Vector Fields — Question 10

Question 10 – Solution

Strategy. Recognize the field as a fixed combination of the radial vector and its $90^\circ$ counterclockwise rotation.

Step 1: Decompose On the unit circle, $$\mathbf e_r=\langle x,y\rangle,\qquad \mathbf e_\theta=\langle-y,x\rangle.$$ Therefore $$\mathbf F =2\langle x,y\rangle+\langle-y,x\rangle =\boxed{2\mathbf e_r+\mathbf e_\theta}.$$ The radial scalar component is $2$ and the counterclockwise tangential scalar component is $1$.

Step 2: Magnitude and angle Because the two unit vectors are perpendicular, $$\boxed{|\mathbf F|=\sqrt{2^2+1^2}=\sqrt 5}.$$ If $\alpha$ is measured counterclockwise from $\mathbf e_r$, then $$\boxed{\alpha=\tan^{-1}\!\left(\frac 12\right)}.$$

Step 3: Point check and zeros At $(0,1)$, $$\mathbf e_r=\langle 0,1\rangle,\quad \mathbf e_\theta=\langle-1,0\rangle,\quad \mathbf F=\boxed{\langle-1,2\rangle}.$$ For the full plane, $\mathbf F=\mathbf 0$ gives $2x-y=0$ and $x+2y=0$. Their coefficient determinant is $5\ne 0$, so only the origin is a zero; in particular, there are none on the unit circle.

Verification The sample vector $\langle-1,2\rangle$ has length $\sqrt 5$ and equals $2\mathbf e_r+\mathbf e_\theta$, confirming all parts.