The given function is $$f(x,y,z) = \frac{1}{\sqrt{x^2+y^2+z^2}}$$
and I need to find gradient and draw the picture of this vector field.
Gradient that I calculate is:
grad $f(x,y,z) = \left(\frac{-x}{(x^2+y^2+z^2)^{\frac{3}{2}}} , \frac{-y}{(x^2+y^2+z^2)^{\frac{3}{2}}} , \frac{-z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\right)$
Could anyone help me to draw the vector field?. Thanks :)
The gradient of any $C^1$ real-valued function, is orthogonal to the levelsets of the function, and points in the direction of growth of the function. Its amplitude is proportional to how fast the function grows.
For this function, the level sets are concentric spheres centered at the origin, and the function grows as you move closer to the origin. Thus, the gradient points inward, orthogonally to the spheres. Its amplitude is its length, which equals, according to your calculation, and Simon's calculation, $\frac1{r^2}$. So the outward arrows grow shorter and shorter as you move away from the origin.
Notice that $\sqrt{x^2 + y^2 + z^2}$ is the distance $r$ from the origin. The vector $(x,y,z)$ points away from the origin. So your $$\operatorname{grad} f = -\frac{1}{r^3}(x,y,z)$$ points in towards the origin. But what happens with the magnitude of $\operatorname{grad} f$?
