VC1.7

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u\cdot v=(t^2 i -tj+(2t+1)k)\cdot ((2t-3)i+j-t k)=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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