The problem I am working on is this:

Find ${\nabla}f(a,b)$ for the differential function $f(x,y)$ given the directional derivatives:

$D_{(i+j)/2}f(a,b) = 3\sqrt2$ , $D_{(3i-4j)/5}f(a,b) = 5$

I've done some reading on Directional Derivatives to see if I could then solve them, but I can't make head or tails of this problem. Could someone guide me through this or point me in the right direction?

Thanks in advance.

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  • $\begingroup$ Perhaps the formula $D_af=(\nabla f)\cdot a$ might help. $\endgroup$ – user137731 Mar 23 '15 at 21:34
  • $\begingroup$ The thing that is confusing me is the $(i+j)/2$ part. Where do I use that? Oh sorry, didn't see the $a$. Okay... so I dot product the function with what exactly? $\endgroup$ – Kommander Kitten Mar 23 '15 at 21:36
  • $\begingroup$ That's your $a$ in my formula. $\endgroup$ – user137731 Mar 23 '15 at 21:36

Hints: $\nabla f(a.b) = (D_i f(a,b), D_j f(a,b))$, write $i$ and $j$ in terms of $(i+j)/2$ and $(3i-4j)/5$, and: $$i = \alpha \cdot \frac{i+j}{2} + \beta \cdot\frac{3i-4j}{5} \implies D_i f(a,b) = \alpha\cdot(3\sqrt{2})+\beta\cdot (5),$$and similarly for the other one.

I'll do the first one for you. Write: $$(1,0) = \alpha \left(\frac{1}{2},\frac{1}{2}\right)+\beta \left(\frac{3}{5},-\frac{4}{5}\right) \implies \begin{cases} \frac{\alpha}{2}+\frac{3\beta}{5} = 1 \\ \frac{\alpha}{2} - \frac{4\beta}{5} = 0\end{cases},$$so that $\beta = 5/7$ and $\alpha = 4/7$. Hence: $$D_i f(a,b) = D_{\frac{4}{7}\left(\frac{i+j}{2}\right)+\frac{5}{7}\left(\frac{3i-4j}{5}\right)}f(a,b) = \frac{4}{7}D_{\frac{i+j}{2}}f(a,b)+\frac{5}{7}D_{\frac{3i-4j}{5}} f(a,b).$$

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  • $\begingroup$ I don't understand what you mean by "write $i$ and $j$ in terms of $(i+j)/2$ and $(3i-4j)/5$ $\endgroup$ – Kommander Kitten Mar 23 '15 at 21:48
  • $\begingroup$ If I told you to write $(1,0)$ as a combination of $(1/2,1/2)$ and $(3/5,-4/5)$ would you understand? There is more than one possible combination, but the final outcome will be the same. $\endgroup$ – Ivo Terek Mar 23 '15 at 21:49
  • $\begingroup$ A bit clearer, but still unsure of myself. Could you explain further? I feel like I'm missing something fairly obvious. $\endgroup$ – Kommander Kitten Mar 23 '15 at 21:54
  • $\begingroup$ I'll add an edit. By the way, I just noticed that we will have a unique combination, I looked too fast. $\endgroup$ – Ivo Terek Mar 23 '15 at 21:56
  • $\begingroup$ @Kommander see if you can do the second one $\endgroup$ – Ivo Terek Mar 23 '15 at 22:01

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