Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 15 - Topics In Vector Calculus - 15.3 Independence Of Path; Conservative Vector Fields - Exercises Set 15.3 - Page 1121: 31

Answer

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Work Step by Step

Given: \[ \mathbf{F} = \nabla \phi(x, y, z) \] where \[ \begin{align*} \mathbf{a} &= \frac{\partial \phi}{\partial x} \\ \mathbf{b} &= \frac{\partial \phi}{\partial y} \\ \mathbf{c} &= \frac{\partial \phi}{\partial z} \end{align*} \] Differentiating the first equation with respect to \(y\) and \(z\): \[ \begin{align*} \frac{\partial a}{\partial y} &= \frac{\partial^2 \phi}{\partial x \partial y} \quad (1) \\ \frac{\partial b}{\partial z} &= \frac{\partial^2 \phi}{\partial y \partial z} \quad (2) \end{align*} \] Analogously, \[ \begin{align*} \frac{\partial a}{\partial x} &= \frac{\partial^2 \phi}{\partial y \partial x} \quad (3) \\ \frac{\partial b}{\partial y} &= \frac{\partial^2 \phi}{\partial z \partial y} \quad (4) \end{align*} \] And \[ \begin{align*} \frac{\partial a}{\partial z} &= \frac{\partial^2 \phi}{\partial x \partial z} \quad (5) \\ \frac{\partial b}{\partial x} &= \frac{\partial^2 \phi}{\partial z \partial x} \quad (6) \end{align*} \] The right sides in equations (1) and (3) are equal, then: \[ \frac{\partial^2 \phi}{\partial x \partial y} = \frac{\partial^2 \phi}{\partial y \partial x} \equiv \frac{\partial a}{\partial y} = \frac{\partial b}{\partial x} \equiv \frac{\partial^2 \phi}{\partial y \partial x} = \frac{\partial^2 \phi}{\partial x \partial y} \equiv \frac{\partial^2 \phi}{\partial y \partial x} = \frac{\partial a}{\partial x} = \frac{\partial b}{\partial y} \] Analogously, from (2) and (5): \[ \frac{\partial^2 \phi}{\partial y \partial z} = \frac{\partial^2 \phi}{\partial z \partial y} \equiv \frac{\partial a}{\partial z} = \frac{\partial c}{\partial y} \equiv \frac{\partial^2 \phi}{\partial z \partial y} = \frac{\partial b}{\partial z} = \frac{\partial c}{\partial y} \] Finally, from (4) and (6): \[ \frac{\partial^2 \phi}{\partial z \partial x} = \frac{\partial^2 \phi}{\partial x \partial z} \equiv \frac{\partial b}{\partial x} = \frac{\partial c}{\partial x} \equiv \frac{\partial^2 \phi}{\partial x \partial z} = \frac{\partial a}{\partial z} = \frac{\partial c}{\partial x} \] This summarizes the relationships between the second partial derivatives of \(\phi\) and the components of \(\mathbf{F}\) in terms of \(a\), \(b\), and \(c\) for each pair of variables \(x\), \(y\), and \(z\).
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