/*! 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 75 Let \(z=3 x^{2}-y^{2}\). Find al... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 75

Let \(z=3 x^{2}-y^{2}\). Find all points at which \(\|\nabla z\|=6\).

Short Answer

Expert verified
The points lie on the ellipse \(9x^2 + y^2 = 9\).

Step by step solution

01

Find the Gradient

First, calculate the gradient of the function \(z = 3x^{2} - y^{2}\). The gradient \(abla z\) is given by \(\left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right)\). Calculate each partial derivative:\[\frac{\partial z}{\partial x} = 6x\]\[\frac{\partial z}{\partial y} = -2y\]Thus, \(abla z = (6x, -2y)\).
02

Calculate the Magnitude of the Gradient

To find \(\|abla z\|\), calculate the magnitude of the gradient vector \(abla z = (6x, -2y)\):\[\|abla z\| = \sqrt{(6x)^{2} + (-2y)^{2}}\]Simplify the expression:\[\|abla z\| = \sqrt{36x^{2} + 4y^{2}}\]
03

Set Magnitude Equal to Given Value and Solve

Set the calculated magnitude of the gradient equal to 6:\[\sqrt{36x^{2} + 4y^{2}} = 6\]Square both sides to eliminate the square root:\[36x^{2} + 4y^{2} = 36\]Divide through by 4:\[9x^{2} + y^{2} = 9\]
04

Identify and Describe Solution

The equation \(9x^{2} + y^{2} = 9\) represents an ellipse centered at the origin. Solve for points on this ellipse by rearranging if needed to highlight the relationship:\[\frac{x^{2}}{1} + \frac{y^{2}}{9} = 1\]This describes an ellipse with semi-major axis along the \(y\)-axis and semi-minor axis along the \(x\)-axis, where the semi-minor axis length is 1, and the semi-major axis length is 3.

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.

Partial Derivative
The partial derivative is a fundamental concept in calculus used to measure how a function changes as its variables change. When dealing with a multivariable function like \(z = 3x^2 - y^2\), the partial derivatives tell us how \(z\) changes with slight changes in just one variable while keeping the other variable constant.
  • Partial Derivative with respect to \(x\): For \(z = 3x^2 - y^2\), the partial derivative with respect to \(x\) is found by differentiating the function treating \(y\) as a constant. Thus we get \(\frac{\partial z}{\partial x} = 6x\).
  • Partial Derivative with respect to \(y\): Similarly, we find the partial derivative with respect to \(y\) by treating \(x\) as a constant, resulting in \(\frac{\partial z}{\partial y} = -2y\).
These derivatives form a vector known as the gradient, pointing in the direction of the steepest increase of the function.
Magnitude of Gradient
The magnitude of a gradient vector \(abla z\) provides insight into how rapidly a function is changing. For the function \(z = 3x^2 - y^2\), its gradient is given by the vector \((6x, -2y)\).
  • Magnitude Calculation: To find the magnitude \(\|abla z\|\), we use the formula for the magnitude of a vector: \(\sqrt{(6x)^2 + (-2y)^2}\).
  • Simplification: Simplifying this expression gives \(\|abla z\| = \sqrt{36x^2 + 4y^2}\).
  • Interpretation: Setting this equal to a given value, like 6 in the exercise, allows us to solve for points where the change represented by the gradient is exactly that value.
The gradient's magnitude can tell us about the slope's steepness for a given surface, indicating critical changes.
Ellipse
An ellipse is a geometric figure representing the set of points where the sum (or difference) of distances from two fixed points, called foci, is constant. In this problem, the equation transformed into \(9x^2 + y^2 = 9\) represents an ellipse.
  • Rewriting the Equation: To make it clear, express it as \(\frac{x^2}{1} + \frac{y^2}{9} = 1\), which showcases its elliptical shape.
  • Axis Orientation & Lengths: This equation tells us that the ellipse is centered at the origin, with its semi-major axis along the y-axis (length 3) and semi-minor axis along the x-axis (length 1).
  • Geometric Meaning: The orientation and lengths indicate how stretched the ellipse is in each direction.
Understanding the equation of an ellipse allows us not only to recognize its shape but also to determine critical points and their properties.

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

Find a unit vector in the direction in which \(f\) decreases most rapidly at \(P\), and find the rate of change of \(f\) at \(P\) in that direction. $$ f(x, y, z)=\frac{x+z}{z-y} ; P(5,7,6) $$

The length, width, and height of a rectangular box are measured with errors of at most \(r \%\) (where \(r\) is small). Use differentials to approximate the maximum percentage error in the computed value of the volume.

Find \(\nabla z\) or \(\nabla w\). $$ w=\ln \sqrt{x^{2}+y^{2}+z^{2}} $$

Locate all relative maxima, relative minima, and saddle points, if any. $$ f(x, y)=x e^{y} $$

Find the directional derivative of \(f\) at \(P\) in the direction of a. $$ f(x, y, z)=e^{x+y+3 z} ; \quad P(-2,2,-1) ; \mathbf{a}=20 \mathbf{i}-4 \mathbf{j}+5 \mathbf{k} $$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

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.