Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.5 Curl and Divergence - 16.5 Exercises - Page 1150: 31

Answer

a ) $\nabla r=\dfrac{\bf{r}}{r}$ b) $\nabla \times r=0$ c) $\nabla (1/r)=\dfrac{-r}{r^3}$ d) $\nabla \ln r=\dfrac{r}{r^2}$

Work Step by Step

a) $\nabla r=\dfrac{\partial (r) }{\partial x}i+\dfrac{\partial (r) }{\partial y}j+\dfrac{\partial (r) }{\partial z}k=\dfrac{x }{\sqrt{x^2+y^2+z^2}} i+\dfrac{y}{\sqrt{x^2+y^2+z^2}} j+\dfrac{z}{\sqrt{x^2+y^2+z^2}} k=\dfrac{(xi+yj+zk)}{\sqrt{x^2+y^2+z^2}} =\dfrac{\bf{r}}{r}$ b) $\nabla \times r=(\dfrac{\partial z}{\partial y}-\dfrac{\partial y}{\partial z})i-(\dfrac{\partial z}{\partial x}-\dfrac{\partial x}{\partial z})j+(\dfrac{\partial y}{\partial x}-\dfrac{\partial x}{\partial y})k=0$ c) $\nabla (1/r)=\begin{vmatrix}i&j&k\\\dfrac{\partial}{\partial x}&\dfrac{\partial }{\partial y}&\dfrac{\partial }{\partial z}\\x&y&z\end{vmatrix}=\dfrac{\partial (1/r) }{\partial x}i+\dfrac{\partial (1/r) }{\partial y}j+\dfrac{\partial (1/r)}{\partial z}k=(\dfrac{-1}{r^2}) \nabla r=\dfrac{-r}{r^3}$ d) $\nabla \ln r=\dfrac{\partial (\ln r) }{\partial x}i+\dfrac{\partial (\ln r) }{\partial y}j+\dfrac{\partial (\ln r)}{\partial z}k=(\dfrac{1}{r}) \nabla r=\dfrac{r}{r^2}$
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