I am really stuck in the following problem:
consider $\Bbb R^2$ under the usual norm . let (x,y)$\in\Bbb R^2-\bar{B}$ ,where B=B(0,1) is open unit ball in $\Bbb R^2$.
Find f(x,y)=distance ((x,y), $\bar{B}$) and it's gradient.
any help would be appericiated..
The origin, $0$, the projection point $P$ and point $M(x,y)$ being aligned:
$$f(x,y)=PM=OM-OP=\sqrt{x^2+y^2}-1$$
thus
$$\vec{grad}f=\binom{\partial f/\partial x}{\partial f/\partial y}=\binom{x/\sqrt{x^2+y^2}}{y/\sqrt{x^2+y^2}}=\frac{1}{\|OM\|}\vec{OM}$$
This last vector expression is important because it connects the gradient to its main meaning: it indicates the direction of greatest increase of a function, here the direction in which the distance to the sphere increases in the fastest way.
(Thanks to Mathemagic who spotted an error that I have corrected.)
Here $f(x,y)=\sqrt{x^2+y^2}-1$
$\nabla f(x,y)=\frac{\Delta f}{\Delta x}i + \frac{\Delta f}{\Delta y}j=\frac{x}{\sqrt{x^2+y^2}}i+ \frac{y}{\sqrt{x^2+y^2}}j$
Interstingly,
$||\nabla f(x,y)||=\sqrt{\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}}=1$