/*! 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 57 Consider the following equations... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the following equations of quadric surfaces. a. Find the intercepts with the three coordinate axes, when they exist. b. Find the equations of the x y-, x z^{-}, \text {and } y z-\text {traces, when they exist. c. Sketch a graph of the surface. $$\frac{y^{2}}{16}+36 z^{2}-\frac{x^{2}}{4}-9=0$$

Short Answer

Expert verified
Based on the given information and the steps taken to analyze and solve the problem, the quadric surface represents a hyperboloid of one sheet with no x-intercept, y-intercepts at (0, ±12, 0), z-intercepts at (0, 0, ±0.5), a hyperbola in the xy-plane and xz-plane, and an ellipse in the yz-plane.

Step by step solution

01

Find the intercepts with the coordinate axes

Since there are three axes, we will solve for each axis. 1. x-axis: Set y = 0 and z = 0: $$\frac{0}{16} + 36(0)^2 - \frac{x^2}{4} - 9 = 0$$ $$\frac{x^2}{4} = -9$$ There is no real value of x that would result in x-intercept. 2. y-axis: Set x = 0 and z = 0: $$\frac{y^2}{16} + 36(0)^2 - \frac{0}{4} - 9 = 0$$ $$\frac{y^2}{16} = 9$$ $$y^2 = 144$$ $$y = \pm 12$$ So, y-intercepts are (0, 12, 0) and (0, -12, 0). 3. z-axis: Set x = 0 and y = 0: $$\frac{0}{16} + 36z^2 - \frac{0}{4} - 9 = 0$$ $$36z^2 = 9$$ $$z^2 = \frac{1}{4}$$ $$z = \pm \frac{1}{2}$$ So, z-intercepts are (0, 0, 0.5) and (0, 0, -0.5).
02

Find the equations of xy-, xz-, and yz- traces

1. xy-trace: Set z = 0: $$\frac{y^2}{16} + 36(0)^2 - \frac{x^2}{4} - 9 = 0$$ $$\frac{y^2}{16} - \frac{x^2}{4} = 9$$ This is the equation of an hyperbola in the xy-plane. 2. xz-trace: Set y = 0: $$\frac{0}{16} + 36z^2 - \frac{x^2}{4} - 9 = 0$$ $$36z^2 - \frac{x^2}{4} = 9$$ This is the equation of an hyperbola in the xz-plane. 3. yz-trace: Set x = 0: $$\frac{y^2}{16} + 36z^2 - \frac{0}{4} - 9 = 0$$ $$\frac{y^2}{16} + 36z^2 = 9$$ This is the equation of an ellipse in the yz-plane.
03

Sketch the graph

We've found the following information: Intercepts: - No x-intercept - y-intercepts at (0, ±12, 0) - z-intercepts at (0, 0, ±0.5) Traces: - Hyperbola in the xy-plane - Hyperbola in the xz-plane - Ellipse in the yz-plane Now, we can sketch the graph of the quadric surface with the given information. Using a graphing tool or manual drawing, plot the points and traces to see the surface. The surface is a hyperboloid of one sheet.

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.

Coordinate Axis Intercepts
Understanding the coordinate axis intercepts is crucial when studying quadric surfaces. These intercepts are the points where the surface crosses the x, y, and z-axes. To find them, we solve the given equation by setting two variables to zero.

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 \(h\) be continuous for all real numbers. a. Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=\int_{x}^{y} h(s) d s\). b. Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=\int_{1}^{x y} h(s) d s\).

Find the points (if they exist) at which the following planes and curves intersect. $$y=2 x+1 ; \quad \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$

An important derivative operation in many applications is called the Laplacian; in Cartesian coordinates, for \(z=f(x, y),\) the Laplacian is \(z_{x x}+z_{y y} .\) Determine the Laplacian in polar coordinates using the following steps. a. Begin with \(z=g(r, \theta)\) and write \(z_{x}\) and \(z_{y}\) in terms of polar coordinates (see Exercise 64). b. Use the Chain Rule to find \(z_{x x}=\frac{\partial}{\partial x}\left(z_{x}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. c. Use the Chain Rule to find \(z_{y y}=\frac{d}{\partial y}\left(z_{y}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. d. Combine parts (b) and (c) to show that $$z_{x x}+z_{y y}=z_{r r}+\frac{1}{r} z_{r}+\frac{1}{r^{2}} z_{\theta \theta}$$

Suppose \(n\) houses are located at the distinct points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) .\) A power substation must be located at a point such that the sum of the squares of the distances between the houses and the substation is minimized. a. Find the optimal location of the substation in the case that \(n=3\) and the houses are located at \((0,0),(2,0),\) and (1,1) b. Find the optimal location of the substation in the case that \(n=3\) and the houses are located at distinct points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right),\) and \(\left(x_{3}, y_{3}\right)\) c. Find the optimal location of the substation in the general case of \(n\) houses located at distinct points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots\) \(\left(x_{n}, y_{n}\right)\) d. You might argue that the locations found in parts (a), (b), and (c) are not optimal because they result from minimizing the sum of the squares of the distances, not the sum of the distances themselves. Use the locations in part (a) and write the function that gives the sum of the distances. Note that minimizing this function is much more difficult than in part (a).

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter 14 .) The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(\varphi=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$ \mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle $$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=k Q / r^{2} .\) Explain why this relationship is called an inverse square law.

See all solutions

Recommended explanations on Math Textbooks

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.