/*! 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 24 Find the linearization of \(f(x,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right) .\) \(f(x, y)=x^{2} e^{y} ;(1,0)\)

Short Answer

Expert verified
The linearization is \( L(x, y) = 2x + y - 1 \).

Step by step solution

01

Recall the Formula for Linearization

The linearization of a function at a given point is given by: \[ L(x, y) = f(x_0, y_0) + f_x(x_0, y_0) (x - x_0) + f_y(x_0, y_0) (y - y_0) \] where \( f_x \) and \( f_y \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \) respectively.
02

Evaluate the Function at the Given Point

Evaluate the function \( f(x, y) = x^2 e^y \) at the point \((1,0)\):\[ f(1,0) = (1)^2 e^0 = 1 \]
03

Compute the Partial Derivative with Respect to x

Differentiate \( f(x, y) = x^2 e^y \) with respect to \( x \):\[ f_x = \frac{\partial}{\partial x}(x^2 e^y) = 2x e^y \]Evaluate it at \( (1, 0) \):\[ f_x(1, 0) = 2(1) e^0 = 2 \]
04

Compute the Partial Derivative with Respect to y

Differentiate \( f(x, y) = x^2 e^y \) with respect to \( y \):\[ f_y = \frac{\partial}{\partial y}(x^2 e^y) = x^2 e^y \]Evaluate it at \( (1, 0) \):\[ f_y(1, 0) = (1)^2 e^0 = 1 \]
05

Construct the Linearization

Substitute all calculated values into the linearization formula:\[ L(x, y) = 1 + 2(x - 1) + 1(y - 0) \]Simplify:\[ L(x, y) = 2x + y - 1 \]

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.

Partial Derivatives
In multivariable calculus, partial derivatives are important tools that help us understand the nature of functions with more than one variable. When we take a partial derivative, we differentiate with respect to one variable while treating other variables as constants.
This is particularly useful in the context of functions like our exercise's function, where we have variables \(x\) and \(y\).
  • The partial derivative with respect to \(x\), denoted \(f_x\), gives us the rate of change of the function as \(x\) changes, holding \(y\) constant.
  • Similarly, the partial derivative with respect to \(y\), denoted \(f_y\), shows how the function changes as \(y\) changes, with \(x\) held constant.
For our given function \(f(x, y) = x^2 e^y\):
  • Finding \(f_x\) involves differentiating \(x^2 e^y\) in terms of \(x\), resulting in \(2x e^y\).
  • Calculating \(f_y\), we differentiate in terms of \(y\), resulting in \(x^2 e^y\).
These partial derivatives are crucial for constructing the linearization of a function, as seen in the solution steps.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus, such as differentiation, into functions of several variables rather than just one.
It allows us to analyze and approximate the behavior of complex systems that depend on multiple factors. In our exercise, we see a function \(f(x, y) = x^2 e^y\) that depends on two variables, \(x\) and \(y\). Understanding the behavior of such a function requires tools from multivariable calculus, including partial derivatives and linearization.
One of the key aspects of multivariable calculus is the ability to approximate functions using linearization. This is similar to using tangent lines in single-variable calculus.
The linearization formula allows us to create a tangent plane that approximates the function near a specific point. This is particularly useful when evaluating functions in contexts where exact calculations are difficult or unnecessary.
Differentiation
Differentiation, a fundamental technique in calculus, involves finding the rate at which things change. For multivariable calculus, we deal with partial derivatives as a form of differentiation, examining how individual changes in variables affect a function.
In the context of our exercise, differentiation helped us calculate the partial derivatives \(f_x\) and \(f_y\). Differentiating a function gives us insights into its behavior, showing exactly how changes in one variable influence the overall output.
  • When differentiating \(f(x, y) = x^2 e^y\) with respect to \(x\), we obtained \(f_x = 2x e^y\).
  • Upon differentiating with respect to \(y\), we calculated \(f_y = x^2 e^y\).
This differentiation tells us:
  • For \(f_x\), how the function's value changes as \(x\) moves.
  • For \(f_y\), how the function's value changes with \(y\).
These derivatives play a critical role in linearization, used to determine the best linear approximation of a function near a specific point, as demonstrated at the given point \((1,0)\). This approximation is extremely useful in various fields such as economics, physics, and engineering, where understanding and predicting change is key.

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 \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\ln \left(\frac{x y}{x^{2}+x y}\right)\)

Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(1.0)}\left(x^{2}-3 y^{2}\right)\)

Use nine evenly spaced points and five colors to draw heat maps of the following functions, defined on their specified domains. \(f(x, y)=x^{2}-y^{2}\) on \(D=\\{(x, y):-1 \leq x \leq 1,-1 \leq y \leq 1\\}\)

Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(-1,3)} x^{2}\left(y^{2}-3 x y\right)\)

A chemical diffuses in a container that occupies the interval \(0 \leq x \leq 1\). The concentration of the chemical at time \(t\) and at a point \(x\) is given by the diffusion equation: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} $$ (a) Suppose that the chemical is allowed to diffuse through the entire container until the concentration reaches an equilibrium value where \(c\) does not change any more with time, that is, \(\partial c / \partial t=0 .\) Suppose that chemical that touches the walls of the container is removed so that $$ c(0, t)=c(1, t)=0 . $$ The steady state concentration of chemical will be a function \(C(x)\) with $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ and \(C(0)=C(1)=0\). Show that \(C(x)=0\) satisfies this differential equation and the constraints as the points \(x=0\) and \(x=1\). (b) Now suppose that chemical is added to the container by a reaction that occurs at the wall \(x=0 .\) This reaction keeps the concentration of chemical at this wall equal to \(c(0, t)=1 . \mathrm{Un}\) der these conditions the steady state distribution of chemical will obey a differential equation: $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ with \(C(0)=1\) and \(C(1)=0 .\) Show that \(C(x)=1-x\) satisfies both the differential equation and the boundary conditions at \(x=0\) and \(x=1\). (c) Notice that the steady state distributions in (a) and (b) do not depend on \(D .\) Can you explain why?

See all solutions

Recommended explanations on Biology 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.