/*! 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 56 Find the points on the surface \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the points on the surface \(y=4+x^{2}+z^{2}\) where the gradient is parallel to \(\vec{i}+\vec{j}+\vec{k}.\)

Short Answer

Expert verified
The point is \((-\frac{1}{2}, \frac{9}{2}, -\frac{1}{2})\).

Step by step solution

01

Understand the Problem

We are given a surface defined by the equation \(y = 4 + x^2 + z^2\) and are asked to find points where the gradient of this surface is parallel to the vector \(\vec{i} + \vec{j} + \vec{k}\). The gradient being parallel means that it will be a scalar multiple of this vector.
02

Find the Gradient of the Surface

The gradient of a function \(f(x, y, z)\) is given by \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\). Here, \(f(x, y, z) = 4 + x^2 + z^2 - y\) (by rearranging the given equation to zero). The gradient is: \[ abla f = (2x)\vec{i} - \vec{j} + (2z)\vec{k} \].
03

Set Up the Parallel Condition

The gradient being parallel to \(\vec{i} + \vec{j} + \vec{k}\) implies that \((2x, -1, 2z) = \lambda (1, 1, 1)\) for some scalar \(\lambda\). This gives us the system of equations: \(2x = \lambda\), \(-1 = \lambda\), and \(2z = \lambda\).
04

Solve for the Scalar and Variables

From \(-1 = \lambda\), we have \(\lambda = -1\). Substituting \(\lambda = -1\) into the other equations, we get:\[ 2x = -1 \Rightarrow x = -\frac{1}{2} \] \[ 2z = -1 \Rightarrow z = -\frac{1}{2} \].
05

Determine the Corresponding y-value

Substitute \(x = -\frac{1}{2}\) and \(z = -\frac{1}{2}\) into the surface equation \(y = 4 + x^2 + z^2\) to find \(y\). Calculating gives: \[ y = 4 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = 4 + \frac{1}{4} + \frac{1}{4} = 4 + \frac{1}{2} = \frac{9}{2} \].
06

Conclude the Solution

The point on the surface where the gradient is parallel to \(\vec{i} + \vec{j} + \vec{k}\) is \( \left(-\frac{1}{2}, \frac{9}{2}, -\frac{1}{2} \right)\).

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.

Parallel Vectors
Parallel vectors are vectors that have the same or exact opposite direction. To check if two vectors are parallel, we can see if one is a scalar multiple of the other. This means that there exists a number, known as a scalar, that when multiplied with one vector gives out the other. For example, consider vectors \( \vec{a} \) and \( \vec{b} \). If \( \vec{a} = k \cdot \vec{b} \), where \( k \) is a scalar, then these vectors are parallel. This often tells us that they point along the same line in space, although they may differ in magnitude or length.
In the context of the exercise, for the gradient of the surface to be parallel to the vector \( \vec{i} + \vec{j} + \vec{k} \), the gradient should become a scalar multiple of \( (1, 1, 1) \). That is, each component of the gradient should equate to a multiple of the components of \( \vec{i} + \vec{j} + \vec{k} \).
Understanding parallel vectors helps in identifying the direction and relation between various vectors in multivariable calculus.
Gradient Vector
The gradient vector \( abla f \) is a fundamental concept in multivariable calculus that conveys the direction and rate of greatest increase of a function of several variables. Imagine a mountain landscape, with the surface defined by a function \( f(x, y, z) \). The gradient tells us which path leads most steeply uphill from any point on this surface.
The gradient of a function \( f(x, y, z) \) is calculated by differentiating with respect to each variable: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
For the exercise, the surface equation was rearranged to \( f(x, y, z) = x^2 + z^2 - y + 4 \), resulting in a gradient \( abla f = (2x, -1, 2z) \). This gradient vector gives us the direction along which the value of the function increases most rapidly. The exercise seeks points where the gradient is parallel to a specific vector, indicating that the direction of steepest increase aligns with this vector.
Surface in Three Dimensions
A surface in three dimensions is a two-dimensional shape existing in three-dimensional space. This can represent anything from a simple plane to complex, undulating shapes like the surface of a mountain. In mathematics, such surfaces are often described by an equation in three variables, such as \( z = f(x, y) \), or, as in the exercise, \( y = 4 + x^2 + z^2 \).
This particular surface describes a paraboloid, which opens upward with its vertex at the minimum point where the three variables equate to minimize and satisfy the equation. To find where certain properties, such as parallel gradients, occur on this surface, we examine the function's equation and its derivatives.
By analyzing derivatives and gradients, students can identify key attributes of the surface. In the context of the problem, points were needed where the gradient aligns parallel to a specified vector, giving insights into where this surface trends in particular directions. Understanding the nature of these surfaces helps comprehend more intricate behaviors, their points, and properties in three-dimensional space.

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

In Problems \(58-64,\) find the quantity. Assume that \(g\) is a smooth function and that $$ \nabla g(2,3)=-2 \vec{i}+\vec{j} \quad \text { and } \quad \nabla g(2.4,3)=4 \vec{i} $$ The slope of the graph of \(g\) at the point (2,3) in the direction of the vector \(\vec{i}+3 \vec{j}\)

A student was asked to find the directional derivative of \(f(x, y)=x^{2} e^{y}\) at the point (1,0) in the direction of \(\vec{v}=4 \vec{i}+3 \vec{j} .\) The student's answer was $$ f_{i i}(1,0)=\operatorname{grad} f(1,0) \cdot \vec{u}=\frac{8}{5} \vec{i}+\frac{3}{5} \vec{j} $$ (a) At a glance, how do you know this is wrong? (b) What is the correct answer?

You are standing at the point (1,1,3) on the hill whose equation is given by \(z=5 y-x^{2}-y^{2}\) (a) If you choose to climb in the direction of steepest ascent, what is your initial rate of ascent relative to the horizontal distance? (b) If you decide to go straight northwest, will you be ascending or descending? At what rate? (c) If you decide to maintain your altitude, in what directions can you go?

the point is on the surface in 3 -space. (a) Find the differential of the equation (that is, of each side). (b) Find \(d z\) at the point. (c) Find an equation of the tangent plane to the surface at the point. $$2 x^{2}+13=y^{2}+3 z^{2},(2,3,2)$$

(a) Check the local linearity of \(f(x, y)=e^{-x} \sin y\) near \(x=1, y=2\) by making a table of values of \(f\) for \(x=0.9,1.0,1.1\) and \(y=1.9,2.0,2.1\) Express values of \(f\) with 4 digits after the decimal point. Then make a table of values for \(x=\) 0.99,1.00,1.01 and \(y=1.99,2.00 .2 .01,\) again showing 4 digits after the decimal point. Do both tables look nearly linear? Does the second table look more linear than the first? (b) Give the local linearization of \(f(x, y)=e^{-x} \sin y\) at \((1,2),\) first using your tables, and second using the fact that \(f_{x}(x, y)=-e^{-x} \sin y\) and \(f_{y}(x, y)=\) \(e^{-x} \cos y\)

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.