Vector Fields — Question 3

Question 3 – Solution

Strategy. Factor the position vector into its length and radial unit vector.

Step 1: Polar form With $r=\sqrt{x^2+y^2}$ and $\mathbf e_r=\langle x/r,y/r\rangle$, $$\mathbf F=\frac{r\mathbf e_r}{r^2}=\frac 1r\mathbf e_r.$$ Hence the domain is $\boxed{\mathbb R^2\setminus\{(0,0)\}}$, the direction is radially outward, and $$\boxed{|\mathbf F|=\frac 1r}.$$

Step 2: Samples $$\mathbf F(1,0)=\langle 1,0\rangle,\qquad \mathbf F(2,0)=\left\langle\frac 12,0\right\rangle,\qquad \mathbf F(0,-2)=\left\langle 0,-\frac 12\right\rangle.$$ Doubling the radius halves the arrow length while preserving radial direction.

Step 3: Limiting behavior As $r\to 0^+$, $|\mathbf F|=1/r\to\infty$, explaining the missing origin. As $r\to\infty$, the magnitude tends to $0$.

Verification Multiplying any vector by its radius gives $\mathbf e_r$, a unit vector, so both the direction and inverse-radius magnitude are confirmed.