(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.

Step-by-step 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:

a

step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: