(Power)
(Power)
Line 30: Line 30:
  
  
<math>=\frac{1}{4\pi}(2\pi+0-0-0)</math>
+
<math>=\frac{1}{4\pi}(\pi+0-(0-\pi))</math>
  
  
 
<math>=\frac{1}{2}</math>
 
<math>=\frac{1}{2}</math>

Revision as of 05:42, 5 September 2008

cos(t-2)

Energy

$ u = (t-\pi) $

$ E=\int_0^{2\pi}{|cos(u)|^2du} $


$ E=\frac{1}{2}\int_0^{2\pi}(1+cos(2(u)))du $


$ E=\frac{1}{2}((u+\frac{1}{2}sin(2(u)))|_{u=-\pi}^{u=2\pi-\pi} $


$ E=\frac{1}{2}(\pi + 0 -(-\pi - 0)) $


$ E=\pi $

Power

$ E=\frac{1}{2\pi-0}\int_0^{2\pi}{|cos(u)|^2du} $


$ =\frac{1}{2\pi-0} *{\frac{1}{2}}\int_0^{2\pi}(1+cos(2u))du $


$ =\frac{1}{4\pi}((u)+\frac{1}{2}sin(2u))|_{u=-\pi}^{u=2\pi-\pi} $


$ =\frac{1}{4\pi}(\pi+0-(0-\pi)) $


$ =\frac{1}{2} $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett