Determine an interval on which the graph is smooth Problem:
Given $\vec r(t) = [\frac 12 \sin{2t}] \hat i + [-\frac 14\cos{4t}]\hat j + [-\frac 14\cos{4t}]\hat k$ determine an interval on which the graph is smooth.
Solution:
My understanding of a smooth graph is that it is continuous. Based on this understanding, I assert that this particular graph is continuous on the interval $(-\infty, \infty)$. 
I suspect my assertion to be incorrect as I did not receive any credit during my exam for stating this interval. Will someone point me in the right direction?
Any advice is greatly appreciated.
 A: I believe that the source of "problems" in your exercise are of two classes:


*

*self-intersections: the curve is not simple

*Values of $t$, s.t. $r'(t)=0$: the curve is not regular
The curve under exam is 
$$r(t)=(\frac{1}{2}\sin(2t),-\frac{1}{4}\cos(4t),-\frac{1}{4}\cos(4t)),$$
with tangent vector
$$r'(t)=(\cos(2t),\sin(4t),\sin(4t)),$$
for all $t\in(-\infty,\infty)$.


*

*Analysis of injectivity and regularity


The curve $r:t\mapsto r(t)$ is simple if it is injective, i.e. there exists no pair $(t_1,t_2)$ with $t_1\neq t_2$ s.t. $r(t_1)=r(t_2)$.
In our case, 
$$\forall t\in (-\infty,\infty)\Rightarrow  r(t+k\pi)=r(t),~~k\in\mathbb Z;$$
in other words, the given curve is not simple.
However, an easy computation shows that
$$\forall t\in (-\infty,\infty)\Rightarrow  r'(t+k\pi)=r'(t),~~k\in\mathbb Z,$$
i.e. at the points of self-intersection the tangent vectors are well defined (returning to the self intersection point after a "time" $+k\pi$ we have the same "velocity" vector). This is not true, in general, for all self intersecting curves (can you visualize an example)?
If we accept self intersecting curves with regular tangent vectors, then $r$ is well defined for all $t\in(-\infty,\infty)$.
We are left with the second source of problems, i.e. values of $t$ s.t. $r'(t)=0$. If the tangent vector $r'(t)$ is equal to the zero vector $0$ for some $t=t_1$, then the curve is not regular and quantities like the curvature of $r$ at $r(t_1)$ cannot be computed. We want to avoid such possibility.
An easy computation shows that $r'(t)=0$ at $t=\pm\frac{\pi}{4}$ and $r'(t)\neq 0$ for all $t\in (-\frac{\pi}{4},\frac{\pi}{4})$.
In summary, for all intervals $(a,b)$ s.t.
$$(a,b)\subset (-\frac{\pi}{4},\frac{\pi}{4}) $$
the curve $r: (a,b)\rightarrow\mathbb R^3$, $t\mapsto r(t)$ is smooth, simple and regular.
