# Continuity of parametric and geometric continuity

We know that in parametric continuity, $$C^1$$ continuity is two successive curve section $$C_1$$ and $$C_2$$ has first parametric derivative is same. That means tangent vector $$t_1$$ is same for both $$C_1$$ and $$C_2$$ and has same direction and magnitudes are same. Like $$C_1'(t=1)=C_2'(t=0).$$

Confusion:1 My question is the above image is right for $$C^1$$ continuity w. r. t tangent $$t_1$$ of $$C_1$$ and $$C_2?$$

But geometric continuity $$G^1$$ continuity is two successive curve section $$C_3$$ and $$C_4$$ has first parametric derivative is proportional to each other. That means tangent vector $$t_2$$ and $$t_3$$ has same direction for both curve sections $$C_3$$ and $$C_4$$ and their magnitudes may or may not be same. Like: $$C_3'(t=1)=(a,b,c),C_4'(t=0)=(ka,kb,kc).$$

That means two tangents $$t_2$$ and $$t_3$$ are parallel to each other. Like:

Confusion:2 My question is above image is right, to show the parallelity of the tangents $$t_2$$ and $$t_3$$ for curve sections $$C_3$$ and $$C_4?$$

We know that $$C^1$$ and $$G^1$$ continuity means $$C^0$$ and $$G^0$$ respectively. But read from (shown below)internet which showing $$C^1$$ and $$G^1$$ continuity, but $$C^0$$ and $$G^0$$ continuity not holding because $$r(t=1) =(-1,1) {\neq}n(t=0)=(1,1).$$

Confusion:3 My question is how can I say $$C^1$$ or $$G^1$$ continuity hold inspite of $$C^0$$ and $$G^0$$ not holding?

• Closing here since it is a cross-site duplicate: math.stackexchange.com/questions/4302092/… Nov 10 at 15:45
• @wychmaster please open the post, in "Math stack exchange" I didn't get any answer till now.
– user17488
Nov 10 at 18:09
• Reopening the question, since the duplication has been removed from Math SE Nov 12 at 8:08

I would rather draw the $$G_1$$ example something like this:
This makes it clear that $$t_2$$ and $$t_3$$ are parallel, but have different lengths in general. (They both start at the same point, but $$t_3$$ is longer.) The way you have drawn it, they have the same length but slightly different directions—they don't quite look parallel.
The standard definition of the $$C^k$$ smoothness classes implies that any $$C^k$$ is contained in all the lower-numbered classes $$C^{k-1}, C^{k-2}, \ldots, C^0$$. So if a curve is in $$C^k$$ then it is also in all the lower classes down to $$C^0$$. The book you found is giving a bad example. Its notation also looks inconsistent/wrong from what you posted, so I would not trust that particular book overly much.
Yes, your understanding of $$C1$$ and $$G1$$, as shown in your drawings, is roughly correct: $$C1$$ means equal derivative vectors, and $$G1$$ means parallel derivative vectors.
Regarding your last question: people who write books and papers and internet articles are, of course, free to make whatever definitions they want. But I think most people would define a curve joint to be $$C1$$ if both positions and first derivative vectors match across the join. In other words, most people would define $$C1$$ to include $$C0$$.