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Chapter 9: Problem 18

Show that the equation represents a sphere, and find its center and radius. $$x^{2}+y^{2}+z^{2}=14 y-6 z$$

Short Answer

Expert verified
The sphere has center \((0, 7, 3)\) and radius \(\sqrt{58}\).

Step by step solution

01

Rearrange the equation

First, we need to rearrange the given equation to make it resemble the standard form of a sphere. Starting with the equation \( x^2 + y^2 + z^2 = 14y - 6z \), move all terms to one side to get \( x^2 + y^2 - 14y + z^2 - 6z = 0 \).
02

Complete the square for y-term

To complete the square for the \( y \) term, focus on \( y^2 - 14y \). Rewrite this in a completed square form: \( (y - 7)^2 - 49 \). Therefore, substitute \( y^2 - 14y \) with \( (y - 7)^2 - 49 \) in the equation.
03

Complete the square for z-term

Next, complete the square for the \( z \) term, \( z^2 - 6z \). Rewrite this as \( (z - 3)^2 - 9 \). Substitute \( z^2 - 6z \) with \( (z - 3)^2 - 9 \) in the equation.
04

Simplify the equation

Now, substitute these completed squares back into the equation: \( x^2 + (y - 7)^2 - 49 + (z - 3)^2 - 9 = 0 \). Simplify it to get: \( x^2 + (y - 7)^2 + (z - 3)^2 = 58 \).
05

Identify the center and radius

The simplified equation \( x^2 + (y - 7)^2 + (z - 3)^2 = 58 \) corresponds to the standard form of a sphere \( (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \), where the center is \((h, k, l) = (0, 7, 3)\) and the radius \( r = \sqrt{58} \).

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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 help transform quadratic expressions into a perfect square trinomial, making them easier to manage and solve. This is crucial when dealing with the equation of a sphere because it helps us simplify and rearrange terms into a recognizable format. To complete the square, you follow these steps:
  • Take the quadratic term and linear term you want to simplify, for instance, the terms involving \( y \) are \( y^2 - 14y \).
  • Identify the coefficient of the linear term, which is \( -14 \) in this case, divide it by 2, and then square the result. Specifically, \( \left(\frac{-14}{2}\right)^2 = 49 \).
  • Add and subtract this square (49 in this case) to transform the expression into a perfect square trinomial. This gives us \( (y-7)^2 - 49 \).
Completing the square reshapes the expression, making such problems more straightforward and preparing them to be inserted into a standard equation format.
Center of a Sphere
In the equation of a sphere, the standard form is \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] where \((h, k, l)\) are coordinates of the sphere's center. Recognizing the center of a sphere involves identifying these values from the equation.After completing the square in the given problem, the equation \[ x^2 + (y - 7)^2 + (z - 3)^2 = 58 \] clearly reveals the center's coordinates:
  • \( h = 0 \), as there is no \( (x - )^2 \) term, meaning the \( x \)-coordinate is 0.
  • \( k = 7 \), derived from \((y-7)^2\).
  • \( l = 3 \), derived from \((z-3)^2\).
Thus, the center of the sphere is at the point \((0, 7, 3)\). Understanding how to extract these from the equation is key to solving sphere-related problems.
Radius of a Sphere
A sphere's radius in its equation determines how far each point on the surface is from its center. In the standard form for a sphere's equation, \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] \( r \) represents the radius.From the equation we derived through completing the square, \[ x^2 + (y - 7)^2 + (z - 3)^2 = 58 \], we identify that \( r^2 = 58 \).To find the radius \( r \), simply take the square root: \( r = \sqrt{58} \).The radius is a measure of size for the sphere, and knowing how to retrieve it will help you in various applications, from geometry problems to practical real-world measurements.

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Most popular questions from this chapter

Equilibrium of Forces The forces \(\mathbf{F}_{1}, \mathbf{F}_{2}, \ldots, \mathbf{F}_{n}\) acting at the same point \(P\) are said to be in equilibrium if the resultant force is zero, that is, if \(\mathbf{F}_{1}+\mathbf{F}_{2}+\cdots+\mathbf{F}_{n}=0 .\) Find (a) the resultant forces acting at \(P,\) and (b) the additional force required (if any) for the forces to be in equilibrium. $$\begin{aligned} &\mathbf{F}_{1}=4 \mathbf{i}-\mathbf{j}, \quad \mathbf{F}_{2}=3 \mathbf{i}-7 \mathbf{j}, \quad \mathbf{F}_{3}=-8 \mathbf{i}+3 \mathbf{j}\\\&\mathbf{F}_{4}=\mathbf{i}+\mathbf{j}\end{aligned}$$

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Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) be vectors, and let \(a\) be a scalar. Prove the given property. $$(\mathbf{u}-\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}$$

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