=-7$

Now just divide by $|\vec u|=\sqrt{5}$ and find the directional derivative is $-7\sqrt{5}/5$.

  • $\begingroup$ Why is $|\vec u| = 5$? what if $\vec i$ is the zero vector and $\vec j =\begin{pmatrix} 1 \\ 1 \end{pmatrix}$? then $|\vec u|=\sqrt{8}$ $\endgroup$ – Oria Gruber May 6 '15 at 22:47
  • 1
    $\begingroup$ I think the author intended for $\vec i$ and $\vec j$ to be interpreted as Cartesian unit vectors $(1,0)$ and $(0,1)$ respectively. $\endgroup$ – WW1 May 6 '15 at 22:59
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    $\begingroup$ @OriaGruber I interpreted $\vec i$ and $\vec j$ to be Cartesian unit vectors. And if not, one can still use $\vec u=-\vec a+\frac12 \vec b$. The calculation would then proceed identically, but we would need to know the magnitudes of the vectors $\vec i$ and $\vec j$ to give a numerical answer. I really do believe that these are Cartesian unit vectors. $\endgroup$ – Mark Viola May 6 '15 at 23:05
  • $\begingroup$ @Dr.MV I think we would need the magnitudes of the vectors and the angle between them, right? The normalization factor $|u|$ depends on $a\cdot b$ in addition to $|a|$ and $|b|$. $\endgroup$ – Josh Burby May 7 '15 at 0:26
  • $\begingroup$ @JoshBurby If we knew the magnitudes of $\vec a$, $\vec b$, and $\vec u$, then that would suffice. $\endgroup$ – Mark Viola May 7 '15 at 0:31

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