Vector Fields — Question 5

Question 5 – Solution

Strategy. A circle’s radial vector is normal to the circle, so orthogonality to that vector proves tangency.

Step 1: Tangency At $(x,y)$ the radial normal is $\mathbf r=\langle x,y\rangle$. Since $$\mathbf F\cdot\mathbf r =\langle-y,x\rangle\cdot\langle x,y\rangle =-xy+xy=0,$$ .

Step 2: Magnitude and orientation $$|\mathbf F|=\sqrt{(-y)^2+x^2}=\sqrt{x^2+y^2}=r.$$ At $(r,0)$ the vector is $\langle 0,r\rangle$, which points upward; therefore the orientation is counterclockwise.

Step 3: The point $(3,4)$ $$\mathbf F(3,4)=\boxed{\langle-4,3\rangle},\qquad |\mathbf F(3,4)|=5,$$ so the counterclockwise unit tangent is $$\boxed{\left\langle-\frac 45,\frac 35\right\rangle}.$$

Verification The dot product $\langle-4,3\rangle\cdot\langle 3,4\rangle=-12+12=0$, confirming tangency at $P$.