Parametrization of a Curve

Question

Consider the paraboloid z= x^2 +y^2. The plane 9x -5y +z -7 =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 parametrization 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.

Find x(t), y(t), and z(t).

Answer 1

** I think this is free of important errors. Comment if you find any (important or not). **

The intersection of the surface where z= x^2 +y^2 and the surface (which happens to be a plane) where 9x -5y +z -7 =0 is the set of points (x,y,z) where z = x^2 + y^2 and also 9x - 5y + z - 7 = 0.

If z =x^2 + y^2 and also (solving for z:) z = -9x + 5y + 7, then x^2 + y^2 = -9x + 5y + 7 . So the points on our curve all lie directly above or below points on the curve where x^2 + y^2 = -9x + 5y + 7 on the (x,y) plane.

Rewrite x^2 + y^2 = -9x + 5y + 7 as x^2 + 9x + y^2 - 5y = 7, and then complete the squares to find out what curve has this equation:

Add and subtract squares of halves of coefficients of x and y:

Rewrite as: x^2 + 9x + (9/2)^2 - (9/2)^2 + y^2 - 5y + (-5/2)^2 - (-5/2)^2 = 7

Then as: (x + 9/2)^2 + (y - 5/2)^2 = 28/4 + 81/4 + 25/4

Which is equivalent to (x + 9/2)^2 + (y - 5/2)^2 = 134/4

which is the circle on the xy plane centered at (-9/2, 5/2) with radius (√134)/2

A parametrization of that curve is (from x = xc + r cos t and y = yc + r sin t, where r is the radius and (xc,yc) is the center of the circle, and t goes from 0 to 2π, so:

x = -9/2 + ((√134)/2) cos t

y = 5/2 + ((√134)/2) sin t

where 0 ≤ t ≤ 2π .

Now just finish the parametrization to give the corresponding z coordinates in addition the x and y coordinates:

x = -9/2 + ((√134)/2) cos t

y = 5/2 + ((√134)/2) sin t

and

z = z = x^2 + y^2 = (-9/2 + ((√134)/2) cos t)^2 + (5/2 + ((√134)/2) sin t)^2

or else

z = -9x + 5y + 7 = -9(-9/2 + ((√134)/2) cos t) + 5(5/2 + ((√134)/2) sin t)

where 0 ≤ t ≤ 2π .