# Can a curve with singular point be a regular curve?

1. Feb 5, 2009

### Cauchy1789

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

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

2. Relevant equations

I know that according to the defintion that a parameterized curve $$\alpha: I \rightarrow \mathbb{R}^3$$ is said to be regular if $$\alpha'(t) \neq 0$$ $$\forall t \in I.$$

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

2. Feb 5, 2009

### Office_Shredder

Staff Emeritus

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

3. Feb 5, 2009

### Cauchy1789

Hi

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

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

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

Sincerrely

Cauchy

4. Feb 5, 2009

### Office_Shredder

Staff Emeritus
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.