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I'm working from Do Carmo's book Differential Geometry, and I was a bit confused about one question - 1.3.5 in particular.

The question asks:

Let $\alpha:(-1,\infty) \to \mathbb{R}^2$ be a parametrized curve given by

$\alpha(t) = (\frac{3at}{1+t^3},\frac{3at^2}{1+t^3})$ (presumably $a \in \mathbb{R}$)

a) Prove that, at $t=0$, $\alpha$ is tangent to the $x$-axis

At first, I thought this means that we had to show that $\alpha(0)$ was on the $x$-axis and $\alpha'(0) \neq 0$, but I realized that $\alpha(t)$ is the tangent curve here, not $\alpha'(t)$. Now, I'm not sure what exactly I should do. $\alpha(0) = (0,0)$ so doesn't that disqualify it from being a tangent line at $t=0$ according to Do Carmo? I assume I'm supposed to integrate $\alpha(t)$ and check that the value attained at $t=0$ lies on the $x$-axis?

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    $\begingroup$ No, you need to compute $\alpha'(0)$. $\endgroup$ – Ted Shifrin Jan 27 '14 at 4:13
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    $\begingroup$ Really you only need to show that $\alpha_{2}'(t) = (\frac{3at^2}{1+t^3})' = 0$. Why? (one should also show that $\alpha_{1}'(t) \neq 0$, if you want to be precise). $\endgroup$ – Chris K Jan 27 '14 at 4:28
  • $\begingroup$ @TedShifrin what confuses me here is that, according to the book, $\beta'(t)$ is said to be the tangent curve to $\beta(t)$ But here, $\alpha(t)$ itself is said to be the tangent curve. I found that $\alpha'(0) = (3a,0)$ $\endgroup$ – Lost Jan 27 '14 at 4:41
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    $\begingroup$ You are confusing "alpha is tangent" and "alpha is the tangent". As others have told you, all you have to do is show that $\alpha(0) = (0,0)$ and that $\alpha'(0)$ is in the direction of the $x$-axis. $\endgroup$ – bubba Jan 27 '14 at 6:28
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    $\begingroup$ English is a tricky language. If I say "$\alpha$ is tangent to the $x$-axis at the point $P$" I mean that the the tangent line of $\alpha$ at the point $P$ is the $x$-axis. It's a statement about the shape of the curve $\alpha$ at the point $P$. If I say "$\beta$ is the tangent curve of $\alpha$", then (according to the definition you gave) I mean that $\beta(t) = \alpha'(t)$ for all $t$. This is a statement about the relationship between entire curves $\alpha$ and $\beta$. Different, right? $\endgroup$ – bubba Jan 27 '14 at 6:59
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senter image description here

The curve looks to be the Folium of Descartes. Although it is parametrized so that a self-intersection is avoided, I've shown the whole curve. It does appear to be tangent to the x-axis. It's also tangent to the y-axis at the origin but that part of the curve would not be present with the parametrization you presented.

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  • $\begingroup$ That is the picture the book provided, but I'm looking to prove it rigorously. $\endgroup$ – Lost Jan 27 '14 at 4:39
  • $\begingroup$ alpha(t) = (x(t),y(t)) , alpha'(t) = (x(t)' , y(t)') . $\endgroup$ – Alan Jan 27 '14 at 5:26

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