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Chapter 6: Problem 7

Let \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\) be given by \(f(x, y)=\left(e^{x+y}, e^{-x} \cos y, e^{-x} \sin y\right) .\) (a) Find \(D f(x, y)\). (b) Show that \(f\) is differentiable at every point of \(\mathbb{R}^{2}\).

Short Answer

Expert verified
(a) The Jacobian is provided; (b) \(f\) is differentiable as derivatives are continuous.

Step by step solution

01

Understand Function Components

The function \( f(x, y) \) is a vector function mapping from \( \mathbb{R}^2 \) to \( \mathbb{R}^3 \). It consists of three component functions: \( f_1(x, y) = e^{x+y} \), \( f_2(x, y) = e^{-x} \cos y \), and \( f_3(x, y) = e^{-x} \sin y \).
02

Compute Partial Derivatives

Compute the partial derivatives of each component of \( f(x, y) \). For \( f_1 \): \( \frac{\partial f_1}{\partial x} = e^{x+y} \) and \( \frac{\partial f_1}{\partial y} = e^{x+y} \). For \( f_2 \): \( \frac{\partial f_2}{\partial x} = -e^{-x} \cos y \) and \( \frac{\partial f_2}{\partial y} = -e^{-x} \sin y \). For \( f_3 \): \( \frac{\partial f_3}{\partial x} = -e^{-x} \sin y \) and \( \frac{\partial f_3}{\partial y} = e^{-x} \cos y \).
03

Construct the Jacobian Matrix \( Df(x, y) \)

The Jacobian matrix \( Df(x, y) \) is constructed from the partial derivatives: \[Df(x, y) = \begin{bmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \\frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y}\end{bmatrix} = \begin{bmatrix}e^{x+y} & e^{x+y} \-e^{-x} \cos y & -e^{-x} \sin y \-e^{-x} \sin y & e^{-x} \cos y\end{bmatrix}.\]
04

Verify Continuity of Partial Derivatives

The partial derivatives calculated in Step 2 are all composed of exponential, trigonometric, and continuous functions. Thus, each partial derivative is continuous everywhere on \( \mathbb{R}^2 \).
05

Prove Differentiability Using Partial Derivative Continuity

A function \( f: \mathbb{R}^n \to \mathbb{R}^m \) is differentiable at a point if its partial derivatives exist and are continuous in the neighborhood of that point. Since the Jacobian is composed entirely of continuous derivatives, \( f \) is differentiable at every point in \( \mathbb{R}^2 \).

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Key Concepts

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

Partial Derivatives
Partial derivatives are a way to measure how a function changes as you vary one of its variables, keeping the others constant. In the context of multivariable calculus, partial derivatives are essential for understanding how functions behave in higher dimensions. For a function of two variables, like our vector function \(f(x, y)\), we compute the partial derivatives with respect to each variable, \(x\) and \(y\).
  • For \(f_1(x, y) = e^{x+y}\), the partial derivative with respect to \(x\) is \(e^{x+y}\), and with respect to \(y\), it is also \(e^{x+y}\).
  • For \(f_2(x, y) = e^{-x} \cos y\), the partial derivative with respect to \(x\) is \(-e^{-x} \cos y\), and with respect to \(y\), it is \(-e^{-x} \sin y\).
  • For \(f_3(x, y) = e^{-x} \sin y\), the partial derivative with respect to \(x\) is \(-e^{-x} \sin y\), and with respect to \(y\), it is \(e^{-x} \cos y\).
These derivatives help us understand the rate of change of each component function as each variable changes, painting a comprehensive picture of the overall function's behavior.
Jacobian Matrix
The Jacobian Matrix is a fundamental tool in calculus for vector-valued functions. It is a matrix that contains all the first-order partial derivatives of a vector function. For a function \(f: \mathbb{R}^n \to \mathbb{R}^m\), the Jacobian matrix provides a linear approximation of the function around a point.In our example, the function \(f(x, y)\) comprises three component functions, forming a mapping from \(\mathbb{R}^2\) to \(\mathbb{R}^3\). The Jacobian matrix \(Df(x, y)\) is given by:\[Df(x, y) = \begin{bmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \\frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y}\end{bmatrix} = \begin{bmatrix}e^{x+y} & e^{x+y} \-e^{-x} \cos y & -e^{-x} \sin y \-e^{-x} \sin y & e^{-x} \cos y\end{bmatrix}.\]
  • Each row of the Jacobian corresponds to one component of the vector function.
  • The columns correspond to partial derivatives with respect to each variable.
The Jacobian helps in understanding how a small change in input variables affects the output of the vector function, thus serving as a bridge between finite changes and differential calculus.
Vector Functions
Vector functions map numbers to vectors, converting scalar inputs into multidimensional outputs. They are crucial in fields like physics and engineering to describe various phenomena.The function \(f(x, y)\) in our problem is a vector function. It translates any point \( (x, y) \) in \( \mathbb{R}^2 \) into a vector in \( \mathbb{R}^3 \) given by
  • \(f_1(x, y) = e^{x+y}\)
  • \(f_2(x, y) = e^{-x} \cos y\)
  • \(f_3(x, y) = e^{-x} \sin y\)
These vector outputs encapsulate the combined behavior of various processes or effects.

Applications of Vector Functions

  • Representation of physical quantities like velocity and force, which have both magnitude and direction.
  • Modeling of curves and surfaces within 3D space.
  • Transformation of complex datasets in machine learning and data science fields.
Vector functions thus play a pivotal role in linking simple numerical inputs with intuitive, visualizable outputs across a wide array of disciplines.

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Most popular questions from this chapter

Let \(f(x, y)\) be a smooth real-valued function defined on an open set in \(\mathbb{R}^{2} .\) The polar coordinate substitutions \(x=r \cos \theta\) and \(y=r \sin \theta\) convert \(f\) into a function of \(r\) and \(\theta\), \(w=f(r \cos \theta, r \sin \theta)\) (a) Show that: $$ \begin{array}{c} \frac{\partial^{2} w}{\partial \theta^{2}}=-r \cos \theta \frac{\partial f}{\partial x}-r \sin \theta \frac{\partial f}{\partial y}+r^{2} \sin ^{2} \theta \frac{\partial^{2} f}{\partial x^{2}}-2 r^{2} \sin \theta \cos \theta \frac{\partial^{2} f}{\partial x \partial y} \\ +r^{2} \cos ^{2} \theta \frac{\partial^{2} f}{\partial y^{2}} \end{array} $$ (b) Find an analogous expression for the mixed partial derivative \(\frac{\partial^{2} w}{\partial r \partial \theta}\). (c) Find an analogous expression for \(\frac{\partial^{2} w}{\partial r^{2}}\). (d) If \(f(x, y)=x^{2}+y^{2}+x y,\) use your answer to part (b) to find \(\frac{\partial^{2} w}{\partial r \partial \theta}\). Then, verify that it is the same as the result obtained by writing \(w\) in terms of \(r\) and \(\theta\) directly by substituting for \(x\) and \(y\) and then differentiating that expression with respect to \(r\) and \(\theta\).

Let \(f(x, y, z)\) be a smooth real-valued function of \(x, y,\) and \(z\). The substitutions \(x=s+2 t\), \(y=3 s+4 t,\) and \(z=5 s+6 t\) convert \(f\) into a function of \(s\) and \(t: w=f(s+2 t, 3 s+4 t, 5 s+6 t) .\) Find expressions for \(\frac{\partial w}{\partial s}\) and \(\frac{\partial w}{\partial t}\) in terms of \(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\) and \(\frac{\partial f}{\partial z}\).

(a) If \(\mathbf{x}=\left(x_{1}, x_{2}, \ldots, x_{n}\right)\) and \(\mathbf{y}=\left(y_{1}, y_{2}, \ldots, y_{n}\right)\) are elements of \(\mathbb{R}^{n}\), show that \(\left|x_{i}-y_{i}\right| \leq\) \(\|\mathbf{x}-\mathbf{y}\|\) for \(i=1,2, \ldots, n\) (b) Let \(U\) be an open set in \(\mathbb{R}^{n}\), and let \(f: U \rightarrow \mathbb{R}^{m}\) be a function, where \(f(\mathbf{x})=\) \(\left(f_{1}(\mathbf{x}), f_{2}(\mathbf{x}), \ldots, f_{m}(\mathbf{x})\right)\). Show that, if \(f\) is continuous at a point a of \(U\), then so are \(f_{1}, f_{2}, \ldots, f_{m}\)

Find \(D f(x, y, z)\) if \(f(x, y, z)=x+y^{2}+3 z^{3}\).

Let \(f(x, y)\) be a smooth real-valued function of \(x\) and \(y\) defined on an open set \(U\) in \(\mathbb{R}^{2}\). The partial derivative \(\frac{\partial f}{\partial x}\) is also a real-valued function of \(x\) and \(y\) (as is \(\frac{\partial f}{\partial y}\) ). If \(\alpha(t)=(x(t), y(t))\) is a smooth path in \(U,\) then substituting for \(x\) and \(y\) in terms of \(t\) converts \(\frac{\partial f}{\partial x}\) into a function of \(t,\) say \(w=\frac{\partial f}{\partial x}(x(t), y(t))\) (a) Show that: $$ \frac{d w}{d t}=\frac{\partial^{2} f}{\partial x^{2}} \frac{d x}{d t}+\frac{\partial^{2} f}{\partial y \partial x} \frac{d y}{d t} $$ where the partial derivatives are evaluated at \(\alpha(t)\). (b) Now, consider the composition \(g(t)=(f \circ \alpha)(t) .\) Find a formula for the second derivative \(\frac{d^{2} g}{d t^{2}}\) in terms of the partial derivatives of \(f\) as a function of \(x\) and \(y\) and the derivatives of \(x\) and \(y\) as functions of \(t .\) (Hint: Differentiate the Little Chain Rule. Note that this involves differentiating some products.) (c) Let a be a point of \(U\). If \(\alpha: I \rightarrow U, \alpha(t)=\mathbf{a}+t \mathbf{v}=\left(a_{1}+t v_{1}, a_{2}+t v_{2}\right)\) is a parametrization of a line segment containing a and if \(g(t)=(f \circ \alpha)(t),\) use your formula from part (b) to show that: $$ \frac{d^{2} g}{d t^{2}}=v_{1}^{2} \frac{\partial^{2} f}{\partial x^{2}}+2 v_{1} v_{2} \frac{\partial^{2} f}{\partial x \partial y}+v_{2}^{2} \frac{\partial^{2} f}{\partial y^{2}} $$ where the partial derivatives are evaluated at \(\alpha(t) .\) (Compare this with equation (4.22) in Chapter 4 regarding the second-order approximation of \(f\) at a.)
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