Problem 89 - Calculus Early Transcendentals

Question

Identify and briefly describe the surfaces defined by the following equations. $$x^{2} / 4+y^{2}-2 x-10 y-z^{2}+41=0$$

Step-by-Step Solution

Answer

Answer: The equation represents a hyperboloid of two sheets centered at (1,5,0) with semi-axes along the x-direction of length 2 and along the y-direction of length 1. However, it cannot be geometrically represented in real space since its z-direction semi-axis is imaginary.

Step 1: Reorder the given equation

Write the equation in a more familiar form: $$\frac{x^2}{4} + y^2 - z^2 - 2x - 10y + 41 = 0$$

Step 2: Complete the square for x and y terms

To better understand the equation, we will complete the square for the x and y terms. For x, let's add and subtract \((2/2)^2 = 1\): $$\frac{x^2}{4} - 2x + 1$$ For y, let's add and subtract \((10/2)^2 = 25\): $$y^2 - 10y + 25$$ Now, rewrite the equation with the completed squares: $$\frac{(x-1)^2}{4} + (y-5)^2 - z^2 + 15 = 0$$

Step 3: Rewrite the equation in standard form

We will rewrite the equation in its standard form, which might resemble the equation of a familiar surface: $$\frac{(x-1)^2}{4} + (y-5)^2 - z^2 = -15$$

Step 4: Identify the surface and describe it

Upon closely examining the equation, we notice that the equation is of the form: $$\frac{(x-a)^2}{A^2} + \frac{(y-b)^2}{B^2} - \frac{z^2}{C^2}= k$$ where \(a = 1\), \(b = 5\), \(A = 2\), \(B = 1\), \(C^2=-15\), and \(k=-1\). The general equation for a hyperboloid of two sheets is: $$\frac{(x-a)^2}{A^2} + \frac{(y-b)^2}{B^2} - \frac{(z^2)}{C^2} = -1$$ Since the given equation matches this form, we can conclude that the surface is a hyperboloid of two sheets, centered at \((1,5,0)\) with semi-axes along the x-direction of length \(2\) and along the y-direction of length \(1\). However, since \(C^2\) is negative, we cannot define the semi-axis corresponding to the z-direction. This indicates that the surface is imaginary, meaning it cannot be represented geometrically in real space.