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.
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.
To find the parametrization of the curve formed by the intersection of the given plane and the paraboloid, we need to first determine the equation of the circle in the xy-plane that lies above the paraboloid.
Given the paraboloid equation z = x^2 + y^2 and the plane equation 3x - 2y + z - 7 = 0, we can substitue the paraboloid equation into the plane equation to eliminate z:
3x - 2y + (x^2 + y^2) - 7 = 0
Rearranging the terms, we have:
x^2 + y^2 + 3x - 2y - 7 = 0
Completing the square for x and y terms:
(x^2 + 3x) + (y^2 - 2y) = 7
To complete the square, we need to add and subtract the necessary terms. For x^2 + 3x, we add (3/2)^2 = 9/4, and for y^2 - 2y, we add (-2/2)^2 = 1:
(x^2 + 3x + 9/4) + (y^2 - 2y + 1) = 7 + 9/4 + 1
(x + 3/2)^2 + (y - 1)^2 = 33/4
This equation describes a circle in the xy-plane centered at (-3/2, 1) with a radius of sqrt(33)/2.
Now, let's parametrize this circle in terms of the variable t:
x = (-3/2) + (sqrt(33)/2) * cos(t)
y = 1 + (sqrt(33)/2) * sin(t)
For t ∈ [0, 2π], as t goes from 0 to 2π, these parametric equations traverse the circle counterclockwise exactly once, starting at the point on the circle with the largest x-coordinate.
To find the parametrization of the curve on the surface, we substitute the parametric equations of x and y into the paraboloid equation:
z = x^2 + y^2
= [(-3/2) + (sqrt(33)/2) * cos(t)]^2 + [1 + (sqrt(33)/2) * sin(t)]^2
Expanding and simplifying the expression:
z = 9/4 - 3sqrt(33)/2 * cos(t) + 33/4 * cos^2(t) + 1 + sqrt(33) * sin(t) + 33/4 * sin^2(t)
Combining like terms, we have:
z = 49/4 + sqrt(33)/2 * cos(t) + sqrt(33)/2 * sin(t) + 33/4 * (cos^2(t) + sin^2(t))
The equation z = 49/4 + sqrt(33)/2 * cos(t) + sqrt(33)/2 * sin(t) simplifies the parametrization of the curve on the surface.