Answer to Question #135326 in Differential Geometry | Topology for tonmoy

Question

If u=3t2i+et j−t k and v=sinti−costj+e2tk then, at t=0, ddt(u×v)=

Answer 1

$\mathbf{We \;have-\;\;u=3t^2\;i+e^t\;j-t\;k\;\;and\;v=sin(t)\;i-cos(t)\;j+e^{2t}\;k}$

$\mathbf{To\;find- \;at\;t=0\;,\;\dfrac{d}{dt}(u \times v)= \;?}$

$\mathbf{So,\;u \times v= \begin{vmatrix} i & j & k\\ 3t^2 & e^t &-t\\ \sin(t) & \cos(t) & e^{2t} \end{vmatrix}=i\;\begin{vmatrix} e^t & -t \\ \cos(t) & e^{2t} \end{vmatrix}-j\;\begin{vmatrix} 3t^2 & -t \\ \sin(t) & e^{2t} \end{vmatrix}}$

$\mathbf{+k\;\begin{vmatrix} 3t^2 & e^t \\ \sin(t) & \cos(t) \end{vmatrix}}$

$\mathbf{\implies u \times v=i\;(e^t.e^{2t}-(-t).\cos(t))-j(3t^2.e^{2t}-(-t).\sin(t))}$

$\mathbf{+k\;(3t^2.\cos(t)-e^t.\sin(t))}$

$\mathbf{=i\;(e^{3t}+t\cos(t))-j(3t^2e^{2t}+t\sin(t))+k(3t^2\cos(t)-e^t\sin(t))}$

$\mathbf{Now,\;\dfrac{d}{dt}(u \times v)=\dfrac{d}{dt}\bigg[i\;(e^{3t}+t\cos(t))-j(3t^2e^{2t}+t\sin(t))}$

$\mathbf{+k(3t^2\cos(t)-e^t\sin(t))\bigg]}$

$\mathbf{=i\;\bigg(\dfrac{d}{dt}(e^{3t}+t\cos(t))\bigg)-j\;\bigg(\dfrac{d}{dt}(3t^2e^{2t}+t\sin(t))\bigg)}$

$\mathbf{+k\;\bigg(\dfrac{d}{dt}(3t^2\cos(t)-e^t\sin(t))\bigg)}$

$=$ $\mathbf{i\;(3e^{3t}+\cos(t)-t\sin(t))-j\;(6t^2e^{2t}+6te^{2t}+\sin(t)+t\cos(t))}$

$\mathbf{+k\;(6t\cos(t)-3t^2\sin(t)-e^t\sin(t)-e^t\cos(t))}$

$\mathbf{At\;t=0,\; \dfrac{d}{dt}(u \times v)=i\;(3+1-0)-j\;(0+0+0+0)+k\;(0-0-0-1)}$

$\mathbf{=4\;i-0\;j+(-1)\;k=4\;i-k}$ ......................Ans.