Problem 51 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The surface has horizontal tangent planes at the points $(1, -1, 2)$ and $(1, -1, -2)$.

Step 1: Compute the gradient of the surface

To find the gradient of the surface, we must first rewrite the given equation as a function of z: $$z^2 = x^2 + y^2 - 2x + 2y + 3.$$ Now, finding the partial derivatives of z with respect to x and y. $$\frac{\partial z^2}{\partial x} = 2x - 2,$$ $$\frac{\partial z^2}{\partial y} = 2y + 2.$$

Step 2: Set the partial derivatives equal to zero

Equate the partial derivative with respect to x and y to zero. $$2x - 2 = 0$$ $$2y + 2 = 0$$

Step 3: Solve the equations for x and y

Solve these equations for x and y: From the first equation: $$2x = 2 \implies x = 1.$$ From the second equation: $$2y = -2 \implies y = -1.$$

Step 4: Find the corresponding z-coordinate

Now that we have x and y, we can plug these values back into the equation for z^2: $$z^2 = (1)^2 + (-1)^2 - 2(1) + 2(-1) + 3 = 1 - 2 + 2 + 3 = 4.$$ Then, we can take the square root of both sides to find z: $$z = \pm \sqrt{4} = \pm 2.$$

Step 5: Write the points with horizontal tangent planes

Finally, we have the points with horizontal tangent planes as: $$(1, -1, 2), (1, -1, -2).$$