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Chapter 12: Problem 51

Find the points at which the following surfaces have horizontal tangent planes. $$x^{2}+y^{2}-z^{2}-2 x+2 y+3=0$$

Short Answer

Expert verified
Answer: The set of points with horizontal tangent planes is given by $$\{(x, y, 0) \mid (x-1)^{2} + (y+1)^{2} = 1\}$$. There isn't a unique solution, as the points lie on a circle in the xy-plane.

Step by step solution

01

Find the gradient of the surface

The gradient of the given implicit surface can be found by taking the partial derivatives of the equation with respect to x, y, and z. Let $$F(x, y, z) = x^{2}+y^{2}-z^{2}-2 x+2 y+3$$, so the gradient is given by: $$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)$$ Computing the partial derivatives, we have: $$\frac{\partial F}{\partial x} = 2x - 2$$ $$\frac{\partial F}{\partial y} = 2y + 2$$ $$\frac{\partial F}{\partial z} = -2z$$ So the gradient of the given surface is: $$\nabla F = (2x - 2, 2y + 2, -2z)$$
02

Determine when the gradient is parallel to the xy-plane

A gradient being parallel to the xy-plane implies that its z-component must be zero. From the gradient we found in step 1, we can set the z-component to zero: $$-2z = 0$$ $$z = 0$$
03

Find the corresponding x and y coordinates for z = 0

With the obtained value of z, substitute it back into the equation of the surface: $$x^{2}+y^{2}-0^{2}-2 x+2 y+3=0$$ $$x^{2}+y^{2}-2 x+2 y+3=0$$ Now we want to find the x and y coordinates that satisfy this equation. To do this, we can complete the square for the x and y terms: $$(x-1)^{2} - 1 + (y+1)^{2} - 1 + 3 = 0$$ $$(x-1)^{2} + (y+1)^{2} = 1$$ This is the equation of a circle with center \((1,-1)\) and radius 1.
04

Identify the points with horizontal tangent planes

The points on the surface with horizontal tangent planes have a z-coordinate of 0 and lie on the circle we found in step 3. Therefore, the points are: $$\{(x, y, 0) \mid (x-1)^{2} + (y+1)^{2} = 1\}$$ There isn't a unique solution for this problem as the points with horizontal tangent planes lie on a circle. So the final answer is given by the set of points in the form of \((x, y, 0)\), where x and y satisfy $$(x-1)^{2} + (y+1)^{2} = 1$$.

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