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?