Problem 27 - Essential Calculus Early Transcendentals

Question

Reduce the equation to one of the standard forms, classify the surface, and sketch it. \(x^{2}-y^{2}+z^{2}-4 x-2 y-2 z+4=0\)

Step-by-Step Solution

Answer

The equation is a hyperboloid of one sheet centered at (2, -1, 1).

Step 1: Group Quadratic Terms

First, group the quadratic terms with similar variables together in the equation: \[ x^2 - 4x - y^2 - 2y + z^2 - 2z + 4 = 0 \]

Step 2: Complete the Square for x

For the terms involving \(x\), \(x^2 - 4x\), we complete the square. - Take half the coefficient of \(x\) (which is 4), square it to get 4, and add and subtract it inside the equation.This gives us: \[ x^2 - 4x = (x-2)^2 - 4 \]

Step 3: Complete the Square for y

Next, complete the square for \(y\), the terms are \(-y^2 - 2y\). - Take half the coefficient of \(y\) (which is -1), square it to get 1, and add and subtract it inside the equation.This gives us: \[ -(y^2 + 2y) = -( (y+1)^2 - 1 ) = -(y+1)^2 + 1 \]

Step 4: Complete the Square for z

Now, complete the square for \(z\), the terms are \(z^2 - 2z\). - Take half the coefficient of \(z\) (which is 1), square it to get 1, and add and subtract it inside the equation.This gives us: \[ z^2 - 2z = (z-1)^2 - 1 \]

Step 5: Substitute Back the Completed Squares

Substitute the completed squares back into the equation:\[ (x-2)^2 - 4 - ((y+1)^2 - 1) + (z-1)^2 - 1 + 4 = 0 \] Simplify further by gathering constant terms.\[ (x-2)^2 - (y+1)^2 + (z-1)^2 - 1 = 0 \]

Step 6: Simplify to a Standard Form

Move the constant to the other side to get the final standard form:\[ (x-2)^2 - (y+1)^2 + (z-1)^2 = 1 \]This is the standard equation of a one-sheet hyperboloid.

Step 7: Classify the Surface

The given equation in its simplified form represents a one-sheet hyperboloid. It is characterized by one variable having a negative square term in a standard hyperboloid equation \((x-h)^2/a^2 - (y-k)^2/b^2 + (z-l)^2/c^2 = 1\).

Step 8: Sketch the Surface

To sketch a hyperboloid of one sheet, remember that it resembles an hourglass shape or a fan. The center is shifted to (2, -1, 1). Sketch the central ellipse at the point and open hyperbolically outwards from this center along the y-axis due to its negative coefficient.

Key Concepts

Complete the Square

Completing the square is a crucial method to solve quadratic equations and simplify expressions. It is particularly useful to transform equations into a form that is easier to graph or analyze. To complete the square:

• Look at the quadratic term and the linear term associated with a specific variable.
• Take half of the linear term's coefficient, square it, and then add and subtract this square in the equation.

By reorganizing our terms, we convert complex expressions into perfect square binomials. For example, for the expression \(x^2 - 4x\), you take half of -4, which is -2, square it to get 4, and then rewrite the expression as \((x-2)^2 - 4\). This method illuminates relationships between variables and constants and simplifies the process of sketching surfaces, like hyperboloids.

Quadratic Equation

A quadratic equation is any equation that can be rewritten in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Understanding how to manipulate and solve these equations is essential as they frequently appear in algebra and calculus. Quadratic equations can be solved by:

• Factoring,
• Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\),
• Or by completing the square, which was employed in our original problem.

These techniques aid in transforming equations into a graphable form. In our case, by completing squares for each variable, we simplified the complex equation into the standard hyperboloid form, which is critical for understanding its geometry and features.

Sketching Surfaces

Sketching surfaces involves visualizing equations in three-dimensional space to understand their behaviors and characteristics. In particular, a hyperboloid of one sheet looks like a central hourglass shape.To accurately sketch:

• Identify the center of the surface from the transformed equation. For example, our hyperboloid's center is identified by \((h, k, l) = (2, -1, 1)\).
• Recognize the orientation of the surface. In our equation, the negative square term suggests the surface opens along the y-axis.
• Start by sketching the central cross-sectional ellipse and build outward hyperbolically from it.

Visualizations assist in understanding complex surface interactions in multivariable functions and are especially important in fields like physics and engineering.