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Chapter 10: Problem 36

Find the Jacobi matrix for each given function. \(\mathbf{f}(x, y)=\left[\begin{array}{c}\sqrt{x^{2}+y^{2}} \\\ e^{-x^{2}}\end{array}\right]\)

Short Answer

Expert verified
The Jacobi matrix is \( \begin{bmatrix} \frac{x}{\sqrt{x^2 + y^2}} & \frac{y}{\sqrt{x^2 + y^2}} \\ -2xe^{-x^2} & 0 \end{bmatrix} \).

Step by step solution

01

Identify the Function Components

The function \( \mathbf{f}(x, y) \) is a vector-valued function with two components: 1. \( f_1(x, y) = \sqrt{x^2 + y^2} \)2. \( f_2(x, y) = e^{-x^2} \). Our goal is to find the Jacobian matrix, which involves computing the partial derivatives of each function with respect to \( x \) and \( y \).
02

Compute Partial Derivatives of \(f_1(x, y)\)

Find the partial derivatives:\- \( \frac{\partial f_1}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \)\- \( \frac{\partial f_1}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \).
03

Compute Partial Derivatives of \(f_2(x, y)\)

Find the partial derivatives: \- \( \frac{\partial f_2}{\partial x} = -2xe^{-x^2} \) (using the chain rule on the exponent)\- \( \frac{\partial f_2}{\partial y} = 0 \) (since \( f_2 \) does not depend on \( y \)).
04

Form the Jacobi Matrix

The Jacobi Matrix \( J \) is constructed by placing the partial derivatives in a matrix where each row corresponds to the partials of each component of the function:\[J = \begin{bmatrix} \frac{x}{\sqrt{x^2 + y^2}} & \frac{y}{\sqrt{x^2 + y^2}} \-2xe^{-x^2} & 0 \end{bmatrix}\]
05

Review the Matrix

The final Jacobi matrix represents the rate of change of each component of \( \mathbf{f} \) with respect to \( x \) and \( y \) variables. It can be used to analyze the behavior of the function near a given point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives help us understand how a function changes when we tweak one of its variables, holding the others constant. When dealing with functions of several variables, such as those in multivariable calculus, partial derivatives become indispensable.
  • Imagine having a function of two variables, like the ones in this exercise: \( f_1(x, y) = \sqrt{x^2 + y^2} \) and \( f_2(x, y) = e^{-x^2} \).
  • Sensitivity: Partial derivatives tell us how sensitive a function is to changes in one independent variable, while keeping the other variable unchanged.
  • Components: For \( f_1(x, y) \), the partial derivative with respect to \( x \) is \( \frac{x}{\sqrt{x^2 + y^2}} \), showing how \( f_1 \) changes when \( x \) changes slightly.
  • Similarly, \( \frac{\partial f_1}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \) expresses the change in \( f_1 \) as \( y \) changes.
  • For \( f_2(x, y) \), notice the partial with respect to \( y \) is zero. This implies that \( f_2 \) doesn't change as \( y \) changes, quite insightful for understanding the depth of such a function.
By gaining clarity on partial derivatives, students will strengthen their foundation for more complex calculus concepts.
Vector-Valued Functions
Vector-valued functions are functions that output vectors instead of simple real numbers. They are particularly useful in multivariable calculus and physics.
  • Representation: As shown in the exercise, a vector-valued function like \( \mathbf{f}(x, y) \) outputs a vector, here seen as \[\begin{align*}\mathbf{f}(x, y) = \begin{bmatrix} \sqrt{x^2 + y^2} \ e^{-x^2} \end{bmatrix}.\end{align*}\]
  • Components: Each entry of the vector is a function of \( x \) and \( y \), representing different physical quantities or dimensions.
  • Understanding: They make it easier to handle problems where multiple interdependent relationships between variables exist.
  • Applications: These functions appear frequently in physics, such as in describing motion along a path where time is the parameter.
The comprehension of vector-valued functions expands the scope of solving real-world problems where multiple outcomes depend on several input variables.
Chain Rule
The chain rule is a fundamental differentiation tool within calculus. It describes how to take the derivative of a composite function and is crucial when dealing with functions where one variable depends on another.
  • Application: In the context of this exercise, it's necessary to use the chain rule while differentiating, especially for functions like \( f_2(x, y) = e^{-x^2} \).
  • Formula: The rule is mathematically represented as \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).
  • Clarity: In our exercise, when differentiating \( f_2 \), its \( x \) derivative involved applying the chain rule: \(-2xe^{-x^2}\), by recognizing the internal function \( g(x) = -x^2 \) and the outer function \( f(g) = e^g \).
  • Complexity: The chain rule simplifies the differentiation process in complex layered functions, enhancing understanding as calculations get intricate.
Grasping how the chain rule intertwines different function layers enables breaking down complexity, a vital skill in calculus.

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Most popular questions from this chapter

Show that $$f(x, y)=\left\\{\begin{array}{cl}\frac{3 r(y+x)}{x^{2}+y^{3}} & \text { for }(x, y) \neq(0,0) \\\0 & \text { for }(x, y)=(0,0)\end{array}\right.$$ is discontinuous at \((0,0)\).

Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(0,1)} \frac{2 x y-3}{x^{2}+y^{2}+1}\)

Find the range of each function \(f(x, y)\), when defined on the specified domain \(D\). \(f(x, y)=\frac{x}{y} ; D=\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq 1, y>x]\)

Find the indicated partial derivatives. \(f(x, y)=x^{1 / 3} y-x y^{-1 / 3} ; f_{y}(1,1)\)

Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(2,0)} \frac{2 x+4 y^{2}}{y^{2}+3 x}\)
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