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Given the function: $$ f(x,y,z) = \frac{1}{x+y+z} $$ What's the directional derivative in the direction of the gradient at $(x,y,z) =(1,1,1)$?

I calculated that: $$ \overrightarrow{\triangledown} f \;\biggr\rvert_{(1,1,1)} = -\frac{1}{9}(1,1,1) $$

Thus the directional derivative in the direction of the gradient at $(1,1,1)$ is: $$ \left(\overrightarrow{\triangledown} f \;\biggr\rvert_{(1,1,1)}\right) \cdot \widehat{\left(\overrightarrow{\triangledown} f \;\biggr\rvert_{(1,1,1)}\right)} = -\frac{1}{9}(1,1,1) \cdot \frac{1}{3\sqrt{3}}(1,1,1) = - 3\sqrt{3} $$

But that answer was marked as false. Why?

Dor
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  • I think you wrote some wrong computations. The last equality is wrong. $-\frac{1}{9}(1,1,1) \cdot \frac{1}{3\sqrt{3}}(1,1,1) = - \frac{1}{27}\sqrt{3}$. – Crostul Feb 10 '15 at 23:06
  • Oh, you're right! I don't understand how I missed that! I blame the late hour..! But I think that you're almost correct - shouldn't it be $-\frac{1}{9\sqrt{3}}$ ?. – Dor Feb 10 '15 at 23:09
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    Typically directional derivatives are defined for unitary vectors, then you must divide the gradient by its norm, but do not change the sign of the vector: the result is always positive. This I think is the main point. – Jesus RS Feb 10 '15 at 23:13
  • Jesus RS: I don't understand - the norm of the gradient is positive, so diving the gradient (which is negative) by its norm (positive), will stay negative. – Dor Feb 10 '15 at 23:17
  • @Dor You might say $-\frac{1}{9\sqrt{3}}$, you might say $-\frac{1}{27}\sqrt3$. They're the same. To see this, expand the first fraction by $\sqrt{3}$. – Arthur Feb 11 '15 at 00:04
  • Back to the sign, you put only a minus, and there must be two, and teh final result positive. – Jesus RS Feb 11 '15 at 13:17

1 Answers1

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It's probably worth while doing a general case first. The directional derivative of $f$ at $P$ in the direction $\bf u$ is $$\nabla f(P)\cdot\widehat{\bf u}\ .$$ For the question you asked we have ${\bf u}=\nabla f(P)$ so this becomes $$\nabla f(P)\cdot\widehat{\nabla f(P)}\ .$$ But for any vector ${\bf v}$ we have $${\bf v}=\|{\bf v}\|\widehat{\bf v}\quad\hbox{so}\quad {\bf v}\cdot\widehat{\bf v} =\|{\bf v}\|(\widehat{\bf v}\cdot\widehat{\bf v})=\|{\bf v}\|$$ (note that this is always positive, as mentioned in the comment by Jesus RS) and the final answer, using your calculation, is just $$\|\nabla f(P)\|=\frac1{3\sqrt3}\ .$$

David
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  • Why $|\nabla f(P)|=\frac1{3\sqrt3}\ $ ? It should be: $$|\nabla f(P)| = |-\frac{1}{9}(1,1,1)| = |(-\frac{1}{9},-\frac{1}{9},-\frac{1}{9})| = \sqrt{(-\frac{1}{9},-\frac{1}{9},-\frac{1}{9}) \cdot (-\frac{1}{9},-\frac{1}{9},-\frac{1}{9})} = 3 \cdot \frac{1}{9^2} = \frac{1}{27}$$ – Dor Feb 10 '15 at 23:32
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    You didn't take the $\sqrt{\ }$. – David Feb 10 '15 at 23:44