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Let $z=4-2x-2y$ be a plane having a curve $\gamma$ on it. The projection of $\gamma$ on $z=0$ is the circle $x^2 + y^2 =1$ .

Find a parameterization of $\gamma$ .

How can I do it ?

I know that the surface is $ x(u,v) = (u,v, 4-2u-2v) $ , and that our curve must be of the form $ \gamma(t) = (u(t), v(t) , 4-2u(t)-2v(t) ) $ . After taking $ z=0, x^2+y^2=1 $ we get $3-8v+5v^2 =0 $ and I can't understand how it helps ...

Will you please help ?

Thanks

criticism
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1 Answers1

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The way I interpret the statement

The projection of $\gamma$ on $z=0$

is the map $P: \mathbb{R}^3 \to \mathbb{R}^3$ given by $P(x,y,z) = (x,y,0)$. Given it is a circumference, we can parametrize it by

$$(x(t),y(t)) = (\cos (t), \sin(t)), \quad t \in \mathbb{R}.$$

Since we want it to be in the plane, just substitute $x(t),y(t)$ in the plane equation.

Therefore, the curve $\gamma$ is parametrized by

$$\gamma(t) = (x(t),y(t),z(t)) = (\cos (t), \sin (t), 4 - 2 \cos(t) - 2 \sin (t)), \quad t \in \mathbb{R}.$$

This way, when we compute $P(\gamma(t))$ we find the unit circumference in the $xy$ plane.

Mark Fantini
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