$\frac{dy/dt}{dx/dt} \text{ at } t = a \text{ or } \lim_{t \to a} \frac{dy/dt}{dx/dt} \text{?}$

Question

Take an example of parametric equation:

\begin{cases} x = t^3\\ y = t^6 \end{cases}

Obviously the formula $\displaystyle \left. \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \right.$ does not work at $t=0 \Big(\displaystyle \frac{dy/dt}{dx/dt} = \frac{0}{0}\Big)$.

But at $t = 0 \text{, } (x, y) = (0, 0)$, and $\displaystyle \frac{dy}{dx} = 0.$

Answer 1

The derivative is not defined at x=0, but the derivative exists in terms of a limit. The point (0,0) is what's called a "cusp". It is a point where the particle momentarily comes to a standstill and then picks up speed again.