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Chapter 12: Problem 52

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

Expert verified
Solution: The surface has a horizontal tangent plane at point (1, 0, 1).

Step by step solution

01

Determine the gradient of the given surface.

The surface is given by $$F(x,y,z) = x^2 + 2y^2 + z^2 - 2x - 2z - 2 = 0$$ We need to find the gradient, which is a vector formed by the partial derivatives of F with respect to x, y, and z: $$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)$$ Calculate the partial derivatives: $$\frac{\partial F}{\partial x} = 2x - 2$$ $$\frac{\partial F}{\partial y} = 4y$$ $$\frac{\partial F}{\partial z} = 2z - 2$$ So, the gradient of F is: $$\nabla F = (2x - 2, 4y, 2z - 2)$$
02

Find the points where the gradient of F is zero.

A horizontal tangent plane occurs when the gradient of F is zero in all three directions. So, we need to solve the following system of equations: $$2x - 2 = 0$$ $$4y = 0$$ $$2z - 2 = 0$$ To solve this system of equations, we can use substitution or elimination. Here, we'll simply use substitution: $$x = 1$$ $$y = 0$$ $$z = 1$$
03

Write the final solution.

Now that we have found the values for x, y, and z that make the gradient of F zero in all three directions, we can say that the surface has a horizontal tangent plane at point: $$(x,y,z) = (1, 0, 1)$$

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