(cos(t-2))
(Energy)
Line 4: Line 4:
 
u = (t-2)
 
u = (t-2)
  
<math>E=\int_0^{2\pi}{|cos(u)|^2dt}</math>
+
<math>E=\int_0^{2\pi}{|cos(u)|^2du}</math>
  
  
<math>E=\frac{1}{2}\int_0^{2\pi}(1+cos(2(u)))dt</math>
+
<math>E=\frac{1}{2}\int_0^{2\pi}(1+cos(2(u)))du</math>
  
  
<math>E=\frac{1}{2}((u+\frac{1}{2}sin(2(u)))|_{u=-2}^{u=2\pi-2} + C</math>
+
<math>E=\frac{1}{2}((u+\frac{1}{2}sin(2(u)))|_{u=-2}^{u=2\pi-2}</math>
  
  
<math>E=\frac{1}{2}(2\pi -2 + .0744 -(-2 - 0.0349)) +C </math>
+
<math>E=\frac{1}{2}(2\pi -2 + .0744 -(-2 - 0.0349))</math>
  
  

Revision as of 05:33, 5 September 2008

cos(t-2)

Energy

u = (t-2)

$ 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=-2}^{u=2\pi-2} $


$ E=\frac{1}{2}(2\pi -2 + .0744 -(-2 - 0.0349)) $


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


$ =\frac{1}{4\pi}(2\pi+0-0-0) $


$ =\frac{1}{2} $

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