I have the curve $\alpha(t):(-1,\infty) \rightarrow R^2$ given by $\alpha(t)= ((\frac{nat}{1+t^3}), (\frac{nat^2}{1+t^3}))$ with $n$ a natural an $a$ a constant both of them fixed. I need to prove that this curve is tangent to the $x$ axis in $t=0$.

What I did:

It's evident that the only point where the curve touch the $x$ axis is in t=0, so $\alpha'(t) = (\frac{na(1+t^3 - nt^2(nat)}{(1+t^3)^2},\frac{2nat(1+t^3)-nt^2(nat^2)}{(1+t^3)^2})$ and $\alpha'(0) = (na,0)$, so the tangent of the curve is parallel to the $x$ axis and then the curve is tangent to the $x$ axis in $t=0$.

It's this all right?

  • $\begingroup$ Yes, that's right $\endgroup$ – Sebastian Schulz Mar 2 at 19:09

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