A particle move so that its position verctor varies with time as `vec r=A cos omega t hat i + A sin omega t hat j`. Find the
a. initial velocity of the particle,
b. angle between the position vector and velocity of the particle at any time, and
c. speed at any instant.

Explanation

a. Position at time `t`
`vec r=A cos omega t hat i+ A sin omega that j`
Instantaneous velocity, `vec v=(dvec r)/(dt)`
We have `vec v=A(d)/(dt)(cos omega t)hat i+A (d)/(dt)(sin omega t)hat j`
`=-A omega sin 0 hat i+A omega cos t hat j`
At `t=0,v=-A omega sin 0 hat i+A omega cos 0 hat i=A omega hat j`
b. For calculating the angle between two vectors. The angle `theta` between `vec r` and `vec v` can be given as
`the=cos^(-1)(vec r.vec v)/(|vec r||vec v|)`
Where
`vec r.vec v=(A cos omega t hat i+A sin omega hat j).(-A omega sin omega t hat i+A omega cos omega t hat j)`
`=omega A^(2)(-cos omega t sin omega t+sin omega t cos omega t)=0`
Hence, `theta=cos t sin omega t+ sin omega t=0`
Hence, `theta=cos^(-1) 0=pi //2`
That means `vec v_|_ vec r`.
c. Speed at any time is the magnitude of instantaneous
velocity, i.e.,
`v=|vec v|=sqrt((-A omega sin omega t)^2+(A omega cos omega t)^2) = A omega`.