Differentiability of curve vs. linear approximation of curve's image set I'm currently trying to understand why in differential geometry it's usually required for a curve $c:I \subset \mathbb{R} \to \mathbb{R}^n$ to be regular, that is $\|c'(t)\| \neq 0$ for all $t \in I$. I was thinking about the image of the smooth curve $c: [0, 2\pi] \to \mathbb{R}^2$ defined by $t \mapsto (\sin t,\; \sin t)$ and I think it shows that we need some extra conditions on $c$ other than being differentiable (this curve is not regular, of course). Wolfgang Kühnel motivates the requirement in his book "Differential Geometry: Curves - Surfaces - Manifolds" as follows:

[...] Differentiability of a map just means that it can be linearly
  approximated. For the image set, however, this no longer needs to be
  the case. From a geometrical point of view it makes sense to require
  that the image curve can be approximated by a line at each point,
  i.e., to require that the image curve has a tangent as a geometrical
  linearization at every point. This means that the derivative of c must
  be non-vanishing.

This confuses me a lot. Where's the difference of linearly approximating $c$ or its image? Let's consider the case of plane curves for a moment. If $f: I \to \mathbb{R}$ is differentiable at $x_0$ there exists a unique tangent at $x_0$. Visually, I'm thinking about the tangent at $(x_0, f(x_0))$ with respect to the graph $G_f = \{\;(x, f(x)\,)\;|\;x \in I\; \}$, which is a subset of $\mathbb{R}^2$. The graph of $f$ determines a regular plane curve, since $\|\frac{d}{dx} (x, f(x))\| = \sqrt{1 + f'(t)^2} \neq 0$. I don't really see in which way this fails if we have different coordinate functions.
Can someone please elaborate a bit on what Kühnel is talking about?
EDIT: To clarify a bit: As seen in Aaron's example, we can have curves that are smooth but also have a cusp (something a real-valued differentiable function of one variable cannot have). Thus it's only natural to seek for conditions that produce "well-behaved" curves. Apparently regularity is the property we're looking for in this case. The question is: WHY and HOW does $c'(t) \neq 0$ prevent this pathological behavior? What is the intuition behind this condition. It's easy to look at a curve like $c(t) = (t^3, t^2)$ and conclude that it is not regular at $t = 0$, but it's not as easy to understand why regularity alone prevents this kind of behavior.
 A: Your definition of regular curve seems off. 
Definition: 

A differentiable curve $c : I \to \mathbb R^n$ is regular when $c'(t) \neq 0$ for every $t \in I$. 

An example where this fails is the curve $c : I  \to \mathbb R^2$ defined by $c(t) = (t^3,t^2)$. Then $c'(t) = (3t^2,2t)$ and $c'(0) = 0$ this curve is not regular and it looks like this

Roughly speaking, a regular curve doesn't have any pointy edges or unexpected bumps.
And perhaps the nature of your misunderstanding lies on distinguishing the trace of a curve with its graph. What is shown in the picture above is called the trace of $c$, that is, the image $c(I) \subset \mathbb R^2$. The graph of $c$ would be the collection of all pairs $(t,c(t))$ and therefore a subset of $\mathbb R^3$. 
A: Curves are taken to regular because we get nice forms for formulas when we can parametrize curves by arc-length. Recall that if $\gamma(t)$ is regular,then we can parametrize $\gamma$ with the unit-speed parameter $s$, $(\|\gamma'(s)\| = 1$ for all $s$). 
