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Using differential I want to find the value of the function $g(x,y,z) = \ln \sqrt{x^2+y^2+z^2}$ if we move from (3,4,12) to a distance of 0.1 in the direction of vector $u = 3i+6j-2k$.

I first found $$|u| = \sqrt{9+36+4} = 7$$ so $$u_o = \frac 3 7 i+ \frac 6 7j - \frac 2 7k$$ and $$\nabla g(3,4,12) = \frac 6 {29}i + \frac 8{29}j+ \frac {24}{29}k$$

Then I don't know what to do. Any ideas?

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Can you explain it a bit more? I think I didn't understand what is going o and that's why I cannot find the solution myself. –  fickonianP Mar 20 '12 at 19:35
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The change in $g$ is approximately the dot product of $\nabla g(3,4,12)$ with $\Delta {\bf x}=.1u_0$. Add the change in $g$ to the "starting value" $g(3,4,12)$. –  David Mitra Mar 20 '12 at 19:36
    
(3,4,12) or(3,4,2)? –  Tpofofn Mar 20 '12 at 20:33
    
@Tpofofn Typo. I think it is ok now. –  fickonianP Mar 20 '12 at 20:53
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1 Answer

You can compute the value of $g$ at the the point ${\bf x}_0=(3,4,12)$. It's $$ g(3,4,12)=\ln(13) $$ You want to estimate the value of $g$ at another point. From ${\bf x}_0$, you move to the the point ${\bf x}_1=(3,4,12)+(.1)u_0$ and you wish to estimate the value of $g$ there.

To do this, you can first consider the change in the value of $g$ between the two points. You can find an approximation to the change in the value of $g$, denoted $\Delta g$, between those two points by using use the formula $$\tag{1} \Delta g\approx \nabla g({\bf x}_0)\cdot \Delta {\bf x},$$ where $\Delta {\bf x}= {\bf x}_1-{\bf x_0} = (.1)u_0$.

So, to approximate the value of $g$ at the point ${\bf x}_1$, add the change in the value of $g$ from ${\bf x}_0$ to ${\bf x}_1$ to the value of $g$ at ${\bf x}_0$: $$\eqalign{ g({\bf x}_1)&= \strut g({\bf x}_0)+ \Delta g\cr &\tag{2}\approx \strut g({\bf x}_0)+\nabla g({\bf x}_0)\cdot \Delta {\bf x} } $$

So you're almost there: compute $(1)$ with ${\bf x}_0=(3,4,12)$ and $\Delta{\bf x}=(.1)u_0 $ and then add to $g(3,4,12)=\ln (13)$. (You may want to check your computation of the gradient, it seems off (you need $z=12$))


Note this is similar to the one variable case with the gradient playing the role of the derivative $$ f(x_1)\approx f(x_0)+(x_1-x_0)f'(x_0). $$

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Can you remove your answer for a while? I don't won't to look at it until I got something by myself. –  fickonianP Mar 20 '12 at 20:28
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@fickonianP Umm...I don't think anyone is making you look at it. –  ItsNotObvious Mar 20 '12 at 20:35
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