Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.3 Exercises - Page 937: 77

Answer

The function $u=1/\sqrt{x^2+y^2+z^2}$ is a solution of the equation $u_{xx}+u_{yy}+u_{zz}=0$.

Work Step by Step

Here, $u_x=\dfrac{-x}{(x^2+y^2+z^2)^{3/2}}$ Now, take the second derivative of $u$ with respect to $x$. $u_{xx}=\dfrac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ ...(1) Now, take the second derivative of $u$ with respect to $y$. $u_{yy}=\dfrac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ ...(2) Now, take the second derivative of $u$ with respect to $z$. $u_{zz}=\dfrac{2z^2-y^2-x^2}{(x^2+y^2+z^2)^{5/2}}$ ...(3) From equations (1), (2) and (3) we have $u_{xx}+u_{yy}+u_{zz}=\dfrac{(2x^2-y^2-z^2)+(2y^2-x^2-z^2)+(2z^2-y^2-x^2)}{(x^2+y^2+z^2)^{5/2}}=0$ Hence, it has been proved that the function $u=1/\sqrt{x^2+y^2+z^2}$ is a solution of the equation $u_{xx}+u_{yy}+u_{zz}=0$.
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