Can a curve with singular point be a regular curve?

**Cauchy1789** · Feb5-09, 04:14 PM

1. The problem statement, all variables and given/known data

Given a parameterized curve $LaTeX Code: \\alpha:(a,b)\\rightarrow \\mathbb{R}^2$ , show that this curve is regular except at t = a.

2. Relevant equations

I know that according to the defintion that a parameterized curve $LaTeX Code: \\alpha: I \\rightarrow \\mathbb{R}^3$ is said to be regular if $LaTeX Code: \\alphasingle-quote(t) \\neq 0$ $LaTeX Code: \\forall t \\in I.$

3. The attempt at a solution

I have read that any curve which has a point where the tangent vector is zero cannot be a regular curve, so how is it even possible to just forget about that singular point in such a proof?

Best regards
Cauchy

**Office_Shredder** · Feb5-09, 04:27 PM

Do you have more information about alpha? Like, what the formula is?

Also, a isn't in your interval, so you're fine anyway?

**Cauchy1789** · Feb5-09, 05:16 PM

Originally Posted by Office_Shredder

Do you have more information about alpha? Like, what the formula is?

Also, a isn't in your interval, so you're fine anyway?

Hi

First of all its suppose be t = p and $LaTeX Code: p \\in I$ and the curve is defined as

$LaTeX Code: \\alpha(t) = (x(t),y(t))$ a parameter curve.

having a singular point on a regular curve isn't that a contradiction?

Sincerrely

Cauchy

**Office_Shredder** · Feb5-09, 05:55 PM

show that this curve is regular except at t = a.

Where a is not in (a,b) so it doesn't even fail the regularity condition.

Anywho, if someone says "Show a curve is regular everywhere except point p" it's like if someone said "show f(x)=|x| is differentiable except at 0" By definition, a differentiable function is differentiable everywhere, but you understand what they mean anyway. Same principle applies here.