Question

Consider the paraboloid z=x^2+y^2. The plane 3x-2y+z-7=0 cuts the paraboloid, its intersection being a curve.
What is the "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

To find the "natural" parametrization of the curve, let's first find the equation of the circle in the xy-plane. We can do this by eliminating the z variable from the given equation of the plane and the equation of the paraboloid.

We have the paraboloid equation: z = x^2 + y^2, and the plane equation: 3x - 2y + z - 7 = 0.

Substitute the expression for z in terms of x and y from the paraboloid equation into the plane equation:

3x - 2y + (x^2 + y^2) - 7 = 0.

Now we need to find the equation of the circle in the xy-plane. Let's rewrite the above equation in the form of a circle equation:

x^2 + y^2 + 3x - 2y = 7.

We can complete the squares for x and y to find the circle's equation in the form (x - a)^2 + (y - b)^2 = r^2:

(x^2 + 3x) + (y^2 - 2y) = 7
(x^2 + 3x + (3/2)^2) + (y^2 - 2y + 1^2) = 7 + (3/2)^2 + 1^2
(x + 3/2)^2 + (y - 1)^2 = 53/4.

Now we have the equation of the circle, which is in the form (x - a)^2 + (y - b)^2 = r^2, where a = -3/2, b = 1, and r^2 = 53/4.

We can use the hint given and parametrize this circle as a function of the variable t. The hint suggests that we start at the point with the largest x-coordinate, which happens to be the point where t = 0. We can parametrize the circle as follows:

x(t) = -3/2 + sqrt(53/4) * cos(t)
y(t) = 1 + sqrt(53/4) * sin(t)

Substitute x(t) and y(t) into the equation of the paraboloid to get the z-coordinate:

z(t) = x(t)^2 + y(t)^2
z(t) = (-3/2 + sqrt(53/4) * cos(t))^2 + (1 + sqrt(53/4) * sin(t))^2.

So the natural parametrization of the curve on the surface is given by:

x(t) = -3/2 + sqrt(53/4) * cos(t)
y(t) = 1 + sqrt(53/4) * sin(t)
z(t) = (-3/2 + sqrt(53/4) * cos(t))^2 + (1 + sqrt(53/4) * sin(t))^2,
for 0 ≤ t ≤ 2π.