I have to parametrize the curve of intersection of two surfaces. The surfaces are:

$$y^2 = z \text{ and } x + y = 4$$

Could someone please show me how to do this step by step? Thanks.

  • $\begingroup$ Where are you stuck? $\endgroup$
    – Pedro
    Nov 9, 2013 at 0:46
  • 3
    $\begingroup$ My guess isyou don't even know what you're supposed to do. You need to find a map $\gamma\colon \Bbb R\to \Bbb R^3$ such that $\{\gamma(t)\colon t\in \Bbb R\}=\{(x,y,z)\in \Bbb R^3\colon y^2=z\land x+y=4\}$. Can you now do this? $\endgroup$
    – Git Gud
    Nov 9, 2013 at 0:47

1 Answer 1


Let $y=t$. Then $x=4-t$ and $z=t^2$ and so $\vec r(t)= (x(t), y(t), z(t))= (4-t, t, t^2)$, $t\in\mathbf R$.


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