Question

(1 point) Consider the paraboloid z=x2+y2. The plane 2x−2y+z−8=0 cuts the paraboloid, its intersection being a curve. Find "the natural" parametrization of this curve. Hint: The curve which is cut lies above a circle in the xy-plane which you should parametrize as a function of the variable t so that the circle is traversed counterclockwise exactly once as t goes from 0 to 2*pi, and the paramterization starts at the point on the circle with largest x coordinate. Using that as your starting point, give the parametrization of the curve on the surface.

Answer 1

d on edge

Explanation:

just did it

Answer 2

`\left \{ {{x(t)=-1+√(10)Sin(t) } \atop {y(t)=1+√(10)Sin(t)}} \right.`

Explanation:

We have the plane 2x-2y+z-8=0 and the paraboloid z=x²+y².

We match the equations:

z=8-2x+2y

z=x²+y²

x²+y²=8-2x+2y

Then, we get the equation of the curve:

x²+2x+y²-2y=8 (adding +2x and -2y in both members of the equation)

(x²+2x+1)+(y²-2y+1)=8+1+1

Trinomials can be reduced as squared binomials

(x+1)²+(y-1)²=10

This is a circle with center in (-1,1) and radius √10

The generic

The parametric expression of the circle is:

`\left \{ {{x(t)=-1+√(10)Sin(t) } \atop {y(t)=1+√(10)Sin(t)}} \right.`