# Third derivative and polar coordinates

## Main Question or Discussion Point

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)er + (r$$\vartheta$$''+2r'$$\vartheta$$')eo. 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
Homework Helper
Hi squenshl! (have a theta: θ )

Hint: what is (er)' ? what is (eθ)' ? er' = $$\vartheta$$'e($$\vartheta$$)
e($$\vartheta$$)' = -$$\vartheta$$'(er)

tiny-tim
Homework Helper
(what happened to that θ i gave you? )

ok, so (rer)' = r'er + rθ'eθ,

and similarly, (r'er + rθ'eθ)' = … ?

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

Last edited:
tiny-tim
Homework Helper
Yes. (But if you keep getting something wrong, it's a good idea to find out why, so show us what you got anyway )

I got R'''(t) = (r'' - r($$\vartheta'$$)2)'er + er'(r''-r($$\vartheta'$$)2) + (r$$\vartheta$$'' + 2r'$$\vartheta'$$)'etheta + etheta'(r$$\vartheta$$'' + 2r'$$\vartheta$$') = (r'' - r($$\vartheta'$$)2)'er + $$\vartheta'$$etheta(r''-r($$\vartheta'$$)2) + (r$$\vartheta$$'' + 2r'$$\vartheta'$$)'etheta - $$\vartheta'$$er(r$$\vartheta$$'' + 2r'$$\vartheta$$')

tiny-tim
Homework Helper
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 = er That's handy, never seen it before.
R'''(t) = r'''er + 3r''er' 3r'er'' + rer'''.
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
Homework Helper
You still use er' = θeθ' and eθ' = -θer' , with the particlar formulas for (t2)' and (t2)''

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

R'''(t) = 0er +3(2)er + 3(2t)er' + t2er
= 6er + 6ter' + t2er

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

Last edited:
tiny-tim
No, 6er' + 6ter'' + t2er''' 