continuitycurves
What is it that distinguishes C continuity (parametric) from G continuity (geometric)?
Give an example curve that is C1 continuous and NOT G1 continuous. Or one which is G1 continuous and not C1 continuous.
I couldn't interpret the different from this :/
As Jose27 noted, uniformly continuous functions need not be differentiable even at a single point.
It is true that if $f$ is defined on an interval in $\mathbb R$ and is everywhere differentiable with bounded derivative, then $f$ is uniformly continuous. In fact, it follows from the Mean Value Theorem that such an $f$ is Lipschitz, which is much stronger.
However, if $f$ is uniformly continuous and everywhere differentiable, then $f$ need not have bounded derivative. Jose27 mentions $\sqrt x$, which would work as an example on the interval $(0,\infty)$. The function $$f\left(x\right)=\begin{cases}x^2\sin\left(\frac{1}{x^2}\right) & :x\neq 0\\ 0 &:x=0\end{cases}$$ is uniformly continuous on any bounded interval such as $(-1,1)$, but has unbounded derivative near $0$. There are also examples where $f'$ is bounded on bounded intervals, but unbounded on $\mathbb R$, while $f$ is uniformly continuous. You can show that any continuous function $f$ on $\mathbb R$ such that $\lim\limits_{|x|\to \infty}f(x)=0$ is uniformly continuous, and using this fact you can see that Nate Eldredge's example here of $\sin(x^4)/(1+x^2)$ provides such an example. Another source of examples is the question
Why if $f'$ is unbounded, then $f$ isn't uniformly continuous?
Consider the parameterized curve $\gamma(t)=(t^3,t^2)$. This parameterization is clearly $C^1$, because each component is (infinitely!) differentiable.
However, the graph of the curve looks like this:
As you can see, the left and right tangent vectors at the origin (where $t=0$) do not match, so the curve is not $G^1$.
This is possible because $\gamma'(0)=\left<0,0\right>$. That is, the parameterization comes gradually to a stop at $t=0$, which allows it to smoothly traverse a geometrically unsmooth curve.
Best Answer
G1 continuous
geometrically continuous, tangents are collinear, may not have same magnitude
C1 continuous
parametric continuity ensured, notice that the magnitude of the vectors are equal
Note: I'm using 'Bezier' curves, so tangents will not overlap (2nd tangent is negative in Bezier), instead face in the opposite direction (as opposed to overlapping tangents in Hermite)