What is $N(t)$ in the definition $E(t) = z(t)+ \frac{N(t)}{\kappa} $ for an evolute?

Question

I am currently reading up on Evolutes in Differential Geometry and in the definition it gives the equation:

$$ E(t) = z(t)+ \frac{N(t)}{\kappa} $$

Where $\kappa$ is the curvature. However, in the specific book that I am reading, for some reason there is no definition for N(t). By searching stack exchange I have found the formula to be

$$ N(t)= \frac{ z'(t) \times z''(t) \times z'(t)}{ \Vert z'(t) \times z''(t) \times z'(t) \Vert } $$

I was wondering if someone could clear up exactly what this N(t) formula is doing for me? My own intuition says that it is simply a normalised tangent vector of the curve but I am hoping to clarify.

Many thanks

score 2 · Accepted Answer · answered Apr 15 '21 at 06:34

2

$N(t):= T'/\kappa$ is the unit normal to the given curve (see here ).

If you have Elementary Differential Geometry by O'Neil (revised 2nd edition) then you'll see the evolute appears in question 13 of section 2.4 (page 79), and $N(t)$ is defined at the start of section 2.3 (page 59). Also see a brief discussion here

answered Apr 15 '21 at 06:34

Theo C.

1,282

2

...With a special caution taken for the resp. sign of $\kappa$ vs. orientation of $N(t)$. – Jean Marie Apr 15 '21 at 06:43
Thank you very much Theo C! I do not have that book but it is available on my university database so I will take a look – T.Omalley Apr 15 '21 at 06:55

What is $N(t)$ in the definition $E(t) = z(t)+ \frac{N(t)}{\kappa} $ for an evolute?

1 Answers1