Question

Find the points at which the following surfaces have horizontal tangent planes. $$x^{2}+2 y^{2}+z^{2}-2 x-2 z-2=0$$

Short answer

Answer: The surface has horizontal tangent planes at points \(P_1=(1, 0, 1+\sqrt{2})\) and \(P_2=(1, 0, 1-\sqrt{2})\).

Step-by-step solution

Recognize the equation of the surface

The given equation is: $$x^{2}+2 y^{2}+z^{2}-2 x-2 z-2=0$$ Rearrange the equation to make it more recognizable: $$\left(x^{2}-2x\right)+2y^{2}+\left(z^{2}-2z\right)-2=0$$ Now complete the squares for the x and z terms: $$\left(x-1\right)^{2}-1+2y^2+\left(z-1\right)^2-1=0$$ Combine the constants and get the equation for the surface: $$\left(x-1\right)^{2}+2y^{2}+\left(z-1\right)^2=2$$ This is the equation of an elliptical paraboloid centered at the point (1, 0, 1).

Find the Gradient of the Surface

To find the gradient (partial derivatives) of the surface, we need to differentiate with respect to x, y, and z. Considering the implicit form of the given surface equation: $$F(x,y,z) = x^{2}+2 y^{2}+z^{2}-2 x-2 z-2=0$$ Calculate the partial derivatives: $$F_x=\frac{\partial F}{\partial x}=2x-2$$ $$F_y=\frac{\partial F}{\partial y}=4y$$ $$F_z=\frac{\partial F}{\partial z}=2z-2$$ The gradient vector is: $$\nabla F = (2x-2, 4y, 2z-2)$$

Determine where the Gradient is equal to 0

The gradient vector \(\nabla F\) is perpendicular to the tangent plane of the surface. A horizontal tangent plane has a normal vector parallel to the z-axis, i.e., it has no x and y components. Thus, we need to find the points on the surface where \(\nabla F\) has no x and y components: $$2x-2 = 0 \Rightarrow x = 1$$ $$4y = 0 \Rightarrow y = 0$$ Now that we found the values of x and y where the gradient has no x and y components, we can find the corresponding z value using the surface equation: $$\left(1-1\right)^{2}+2(0)^{2}+\left(z-1\right)^2=2$$ $$\left(z-1\right)^2=2$$ $$z-1=\pm \sqrt{2}$$ $$z=1\pm \sqrt{2}$$ Thus, there are two points on the surface where the tangent plane is horizontal:

Find the Corresponding Points on the Surface

Using the coordinates we found above, the two points (x, y, z) where the given surface has horizontal tangent planes are: $$P_1=(1, 0, 1+\sqrt{2})$$ $$P_2=(1, 0, 1-\sqrt{2})$$