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 \theta
r = a(4-cos\theta) and d\theta/dt = 6

Im pretty sure to take the derivative:

r' = a(4+sin\theta)
r" = a(4+cos\theta)

what should i do next?

 

[HallsofIvy]

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

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

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

You were given d\theta/dt before and I gave you dr/dt 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?