Let $\boldsymbol{X}:\boldsymbol{R}^2\to \boldsymbol{R}^3$ be the paramtrized surface given by$$\boldsymbol{X}(s,t)=(s^2-t^2,s+t,s^2+3t)$$ I'm trying to describe the parametrized surface by an equation of the form $z=f(x,y)$

  • 1
    $\begingroup$ Haven't got any idea $\endgroup$
    – user202080
    Dec 18 '14 at 17:08

You have three equations: $x=s^2-t^2$, $y=s+t$ and $z=s^2+3t$. You need to eliminate $s$ and $t$.

Rearranging $y=s+t$ gives $t=y-s$. Substituting this into the other two equations gives

$$\begin{eqnarray*} x &=& 2ys-y^2 \\ z&=& s^2-3s+3y \end{eqnarray*}$$

Now we can rearrange $x = 2ys-y^2$ to solve for $s$. We get $\displaystyle{s=\frac{x+y^2}{2y}}$.

Finally, substituting this into $z= s^2-3s+3y$ gives an equation in $x$, $y$ and $z$:

$$z = \left(\frac{x+y^2}{2y}\right)^{\!2} - 3\left(\frac{x+y^2}{2y}\right) + 3y$$

This assumes that $y \neq 0$. If $y=0$ then $s=-t$, meaning that $x=0$ and $z=t^2+3t$. Here $t$ is arbitrary. Completing the square gives $$z=\left(t+\frac{3}{2}\right)^{\!2} - \frac{9}{4}$$ This tells is that for al $t$, $z \ge -\frac{9}{4}$. So, when $x=y=0$, we have to add the semi-axis $z \ge -\frac{9}{4}$.


Mathematica says this:

In[1]:=Reduce[Exists[{s, t}, x == s^2 - t^2 && y == s + t && z == s^2 + 3 t], {z}]

Out[1] = (y == 0 && x == 0) || (y != 0 && z == (x^2 - 6 x y + 2 x y^2 + 6 y^3 + y^4)/(4 y^2))

So, the answer seems to be $$z=\frac{x^2+2 x y^2-6 x y+y^4+6 y^3}{4 y^2}$$ when $y \neq 0$. When $y = 0$ the value of $z$ is arbitrary and $x = 0$.

  • $\begingroup$ Hmm, this can't be right. $\endgroup$ Dec 18 '14 at 17:16
  • $\begingroup$ I mean that I disagree with my answer. $\endgroup$ Dec 18 '14 at 17:18

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