# Parametrized curve tangent to a line

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

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?

• No, you need to compute $\alpha'(0)$. – Ted Shifrin Jan 27 '14 at 4:13
• 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). – Chris K Jan 27 '14 at 4:28
• @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 $\alpha'(0) = (3a,0)$ – Lost Jan 27 '14 at 4:41
• 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. – bubba Jan 27 '14 at 6:28
• 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? – bubba Jan 27 '14 at 6:59