VC2.7

Given, $r=2t^2i+tj-3t^3k\,$

$\frac{dr}{dt}=4ti+j-9t^2k\,$

$\frac{d^r}{dt^2}=4i+0j-18tk\,$

Therefore, $r\times \frac{d^2 r}{dt^2}=(2t^2i+tj-3t^2k)\times (4i+0j-18tk)\,$

= $\begin{vmatrix} i & j & k \\ 2t^2 & t & -3t^2 \\ 4 & 0 & -18t \end{vmatrix}\,$

= $-18t^2i-(-36t^3+12t^3)j-4tk=-18t^2i+24t^3j-4tk\,$

Hence, $\int_1^2 r\times \frac{d^2 r}{dt^2}\,dt=\int_1^2 (-18t^2i+24t^3j-4tk)\,dt\,$

= $-18i\int_1^2 t^2\,dt+24j\int_1^2 t^3\,dt-4k\int_1^2 t\,dt\,$

= $-18i[\frac{t^3}{3}]+24j[\frac{t^4}{4}]-4k[\frac{t^2}{2}]\,$ at t=[1,2]

= $-6(8-1)i+6(16-1)j-2(4-1)k=-42i+90j-6k\,$