/*! 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 1 Consider the ellipsoid \(x^{2}+y... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the ellipsoid \(x^{2}+y^{2}+4 z^{2}=12 .\) (a) Find an equation of the tangent plane to the ellipsoid at the point \((2,2,1) .\) (b) Find parametric equations of the line that is normal to the ellipsoid at the point \((2,2,1) .\) (c) Find the acute angle that the tangent plane at the point \((2,2,1)\) makes with the \(x y\) -plane.

Short Answer

Expert verified
(a) The tangent plane is \(4x + 4y + 8z = 24\). (b) Parametric equations are \(x = 2 + 4t\), \(y = 2 + 4t\), \(z = 1 + 8t\). (c) The acute angle is \(\cos^{-1}\left( \frac{2}{\sqrt{6}} \right)\).

Step by step solution

01

Identify the Gradient

To find the tangent plane to the ellipsoid at a given point, we need the gradient of the ellipsoid's equation. The equation of the ellipsoid is \(x^2 + y^2 + 4z^2 = 12\). The gradient vector \(abla f(x, y, z)\) is computed as the partial derivatives: \(\frac{\partial f}{\partial x} = 2x\), \(\frac{\partial f}{\partial y} = 2y\), \(\frac{\partial f}{\partial z} = 8z\). Thus, \[ abla f(x, y, z) = (2x, 2y, 8z) \].
02

Evaluate the Gradient at the Point

Now, evaluate the gradient at the point \((2,2,1)\):\[ abla f(2, 2, 1) = (2(2), 2(2), 8(1)) = (4, 4, 8) \].
03

Write the Equation of the Tangent Plane

Using the gradient, the equation of the tangent plane at \((x_0, y_0, z_0)\) is \[2(x_0)(x-x_0) + 2(y_0)(y-y_0) + 8(z_0)(z-z_0) = 0. \]Substitute \((x_0, y_0, z_0) = (2, 2, 1)\) and \(abla f(2, 2, 1) = (4, 4, 8)\) into the formula:\[ 4(x - 2) + 4(y - 2) + 8(z - 1) = 0, \]which simplifies to \[ 4x + 4y + 8z = 24. \]
04

Parametrize the Normal Line

A normal line can be described by a direction vector through a point. Here, the direction vector is the gradient vector at \((2, 2, 1)\), i.e., \((4, 4, 8)\). The parametric equations starting from \((2, 2, 1)\) are: \[ x = 2 + 4t, \]\[ y = 2 + 4t, \]\[ z = 1 + 8t, \]where \(t\) is the parameter.
05

Calculate the Angle Between the Tangent Plane and XY-Plane

The angle between two planes can be found using the normal vectors of the planes and their dot product. The normal vector to the XY-plane is \((0, 0, 1)\), and the normal vector to the tangent plane is \((4, 4, 8)\). The dot product is:\[ (4, 4, 8) \cdot (0, 0, 1) = 8. \]The magnitudes are:\[ ||(4, 4, 8)|| = \sqrt{4^2 + 4^2 + 8^2} = \sqrt{96} = 4\sqrt{6}, \]\[ ||(0, 0, 1)|| = 1. \]Thus, the cosine of the angle \(\theta\) is given by:\[ \cos \theta = \frac{8}{4\sqrt{6}} = \frac{2}{\sqrt{6}}. \]The acute angle \(\theta\) is:\[ \theta = \cos^{-1}\left( \frac{2}{\sqrt{6}} \right). \]

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.

Gradient Vector
A gradient vector is crucial when finding the equation of a tangent plane to a surface defined by a scalar function. In simple terms, the gradient vector provides the direction of the steepest ascent for a multivariable function. For the ellipsoid given by the equation \(x^2 + y^2 + 4z^2 = 12\), the gradient vector is found by computing the partial derivatives of each variable in the function:
  • \(\frac{\partial f}{\partial x} = 2x\)
  • \(\frac{\partial f}{\partial y} = 2y\)
  • \(\frac{\partial f}{\partial z} = 8z\)
Evaluating these at point \((2, 2, 1)\), we determine the gradient vector as \((4, 4, 8)\). This vector not only helps in defining the tangent plane but also plays a key role in constructing the equation of any normal line to the surface at a specific point.
Parametric Equations
Parametric equations are helpful in expressing lines and curves by setting each spatial variable as a function of one or more parameters. In the context of the normal line to an ellipsoid, the parametric equations allow us to describe the line in terms of a single variable, often denoted by \(t\). For instance, given a point \((2, 2, 1)\) on our ellipsoid and a direction vector \((4, 4, 8)\), the parametric equations of the normal line are derived as follows:
  • \(x = 2 + 4t\)
  • \(y = 2 + 4t\)
  • \(z = 1 + 8t\)
Here, \(t\) varies over all real numbers, and each equation represents one dimension of space. As \(t\) changes, the equations describe a line traveling through the point \((2, 2, 1)\) along the direction of the gradient vector \((4, 4, 8)\) that is normal to the surface.
Angle Between Planes
Determining the angle between two planes involves understanding their normal vectors. For two planes with normal vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\), the cosine of the angle \(\theta\) between them is derived from the dot product of the vectors. The formula goes:\[\cos\theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{||\mathbf{n}_1|| \cdot ||\mathbf{n}_2||}\]In our example, the normal vector to the tangent plane at \((2, 2, 1)\) is \((4, 4, 8)\) and for the \(xy\)-plane, it is \((0, 0, 1)\). The dot product is 8, and their magnitudes are \(4\sqrt{6}\) and 1 respectively. Thus, the acute angle \(\theta\) between the planes is:\[\theta = \cos^{-1}\left( \frac{2}{\sqrt{6}} \right)\]This angle measures how steeply the tangent plane to the ellipsoid inclines relative to the horizontal \(xy\)-plane.
Normal Line
A normal line to a surface at a given point is perpendicular to the tangent plane at that point. Constructing a normal line can be approached similarly to parametrizing a line, where we use a point on the surface and the vector normal to the surface at that point (from the gradient). For the ellipsoid in our case, where the point is \((2, 2, 1)\) and the normal vector \((4, 4, 8)\), the normal line is defined by parametric equations:
  • \(x = 2 + 4t\)
  • \(y = 2 + 4t\)
  • \(z = 1 + 8t\)
These are created by stepping from the point \((2, 2, 1)\) in the direction \((4, 4, 8)\). This method ensures the line remains perpendicular to the tangent plane, thus maintaining the geometric definition of a normal line.

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

Confirm that the mixed second-order partial derivatives of \(f\) are the same. $$ f(x, y)=e^{x} \cos y $$

Solve using Lagrange multipliers. Find a vector in 3 -space whose length is 5 and whose components have the largest possible sum.

The length, width, and height of a rectangular box are \(l=5, w=2,\) and \(h=3,\) respectively. (a) Find the instantaneous rate of change of the volume of the box with respect to the length if \(w\) and \(h\) are held constant. (b) Find the instantaneous rate of change of the volume of the box with respect to the width if \(l\) and \(h\) are held constant. (c) Find the instantaneous rate of change of the volume of the box with respect to the height if \(l\) and \(w\) are held constant.

The volume \(V\) of a right circular cylinder is given by the formula \(V=\pi r^{2} h\), where \(r\) is the radius and \(h\) is the height. (a) Find a formula for the instantaneous rate of change of \(V\) with respect to \(r\) if \(r\) changes and \(h\) remains constant. (b) Find a formula for the instantaneous rate of change of \(V\) with respect to \(h\) if \(h\) changes and \(r\) remains constant. (c) Suppose that \(h\) has a constant value of 4 in, but \(r\) varies. Find the rate of change of \(V\) with respect to \(r\) at the point where \(r=6 \mathrm{in}\). (d) Suppose that \(r\) has a constant value of \(8 \mathrm{in},\) but \(h\) varies. Find the instantaneous rate of change of \(V\) with respect to \(h\) at the point where \(h=10\) in.

Confirm that the mixed second-order partial derivatives of \(f\) are the same. $$ f(x, y)=\left(x^{2}-y^{2}\right) /\left(x^{2}+y^{2}\right) $$

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.