Question

At what point on the surface z = 2 + x2 + y2 is its tangent plane parallel to the following planes?

(a) z = 6
(x, y, z) =
(b) z = 6 + 4x − 12y
(x, y, z) =

Answer 1

The point where the surface z = 2 + x^2 + y^2 has a tangent plane parallel to z = 6 is (0, 0, 2). For the plane z = 6 + 4x - 12y, the point is (2, -6, 24). We find these by equating the normal vectors of the planes to the gradient of z given by its partial derivatives.

Step-by-step explanation:

To find the point on the surface z = 2 + x^2 + y^2 where its tangent plane is parallel to a given plane, we need to determine the gradient of the surface, as this gradient will be perpendicular to the tangent plane. For two planes to be parallel, their normal vectors must be scalar multiples of each other.

Finding the normal vector:

We calculate the partial derivatives of z = 2 + x^2 + y^2 concerning x and y to get the normal vector of the tangent plane:

∂z/∂x = 2x

∂z/∂y = 2y

Therefore, the normal vector to the surface at any point (x, y) is (2x, 2y, -1) since we take the negative of the gradient for the z-component to match the standard form of the plane equation z = ax + by + c where the vector (a, b, -1) is normal to the plane.

(a) Case:

If the tangent plane is parallel to the plane z = 6, its normal vector is (0, 0, 1), which is parallel to (2x, 2y, -1) only if both x and y are zero. Hence, the point is (0, 0, 2).

(b) Case:

If the tangent plane is parallel to the plane z = 6 + 4x - 12y, the normal vector is (4, -12, 1), which needs to be a scalar multiple of (2x, 2y, -1). Solving the equations 4 = 2x and -12 = 2y gives us x = 2 and y = -6. Therefore, the point is (2, -6, 24).

Answer 2

a. This equation has no solution, which means that there is no point on the surface
`\(z = 1 + x^2 + y^2\)` where the tangent plane is parallel to
`\(z = 6\)`.

b. The correct points on the surface
`\(z = 1 + x^2 + y^2\)` where the tangent plane is parallel to
`\(z = 6\)` are
`\((√(2), √(2), 6)\)` and
`\((- √(2), - √(2), 6)\)`.

To find the point on the surface
`\(z = 1 + x^2 + y^2\)` where the tangent plane is parallel to a given plane, we need to consider the normal vector of the surface and the normal vector of the given plane.

The surface
`\(z = 1 + x^2 + y^2\)` can be expressed as
`\(F(x, y, z) = 1 + x^2 + y^2 - z = 0\)`. The normal vector of the surface is then given by the gradient of \(F\):

`\[ \\abla F = \langle (\partial F)/(\partial x), (\partial F)/(\partial y), (\partial F)/(\partial z) \rangle \]`

So, compute the gradient:

`\[ \\abla F = \langle 2x, 2y, -1 \rangle \]`

Now, let's consider the given planes:

(a)
`$z=6$` : The normal vector of this plane is
`$\langle 0,0,1\rangle$`.

(b)
`$z=6+6 x-10 y$`: The normal vector of this plane is
`$\langle 6,-10,-1\rangle$`.

For the tangent plane to be parallel to the given planes, the normal vectors of the surface and the planes must be proportional. This implies that:

`\[ \\abla F = \lambda \langle 0, 0, 1 \rangle \quad \text{for (a)}`

`\[ \\abla F = \lambda \langle 6, -10, -1 \rangle \quad \text{for (b)}`

For (a), since the
`\(z\)`-component of
`\(\\abla F\)` is -1, we can set
`\(\lambda = 1` to satisfy the condition. This leads to the equation:

`\[ -1 = 1 \]`

This equation has no solution, which means that there is no point on the surface
`\(z = 1 + x^2 + y^2\)` where the tangent plane is parallel to
`\(z = 6\)`.

For (b), set the components proportional:

`\[ (2x)/(6) = (2y)/(-10) = (-1)/(-1) \]`

Solve these equations to find
`\(x\)` and
`\(y\)`. In this case, the solution is
`\(x = \pm √(2)\)` and
`\(y = \pm √(2)\)`. Now, substitute these values into the equation of the surface to find the corresponding
`\(z\)` values:

`\[ z = 1 + x^2 + y^2 \]`

`\[ z = 1 + (√(2))^2 + (√(2))^2 = 6 \]`

`\[ z = 1 + (-√(2))^2 + (-√(2))^2 = 6 \]`

Therefore, the correct points on the surface
`\(z = 1 + x^2 + y^2\)` where the tangent plane is parallel to
`\(z = 6\)` are
`\((√(2), √(2), 6)\)` and
`\((- √(2), - √(2), 6)\)`. The provided solution is correct for the second case.