/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Find the standard equation of th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 35

Find the standard equation of the sphere. $$ \text { Center: }(1,1,5) ; \text { radius: } 3 $$

Short Answer

Expert verified
The standard equation of the sphere with center at (1,1,5) and radius 3 is \((x-1)^2 + (y-1)^2 + (z-5)^2 = 9\).

Step by step solution

01

Identify the Components of the Equation

Given that the center of the sphere is at (1,1,5) and radius \(r\) is equal to 3. The formula for the standard equation of a sphere is \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\). In this equation (a,b,c) stand for the coordinates of the center and \(r\) for the radius of the sphere.
02

Substitute Given Values Into the Standard Equation

Substitute a=1, b=1, c=5 and \(r\) =3 into the formula. Doing this we get, \((x-1)^2 + (y-1)^2 + (z-5)^2 = 3^2\)
03

Simplify the Equation

The equation becomes \((x-1)^2 + (y-1)^2 + (z-5)^2 = 9\).

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 91影视!

Key Concepts

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

Sphere
A sphere is a perfectly round three-dimensional shape, much like a ball or globe. Its defining characteristic is that every point on its surface is equidistant from its center. This property makes it one of the most symmetrical objects in geometry.

In mathematics, spheres are important shapes, and they can be described using equations. The standard equation of a sphere helps us define it in a coordinate system, providing a way to calculate positions and distances related to the sphere. This equation reflects the balance and symmetry inherent in its shape.
  • The formula \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\) represents the sphere in a three-dimensional space.
  • Every point \(x, y, z\) that satisfies this equation falls on the sphere's surface.
  • The center \((a, b, c)\) and radius \(r\) are crucial in defining the exact size and position of the sphere.
Radius
The radius of a sphere is the distance from its center to any point on its surface. This is a constant measure because all points on the surface are the same distance from the center. The radius is half the diameter of the sphere and is a key factor in calculating many of its properties.

In the standard sphere equation, the radius \(r\) appears squared on the right side of the equation as \(r^2\). This is because in geometry, distances are often squared鈥攃alculating squared deviations is a fundamental part of many geometric relationships.
  • Mathematicians use the radius to calculate the sphere's surface area, using the formula \(4\pi r^2\).
  • It's also used to find the volume of the sphere, calculated by \(\frac{4}{3}\pi r^3\).
  • In our example, the sphere's radius is 3, making its role critical in determining the size and extent of the sphere.
Center of a Sphere
The center of a sphere is a point in space that is exactly in the middle of the sphere. It acts like the heart of the sphere, where if you could insert lines from the surface, they would all meet at this central point.

In the standard equation of a sphere \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\), the coordinates \((a, b, c)\) specify the center of the sphere. Each of these values adjusts the position of the sphere along the respective axis in a 3D coordinate system.
  • The center helps define the sphere's location in space relative to other objects.
  • Changing the center coordinates simply moves the sphere without altering its size or shape.
  • In our example, the center is located at \(1, 1, 5\), positioning the sphere uniquely within the coordinate system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Examine the function for relative extrema and saddle points. $$ f(x, y)=x^{2}-y^{2}+4 x-4 y-8 $$

Evaluate the double integral. $$ \int_{0}^{1} \int_{0}^{x} \sqrt{1-x^{2}} d y d x $$

Evaluate the double integral. $$ \int_{0}^{2} \int_{0}^{2}\left(6-x^{2}\right) d y d x $$

Examine the function for relative extrema and saddle points. $$ f(x, y)=-3 x^{2}-2 y^{2}+3 x-4 y+5 $$

Find the critical points and test for relative extrema. List the critical points for which the Second-Partials Test fails. $$ f(x, y)=x^{3}+y^{3}-3 x^{2}+6 y^{2}+3 x+12 y+7 $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.