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Suppose that there is a surface described by:

$\phi(x,y,z)=c$

And suppose that there is a fixed point A:
$\vec{r_A}=(x_A,y_A,z_A)$

Let $\vec r$ be position vector of any point on the surface so that:

$R=|\vec r-\vec r_A|$

Show that $\nabla R$ is a unit vector whose direction is along $\vec r-\vec r_A$

I tried to write $z=f(x,y)$ and then calculate what is $\nabla R$ but I got that it has no $z$ component which leads to a contradiction...

  • There is something wrong here. As defined in your question, $R$ is a function on your surface (let's call it $M$). As such, it can only be differentiated along $M$, and it's gradient (wrt M) is a tangent vector of $M$.However, $\vec {r}-\vec r_A$ is, in general, not tangent to $M$. – Thomas May 12 '18 at 09:29

2 Answers2

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By denoting $\vec r = (x,y,z)$ you can write $R = \sqrt{(x-x_A)^2+(y-y_A)^2+(z-z_A)^2}$. Then as $\nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz})$ I will just do the calculation for the $x$-coordinate, as it is completely analogous for $y,z$:

$\frac{dR}{dx} = (\frac{1}{2}) R^{-1} 2 (x-x_A) = \frac{x-x_A}{R}$

And by putting all coordinates together that gives you:

$\nabla R = \frac{1}{R} (\vec r - \vec r_a) $ which is just $\vec r - \vec r_a$ divided by its norm and has therefore unit length.

FeWa
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$$R=|\vec{r}-\vec{r}_a|=|(x-x_a)\mathbf{i}+(y-y_a)\mathbf{j}+(z-z_a)\mathbf{k}|=\sqrt{(x-x_a)^2+(y-y_a)^2+(z-z_a)^2}$$

Now take $\nabla R$

mallan
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