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Let $C$ be the curve associated to a regular, simple path $\theta:[0,l]\rightarrow \Bbb R^2 $; also assume that $((x'(s))^2+((y'(s))^2=b^2$ and let $S$ be the surface generated by the circles of radius $b$, orthogonal to, and centered in points of the curve $\rho(s)=(\theta(s),0) $.

I need to obtain a parametrization for $S$. For this, I was suggested to note that the normal vector to the plane that contains the circle is $(y'(s),-x'(s),0)$, so that the vector that passes trough $(x(s),y(s),0)$ and a point in $S$ must be orthogonal to this vector.

I haven't been able to find a parametrization for $S$, mainly because I don't understand the geometric figure being described above. I'd appreciate any help.

The purpose of this exercise is to then use surface integrals to compute the area of $S$, and then, defining a 'torus' as an special case of the latter construction for a circle centered in the origin with radius $a$, with $a>b$ and calculating its area.

Weierstraß Ramirez
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  • Are you sure the normal to the plane is $(y',-x',0)$? If the plane of the circle is orthogonal to the curve, then its normal should be parallel to the tangent $(x',y',0)$, and $(y',-x',0)$ would be a vector along the plane of the circle. –  Dec 04 '14 at 15:47

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Here's a parameterization: writing $$ \theta(s) = (x(s), y(s)) $$ I'm going to define $$ H(s, t) = ( x(s), y(s), 0) + \cos(t) (0, 0, b) + \sin(t) (-y'(s), x'(s), 0) $$ which we can regard as having the form $$ H(s, t) = ( x(s), y(s), 0) + \cos(t) \mathbf v + \sin(t) \mathbf w $$ where $\mathbf v$ and $\mathbf w$ are orthogonal vectors of length $b$, and they evidently span the plane orthogonal to $(x'(s), y'(s), 0)$, the tangent to the "centerline" curve.

For fixed $s$, as $t$ varies from $0$ to $2\pi$, this describes a circle of radius $b$ in the plane orthogonal to $(x'(s), y'(s), 0)$, and containing the point $(x(s), y(s), 0)$.

For $t = 0$, as $s$ varies, $H(s, 0)$ describes a curve parallel to the centerline curve $(x(s), y(s), 0)$, offset by $b$ units in the $z$ direction.

In the event that the curvature $\kappa(s)$ of the original curve $\theta$ is larger than or equal to $1/b$ in absolute value, this surface will have singularities (i.e., the derivative of the map $H$ will not have rank 2 at every point of its domain). If the absolute curvature of $\theta$ is strictly bounded by $1/b$, however, then $H$ will be a nonsingular parameterization.

One more small note about this parameterization: the partials of $H$ at any point $(s, t)$ actually end up being perpendicular, so the length of their cross product (which comes up in various integrals) is just the product of their lengths, which is $b^2$.

John Hughes
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  • Could I verify from this that when $C$ is a circle of radius $a$, then its parametrization is given by $H(u,v)=((a+b \cos{u})\cos v,(a+b \cos{u})\sin v, b \sin u )$ ? – Weierstraß Ramirez Dec 14 '14 at 04:44
  • That looks exactly right to me. – John Hughes Dec 14 '14 at 04:48
  • Alright, I know the $a$ term most respond the fact that the 'centerline' has radius $a$, however I'm not sure how I don't know how to get it from your equations. – Weierstraß Ramirez Dec 14 '14 at 04:52
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    The torus you're looking for is an "offset surface" from a centerline-curve $C$, which is (according to your question) a curve in the $xy$-plane. You want that to be a radius $a$ circle. Placing it at the origin, the standard description of such a curve is $C(t) = (a \cos t, a \sin t)$. If you plug this into my formula for $H$, you get the expression you wrote down three comments before this one. – John Hughes Dec 14 '14 at 13:18