Can a curve with singular point be a regular curve?

[Cauchy1789]

Homework Statement

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

Homework 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.

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]

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]

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 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

 

[Office_Shredder]

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.