Finding the tangent to a parametric curve $(t^3, t^5)$ at $(0,0)$

Question

The curve $(t^3, t^5)$ at that point $(0,0)$ does not have a tangent vector as when you work it out, you will arrive at $(0,0)$.

Question: How can you find a new parametrisation for the curve such that you get a tangent vector at $(0,0)$. Is this even possible.

My attempts:

Changing the curve's position by taking $t \to t+1$ won't work because it just changes the position of the point we're trying to find the tangent on
A trig sub may be possible, however I have had no luck

Any help or guidance would be greatly appreciated!

Answer 1

$\frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}}$

$\frac {5t^4}{3t^2} = \frac 53 t^2$

As noted the derivative appears to not exist when $t=0$ But, the limit exists.

$\lim_\limits{t\to 0} \frac 53 t^2 = 0$

$y = 0$

An alternative would be to say

$y^3 = x^5 \\ y = x^\frac 53\\ y' = \frac 53 x^\frac 23$

Answer 2

$(0,0) $ is a singular point. but

$$t^3=0+0t+0t^2+t^3+0t^4+0t^5+t^5\epsilon(t)$$ and $$t^5=0+0t+0t^2+0t^3+0t^4+t^5+t^5\epsilon(t)$$ so

$(0,0)$ is an inflexion point with $\vec{i}=(1, 0) $ as the tangent vector.