If there is a parametrized curve: $$r(t) = t^2\,i + ln(t)\,j + \frac{1}{t}\,k$$
then as I understand it, to find surfaces that contain the curve, you can solve for $t$ and find three equations, each missing one of the three variables, giving three equations for surfaces that contain the curve.
$$x = t^2\\y = ln(t)\\z = \frac{1}{t}$$
$$1)\,\,t = \frac{1}{z} \Rightarrow x = (\frac{1}{z})^2 = \frac{1}{z^2}$$$$2)\,\,y = ln(t) \Rightarrow t = e^y \Rightarrow x = e^{2y}$$$$3)\,\,z = \frac{1}{t} = \frac{1}{e^y}$$
Above are three surfaces that contain the curve; but are there infinitely many surfaces that contain this curve? Could you not translate that surface infinitely many times and still have the curve lie on the surface?
$\endgroup$ 11 Answer
$\begingroup$There are infinitely many surfaces indeed. An easy way to generate them is to take any three functions with the property $f(0)=g(0)=h(0)=0$. Then the surface$$ r(t,s) = (t^2 + f(s))\mathbf{i} + (\ln t + g(s))\mathbf{j} + (1/t +h(s))\mathbf{k} $$will contain your curve ($s=0$). By the way, three surfaces you have found are:
- $f(s) = s$, $g(s) =0$, $h(s) = 0$
- $f(s) = 0$, $g(s) =s$, $h(s) = 0$
- $f(s) = 0$, $g(s) =0$, $h(s) = s$
