Vector normal to parametrized circle

Question

Suppose there is a circle given by equation c: $(x-3)^2+(y-2)^2=4$ and and another identical circle given by the parametric equation $$\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}3\\2\end{array}\right)+2\left(\begin{array}{c}cos\theta\\sin\theta\end{array}\right)\quad0\le\theta\le2\pi$$So we are given point $A$ on the circle. The norm for point A, using the Cartesian equation is $$\nabla c=\left(\begin{array}{c}2x-6\\2y-4\end{array}\right)$$ So at point $A$, a line orthogonal to the circle can be written as $(x,y)=A+t\cdot\nabla c\quad-\infty\le t\le\infty.$ I would like to know how to write the vector normal, $\mathbf{u}$, to the parametrized circle equation so that I could obtain a similar (as in identical) orthogonal line at $A$. What seems obvious is to use $\mathbf{u}=(dx/d\theta, dy/d\theta) = (-2sin\theta,\,2cos\theta)$, but if I let $A=(5,2)$ which is just a point on the circle 2 units to the right center, then $\nabla c = (4,0)$, but since $\theta = 0$, $\mathbf{u} = (0,2)$. Where have I gone wrong? I would like to be able to apply this to a 3d parametric circle. Thought I should understand the 2d case first.

Answer 1

Your fundamental error here is that $\mathbf u'$ is tangent to the curve, not normal to it. Once you’ve corrected that, you will still find that the gradient of $c$ is not equal to the normal (not “norm”) computed from the parameterization $\mathbf u(\theta)$, but this is to be expected—there’s no such thing as the normal to the curve. The vectors will be scalar multiples of each other, but there’s no guarantee that they’ll be equal. Indeed, you could just as well parameterize the circle as $(3+\cos\theta, 2-\sin\theta)$ or $(3+\cos2\theta,2+\sin2\theta)$, both of which will produce different tangent vectors.