Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 809: 79


The Laplace's equation is satisfied.

Work Step by Step

We need to compute the Laplace's equation and prove that it equals to $0$. In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa: $f_x=-x (x^2+y^2+z^2)^{-3/2} \implies f_{xx}=-x (\dfrac{-3}{2}) (x^2+y^2+z^2)^{-5/2} (2x)+(x^2+y^2+z^2)^{-3/2}(-1)=(x^2+y^2+z^2)^{-5/2} (2x^2-y^2-z^2)$ and $f_y=-y (x^2+y^2+z^2)^{-3/2} \implies f_{yy}=-y (\dfrac{-3}{2}) (x^2+y^2+z^2)^{-5/2} (2y)+(x^2+y^2+z^2)^{-3/2}(-1)=(x^2+y^2+z^2)^{-5/2} (2y^2-x^2-z^2)$ and $f_z=-z (x^2+y^2+z^2)^{-3/2} \implies f_{zz}=-z (\dfrac{-3}{2}) (x^2+y^2+z^2)^{-5/2} (2z)+(x^2+y^2+z^2)^{-3/2}(-1)=(x^2+y^2+z^2)^{-5/2} (2z^2-x^2-y^2)$ Consider the Laplace's equation $\nabla^2 f =\dfrac{\partial^2 f}{\partial x^2}+\dfrac{\partial^2 f}{\partial y^2}$ $\nabla^2 f =(x^2+y^2+z^2)^{-5/2} (2x^2-y^2-z^2+2y^2-x^2-z^2+2z^2-x^2-y^2)=0$ Thus, the Laplace's equation is satisfied.
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