The next question is from Do-Carmo's baby book, page 30 question 3 in section 1-6.

The question goes as follows: Show that the curvature $k(t)\neq 0$ of a regular parametrized curve $\alpha : I\rightarrow \mathbb{R}^3$ is the curvature at $t$ of of the plane curve $\pi \circ \alpha$, where $\pi$ is the normal projection of $\alpha$ over the osculating plane at $t$.

Now I guess I don't understand what is $\pi$ here, I mean I have $$\alpha(s)= (x(s),y(s),z(s))$$ assume it's in arclength parametrization, so basically I want to show that:

$$(\pi \circ \alpha (s))'' = k \cdot n_{\pi \circ \alpha (s)}\;,$$

but I am not sure here what $\pi$ equals to, I mean if its a projection on the osculating plane which is the plane of $xy$ then shouldn't it be the binormal to $\alpha$ (in which case it's $(0,0,z(s))$)?

Any hints?


  • $\begingroup$ You can get the symbol for concatenation using \circ. To put punctuation on a line after a displayed equation (instead of at the beginning of the next line), you need to include it before the closing $$ (preferably preceded by a \; to set it off from the equation). $\endgroup$ – joriki Aug 5 '11 at 8:59

Don't use coordinates to built your intuition! Use them when you need to make calculations.

Given a point $\alpha(t)$ on our curve, consider two points $X$ and $Y$ which move on the curve. The plane determined by these three points approaches a fixed plane as $X$ and $Y$ approach the point $\alpha(t)$. This plane is called the osculating plane.

They also determine a unique circle passing through all three points, and these circles approach a fixed circle as $X$ and $Y$ get closer to $\alpha(t)$. This is the circle of curvature.

It is an easy exercise to check that these are compatible with the usual definitions. With this in mind the problem reduces to showing that projecting the curve to the osculating plane at $\alpha(t)$ does not change the osculating circle. This should be obvious once phrased as follows: Given a sequence of triangles whose limit is a single point, the limit of their circumcircles is the same as the limit of their circumellipses provided the eccentricity of these ellipses goes to zero.

  • $\begingroup$ I am not sure I understand, so basically this projection is onto the plane tn of alpha? $\endgroup$ – MathematicalPhysicist Aug 5 '11 at 11:35
  • $\begingroup$ Ok, I think I have got an answer. let's assume that the osculating circle of $\alpha$, is given by the equation : $$x^2(s)+y^2(s)=\frac{1}{k^2}$$ Now under the projection of $\pi$ onto the osculatin plane we get $$\pi\circ \alpha(s)=(x(s),y(s),0)$$, so we get: $$ |\pi\circ\alpha(s)|^2 =\frac{1}{k^2}$$ So upto a sign the curvatures of both $\alpha$ and $\pi\circ\alpha$, now they have the same sign cause the orientation of both of these curves are the same. Does it need anymore tweaking? $\endgroup$ – MathematicalPhysicist Aug 5 '11 at 11:50

Although the geometric intuition given in the other response is amazing. I encountered the same problem and solved it in two (analytic) ways:

  1. We know that the Osculating Plane in $t\in I$ is the set of points $x$ in $\mathbb{R}^3$ such that $x=\alpha(t)+r_1\hat{t}(t)+r_2\hat{n}(t)$ where $\hat{t}(t)=\frac{\alpha'(t)}{\|\alpha'(t)\|}$,$\hat{n}(t)=\frac{\alpha''(t)}{\|\alpha''(t)\|}$ and $r_1,r_2\in\mathbb{R}$.

    Now for every $y\in\mathbb{R}^3$ define the projection of $x$ over $y$ to be $P_{y}(x):=\frac{x\cdot y}{\|y\|^2}y$.

    To answer your question: the projection of a vector $x$ over a plane $\pi_{_\mathcal{P}}(x)$ is obtained by computing the linear combination of the projection of $x$ to each of the spanning vectors of the plane, in this case: $\pi_{_\mathcal{P}}(x)=P_{\hat{n}(t)}(x)+P_{\hat{t}(t)}(x)=[x\cdot\hat{n}(t)]\hat{n}(t)+[x\cdot\hat{t}(t)]\hat{t}(t)$.

    Finally, if we compose $\pi_{_\mathcal{P}}\circ\alpha$, derivate that function two times and evaluate in $t$ (to get its curvature) we get the desired result.

  2. Another way to get the proyection of a point $u\in\mathbb{R}^3$ over the plane $\mathcal{P}$ is to remember that a plane is the geometric space whose points $x=(x,y,z)$ satisfy that $ax+by+cz+d=0$ where the vector $n=(a,b,c)$ is normal to the plane and $n\cdot x_0=-d$ for some $x_0\in\mathcal{P}$. If you consider the line "with foot" in $u$ and spanned by $n$, i.e., $\mathcal{L}=\{u+tn|t\in\mathbb{R}\}$, that line should intersect the plane $\mathcal{P}$ in the point $\pi_{_\mathcal{P}}(u)$. Doing some computations you should get that: $$\pi_{_\mathcal{P}}(u)=u+t_0n=u-P_{n}(u)-P_n(x_0)$$ Proceding in the same way as in 1. using $\hat{b}(t)=\hat{n}(t)\times\hat{t}(t)$ as the vector normal to $\mathcal{P}$ should give the same result.


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