Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks.

share|improve this question
    
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). –  joriki Aug 5 '11 at 8:59

1 Answer 1

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.

share|improve this answer
    
I am not sure I understand, so basically this projection is onto the plane tn of alpha? –  MathematicalPhysicist Aug 5 '11 at 11:35
    
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? –  MathematicalPhysicist Aug 5 '11 at 11:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.