Parametric curves - derivate at the given point (tangent line)

Question

$x(t) = t^2, $ $y(t) = t^5$

I want to find derivate at point $(0,0$) and tangent line if derivate exists.

I did it following way:

$\frac{dy}{dx}\frac{5}{2}t^3$ if i put in $t=0$ i will get $0$ - does it mean that derivate at $(0,0)$ exists and tangent line is $y=0$?

Answer 1

Here is a graph of your given curve and its derivative. At $t =0$ the derivative does not exist as the left hand derivative and the right hand derivative do not match.

Answer 2

Hint:we have $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{5t^4}{2t}=\frac{5}{2}t^3$$

Answer 3

You’ve simplified the expression for ${dy\over dx}$ too soon. In full, it’s ${5t^4\over2t}$. This is equal to $\frac52t^3$ unless $t=0$, where the expression is undefined. That’s exactly the situation at $(0,0)$, so you can’t use the value of $\frac52t^2=0$ directly.