The vector field (mathbf{F}(x, y)=leftlanglefrac{x}{x^{2}+y^{2}}, frac{y}{x^{2}+y^{2}}ightangle) is defined on the domain (mathcal{D}={(x, y) eq(0,0)}). (a) Is (mathcal{D})

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The vector field \(\mathbf{F}(x, y)=\left\langle\frac{x}{x^{2}+y^{2}}, \frac{y}{x^{2}+y^{2}}ightangle\) is defined on the domain \(\mathcal{D}=\{(x, y) eq(0,0)\}\).
(a) Is \(\mathcal{D}\) simply connected?
(b) Show that \(\mathbf{F}\) satisfies the cross-partials condition. Does this guarantee that \(\mathbf{F}\) is conservative?
(c) Show that \(\mathbf{F}\) is conservative on \(\mathcal{D}\) by finding a potential function.
(d) Do these results contradict Theorem 4?

THEOREM 4 Existence of a Potential Function Let F be a vector field on a simply connected domain D. If F

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Calculus

ISBN: 9781319055844

4th Edition

Authors: Jon Rogawski, Colin Adams, Robert Franzosa

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