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If I've got the surface in $\mathbb{R}^3$ described by: $x(s,t)=s^2-t^2$, $y(s,t)=s+t$, $z(s,t)=s^2+3t$ for $(s,t)\in\mathbb{R}^2$, and I'm told this surface is the graph of a function $f(x,y)$, how can I find $f$?

In general, is there a systematic way to do this sort of problem?

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From $x(s,t)=s^2-t^2$ and $y(s,t)=s+t$,

$$\frac{x(s,t)}{y(s,t)}=\frac{s^2-t^2}{s+t}=\frac{(s-t)(s+t)}{(s+t)}=s-t\\ \implies y-\frac{x}{y}=s+t+t-s=2t\\ \implies t=\frac12\left(y-\frac{x}{y}\right),$$

and

$$\implies y+\frac{x}{y}=s+t+s-t=2s\\ \implies s=\frac12\left(y+\frac{x}{y}\right).$$

Now in order to find $z=f(x,y)$, we use $z(s,t)=s^2+3t$ to find:

$$f(x,y)=z(s(x,y),t(x,y))=s^2+3t=\frac14\left(y+\frac{x}{y}\right)^2+\frac32\left(y-\frac{x}{y}\right).$$

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    $\begingroup$ What happens if $y=0$? $\endgroup$ – Peter May 20 '14 at 5:04
  • $\begingroup$ @Peter Well first of all, if $y=0$, then $s=-t$ and so $x=0$ also. So we only have to worry about things blowing up at the origin. $\endgroup$ – David H May 20 '14 at 5:15

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