Wehave−u=3t2i+etj−tkandv=sin(t)i−cos(t)j+e2tk
Tofind−att=0,dtd(u×v)=?
So,u×v=∣∣i3t2sin(t)jetcos(t)k−te2t∣∣=i∣∣etcos(t)−te2t∣∣−j∣∣3t2sin(t)−te2t∣∣
+k∣∣3t2sin(t)etcos(t)∣∣
⟹u×v=i(et.e2t−(−t).cos(t))−j(3t2.e2t−(−t).sin(t))
+k(3t2.cos(t)−et.sin(t))
=i(e3t+tcos(t))−j(3t2e2t+tsin(t))+k(3t2cos(t)−etsin(t))
Now,dtd(u×v)=dtd[i(e3t+tcos(t))−j(3t2e2t+tsin(t))
+k(3t2cos(t)−etsin(t))]
=i(dtd(e3t+tcos(t)))−j(dtd(3t2e2t+tsin(t)))
+k(dtd(3t2cos(t)−etsin(t)))
= i(3e3t+cos(t)−tsin(t))−j(6t2e2t+6te2t+sin(t)+tcos(t))
+k(6tcos(t)−3t2sin(t)−etsin(t)−etcos(t))
Att=0,dtd(u×v)=i(3+1−0)−j(0+0+0+0)+k(0−0−0−1)
=4i−0j+(−1)k=4i−k ......................Ans.
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