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?

  • 3
    $\begingroup$ No, you need to compute $\alpha'(0)$. $\endgroup$ – Ted Shifrin Jan 27 '14 at 4:13
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.