/*! 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 52 Find an equation of the surface ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 52

Find an equation of the surface consisting of all points \(P(x, y, z)\) that are twice as far from the plane \(z=-1\) as from the point \((0,0,1),\) Identify the surface.

Short Answer

Expert verified
The surface is a hyperboloid of one sheet.

Step by step solution

01

Identify the Distances

To find the equation, we need to express the distances from a point \(P(x, y, z)\) to the plane \(z = -1\) and the point \((0, 0, 1)\). The distance from \(P\) to the plane is \(|z + 1|\), and the distance from \(P\) to the point \((0, 0, 1)\) is \(\sqrt{x^2 + y^2 + (z - 1)^2}\).
02

Set up the Relationship

The problem states that the distance to the plane is twice the distance to the point. Therefore, we write the equation: \[ |z + 1| = 2\sqrt{x^2 + y^2 + (z - 1)^2} \]
03

Eliminate the Absolute Value

Since \(|z + 1|\) is the absolute distance to the plane, we consider both cases: when \(z + 1 \geq 0\) and when \(z + 1 < 0\). However, since it's the absolute value of the distance, \(|z + 1| = z + 1\) can be used as is because it will simplify correctly.
04

Square Both Sides

To remove the square root, we square both sides of the equation from Step 2:\[ (z + 1)^2 = 4(x^2 + y^2 + (z - 1)^2) \]
05

Simplify the Equation

Expand both sides:- Left side: \((z + 1)^2 = z^2 + 2z + 1\)- Right side: \(4(x^2 + y^2 + z^2 - 2z + 1) = 4x^2 + 4y^2 + 4z^2 - 8z + 4\)Equating both sides gives:\[ z^2 + 2z + 1 = 4x^2 + 4y^2 + 4z^2 - 8z + 4 \]
06

Rearrange and Simplify Further

Rearrange the terms:\[ z^2 + 2z + 1 - 4z^2 + 8z - 4 = 4x^2 + 4y^2 \]Simplify:\[ -3z^2 + 10z - 3 = 4x^2 + 4y^2 \]Divide every term by \(-3\) to make it a standard form:\[ z^2 - \frac{10}{3}z + 1 = -\frac{4}{3}(x^2 + y^2) \]
07

Identify the Surface

The final equation \( z^2 - \frac{10}{3}z + 1 = -\frac{4}{3}(x^2 + y^2) \) can be identified as a hyperboloid of one sheet if we consider transformations to standard forms of quadratic surfaces based on coefficients and signs.

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.

Distance Calculation
In this context, calculating the distance involves both a plane and a point. To determine an equation for the given surface, we must assess how far a generic point \( P(x, y, z) \) is from the specific plane \( z = -1 \) and the point \( (0, 0, 1) \).
  • Distance to the Plane: The formula for the distance from a point to the plane \( z = -1 \) is given by \( |z + 1| \). This is because the plane's equation can be rewritten as \( z + 1 = 0 \), and thus the absolute value indicates the perpendicular distance in the \( z \)-axis direction.
  • Distance to the Point: To calculate the distance to the point \( (0, 0, 1) \), utilize the 3D distance formula: \( \sqrt{x^2 + y^2 + (z - 1)^2} \). This expression includes the differences in each coordinate, squared, and summed under a square root, as per Pythagoras' theorem generalized in three dimensions.
By setting these distances relative to one another, we formulate an equation that will help identify the surface.
Hyperboloid of One Sheet
A hyperboloid of one sheet is a fascinating quadric surface characterized by its saddle-like shape. This type of surface can be expressed in several standard forms, generally involving squares of the \( x \), \( y \), and \( z \) coordinates.
  • Typically takes the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \), where one variable has a negative coefficient, typifying the surface鈥檚 hyperbolic nature.
  • It allows slicing through with hyperbolas and ellipses, depending on the plane of intersection, which causes the surface to appear as one continuous sheet.
  • In our problem, the transformed equation resembles a hyperboloid of one sheet by having differing coefficients and a negative sign playing a crucial role in its identification.
Recognizing these surface properties can help not only in solving equations but also in visualizing the geometry involved.
Equations of Surfaces
Manipulating equations is crucial in identifying surfaces described by specific conditions. Transforming expressions to standard forms is one useful technique.
  • Equation Setup: Initially, equations reflect the conditions set by the problem 鈥� in this case, \( |z + 1| = 2\sqrt{x^2 + y^2 + (z - 1)^2} \).
  • Simplifying: Through algebraic transformations such as removing absolute values and squaring both equation sides, we can simplify complex expressions.
  • Reformulating: Rewriting facilitates matching to known surface forms, such as the hyperboloid of one sheet. Adjusting the final form \( z^2 - \frac{10}{3}z + 1 = -\frac{4}{3}(x^2 + y^2) \) reveals familiar properties.
These steps mirror the approach in tackling surface equations and tease apart relationships among geometric elements represented in algebraic forms.

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

Let \(L\) be the line that passes through the point \(\left(x_{0}, y_{0}, z_{0}\right)\) and is parallel to the vector \(\mathbf{v}=\langle a, b, c\rangle,\) where \(a, b,\) and \(c\) are nonzero. Show that a point \((x, y, z)\) lies on the line \(L\) if and only if $$ \frac{x-x_{0}}{a}=\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c} $$ These equations, which are called the symmetric equations of \(L,\) provide a nonparametric representation of \(L .\)

An equation of a surface is given in rectangular coordinates. Find an equation of the surface in (a) cylindrical coordinates and (b) spherical coordinates. $$ z^{2}=x^{2}-y^{2} $$

Convert from cylindrical to spherical coordinates. $$ \begin{array}{ll}{\text { (a) }(\sqrt{3}, \pi / 6,3)} & {\text { (b) }(1, \pi / 4,-1)} \\ {\text { (c) }(2,3 \pi / 4,0)} & {\text { (d) }(6,1,-2 \sqrt{3})}\end{array} $$

Find an equation of the plane that satisfies the stated conditions. The plane through the point \((-1,4,2)\) that contains the line of intersection of the planes \(4 x-y+z-2=0\) and \(2 x+y-2 z-3=0\)

Let \(L_{1}\) and \(L_{2}\) be the lines whose parametric equations are $$ \begin{array}{ll}{L_{1}: x=1+2 t,} & {y=2-t, \quad z=4-2 t} \\ {L_{2}: x=9+t,} & {y=5+3 t, \quad z=-4-t}\end{array} $$ (a) Show that \(L_{1}\) and \(L_{2}\) intersect at the point \((7,-1,-2) .\) (b) Find, to the nearest degree, the acute angle between \(L_{1}\) and \(L_{2}\) at their intersection. (c) Find parametric equations for the line that is perpendicular to \(L_{1}\) and \(L_{2}\) and passes through their point of intersection.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

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.