I have to parametrize the curve of intersection of 2 surfaces.
The surfaces are:
$z=x^2+y^2$ and $2x-4y-z-1=0$
Could someone please show me how to do this step by step?
Thank you.
[Math] Parametrize the curve of intersection of 2 surfaces
calculus
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Best Answer
Well, setting the common value of $z$ equal to itself, we get
$$x^2+y^2=2 x-4 y-1$$
or
$$x^2-2 x + y^2+4 y=-1 \implies (x-1)^2+(y+2)^2=4 $$
This is a circle of radius $2$ centered at $(1,-2)$. A standard parametrization of this is
$$x=1+2 \cos{t}$$ $$y=-2+2 \sin{t}$$