Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.1 Vector Fields - Exercises - Page 919: 48

Answer

$\nabla \varphi = \frac{{{{\bf{e}}_r}}}{r}$

Work Step by Step

We have $\varphi = \ln r$, where $r = \sqrt {{x^2} + {y^2}} $. The gradient of $\varphi $ is $\nabla \varphi = \left( {\frac{{\partial \varphi }}{{\partial x}},\frac{{\partial \varphi }}{{\partial y}}} \right)$. Evaluate $\frac{{\partial \varphi }}{{\partial x}}$ $\frac{{\partial \varphi }}{{\partial x}} = \frac{1}{r}\frac{{\partial r}}{{\partial x}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\frac{\partial }{{\partial x}}\sqrt {{x^2} + {y^2}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\left( {\frac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$ $\frac{{\partial \varphi }}{{\partial x}} = \frac{x}{{{r^2}}}$ Evaluate $\frac{{\partial \varphi }}{{\partial y}}$ $\frac{{\partial \varphi }}{{\partial y}} = \frac{1}{r}\frac{{\partial r}}{{\partial y}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\frac{\partial }{{\partial y}}\sqrt {{x^2} + {y^2}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\left( {\frac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)$ $\frac{{\partial \varphi }}{{\partial y}} = \frac{y}{{{r^2}}}$ Therefore, $\nabla \varphi = \left( {\frac{{\partial \varphi }}{{\partial x}},\frac{{\partial \varphi }}{{\partial y}}} \right) = \frac{1}{r}\left( {\frac{x}{r},\frac{y}{r}} \right)$. Since ${{\bf{e}}_r} = \left( {\frac{x}{r},\frac{y}{r}} \right)$, so $\nabla \varphi = \frac{{{{\bf{e}}_r}}}{r}$.
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