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Chapter 10: Problem 15

Show that the equation represents a sphere, and find its center and radius. \(2 x^{2}+2 y^{2}+2 z^{2}=8 x-24 z+1\)

Short Answer

Expert verified
The sphere's center is (2, 0, -6), and its radius is approximately 6.36.

Step by step solution

01

Reorganize the Equation

First, move all terms to one side of the equation to express the equation in the form \[2x^2 - 8x + 2y^2 + 2z^2 + 24z - 1 = 0.\] The idea is to arrange it such that it can be converted into the standard form of a sphere.
02

Factor Common Terms

Notice that some terms have a common factor. Factor 2 out of all the quadratic terms to simplify the equation:\[2(x^2 - 4x + y^2 + z^2 + 12z) = 1.\]
03

Complete the Square

Complete the square for each variable: - For the \(x\) terms: \(x^2 - 4x = (x-2)^2 - 4\) - For the \(z\) terms: \(z^2 + 12z = (z+6)^2 - 36\). Now substituting back, we get:\[2((x-2)^2 - 4 + y^2 + (z+6)^2 - 36) = 1.\]
04

Simplify and Identify Sphere Equation

Simplify the completed square equation:\[2((x-2)^2 + y^2 + (z+6)^2 - 40) = 1.\]Distribute the factor of 2 inside:\[2(x-2)^2 + 2y^2 + 2(z+6)^2 - 80 = 1.\]Move the 80 over:\[2(x-2)^2 + 2y^2 + 2(z+6)^2 = 81.\] Finally, divide the entire equation by 2 to obtain:\[(x-2)^2 + y^2 + (z+6)^2 = rac{81}{2}.\]
05

Identify Center and Radius

The equation is now in the standard form of a sphere \[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\], where \((h, k, l)\) is the center and \(r\) is the radius. Comparing, the center is \((2, 0, -6)\) and the radius is \( rac{ ext{radius squared = } 81}{2}\).Therefore, the radius \( r \) is \( \sqrt{\frac{81}{2}} = \sqrt{40.5} \approx 6.36.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
Completing the square is a technique used to organize quadratic terms into a perfect square form. This method is essential when transforming an equation to fit the standard equation of a sphere.
Here is how it works:
  • Take a quadratic expression like \( x^2 - 4x \).
  • Find the value needed to make it a perfect square. This value is found by taking half of the coefficient of \( x \), squaring it, and adding it to the equation.
  • In this case, half of \(-4\) is \(-2\), and \((-2)^2\) is \(4\). So, add and subtract \(4\) in the expression: \( x^2 - 4x = (x-2)^2 - 4 \).
Applying this method helps break down complex quadratic forms into simpler, factorable expressions.
For the z terms, such as \( z^2 + 12z \), add and subtract \((12/2)^2 = 36\) to it:
  • You'll obtain \((z+6)^2 - 36\).
This step prepares the given equation for transformation into the standard sphere format by revealing the hidden structure of a perfect square.
Center of Sphere
Once you've completed the square, the equation takes on a recognizable form that allows you to identify the center of the sphere.
For a sphere, the equation often resembles:\[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\]where \((h, k, l)\) signify the center coordinates.- In our converted equation, \((x-2)^2 + y^2 + (z+6)^2 = \frac{81}{2}\), you can see:
  • The \(x\) term's perfect square form is \((x-2)^2\), giving \(h=2\).
  • The \(y\) term's perfect square form is \(y^2\), implying \(k=0\) considering its absence as a squared term.
  • The \(z\) term's perfect square form is \((z+6)^2\), leading to \(l=-6\).
Therefore, the center of the sphere is at the point \((2, 0, -6)\).
This interpretation is vastly simplified by recognizing the equation's form, giving you a direct method to find the sphere's center.
Radius of Sphere
The sphere's equation also shows us how to determine its radius.
We start from the form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). In this form, \(r^2\) is the squared radius.
From our completed square equation, we have:\[(x-2)^2 + y^2 + (z+6)^2 = \frac{81}{2}\]
  • This means the squared radius \(r^2\) is equal to \(\frac{81}{2}\).
  • To find the radius, simply take the square root: \(r = \sqrt{\frac{81}{2}}\).
  • After calculating \(\sqrt{\frac{81}{2}}\), you get approximately \(6.36\).
By understanding this part of the equation, you're able to not only identify but also compute the sphere's radius efficiently. This process reveals the hidden physical size encoded within the equation.

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