If u(t) = < sin(5t), cos(7t), t > and v(t) = < t, cos(7t), sin(5t) >, use the formula below to find the given derivative. d/(dt)[u(t) cross v(t)] = u'\(t\) cross v(t) + u(t) cross v'\(t\)

Answer:[tex]\frac{d(u(t)\times v(t))}{dt}=i(-7 sin (7t)sin5t-cos7t-7tsin7t-5cos5tcos7t)-2j(5sin5tcos5t-t)+k(5cos5tcos7t+7tsin7-7sin5tcos7t-cos7t)[/tex]Step-by-step explanation:We are given that [tex]u(t)=<sin(5t),cos(7t),t>[/tex]and [tex]v(t)=<t,cos(7t),sin(5t)>[/tex]We have to find the value of [tex]\frac{d(u(t)\times v(t))}{dt}[/tex]We are given that [tex]\frac{d(u(t)\times v(t)}{dt}=u'(t)\times v(t)+u(t)\times v'(t)[/tex]We are finding u'(t) and v'(t)[tex]u'(t)=<5cos(5t),-7sin(7t),1>[/tex][tex]v'(t)=<1,-7sin(7t),5cos(5t)>[/tex]We are finding [tex]u'(t)\times v(t)[/tex][tex]u'(t)\times v(t)=\begin{vmatrix}i&j&k\\5cos5t&-7sin7t&1\\t&cos7t&sin5t\end{vmatrix}[/tex][tex]u'(t)\times v(t)=(-7 sin (7t)sin5t-cos7t)i-j(5cos5tsin5t-t)+k(5cos5tcos7t+7tsin7t)[/tex][tex]u(t)\times v'(t)=\begin{vmatrix}i&j&k\\sin5t&cos7t&t\\1&-7sin7t&5cos5t\end{vmatrix}[/tex][tex]u(t)\times v'(t)=i(5cos5tcos7t+7tsin7t)-j(5sin5tcos5t-t)+k(-7sin5tcos7t-cos7t)[/tex]Substitute the values then we get [tex]\frac{d(u(t)\times v(t)}{dt}==(-7 sin (7t)sin5t-cos7t)i-j(5cos5tsin5t-t)+k(5cos5tcos7t-7tsin7t)+i(5cos5tcos7t+7tsin7t)-j(5sin5tcos5t-t)+k(-7sin5tcos7t-cos7t)[/tex][tex]\frac{d(u(t)\times v(t))}{dt}=(-7 sin (7t)sin5t-cos7t)i-j(5cos5tsin5t-t)+k(5cos5tcos7t-7tsin7t)+i(5cos5tcos7t+7tcos7t)-j(5sin5tcos5t-t)+k(-7sin5tcos7t-cos7t)[/tex][tex]\frac{d(u(t)\times v(t))}{dt}=i(-7 sin (7t)sin5t-cos7t+7tsin7t+5cos5tcos7t)-j(5cos5tsin5t-t-t+5cos5tsin5t)+k(5cos5tcos7t+7tsin7-7sin5tcos7t-cos7t)[/tex][tex]\frac{d(u(t)\times v(t))}{dt}=i(-7 sin (7t)sin5t-cos7t-7tsin7t-5cos5tcos7t)-2j(5sin5tcos5t-t)+k(5cos5tcos7t+7tsin7-7sin5tcos7t-cos7t)[/tex]