Finding the number of surfaces containing a curve given its parametrization 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?

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:

*

*$f(s) = s$, $g(s) =0$, $h(s) = 0$

*$f(s) = 0$, $g(s) =s$, $h(s) = 0$

*$f(s) = 0$, $g(s) =0$, $h(s) = s$
