Question #135326
If u=3t2i+et j−t k and v=sinti−costj+e2tk then, at t=0, ddt(u×v)=
1
Expert's answer
2020-09-28T19:56:31-0400

We  have    u=3t2  i+et  jt  k    and  v=sin(t)  icos(t)  j+e2t  k\mathbf{We \;have-\;\;u=3t^2\;i+e^t\;j-t\;k\;\;and\;v=sin(t)\;i-cos(t)\;j+e^{2t}\;k}


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


So,  u×v=ijk3t2ettsin(t)cos(t)e2t=i  ettcos(t)e2tj  3t2tsin(t)e2t\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}}


+k  3t2etsin(t)cos(t)\mathbf{+k\;\begin{vmatrix} 3t^2 & e^t \\ \sin(t) & \cos(t) \end{vmatrix}}


    u×v=i  (et.e2t(t).cos(t))j(3t2.e2t(t).sin(t))\mathbf{\implies u \times v=i\;(e^t.e^{2t}-(-t).\cos(t))-j(3t^2.e^{2t}-(-t).\sin(t))}


+k  (3t2.cos(t)et.sin(t))\mathbf{+k\;(3t^2.\cos(t)-e^t.\sin(t))}


=i  (e3t+tcos(t))j(3t2e2t+tsin(t))+k(3t2cos(t)etsin(t))\mathbf{=i\;(e^{3t}+t\cos(t))-j(3t^2e^{2t}+t\sin(t))+k(3t^2\cos(t)-e^t\sin(t))}


Now,  ddt(u×v)=ddt[i  (e3t+tcos(t))j(3t2e2t+tsin(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))}


+k(3t2cos(t)etsin(t))]\mathbf{+k(3t^2\cos(t)-e^t\sin(t))\bigg]}


=i  (ddt(e3t+tcos(t)))j  (ddt(3t2e2t+tsin(t)))\mathbf{=i\;\bigg(\dfrac{d}{dt}(e^{3t}+t\cos(t))\bigg)-j\;\bigg(\dfrac{d}{dt}(3t^2e^{2t}+t\sin(t))\bigg)}


+k  (ddt(3t2cos(t)etsin(t)))\mathbf{+k\;\bigg(\dfrac{d}{dt}(3t^2\cos(t)-e^t\sin(t))\bigg)}


== i  (3e3t+cos(t)tsin(t))j  (6t2e2t+6te2t+sin(t)+tcos(t))\mathbf{i\;(3e^{3t}+\cos(t)-t\sin(t))-j\;(6t^2e^{2t}+6te^{2t}+\sin(t)+t\cos(t))}


+k  (6tcos(t)3t2sin(t)etsin(t)etcos(t))\mathbf{+k\;(6t\cos(t)-3t^2\sin(t)-e^t\sin(t)-e^t\cos(t))}


At  t=0,  ddt(u×v)=i  (3+10)j  (0+0+0+0)+k  (0001)\mathbf{At\;t=0,\; \dfrac{d}{dt}(u \times v)=i\;(3+1-0)-j\;(0+0+0+0)+k\;(0-0-0-1)}


=4  i0  j+(1)  k=4  ik\mathbf{=4\;i-0\;j+(-1)\;k=4\;i-k} ......................Ans.













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