Find the acceleration vector in terms of u subscript r and u subscript

[physics_88]

Find the acceleration vector in terms of u subscript r and u subscript [tex]\theta[/tex]
r = a(4-cos[tex]\theta[/tex]) and d[tex]\theta[/tex]/dt = 6

Im pretty sure to take the derivative:

r' = a(4+sin[tex]\theta[/tex])
r" = a(4+cos[tex]\theta[/tex])

what should i do next?

 

[HallsofIvy]

Your derivative is not quite right- [itex]dr/d\theta= asin(\theta)[/itex]- the derivative of "4" is 0 since it is a constant. Then, by the chain rule, [itex]dr/dt= asin(\theta) d\theta/dt= 6a sin(\theta)[/itex].

The acceleration vector is [itex]\left<d^2x/dt^2, d^2y/dt^2\right>[/itex]

Of course, [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so that [itex]dx/dt= (dr/dt)cos(\theta)- rsin(\theta)[/itex],
[itex]dy/dt= (dr/dt)sin(\theta)+ rcos(\theta)[/itex],
[itex]d^2x/dt^2= (d^2r/dt^2)cos(\theta)- 2(dr/dt)sin(\theta)- rcos(\theta)[/itex],
and
[itex]d^2y/dt^2= (d^2r/dt^2)sin(\theta)+ 2(dr/dt)cos(\theta)- rsin(\theta)[/itex]

You were given [itex]d\theta/dt[/itex] before and I gave you [itex]dr/dt[/itex] above.

 

[physics_88]

I appreciate that, but is there more to it.

I don't perfer the answer, but can you guide me through it.

What does it mean u subscript r and u sunbscript theta?