Showing that a specific curve is regular. 
Define the curve by $c(t):=(sin(pt)+r)(cos(qt),sin(qt))$ for $p,q \in \mathbb{Q}$ and $r\in \mathbb{R}$. Determine for which $p,q$ is the curve regular, i.e. $c'(t) \neq (0,0)$ for any $t\in \mathbb{R}$, where $c'(t)$ denotes the derivative of $c(t)$ w.r.t. $t$

My ideas: First of all, it applies $$c'(t)=(q\cdot sin(qt)(-sin(pt)-r)+p\cdot cos(pt)cos(qt),  q\cdot cos(qt)(sin(pt)+r)+p\cdot cos(pt)sin(qt))$$
I tried to use the substitution $$x_1:=cos(pt), x_2:=cos(qt), y_1:=sin(qt), y_2:=sin(pt)+r$$ and then I get the following equations $p\cdot x_1 x_2 = q \cdot y_2 y_1$ and $q\cdot x_2 y_1=-p \cdot x_1 y_2$. By using the cases $y_1=0$ and $y_1 \neq 0$, I would have to go through several cases by assumption on $p,q,r$. 
Is there a nicer approach to determine $q,p,r$?
 A: Note that
$$
c'(t)=(\sin(pt)+r)(\cos(qt),\sin(qt))'+(\sin(pt)+r)'(\cos(qt),\sin(qt))
$$
and that, if $q\neq 0$,
$$
(\cos(qt),\sin(qt))' \perp (\cos(qt),\sin(qt)).
$$
Hence, $c'(t)=0$ implies
$$
\sin(pt)+r=0, \quad \sin(pt)'+r=0.
$$
One can see that, if $r, q\ne 0$, then the above equation has a root.
A: With
$c(t) = (\sin(pt) + r)(\cos(qt), \sin(qt)), \tag{1}$
we have
$c'(t) = (p\cos(pt))(\cos(qt), \sin(qt)) + q(\sin(pt) + r)(-\sin(qt), \cos(qt)). \tag{2}$
Note that
$(\cos(qt), \sin(qt)) \cdot (-\sin(qt), \cos(qt)) = 0;$
$(\cos(qt), \sin(qt)) \cdot (\cos(qt), \sin(qt)) =  (-\sin(qt), \cos(qt)) \cdot  (-\sin(qt), \cos(qt)) = 1; \tag{3}$
thus
$\Vert c'(t) \Vert^2 = c'(t) \cdot c'(t) = p^2 cos^2(pt) + q^2(\sin(pt) + r)^2. \tag{4}$
$c(t)$ is regular everywhere (that is, for all $t \in \Bbb R$) if and only if $c'(t) \ne 0$ everywhere, which, in the light of (4), may be taken to mean that $c(t)$ is regular if and only if there is no $t \in \Bbb R$ such that
$p\cos(pt) = 0. \tag{5}$
and
$q(\sin(pt) + r) = 0. \tag{6}$
If $q = 0$, we thus see that $c(t)$ cannot be regular for any $p \in \Bbb R$, since $p\cos(pt) = 0$ whenever $p = 0$ or $pt$ is an odd multiple of $\pi / 2$, i.e. when $t = (2n + 1)\pi/2p$ for $n \in \Bbb Z$.  When $q \ne 0$, (6) yields
$\sin(pt) + r = 0; \tag{7}$
when $pt = (2n + 1)\pi/2$, we have $\sin (pt) = \pm 1$, so there is no $t \in \Bbb R$ at which (5) and (6) hold simultaneously unless $r = \pm 1$; $c(t)$ is in this case regular for $r \ne \pm 1$.   When $r \in \{ -1, 1\}$, then $c(t)$ is not regular at the points $t$ where
$pt = \dfrac{(4n + 1)\pi}{2}, \; \; r =- 1; \tag{8}$
$pt = \dfrac{(4n + 3)\pi}{2}, \; \; r = 1. \tag{9}$
We can summarize these results as follows:  
$q = 0, p = 0: \; \; \text{not regular for any} \; t \in \Bbb R;$
$q = 0, p \ne 0: \; \; \text{not regular for} \; t = \dfrac{(2n + 1)\pi}{2}; $ 
$q \ne 0, p = 0, r = 0: \; \; \text{not regular for any} \; t \in \Bbb R;$
$q \ne 0, p = 0, r \ne 0: \; \; \text{regular for all} \; t \in \Bbb R;$
$q \ne 0, p \ne 0, r \ne \pm 1: \; \; \text{regular for all} \; t \in \Bbb R;$
$q \ne 0, p \ne 0, r = -1: \; \; \text{not regular for} \; t = \dfrac{(4n + 1)\pi}{2p};$
$q \ne 0, p \ne 0, r = 1: \; \; \text{not regular for} \; t = \dfrac{(4n + 3)\pi}{2p}.$
A glance at the above shows that whether $p, q \in \Bbb Q$ or not does not bear significantly on the regularity of $c(t)$.
Lastly, not to put too fine a point on it, there are a couple of erroneous statements in Ma Ming's answer and comment.  The assertion (from the comments) that $q = 0$ implies $c(t)$ is constant is false; for $q = 0$ we have $c(t) = (\sin(pt) + r)(1, 0)$; this "curve" lies entirely within the line $y = 0$ (assuming we use $x$-$y$ coordinates in the plane $\Bbb R^2$), but it is not constant for $p \ne 0$; it's regularity properties are found under the cases $q = 0$ in the above list.  Furthermore, $c(t) = 0$ implies $(\sin(pt) + r)' = p\cos(pt) = 0$, not $(\sin (pt))' + r = p\cos(pt) + r = 0$.  Lastly, the hypothesis $q, r \ne 0$ only implies the equations have a zero it $t$ when $p \ne 0$, $r = \pm 1$; the above list correctly enumerates all cases.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
