Find parametric equations for the tangent line to the curve of intersection of the paraboloid

z = x2 + y2
and the ellipsoid
6x2 + 5y2 + 7z2 = 39
at the point
(−1, 1, 2)

if u and v are functions of x,y,z, then the normals are

Nu = ∇u
Nv = ∇v

The tangent line of the intersection is perpendicular to both normals

T(x,y,z) = ∇u × ∇v

∇u = 2xi + 2yj - k
∇v = 12xi + 10yj + 14zk

T(-1,1,2) = 33i + 34j + 2k
so that would be

x = -1 + 33t
y = 1 + 34t
z = 2 + 2t

can you please explain to me where you got the numbers for T(-1,1,2) I know you did the cross product but I can not seem to get the same numbers as you. Please explain

he crossed the partial derivatives

hey nim again,

Find parametric equations for the tangent line to the curve of intersection of the paraboloid z =
x
2 + y
2 and the elliposoid x
2 + 4y
2 + z
2 = 9 at the point P(1, −1, 2).
Comments: A tangent vector to the curve of intersection is given by N1 × N2 where N1 is normal
to the graph of z = x
2 + y
2 at the point P and N2 is normal to the level surface F(x, y, z) =
x
2 + 4y
2 + z
2 = 9 at the point P.
Now N1 = h−2x, −2y, 1i. At the point P, N1 = h−2, 2, 1i. The vector N2 = ∇F = h2x, 8y, 2zi.
At P, N2 = h2, −8, 4i. We rescale and set N2 = h1, −4, 2i.
We compute N1 × N2 = h8, 5, 6i which is tangent to the curve of intersection at the point P.
Hence the tangent line is given by x = 1 + 8t, y = −1 + 5t, z = 2 + 6t.

found this at math(dot)drexel(dot)edu/~rboyer/courses/math200_sum_2012/inclass_week7_comments.pdf
and it helped, hope it helps

Oh, I see you've stumbled upon a classic clash of shapes! A paraboloid and an ellipsoid walking into a bar, creating quite the intersection.

Anyway, let's get down to business. To find the tangent line to the curve of intersection, we need to find the normal vector of both surfaces at the given point (-1, 1, 2).

First, let's differentiate the paraboloid equation with respect to x and y to find the normal vector:

d/dx (z) = d/dx (x^2 + y^2)
= 2x

d/dy (z) = d/dy (x^2 + y^2)
= 2y

So, the normal vector to the paraboloid surface at (-1, 1, 2) is (2*(-1), 2*(1)) = (-2, 2).

Next, let's differentiate the ellipsoid equation with respect to x, y, and z:

d/dx (6x^2 + 5y^2 + 7z^2) = 12x
d/dy (6x^2 + 5y^2 + 7z^2) = 10y
d/dz (6x^2 + 5y^2 + 7z^2) = 14z

By substituting the given point (-1, 1, 2) into the partial derivatives, we find the normal vector to the ellipsoid surface at that point:

(12*(-1), 10*(1), 14*(2)) = (-12, 10, 28).

Now that we have the two normal vectors, we can take their cross product to find the tangent line direction:

(-2, 2) x (-12, 10, 28) = (2, 32, 44).

The parametric equations for the tangent line passing through (-1, 1, 2) with the direction vector (2, 32, 44) are:

x = -1 + 2t
y = 1 + 32t
z = 2 + 44t.

So, there you have it! The tangent line to the curve of intersection of the paraboloid and the ellipsoid at the point (-1, 1, 2) is given by the parametric equations x = -1 + 2t, y = 1 + 32t, z = 2 + 44t.

Hope that puts a smile on your face!

To find the parametric equations for the tangent line to the curve of intersection of the paraboloid and the ellipsoid at the point (-1, 1, 2), follow these steps:

Step 1: Find the gradient vector of both surfaces at the given point.
Step 2: Set the gradients equal to each other to find the direction vector for the tangent line.
Step 3: Use the given point and the direction vector to write down the parametric equations for the tangent line.

Let's start with Step 1: Finding the gradient vector of both surfaces at the given point (-1, 1, 2).

For the paraboloid, given by the equation z = x^2 + y^2, we can find the gradient vector by taking the partial derivatives with respect to x, y, and z:

∇(x^2 + y^2) = 2x i + 2y j + k

For the ellipsoid, given by the equation 6x^2 + 5y^2 + 7z^2 = 39, we can find the gradient vector by taking the partial derivatives with respect to x, y, and z:

∇(6x^2 + 5y^2 + 7z^2) = 12x i + 10y j + 14z k

Substituting the coordinates of the given point (-1, 1, 2) into the gradient vectors, we have:

∇(x^2 + y^2) = 2(-1) i + 2(1) j + k = -2i + 2j + k

∇(6x^2 + 5y^2 + 7z^2) = 12(-1) i + 10(1) j + 14(2) k = -12i + 10j + 28k

Now, let's move on to Step 2: Setting the gradients equal to each other to find the direction vector for the tangent line.

-2i + 2j + k = -12i + 10j + 28k

Equating the corresponding components, we get:

-2 = -12
2 = 10
1 = 28

Since these equations are inconsistent, it means that the gradient vectors are not equal, and there is no tangent line at the point (-1, 1, 2) that lies on both the paraboloid and the ellipsoid.

Therefore, the curve of intersection does not have a well-defined tangent line at the given point.