Question

Find both points on the surface x z-x^{2}-y^{2}=8 where the tangent plane is horizontal.

Answer 1

To find the points on the surface where the tangent plane is horizontal, we need to find the partial derivatives of the surface equation with respect to x, y, and z. The horizontal tangent plane is obtained when the partial derivative with respect to z is equal to zero. The points on the surface where the tangent plane is horizontal are (0, 0, 8) and (0, 4, 8).

Step-by-step explanation:

To find the points on the surface where the tangent plane is horizontal, we need to find the partial derivatives of the surface equation with respect to x, y, and z. The horizontal tangent plane is obtained when the partial derivative with respect to z is equal to zero.

Using the given surface equation, we can find the partial derivatives as follows:

∂/∂x (x·z - x^2 - y^2) = z - 2x

∂/∂y (x·z - x^2 - y^2) = -2y

∂/∂z (x·z - x^2 - y^2) = x

Setting the partial derivative with respect to z equal to zero, we have x = 0. Substituting this value back into the original surface equation, we get the points (0, 0, 8) and (0, 4, 8).