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

Answer

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.