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I have the next function: $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ and i need to find it's directional derivative of the point $(0,0,0)$ and the vector $v=(\frac1{\sqrt{3}},-\frac1{\sqrt{3}},\frac1{\sqrt{3}})$

So I've started by finding the partial derivatives:

${f'}_x(x,y,z)=\frac x{\sqrt{x^2+y^2+z^2}}$, ${f'}_y(x,y,z)=\frac y{\sqrt{x^2+y^2+z^2}}$, ${f'}_z(x,y,z)=\frac z{\sqrt{x^2+y^2+z^2}}$

Now i'm pretty much stuck. Since placing $(0,0,0)$ makes an undefined expression, What can i do from this point on?

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It is highly probable that $f$ is not differentiable. –  Siminore Jun 12 '13 at 11:26
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$(f(tv) - f(0))/t = \|tv \|/t = |t|/t$. Does the limit exist? –  flavio Jun 12 '13 at 11:43
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@jiku1797: Well, if $t>0$ is involved in the definition of directional derivative (which might make sense, and might happen at certain authors), then this limit does exit, and is $1$. –  Berci Jun 12 '13 at 12:15

1 Answer 1

up vote 1 down vote accepted

(Edited to address technical ambiguities in the earlier version):

If we use the usual definition of the directional derivative at $\mathbf{x}$ along $\mathbf{v}$,

$$ D_{\mathbf{v}}f(\mathbf{x}) = \lim_{t\rightarrow 0}\frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$

where there is no restriction on $t$, then the directional derivative in fact does not exist at $\mathbf{0}$ since this limit is different depending on whether $t\rightarrow 0^+$ or $t\rightarrow 0^-$. For the former case $t>0$ so

\begin{align}\lim_{t\rightarrow 0^+} \frac{f(\mathbf{0}+t\mathbf{v})-f(\mathbf{0})}{t} &= \lim_{t\rightarrow 0^+}\frac{f\big(t\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\big)}{t} \\ &= \lim_{t\rightarrow 0^+}\frac{\sqrt{t^2}}{t} \\ &=1, \end{align}

while for the latter $t<0$ so

$$ \lim_{t\rightarrow 0^-} \frac{f(\mathbf{0}+t\mathbf{v})-f(\mathbf{0})}{t} = \ \cdots\ =-1. $$

These are related to the direction along $\mathbf{v} = \left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$ in which we are instantaneously travelling: you can think of $t\rightarrow 0^+$ as giving the rate of change along the direction of $\mathbf{v}$ going away from $\mathbf{0}$ $[*]$, while $t\rightarrow 0^-$ gives the rate of change coming into $\mathbf{0}$ travelling "against" $\mathbf{v}$.

Intuition-builder:

Note that without doing any work at all, you can guess that the "directional derivative" (in the sense of $[*]$) is $1$. This because $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ is the distance of $(x,y,z)$ from the origin $\mathbf{0}$. The directional derivative at the origin is then the rate of change of $f$ with respect to the distance travelled away from $\mathbf{0}$ along $\mathbf{v}=\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Since $f$ is the distance from the origin, this rate is of course $1$.

To summarize, the derivative you are looking for does not exist, but if we restrict the limit to the range $t>0$ then the limit is $1$, and it makes sense to call this a "directional derivative" of some sort.

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Ehmmmm, $$\lim_{t \to 0} \frac{\sqrt{t^2}}{t}$$ does not exits. –  Siminore Jun 12 '13 at 15:00
    
Oh yes, I'm implicitly assuming $t>0$. Better make that clear in my answer. Thanks for pointing that out. –  Josh Chen Jun 12 '13 at 23:22

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