VC2.6

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\int _{2}^{3}[r\cdot {\frac  {dr}{dt}}]\,dt=[{\frac  {r^{2}}{2}}]=[{\frac  {1}{2}}|r(t)|^{2}]\, at t=2,3

={\frac  {1}{2}}[|r(3)|^{2}-|r(2)|^{2}]\,

Hence,from given results r(2)=2i-j+2k,r(3)=4i-2j+3k\,

Therefore,|r(2)|^{2}=4+1+4=9,|r(3)|^{2}=16+4+9=29\,

So, \int _{2}^{3}[r\cdot {\frac  {dr}{dt}}]\,dt={\frac  {1}{2}}(29-9)=10\,

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