VC1.7

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u\cdot v=(t^{2}i-tj+(2t+1)k)\cdot ((2t-3)i+j-tk)=t^{2}(2t-3)-t-t(2t+1)=2t^{3}-5t^{2}-2t\,

{\frac  {d}{dt}}(2t^{3}-5t^{2}-2t)=6t^{2}-10t-2=-6\, [When t=1]

Now, we have u\times v={\begin{vmatrix}i&j&k\\t^{2}&-t&2t+1\\2t-3&1&-t\end{vmatrix}}=(t^{2}-2t-1)i-[-t^{3}-(2t+1)(2t-3)]j+[t^{2}+t(2t-3)]k\,

(t^{2}-2t-1)i+(t^{3}+4t^{2}-4t-3)j+(3t^{2}-3t)k\,

Therefore,{\frac  {d}{dt}}(u\times v)={\frac  {d}{dt}}(t^{2}-2t-1)i+{\frac  {d}{dt}}(t^{3}+4t^{2}-4t-3)j+{\frac  {d}{dt}}(3t^{2}-3t)k=(2t-2)i+(3t^{2}+8t-4)j+(6t-3)k=7j+3k\, [When t=1]

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