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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!

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6 Answers 6

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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.

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  • $\begingroup$ This doesn't seem right. If you have $x=t^3$ $y=t^3$ you get $x^\prime(0)=0$ and $y^\prime(0)=0$ yet the graph obviously has a tangent. $\endgroup$
    – DRF
    Commented Apr 3, 2018 at 12:46
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    $\begingroup$ Note: I'm not disagreeing with the result but I am disagreeing with the explanation given the confusion the OP seems to have. $\endgroup$
    – DRF
    Commented Apr 3, 2018 at 12:56
  • $\begingroup$ @DRF Yes of course you are right, indeed I should state that I'm referring to a "possible" singularity point at the origin which is indeed is a cusp. $\endgroup$
    – user
    Commented Apr 3, 2018 at 13:14
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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...

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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 $x<0$.

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  • $\begingroup$ It looks to me line the line $y=0$ is tangent to the curve at $(x,y)=(0,0)$. $\endgroup$ Commented Apr 3, 2018 at 12:26
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    $\begingroup$ How does it follow that because you can't use one way to compute the derivative (as it doesn't work if $dx/dt=0$) you get that the derivative doesn't exist. You have the same problem in the case of $x=t^3$, $y=t^3$ yet there the derivative obviously exists. $\endgroup$
    – DRF
    Commented Apr 3, 2018 at 12:50
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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.

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    $\begingroup$ Same counterexample as elsewhere. In this case there is a cusp, but if you have $\alpha(t)=(t^3,t^3)$ you also get $\alpha^\prime(0)=(0,0)$ yet there the parametrized function is a line which does have a tangent. $\endgroup$
    – DRF
    Commented Apr 3, 2018 at 12:52
  • $\begingroup$ @DRF. The problem is it has a tangent line but not a derivative. $\endgroup$ Commented Apr 3, 2018 at 14:18
  • $\begingroup$ That I think is a crucial point to elaborate. The parametric function doesn't have a derivative in either of the cases (the original or my case) since there the derivative IIRC has a different meaning. But the parametrization of my curve can be changed. You can have $\alpha(t)=(t,t)$ which parametrizes the same curve but has a nice derivative everywhere namely $\alpha^\prime(t)=(1,1)$ . On the other hand in the case the OP has the curve doesn't have a tangent at $(0,0)$ and none of it's parametrization will have a derivative there. $\endgroup$
    – DRF
    Commented Apr 3, 2018 at 15:09
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Hint:

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

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

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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:

  1. Find an implicit form of your curve, in this case $y^2-x^5=0$.
  2. Take all terms of lowest degree, in this case $y^2=0$.
  3. The zeros of this equation specify the tagent cone.

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

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