The characteristic equation of a particle moving in a curved path are: `x=e^(-t), y=cos 3t` and z=2 sin 3t, where t stands for time. Find out
(ii) Velocity and acceleration at t=0.

Explanation

`veca=(vec(dv))/(dt)=e^(-t)hati-18 cos3thatj-18sin3thatk=xhati-9yhatj-9zhatk`
(ii) At t=0 , `e^0hati-6sin(3xx0)hatj+6 cos(3xx0)hatk`
`=-hati-6xx0hatj+6xx1hatk=-hati+6hatk`
`v=|vecv|=sqrt((-1)+6^2)=sqrt(37)`
At t=0 , `veca=e^0hati-18cos(3xx0)hatj-18 sin (3xx0)hatk`
`=hati-(18xx1)hatj(18xx0)hatk`
`=hati-18hatj`
`a=|veca|=sqrt(1^2+(-18)^2)=sqrt(325)=5sqrt(13).`