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.
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$\begingroup$ Where are you stuck? $\endgroup$– Pedro ♦Nov 9, 2013 at 0:46
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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 GudNov 9, 2013 at 0:47
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1 Answer
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Let $y=t$. Then $x=4-t$ and $z=t^2$ and so $\vec r(t)= (x(t), y(t), z(t))= (4-t, t, t^2)$, $t\in\mathbf R$.
