Question

Find the space curve of vector function r(t), where the following surfaces intersect:
3x^2 + y^2 + 1 = z, x^2 +3y^2 + z = 5

Answer 1

We can use the method of Lagrange multipliers to find the intersection curve. Let the two surfaces be given by the equations:

f(x,y,z) = 3x^2 + y^2 + 1 - z = 0
g(x,y,z) = x^2 + 3y^2 + z - 5 = 0

We want to find the common solutions, subject to the constraint that these points lie on a curve. Let r(t) = (x(t), y(t), z(t)) be a parameterization of the curve. Then we want to find r(t) such that:

f(x(t), y(t), z(t)) = 0
g(x(t), y(t), z(t)) = 0

We introduce a Lagrange multiplier λ and consider the function:

F(x,y,z,λ) = f(x,y,z) - λg(x,y,z)

We want to find critical points of this function subject to the constraint that r(t) lies on the curve. That is, we want to solve the equations:

∂F/∂x = 6x - 2λx = 0
∂F/∂y = 2y - 6λy = 0
∂F/∂z = -1 - λ = 0
∂F/∂λ = x^2 + 3y^2 + z - 5 = 0

Solving these equations, we obtain:

λ = -1
3x^2 + y^2 = 1
x^2 + 3y^2 + z - 5 = 0

Substituting λ = -1 into the first two equations gives us the equation of an ellipse:

3x^2 + y^2 = 1

Substituting this into the third equation and solving for z, we get:

z = -2x^2 - 3y^2 + 5

Thus, the space curve of the intersection is given by:

r(t) = (x(t), y(t), z(t)) = (cos(t)/sqrt(3), sin(t), -2cos^2(t)/sqrt(3) - 3sin^2(t) + 5)

for t in [0,2π].