/*! 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 22 If \(F(x, y)=\ln \left(x^{2}+x y... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 22

If \(F(x, y)=\ln \left(x^{2}+x y+y^{2}\right)\), find \(F_{x}(-1,4)\) and \(F_{y}(-1,4)\)

Short Answer

Expert verified
\( F_x(-1,4) = \frac{2}{13} \) and \( F_y(-1,4) = \frac{7}{13} \).

Step by step solution

01

Understand the Problem

We need to find the partial derivatives \( F_x(x, y) \) and \( F_y(x, y) \) of the function \( F(x, y)=\ln (x^2+xy+y^2) \), and then evaluate these derivatives at the point \((-1, 4)\).
02

Find Partial Derivative with Respect to x

To find \( F_x(x, y) \), use the chain rule. Start by setting \( u = x^2 + xy + y^2 \). Then \( F(x, y) = \ln u \), so \( F_x = \frac{1}{u} \cdot \frac{\partial u}{\partial x} \). We have \( u = x^2 + xy + y^2 \), so \( \frac{\partial u}{\partial x} = 2x + y \). Therefore, \( F_x(x, y) = \frac{2x + y}{x^2 + xy + y^2} \).
03

Evaluate \( F_x(-1, 4) \)

Substitute \( x = -1 \) and \( y = 4 \) into the expression for \( F_x(x, y) \). Find \( F_x(-1, 4) = \frac{2(-1) + 4}{(-1)^2 + (-1)(4) + 4^2} = \frac{-2 + 4}{1 - 4 + 16} = \frac{2}{13} \).
04

Find Partial Derivative with Respect to y

To find \( F_y(x, y) \), again use the chain rule. Start with \( u = x^2 + xy + y^2 \) so \( F(x, y) = \ln u \), therefore \( F_y = \frac{1}{u} \cdot \frac{\partial u}{\partial y} \). Here, \( \frac{\partial u}{\partial y} = x + 2y \). Thus, \( F_y(x, y) = \frac{x + 2y}{x^2 + xy + y^2} \).
05

Evaluate \( F_y(-1, 4) \)

Substitute \( x = -1 \) and \( y = 4 \) into the expression for \( F_y(x, y) \). Find \( F_y(-1, 4) = \frac{-1 + 2(4)}{(-1)^2 + (-1)(4) + 4^2} = \frac{-1 + 8}{1 - 4 + 16} = \frac{7}{13} \).

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.

Chain Rule
The chain rule is a fundamental differentiation rule used frequently in calculus, especially when dealing with composite functions. It's crucial when working with functions depending on multiple variables, as seen in partial derivatives.

In the context of the exercise, the function is given by \[F(x, y) = \ln(x^2 + xy + y^2)\] and to differentiate this with respect to one variable, we use the chain rule. Here's how it helps:
  • Firstly, identify the inner function \(u = x^2 + xy + y^2\), which means \(F(x, y) = \ln u\).
  • To find the partial derivative \(F_x\) or \(F_y\), differentiate the outer function (\(\ln(u)\)) with respect to \(u\), giving \(\frac{1}{u}\).
  • Next, find the derivative of the inner function with respect to the desired variable: \(\frac{\partial u}{\partial x} = 2x + y\) for \(x\) and \(\frac{\partial u}{\partial y} = x + 2y\) for \(y\).
Using the chain rule, you can then reach the partial derivatives \(F_x\) and \(F_y\), expressed in terms of \(u\). This method simplifies evaluating derivatives for complex multivariable functions.
Evaluation at a Point
Once you've found the partial derivatives, evaluating them at a specific point involves substituting the coordinates into the derivative expressions.

For the problem, we need to evaluate both \(F_x(x, y)\) and \(F_y(x, y)\) at the point \((-1, 4)\). Here's the approach broken down:
  • Start with the partial derivatives found earlier: \(F_x(x, y) = \frac{2x + y}{x^2 + xy + y^2}\) and \(F_y(x, y) = \frac{x + 2y}{x^2 + xy + y^2}\).
  • Substitute \(x = -1\) and \(y = 4\) into these expressions.
  • For \(F_x(-1, 4)\), you calculate \(\frac{-2 + 4}{1 - 4 + 16} = \frac{2}{13}\).
  • Similarly, for \(F_y(-1, 4)\), you compute \(\frac{-1 + 8}{1 - 4 + 16} = \frac{7}{13}\).
Evaluating at a point gives specific insights into the behavior and slope of the function's graph at that particular location.
Logarithmic Functions
Logarithmic functions play a vital role in many areas of mathematics, particularly in calculus for differentiating composite expressions like in our exercise function \(F(x, y) = \ln(x^2 + xy + y^2)\).

Understanding their properties is key:
  • The derivative of \(\ln(u)\) with respect to \(u\) is simply \(\frac{1}{u}\). This fundamental rule makes logarithmic differentiation straightforward.
  • Logarithms are useful because they transform multiplicative relationships into additive ones, simplifying complex expressions before differentiation.
  • In the exercise, since \(F(x, y)\) involves \(\ln(u)\), calculating its partial derivatives is efficient using the chain rule.
Knowing how to handle logarithmic functions allows for precise calculation of derivatives, especially when dealing with products, quotients, or powers embedded within logarithmic expressions.

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 the directional derivative of \(f(x, y, z)=x y+z^{2}\) at \((1,1,1)\) in the direction toward \((5,-3,3)\).

If the temperature of a plate at the point \((x, y)\) is \(T(x, y)=10+x^{2}-y^{2}\), find the path a heat-seeking particle (which always moves in the direction of greatest increase in temperature) would follow if it starts at \((-2,1) .\) Hint: The particle moves in the direction of the gradient $$ \nabla T=2 x \mathbf{i}-2 y \mathbf{j} $$ We may write the path in parametric form as $$ \mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j} $$ and we want \(x(0)=-2\) and \(y(0)=1 .\) To move in the required direction means that \(\mathbf{r}^{\prime}(t)\) should be parallel to \(\nabla T\). This will be satisfied if $$ \frac{x^{\prime}(t)}{2 x(t)}=-\frac{y^{\prime}(t)}{2 y(t)} $$ together with the conditions \(x(0)=-2\) and \(y(0)=1\). Now solve this differential equation and evaluate the arbitrary constant of integration.

Use differentials to find the approximate amount of copper in the four sides and bottom of a rectangular copper tank that is 6 feet long, 4 feet wide, and 3 feet deep inside, if the sheet copper is \(\frac{1}{4}\) inch thick. Hint: Make a sketch.

Least Squares Given \(n\) points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots\), \(\left(x_{n}, y_{n}\right)\) in the \(x y\) -plane, we wish to find the line \(y=m x+b\) such that the sum of the squares of the vertical distances from the points to the line is a minimum; that is, we wish to minimize $$ f(m, b)=\sum_{i=1}^{n}\left(y_{i}-m x_{i}-b\right)^{2} $$ (a) Find \(\partial f / \partial m\) and \(\partial f / \partial b\), and set these results equal to zero. Show that this leads to the system of equations $$ \begin{aligned} m \sum_{i=1}^{n} x_{i}^{2}+b \sum_{i=1}^{n} x_{i} &=\sum_{i=1}^{n} x_{i} y_{i} \\\ m \sum_{i=1}^{n} x_{i}+n b &=\sum_{i=1}^{n} y_{i} \end{aligned} $$ (b) Solve this system for \(m\) and \(b\). (c) Use the Second Partials Test (Theorem C) to show that \(f\) is minimized for this choice of \(m\) and \(b\).

Find the minimum of \(f(x, y)=x^{2}+4 x y+y^{2}\) subject to the constraint \(x-y-6=0\).
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

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.