(Energy)
Line 6: Line 6:
 
<math>=\frac{1}{2}(2\pi-0-0+0)</math><br>
 
<math>=\frac{1}{2}(2\pi-0-0+0)</math><br>
 
'''<math>=\pi</math>'''
 
'''<math>=\pi</math>'''
 +
 +
== Power ==
 +
<math>P=\frac{1}{2\pi-0}\int_0^{2\pi}{|sin(t)|^2dt}</math><br>

Revision as of 19:36, 2 September 2008

x(t) = sin t

Energy

$ E=\int_0^{2\pi}{|sin(t)|^2dt} $
$ =\frac{\int_0^{2\pi}(1-cos(2t))dt}{2} $
$ =\frac{t-\frac{1}{2}sin(2t)}{2}|_{t=0}^{t=2\pi} $
$ =\frac{1}{2}(2\pi-0-0+0) $
$ =\pi $

Power

$ P=\frac{1}{2\pi-0}\int_0^{2\pi}{|sin(t)|^2dt} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett