Does the curve $(t^2, t^5)$ have a tangent at the origin? My question is whether or not the curve $x(t) = t^{2}, \ y(t) = t^{5}$
has a tangent at $(x, y) = (0, 0)$. 
I don't really know what to do if both $dy = dx = 0$, so I tried to take the limit $\lim_{t \to 0} \frac{dy}{dx} = \lim_{t \to 0}\frac{5t^{4}}{2t}$ which is zero. But does this mean the curve has a tangent, with slope $0$? My book is very unclear with this situation. Thanks!
 A: Note that since


*

*$x'(0)=y'(0)=0$


we can conclude that the origin is  a possible singular point and that is indeed precisely the case since $\frac {dy}{dx}$ is not defined at the origin.
A: This is what I did,
From the parametric points, we can derive the actual equation as, $y^2=x^5$,
Now, If we differentiate it,
$$2y\dfrac{dy}{dx}=5x^4$$
$$\implies \dfrac{dy}{dx}=\dfrac{5x^4}{2y}$$
Now, The limit of this function, as $(x,y)\rightarrow(0,0)$ does not exist which you can see from the graph of $y^2=x^5$ graph
You can draw this graph with some substitution of values too, let me give you something to start on,
In, $$y^2=x^5\,\,,\,\, y^2>0 \implies x^5>0 \implies x>0 $$
Using such relations, you roughly draw the curve, 
As you may have drawn, you can see that the function is not differentiable on $(0,0)$ therefore you cannot draw a tangent to the curve at that point.
Hope this answers your query...
A: Note that $$\frac{dx(t)}{dt}=2t;\quad\frac{dy(t)}{dt}=5t^4\implies\frac{dy}{dx}=\frac52t^3$$ is true only when $\frac{dx(t)}{dt}\neq0$, which is not true in this case, so there is no derivative at the origin.
As an example, consider the function $y=x\sqrt x$. It satisfies $x(t)=t^2$ and $y(t)=t^5$, but $\sqrt x$ is not defined when $x<0$.
A: The curve is $\alpha(t) = (t^2, t^5)$. Then $\alpha'(t)=(2t, 5t^4)$. In particular, $\alpha'(0) = (0,0)$. Clearly there is no unit tangent vector at $t=0$ even if there is still a tangent line there.
Notice that near, but not at, $t=0$, the unit tangent vector is
$$T(t) = \dfrac{(2t, 5t^4)}{\sqrt{4t^2+25t^8}} 
       = \dfrac{t}{|t|} \dfrac{(2, 5t^3)}{\sqrt{4+25t^6}} $$
So, approacing $0$ from the right and the left, we get
$$\lim_{t \to 0^+}T(t) = (1,0)
  \qquad \text{and} \qquad 
  \lim_{t \to 0^-}T(t)=(-1,0)$$
This implies that the curve is tangent to the line $y=0$.
What you have is a cusp at $t=0$. Wikipedia has a gobbledygookie definition  here and Wolfram Mathworld has a user-friendly definition here.
A: Hint:
$$\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}$$
$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$$
A: Other answers make clear why there does not exist a tangent at the origin: there is a singularity.
However, from a graphical point of view, we expect that the line $y=0$ has a special meaning. This line is the tangent cone of your curve:  https://en.wikipedia.org/wiki/Tangent_cone
You can compute this tangent cone at the origin as follows:


*

*Find an implicit form of your curve, in this case $y^2-x^5=0$.

*Take all terms of lowest degree, in this case $y^2=0$.

*The zeros of this equation specify the tagent cone.


(The general case can be computed by shifting to the origin.)
