Third derivative and polar coordinates

[squenshl]

I'm studying for a maths test.
I know that the second derivative of the position R(t) of a particle moving in the plane, in polar coordinates, is (r''-r(\vartheta')²)e_r + (r\vartheta''+2r'\vartheta')e_o. o = \vartheta

How to differentiate this to find R'''(t), in polar coordinates and in turn find R'''(1) in polar coordinates if R(t) has polar coordinates r(t) = t², \vartheta(t) = t²

 

[tiny-tim]

Hi squenshl!

(have a theta: θ )

Hint: what is (e_r)' ? what is (e_θ)' ?

 

[squenshl]

e_r' = \vartheta'e_(\vartheta)
e_(\vartheta)' = -\vartheta'(e_r)

 

[tiny-tim]

(what happened to that θ i gave you? )

ok, so (re_r)' = r'e_r + rθ'e_θ,

and similarly, (r'e_r + rθ'e_θ)' = … ?

(r'e_r + rθ'e_θ)'' = … ? ​

 

[squenshl]

I keep getting different answers everytime.
I must be doing something wrong.
Can I just do the fact that e_r' = \vartheta'e_\vartheta & e_\vartheta = -\vartheta'e_r
and then just use the product rule on R''(t) to get R'''(t).

 

[tiny-tim]

Yes.

(But if you keep getting something wrong, it's a good idea to find out why, so show us what you got anyway )

 

[squenshl]

I got R'''(t) = (r'' - r(\vartheta')²)'e_r + e_r'(r''-r(\vartheta')²) + (r\vartheta'' + 2r'\vartheta')'e_theta + e_theta'(r\vartheta'' + 2r'\vartheta') = (r'' - r(\vartheta')²)'e_r + \vartheta'e_theta(r''-r(\vartheta')²) + (r\vartheta'' + 2r'\vartheta')'e_theta - \vartheta'e_r(r\vartheta'' + 2r'\vartheta')

 

[tiny-tim]

Looks ok so far.

But it'll save a lot of repetition if you use the formula (ab)''' = a'''b + 3a''b' + 3a'b'' + ab''', with a = r and b = e_r

 

[squenshl]

That's handy, never seen it before.
R'''(t) = r'''e_r + 3r''e_r' 3r'e_r'' + re_r'''.
Does this help to R'''(1) in polar coordinates if R(t) has polar coordinates r(t) = t² & \vartheta(t) = t².
How do I go about finding R'''(1) then.

 

[tiny-tim]

You still use e_r' = θe_θ' and e_θ' = -θe_r' , with the particlar formulas for (t²)' and (t²)''

 

[squenshl]

Since r = t2 \vartheta = t²
r' = 2t \vartheta' = 2t
r'' = 2 \vartheta'' = 2
r''' = 0 \vartheta''' = 0

R'''(t) = 0e_r +3(2)e_r + 3(2t)e_r' + t²e_r
= 6e_r + 6te_r' + t²e_r

Then R'''(1) = 6e_r + 6(1)e_r + (1)²e_r
= 6e_r + 6e_r' + e_r

 

[tiny-tim]

R'''(t) = 0e_r +3(2)e_r + 3(2t)e_r' + t²e_r
= 6e_r + 6te_r' + t²e_r

No, 6e_r' + 6te_r'' + t²e_r'''