Intersection of surfaces/tangent vector

[Odyssey]

Hi, how do I find a tangent vector to the curve intersection of the surfaces $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2$ which has positive x-direction?

Thanks in advance.

 

[arildno]

Note that your last equation implies that we may write:
$$x=z\cos\theta,y=z\sin\theta,0\leq\theta\leq2\pi$$
Inserting these relations into your first equation, yields:
$$z^{2}=\frac{25}{3}$$
Can you take it from here?

 

[wais]

To find the tangent vector to the curve intersection of the two surfaces, we can use the method of Lagrange multipliers. This method involves finding the gradient vectors of both surfaces and setting them equal to each other, along with the constraint equation for the curve intersection.

First, we can find the gradient vectors of the two surfaces:
∇(2x^2+2y^2-z^2) = (4x, 4y, -2z)
∇(x^2+y^2-z^2) = (2x, 2y, -2z)

Next, we set these two gradient vectors equal to each other, along with the constraint equation x^2+y^2=z^2:
(4x, 4y, -2z) = λ(2x, 2y, -2z)
x^2+y^2-z^2 = 0

Solving this system of equations, we get three possible solutions:
- The first solution is when λ=1 and x=y=√2, z=√3. This is a point on the curve of intersection.
- The second solution is when λ=0 and x=y=z=0. This is the origin, which is not on the curve of intersection.
- The third solution is when λ=-1 and x=y=-√2, z=-√3. This is another point on the curve of intersection.

To determine the tangent vector with positive x-direction, we can choose the first solution, which gives us the point (x,y,z)=(√2,√2,√3).

To find the tangent vector at this point, we can take the partial derivatives of the constraint equation with respect to x, y, and z:
∂x(x^2+y^2-z^2) = 2x
∂y(x^2+y^2-z^2) = 2y
∂z(x^2+y^2-z^2) = -2z

Plugging in the values from the point (x,y,z)=(√2,√2,√3), we get the tangent vector (2√2, 2√2, -2√3).

Therefore, the tangent vector to the curve intersection of the surfaces 2x^2+2