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Q 12.

Expert-verified
Found in: Page 1131

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Give an example of a field with positive divergence at (1, 0, π).

$\mathrm{An}\mathrm{example}\mathrm{of}\mathrm{field}\mathrm{with}\mathrm{positive}\mathrm{divergence}\mathrm{at}\left(1,0,\mathrm{\pi }\right)\mathrm{is},\phantom{\rule{0ex}{0ex}}\mathrm{F}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)={\mathrm{x}}^{2}\mathrm{i}+3\mathrm{sin}\left(\mathrm{y}\right)\mathrm{j}+\mathrm{cos}\left(\mathrm{z}\right)\mathrm{k}$

See the step by step solution

Point is (1,0,π)

## Step 2. Example of field

$\mathrm{If}\mathrm{F}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)={\mathrm{F}}_{1}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{i}+{\mathrm{F}}_{2}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{j}+{\mathrm{F}}_{3}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{k}\mathrm{is}\mathrm{a}\mathrm{vector}\mathrm{field},\mathrm{then}\phantom{\rule{0ex}{0ex}}\mathrm{the}\mathrm{divergence}\mathrm{of}\mathrm{the}\mathrm{vector}\mathrm{field}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{follows}:\phantom{\rule{0ex}{0ex}}\nabla .\mathrm{F}=\frac{\partial {\mathrm{F}}_{1}}{\partial \mathrm{x}}+\frac{\partial {\mathrm{F}}_{2}}{\partial y}+\frac{\partial {\mathrm{F}}_{3}}{\partial z}\phantom{\rule{0ex}{0ex}}\mathrm{Note}\mathrm{that}\mathrm{value}\mathrm{of}\nabla .\mathrm{F}\mathrm{is}\mathrm{scalar}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Consider}\mathrm{the}\mathrm{following}\mathrm{vector}\mathrm{field}:\phantom{\rule{0ex}{0ex}}\mathrm{F}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)={\mathrm{x}}^{2}\mathrm{i}+3\mathrm{sin}\left(\mathrm{y}\right)\mathrm{j}+\mathrm{cos}\left(\mathrm{z}\right)\mathrm{k}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{divergence}\mathrm{of}\mathrm{this}\mathrm{vector}\mathrm{field}\mathrm{will}\mathrm{be},\phantom{\rule{0ex}{0ex}}\nabla .\mathrm{F}=\frac{\partial \left({\mathrm{x}}^{2}\right)}{\partial \mathrm{x}}+\frac{\partial \left(3\mathrm{sin}\left(\mathrm{y}\right)\right)}{\partial y}+\frac{\partial \left(\mathrm{cos}\mathrm{z}\right)}{\partial z}\phantom{\rule{0ex}{0ex}}=2\mathrm{x}+3\mathrm{cosy}-\mathrm{sinz}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{At}\mathrm{the}\mathrm{origin},\mathrm{that}\mathrm{is},\mathrm{at}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(1,0,\mathrm{\pi }\right),\mathrm{the}\mathrm{divergence}\mathrm{of}\mathrm{this}\mathrm{vector}\mathrm{field}\mathrm{will}\mathrm{be},\phantom{\rule{0ex}{0ex}}\nabla .\mathrm{F}=2.1+3\mathrm{cos}0-\mathrm{sin}\mathrm{\pi }\phantom{\rule{0ex}{0ex}}=2+3-0\phantom{\rule{0ex}{0ex}}=5,\phantom{\rule{0ex}{0ex}}\mathrm{which}\mathrm{is}\mathrm{positive}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{an}\mathrm{example}\mathrm{of}\mathrm{field}\mathrm{with}\mathrm{positive}\mathrm{divergence}\mathrm{at}\left(1,0,\mathrm{\pi }\right)\mathrm{is},\phantom{\rule{0ex}{0ex}}\mathrm{F}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)={\mathrm{x}}^{2}\mathrm{i}+3\mathrm{sin}\left(\mathrm{y}\right)\mathrm{j}+\mathrm{cos}\left(\mathrm{z}\right)\mathrm{k}$