/*! 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 the points at which the fol... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 52

Find the points at which the following surfaces have horizontal tangent planes. $$x^{2}+2 y^{2}+z^{2}-2 x-2 z-2=0$$

Short Answer

Expert verified
Answer: The surface has horizontal tangent planes at points \(P_1=(1, 0, 1+\sqrt{2})\) and \(P_2=(1, 0, 1-\sqrt{2})\).

Step by step solution

01

Recognize the equation of the surface

The given equation is: $$x^{2}+2 y^{2}+z^{2}-2 x-2 z-2=0$$ Rearrange the equation to make it more recognizable: $$\left(x^{2}-2x\right)+2y^{2}+\left(z^{2}-2z\right)-2=0$$ Now complete the squares for the x and z terms: $$\left(x-1\right)^{2}-1+2y^2+\left(z-1\right)^2-1=0$$ Combine the constants and get the equation for the surface: $$\left(x-1\right)^{2}+2y^{2}+\left(z-1\right)^2=2$$ This is the equation of an elliptical paraboloid centered at the point (1, 0, 1).
02

Find the Gradient of the Surface

To find the gradient (partial derivatives) of the surface, we need to differentiate with respect to x, y, and z. Considering the implicit form of the given surface equation: $$F(x,y,z) = x^{2}+2 y^{2}+z^{2}-2 x-2 z-2=0$$ Calculate the partial derivatives: $$F_x=\frac{\partial F}{\partial x}=2x-2$$ $$F_y=\frac{\partial F}{\partial y}=4y$$ $$F_z=\frac{\partial F}{\partial z}=2z-2$$ The gradient vector is: $$\nabla F = (2x-2, 4y, 2z-2)$$
03

Determine where the Gradient is equal to 0

The gradient vector \(\nabla F\) is perpendicular to the tangent plane of the surface. A horizontal tangent plane has a normal vector parallel to the z-axis, i.e., it has no x and y components. Thus, we need to find the points on the surface where \(\nabla F\) has no x and y components: $$2x-2 = 0 \Rightarrow x = 1$$ $$4y = 0 \Rightarrow y = 0$$ Now that we found the values of x and y where the gradient has no x and y components, we can find the corresponding z value using the surface equation: $$\left(1-1\right)^{2}+2(0)^{2}+\left(z-1\right)^2=2$$ $$\left(z-1\right)^2=2$$ $$z-1=\pm \sqrt{2}$$ $$z=1\pm \sqrt{2}$$ Thus, there are two points on the surface where the tangent plane is horizontal:
04

Find the Corresponding Points on the Surface

Using the coordinates we found above, the two points (x, y, z) where the given surface has horizontal tangent planes are: $$P_1=(1, 0, 1+\sqrt{2})$$ $$P_2=(1, 0, 1-\sqrt{2})$$

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!

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

Identify and briefly describe the surfaces defined by the following equations. $$z^{2}+4 y^{2}-x^{2}=1$$

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y)=\ln \left(1+x^{2}+y^{2}\right)$$

Production functions Economists model the output of manufacturing systems using production functions that have many of the same properties as utility functions. The family of Cobb-Douglas production functions has the form \(P=f(K, L)=C K^{a} L^{1-a},\) where K represents capital, L represents labor, and C and a are positive real numbers with \(0

Let \(E\) be the ellipsoid \(x^{2} / 9+y^{2} / 4+z^{2}=1, P\) be the plane \(z=A x+B y,\) and \(C\) be the intersection of \(E\) and \(P\). a. Is \(C\) an ellipse for all values of \(A\) and \(B\) ? Explain. b. Sketch and interpret the situation in which \(A=0\) and \(B \neq 0\). c. Find an equation of the projection of \(C\) on the \(x y\) -plane. d. Assume \(A=\frac{1}{6}\) and \(B=\frac{1}{2} .\) Find a parametric description of \(C\) as a curve in \(\mathbb{R}^{3}\). (Hint: Assume \(C\) is described by \(\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle\) and find \(a, b, c, d, e, \text { and } f .)\)

Limits at (0,0) may be easier to evaluate by converting to polar coordinates. Remember that the same limit must be obtained as \(r \rightarrow 0\) along all paths to (0,0) Evaluate the following limits or state that they do not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{(x-y)^{2}}{\left(x^{2}+y^{2}\right)^{3 / 2}}$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Discrete 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.