Ian at 16:43, 27 March 2011

2011-03-27T16:43:45Z

←Older revision		Revision as of 16:43, 27 March 2011
Line 13:		Line 13:
	& V=6 \\		& V=6 \\
	\end{align}</math>		\end{align}</math>
-
-	Note that <math>V</math> is a dimensionless number.

Ian: New page: From part a) $(5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j}$ Consider the $\mathbf{j}$ component. That is...

2010-10-01T16:29:17Z

New page: From part a) <math>(5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j}</math> Consider the <math>\mathbf{j}</math> component. That is...

New page

From part a)

<math>(5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j}</math>

Consider the
<math>\mathbf{j}</math>
component. That is the <math>\mathbf{j}</math> component on the left hand side must be the same as the <math>\mathbf{j}</math> component on the right hand side, that is zero.

<math>\begin{align}
& mV-\lambda mV+2m+2\lambda m=0 \\
& V-2V+2+4=0 \\
& -V+6=0 \\
& V=6 \\
\end{align}</math>

Note that <math>V</math> is a dimensionless number.

Solution 16.10b - Revision history

Ian at 16:43, 27 March 2011

Ian: New page: From part a) (5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j} Consider the \mathbf{j} component. That is...

Ian: New page: From part a) $(5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j}$ Consider the $\mathbf{j}$ component. That is...