/*! 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 65 In Exercises \(63-66,\) use the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 65

In Exercises \(63-66,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$f(x, y)=\sqrt{2 x+3 y-1}, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(-2,3)$$

Short Answer

Expert verified
\( \frac{\partial f}{\partial x} = \frac{1}{2} \) at \((-2, 3)\) and \( \frac{\partial f}{\partial y} = \frac{3}{4} \) at \((-2, 3)\).

Step by step solution

01

Understand the Partial Derivative with Respect to x

The partial derivative of a function \( f(x, y) \) with respect to \( x \) evaluates the rate at which the function changes as \( x \) changes, while \( y \) is held constant. We use the limit definition: \( \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x + h, y) - f(x, y)}{h} \).
02

Apply the Limit Definition for \( \frac{\partial f}{\partial x} \)

Begin with the function \( f(x, y) = \sqrt{2x + 3y - 1} \). According to the limit definition, substitute \( f(x + h, y) \) and \( f(x, y) \):- \( f(x + h, y) = \sqrt{2(x + h) + 3y - 1} = \sqrt{2x + 2h + 3y - 1} \)- \( f(x, y) = \sqrt{2x + 3y - 1} \)Therefore, the expression becomes: \[ \frac{f(x + h, y) - f(x, y)}{h} = \frac{\sqrt{2x + 2h + 3y - 1} - \sqrt{2x + 3y - 1}}{h} \].
03

Simplify Using the Conjugate

Multiply and divide by the conjugate to simplify the expression:\[ \frac{\sqrt{2x + 2h + 3y - 1} - \sqrt{2x + 3y - 1}}{h} \times \frac{\sqrt{2x + 2h + 3y - 1} + \sqrt{2x + 3y - 1}}{\sqrt{2x + 2h + 3y - 1} + \sqrt{2x + 3y - 1}} \].This becomes:\[ \frac{(2x + 2h + 3y - 1) - (2x + 3y - 1)}{h(\sqrt{2x + 2h + 3y - 1} + \sqrt{2x + 3y - 1})} = \frac{2h}{h(\sqrt{2x + 2h + 3y - 1} + \sqrt{2x + 3y - 1})} \].
04

Evaluate the Limit for \( \frac{\partial f}{\partial x} \)

Cancel \( h \) and take the limit:\[ \frac{2h}{h(\sqrt{2x + 2h + 3y - 1} + \sqrt{2x + 3y - 1})} = \frac{2}{\sqrt{2x + 2h + 3y - 1} + \sqrt{2x + 3y - 1}} \].As \( h \to 0 \), the limit is:\[ \frac{2}{2\sqrt{2x + 3y - 1}} \].
05

Compute \( \frac{\partial f}{\partial x} \) at \( (x, y) = (-2, 3) \)

Substitute \( x = -2 \) and \( y = 3 \) into the expression:\[ \frac{2}{2\sqrt{2(-2) + 3(3) - 1}} = \frac{2}{2\sqrt{-4 + 9 - 1}} = \frac{2}{2\sqrt{4}} = \frac{2}{4} = \frac{1}{2} \].
06

Understand the Partial Derivative with Respect to y

Similar to the partial derivative with respect to \( x \), for \( y \), we hold \( x \) constant and use:\( \frac{\partial f}{\partial y} = \lim_{k \to 0} \frac{f(x, y + k) - f(x, y)}{k} \).
07

Apply the Limit Definition for \( \frac{\partial f}{\partial y} \)

Substitute \( f(x, y + k) \) into the function:- \( f(x, y + k) = \sqrt{2x + 3(y + k) - 1} = \sqrt{2x + 3y + 3k - 1} \)- \( f(x, y) = \sqrt{2x + 3y - 1} \)The expression becomes: \[ \frac{\sqrt{2x + 3y + 3k - 1} - \sqrt{2x + 3y - 1}}{k} \].
08

Simplify Using the Conjugate

Multiply and divide by the conjugate:\[ \frac{\sqrt{2x + 3y + 3k - 1} - \sqrt{2x + 3y - 1}}{k} \times \frac{\sqrt{2x + 3y + 3k - 1} + \sqrt{2x + 3y - 1}}{\sqrt{2x + 3y + 3k - 1} + \sqrt{2x + 3y - 1}} \].This becomes:\[ \frac{3k}{k(\sqrt{2x + 3y + 3k - 1} + \sqrt{2x + 3y - 1})} \].
09

Evaluate the Limit for \( \frac{\partial f}{\partial y} \)

Cancel \( k \) and take the limit:\[ \frac{3k}{k(\sqrt{2x + 3y + 3k - 1} + \sqrt{2x + 3y - 1})} = \frac{3}{\sqrt{2x + 3y + 3k - 1} + \sqrt{2x + 3y - 1}} \].As \( k \to 0 \), the limit is:\[ \frac{3}{2\sqrt{2x + 3y - 1}} \].
10

Compute \( \frac{\partial f}{\partial y} \) at \( (x, y) = (-2, 3) \)

Substitute \( x = -2 \) and \( y = 3 \) into the expression:\[ \frac{3}{2\sqrt{2(-2) + 3(3) - 1}} = \frac{3}{2\sqrt{-4 + 9 - 1}} = \frac{3}{2\sqrt{4}} = \frac{3}{4} \].

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.

Limit Definition
The limit definition is a fundamental concept in calculus, particularly when dealing with derivatives. It helps us understand how functions behave as they approach certain points or infinitesimal changes. When we talk about partial derivatives, add a slight twist: we focus on changes with respect to only one of several variables, keeping the others constant.

In the limit definition of a partial derivative, we see:
  • For the function \( f(x, y) \), the partial derivative with respect to \( x \) can be computed using:
\[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x + h, y) - f(x, y)}{h} \]
  • This expression captures how \( f(x, y) \) changes as \( x \) changes by a tiny amount \( h \), while \( y \) remains fixed.
  • The same concept applies to the partial derivative with respect to \( y \), swapping \( x \) and \( y \) in the definition.
Through these definitions, we apply mathematical precision to understand local behaviors of functions in multivariable contexts.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of multiple variables. When dealing with functions like \( f(x, y) \), rather than looking at a single input and output, we explore how changes in multiple inputs affect the output.

Key aspects in multivariable calculus include:
  • Partial Derivatives: Understanding partial derivatives is crucial because they describe how functions change parallel to each coordinate axis.
  • Gradient: Beyond single partial derivatives, the gradient vector combines all partial derivatives, providing a detailed sense of direction for steepest ascent or descent.
In practical situations, functions may depend on several variables, so multivariable calculus helps us break down and analyze their intricate changes.
Think of multivariable calculus as a toolkit to navigate the complex terrains of functions where changes in one variable can uniquely influence changes in others.
Rate of Change
The rate of change in calculus is essentially how one quantity varies with respect to another. When looking at functions, this signals how outputs or dependent variables shift as inputs or independent variables change.

Partial derivatives specifically tell us the rate of change for functions of several variables:
  • For a function \( f(x, y) \), the rate of change in \( f \) as \( x \) changes is given by the partial derivative \( \frac{\partial f}{\partial x} \).
  • Similarly, the change in \( f \) as \( y \) changes, with \( x \) held constant, is captured by \( \frac{\partial f}{\partial y} \).
This concept is helpful in many fields like physics, economics, and engineering where understanding how small changes influence outcomes is necessary. The rate of change gives insight into the behavior of complex systems and dynamics that rely on various inputs.

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

Suppose that the equation \(g(x, y, z)=0\) determines \(z\) as a differentiable function of the independent variables \(x\) and \(y\) and that \(g_{z} \neq 0 .\) Show that $$ \left(\frac{\partial z}{\partial y}\right)_{x}=-\frac{\partial g / \partial y}{\partial g / \partial z} $$

Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes \(y+2 z=12\) and \(x+y=6 .\)

In Exercises \(63-66,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$f(x, y)=4+2 x-3 y-x y^{2}, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(-2,1)$$

In Exercises \(71-76\) , you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$ \begin{array}{l}{f(x, y)=x^{4}+y^{2}-8 x^{2}-6 y+16,-3 \leq x \leq 3} \\ {-6 \leq y \leq 6}\end{array} $$

Consider the function \(f(x, y)=x^{2}+y^{2}+2 x y-x-y+1\) over the square \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\) a. Show that \(f\) has an absolute minimum along the line segment \(2 x+2 y=1\) in this square. What is the absolute minimum value? b. Find the absolute maximum value of \(f\) over the square.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

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.