Chapter 14: Problem 39
Exer. \(39-50:\) Sketch the graph of the equation. $$ x^{2}+y^{2}+z^{2}-14 x+6 y-8 z+10=0 $$
Short Answer
Expert verified
The graph is a sphere centered at (7, -3, 4) with radius 8.
Step by step solution
01
Identify the Type of Surface
The given equation is of the form \(x^2 + y^2 + z^2 + Ax + By + Cz + D = 0\). This indicates the equation of a sphere in 3D space.
02
Rearrange Equation
The original equation is \(x^2 + y^2 + z^2 - 14x + 6y - 8z + 10 = 0\). Rearrange by moving the constant to one side: \(x^2 - 14x + y^2 + 6y + z^2 - 8z = -10\).
03
Complete the Square for Each Variable
Complete the square for each variable:- For \(x\): \(x^2 - 14x\) becomes \((x-7)^2 - 49\)- For \(y\): \(y^2 + 6y\) becomes \((y+3)^2 - 9\)- For \(z\): \(z^2 - 8z\) becomes \((z-4)^2 - 16\)
04
Substitute Completed Squares into Equation
Replace the completed squares back into the equation:\((x-7)^2 - 49 + (y+3)^2 - 9 + (z-4)^2 - 16 = -10\)
05
Simplify the Equation
Add the constants on the left-hand side and simplify:\((x-7)^2 + (y+3)^2 + (z-4)^2 = 64\). This is the equation of a sphere with center \((7, -3, 4)\) and radius \(8\), since \(64 = 8^2\).
06
Sketch the Sphere
To sketch the sphere:- Draw a 3D coordinate system.- Plot the center at \((7, -3, 4)\).- Draw a sphere with radius \(8\) centered at this point. Ensure to depict the symmetry in all directions.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of a Sphere
In three-dimensional geometry, a sphere is a set of points that are all the same distance from a fixed point, known as the center. This distance is called the radius. The general equation of a sphere in three-dimensional space is given by \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\] where - \((h, k, l)\) represent the coordinates of the center of the sphere, - \(r\) is the radius of the sphere.In the problem's original form, we had the equation \(x^2 + y^2 + z^2 - 14x + 6y - 8z + 10 = 0\).To convert this to the standard equation of a sphere, we completed the square for each variable, revealing the sphere's center and radius.
Completing the Square
Completing the square is a mathematical technique used to transform a quadratic expression into a perfect square trinomial. This method is essential for converting the general equation of a circle or sphere into its standard form. In the given problem, the equation had terms like \(x^2 - 14x\), \(y^2 + 6y\),and \(z^2 - 8z\).To complete the square:- For \(x\), find half of \(-14\) (which is \(-7\)) and square it to get \(49\). This makes \(x^2 - 14x\) into \((x-7)^2 - 49\).- For \(y\), half of \(6\) is \(3\), and squaring gives \(9\), so \(y^2 + 6y\) becomes \((y+3)^2 - 9\).- For \(z\), half of \(-8\) is \(-4\), and squaring yields \(16\). Hence, \(z^2 - 8z\) turns into \((z-4)^2 - 16\).These adjustments allow us to simplify and rearrange the equation into a more recognizable form.
Graphing 3D Surfaces
Graphing in three-dimensional space can be a challenging but rewarding task, especially when it involves shapes like spheres. To graph a sphere from its equation, it is essential to first identify the center and radius from the standard form equation.For the sphere determined by our exercise:- Identify the center at \((7, -3, 4)\),- Recognize the radius \(8\), since \(64 = 8^2\).Once these elements are known, you can proceed with sketching:
- Begin by drawing a three-dimensional coordinate system.
- Plot the center of the sphere at the identified coordinates.
- Use the radius to determine the size of the sphere, ensuring symmetry in all directions from the center.