# Parametric equations of a cycloid

Given a parametric equation of a cycloid ($t \in R$):

$$x(t)=r(t-\sin(t)); \\ y(t)=r(1-\cos(t)).$$

A vector $v=(x'(t),y'(t))$ if is not equals to zero then is a tangent vector to the curve at $(x(t),y(t))$. Given that $||v||$ is a vector norm and that a unit tangent vector (a tangent vector with a length $1$) is:

$$T=\frac{v}{||v||}$$ to a curve at the same point.

Are there points on the curve at which we could not construct vector $T$ (means that $T$ could not exists at those points). If so what does the curve look like at those points?

My solution: My guess that the only way for which T could not exists is if the vector norm $||v||$ is $0$. Hence equating $||v||= 0$. And from there solve for the value of $t$. But, I do not know how how to proceed from here.

I got x'(t)=r(1-cos(t)) y'(t)=r(sin(t))

hence ||v||= sqrt(x'(t)^2+y'(t)^2) =0

Which leads to the expression 2-2cos(t)=0

solving it t=0

Is this correct?

Do what the question instructs: given \begin{align*} x(t) &= r(t-\sin t) \\ y(t) &= r(1-\cos t), \end{align*} compute the vector $$\boldsymbol v = ( x'(t), y'(t) ).$$ For what values of $t$ is $\boldsymbol v = (0,0)$?
• Are you sure that you're computing the derivative properly? You ought to be able to show that $t$ is an integer multiple of $2\pi$. – Sammy Black Aug 20 '14 at 8:05
• What about $t = 2\pi$? Does that work too? How about $t = -4\pi$? You are not restricted to $0 \le t < 2\pi$. – heropup Aug 20 '14 at 13:54
This is quite expected at all $t = 2 \pi n$ points which are singular. The tracing point P on a circle radius r rotates around itself at zero velocity at the cusp of cycloid when circle rolls on line y=0 touching it at these instances.