Identify 2 points $(x,y)$ at which the curve is not smooth (Parametric Equation)

Question

Original parametric equations:

$x(t) = 7t - 7 \sin(t)$

$y(t) = 7 - 7 \cos(t)$

1st derivatives:

$x'(t) = 7 - 7 \cos(t)$

$y'(t) = 7 \sin(t)$

I've come to the conclusion that the curve is not smooth when $t = 2k\pi$ ($x$-intercepts). I'm just not entirely sure why though. My best guess for why $t = 2k\pi$ is because as $t$ increases toward $2\pi$, the McDonalds like arches on the graph drop toward $y = 0$ very rigidly/sharply, rather than smooth.

To me my assumption seems a little to literal, but I can't really come up with any other reasons when I analyze the formal criteria of a smooth curve where "a curve is called smooth when f' and g' are continuous on an interval (I) and not simultaneously 0, except possibly at the endpoints of interval (I)."

Considering that $t = 2\pi$ is a full rotation of the unit circle, I would assume that the x-intercept at $t = 2k\pi$ can technically be an endpoint of an interval, although $\cos/\sin$ are indefinitely continuous otherwise. Any thoughts/solutions are appreciated!

Answer 1

Hint: If we have $y=f(t)$ and $x=g(t)$ then: $$\frac {dy}{dt}=f'(t)$$ and $$\frac {dx}{dt}=g'(t)$$ Hence we have- $$\frac {dy}{dx}=\frac {f'(t)}{g'(t)}$$ Now note that a function is said to be differentiable if it's left hand and right hand derivatives are equal, hence to compare it is necessary that they be finite. You will see that- $$\frac {dy}{dx}=\frac {\sin t}{1-\cos t}=\cot(\frac t2)$$ This means that at $t=2k\pi$, we have the derivative tending towards $\infty$. Hence the sharp change.

Also, it might be interesting to consider, that the curve you've traced out here is called a cycloid, and this would be the locus traced by a point on a disc of diameter $7$ units which is pure rolling on the ground(which you can consider the $x$-axis).

Answer 2

For the sake of this discussion, a curve is "smooth" at a point $x$ if $f'(x)$ exists. $f'(x)$ can alternatively be written as $\frac{dy}{dx}$, and we know that $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ Therefore, for this problem, we find the values of $t$ at which $\frac{dy}{dx}$ DNE. That occurs precisely when $\frac{dx}{dt}=0$ but $\frac{dy}{dt} \neq 0$. Those values of $t$, as you discovered, are at $t = 2\pi k$ for $k \in \mathbb{Z}$.