| Version 2 |
Version 1 |
| \textbf{Example.}\, Show that the vector field |
\textbf{Example.}\, Show that the vector field |
| \begin{align} |
\begin{align} |
| \vec{U} \,:=\, y\,\vec{i}+(x+\sin{z})\,\vec{j}+y\cos{z}\,\vec{k} |
\vec{U} \,:=\, y\,\vec{i}+(x+\sin{z})\,\vec{j}+y\cos{z}\,\vec{k} |
| \end{align} |
\end{align} |
| is laminar everywhere in $\mathbb{R}^3$ and determine its scalar potential.\\ |
is laminar everywhere in $\mathbb{R}^3$ and determine its scalar potential.\\ |
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| We have |
We have |
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$\displaystyle\nabla\!\times\!\vec{U} = \left|\begin{matrix}
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$\displaystyle\nabla\times\vec{U} = \left|\begin{matrix}
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| \vec{i} & \vec{j} & \vec{k}\\ |
\vec{i} & \vec{j} & \vec{k}\\ |
| \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\ |
\frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\ |
| y & x\!+\!\sin{z} & y\cos{z} |
y & x\!+\!\sin{z} & y\cos{z} |
| \end{matrix}\right| \\= |
\end{matrix}\right| \\= |
| \left(\frac{\partial(y\cos{z})}{\partial{y}}-\frac{\partial(x\!+\!\sin{z})}{\partial{z}}\right)\vec{i} |
\left(\frac{\partial(y\cos{z})}{\partial{y}}-\frac{\partial(x\!+\!\sin{z})}{\partial{z}}\right)\vec{i} |
| +\left(\frac{\partial{y}}{\partial{z}}-\frac{\partial(y\cos{z})}{\partial{x}}\right)\vec{j} |
+\left(\frac{\partial{y}}{\partial{z}}-\frac{\partial(y\cos{z})}{\partial{x}}\right)\vec{j} |
| +\left(\frac{\partial(x\!+\!\sin{z})}{\partial{x}}-\frac{\partial{y}}{\partial{y}}\right)\vec{k}$,\\ |
+\left(\frac{\partial(x\!+\!\sin{z})}{\partial{x}}-\frac{\partial{y}}{\partial{y}}\right)\vec{k}$,\\ |
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which is identically $\vec{0}$ for all $x$, $y$, $z$.\, Thus $\vec{U}$ is laminar.\\
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which is identically $\vec{0}$ for all $x$, $y$, $z$.\, Thus $\vec{U}$ is laminar. |
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| The scalar potential \,$u = u(x,\,y,\,z)$\, must satisfy the conditions |
The scalar potential \,$u = u(x,\,y,\,z)$\, must satisfy the conditions |
| $$\frac{\partial{u}}{\partial{x}} = y,\quad \frac{\partial{u}}{\partial{y}} = x\!+\!\sin{z},\quad \frac{\partial{u}}{\partial{z}} = y\cos{z}.$$ |
$$\frac{\partial{u}}{\partial{x}} = y,\quad \frac{\partial{u}}{\partial{y}} = x\!+\!\sin{z},\quad \frac{\partial{u}}{\partial{z}} = y\cos{z}.$$ |
| Thus we can write |
Thus we can write |
| $$u = \int y\,dx = xy+C_1,$$ |
$$u = \int y\,dx = xy+C_1,$$ |
| where $C_1$ may depend on $y$ or $z$. Differentiating this result with respect to $y$ and comparing to the second condition, we get |
where $C_1$ may depend on $y$ or $z$. Differentiating this result with respect to $y$ and comparing to the second condition, we get |
| $$\frac{\partial{u}}{\partial{y}} = x+\frac{\partial{C_1}}{\partial{y}} = x+\sin{z}.$$ |
$$\frac{\partial{u}}{\partial{y}} = x+\frac{\partial{C_1}}{\partial{y}} = x+\sin{z}.$$ |
| Accordingly, |
Accordingly, |
| $$C_1 = \int\sin{z}\,dy = y\sin{z}+C_2,$$ |
$$C_1 = \int\sin{z}\,dy = y\sin{z}+C_2,$$ |
| where $C_2$ may depend on $z$.\, So |
where $C_2$ may depend on $z$.\, So |
| $$u = xy+y\sin{z}+C_2.$$ |
$$u = xy+y\sin{z}+C_2.$$ |
| Differentiating this result with respect to $z$ and comparing to the third condition yields |
Differentiating this result with respect to $z$ and comparing to the third condition yields |
| $$\frac{\partial{u}}{\partial{z}} \,=\, y\cos{z}+\frac{\partial{C_2}}{\partial{z}} \,=\, y\cos{z}.$$ |
$$\frac{\partial{u}}{\partial{z}} \,=\, y\cos{z}+\frac{\partial{C_2}}{\partial{z}} \,=\, y\cos{z}.$$ |
| This means that $C_2$ is an arbitrary \PMlinkescapetext{constant}. Thus the required potential function has the form |
This means that $C_2$ is an arbitrary \PMlinkescapetext{constant}. Thus the required potential function has the form |
| $$u = xy+y\sin{z}+C.$$ |
$$u = xy+y\sin{z}+C.$$ |
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