I'm having trouble figuring out this homework question for my Calculus III class:
"Find a field $\vec{G}=P(x,y)\vec{i}+Q(x,y)\vec{j}$ in the $xy$-plane with the property that at any point $(a,b)\neq (0,0)$, $\vec{G}$ is a vector of magnitude $\sqrt{a^2+b^2}$ tangent to the circle $x^2+y^2=a^2+b^2$ and pointing in the clockwise direction. (The field is undefined at $(0,0)$.)"
I have to fill in the answers for $P(x,y)$ and $Q(x,y)$.
I've tried parameterizing the circle with a variable $t$, but that didn't seem to get me anywhere. Once I got the problem wrong, it said to differentiate the circle in terms of $x$ and use the slope to find the vector field. So then I tried implicit differentiation, treating $a$ and $b$ as constants and $y$ as a function of $x$:
$\frac{d}{dx}\left(x^2+y^2=a^2+b^2\right)$
$2x+2y\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\frac{x}{y}$
Now that I got this slope, I'm not sure how to use this to find the vector field. And it makes it even more confusing since this is clockwise. Any help would be appreciated! Thank you! 🙂
Best Answer
For perpendicularity you need $Pa+Qb=0$, and you also need $P^2+Q^2=a^2+b^2$ (at the point $(a,b)$). Solving the system you get $(P,Q)=(-b,a)$ or $(P,Q)=(b,-a)$. Only the second is oriented clockwise. So your answer is $$ P(x,y)=y,\qquad Q(x,y)=-x. $$