Third derivative and polar coordinates

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Discussion Overview

The discussion revolves around finding the third derivative of a position vector R(t) expressed in polar coordinates, particularly focusing on the differentiation process and the application of the product rule. Participants explore the implications of the second derivative and how to compute R'''(1) given specific functions for r(t) and θ(t).

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant states the second derivative of the position R(t) in polar coordinates and seeks guidance on finding the third derivative R'''(t).
  • Another participant suggests considering the derivatives of the unit vectors er and eθ, prompting further exploration of their derivatives.
  • There is a discussion on applying the product rule to the second derivative to derive the third derivative, with one participant expressing confusion over obtaining consistent results.
  • Participants share a formula for differentiating products, which could simplify the computation of R'''(t).
  • One participant provides specific values for r and θ, along with their derivatives, and attempts to compute R'''(t) using these values.
  • Another participant points out a potential error in the computation of R'''(t) and suggests corrections to the terms involved.

Areas of Agreement / Disagreement

Participants generally engage in a collaborative exploration of the differentiation process, but there is no consensus on the final form of R'''(t) or the specific computation of R'''(1), as some participants express uncertainty and propose different approaches.

Contextual Notes

Some participants note the importance of correctly applying the product rule and the derivatives of the unit vectors, indicating that misunderstandings may arise from these steps. There are also unresolved aspects regarding the correctness of the computed derivatives and the application of the formulas.

Who May Find This Useful

This discussion may be useful for students studying polar coordinates and derivatives in the context of physics or mathematics, particularly those preparing for exams or seeking clarification on differentiation techniques in vector calculus.

squenshl
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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
 
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Hi squenshl! :smile:

(have a theta: θ :wink:)

Hint: what is (er)' ? what is (eθ)' ? :smile:
 
er' = \vartheta'e(\vartheta)
e(\vartheta)' = -\vartheta'(er)
 
(what happened to that θ i gave you? :confused:)

ok, so (rer)' = r'er + rθ'eθ,

and similarly, (r'er + rθ'eθ)' = … ?

(r'er + rθ'eθ)'' = … ? :smile:
 
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:
Yes. :smile:

(But if you keep getting something wrong, it's a good idea to find out why, so show us what you got anyway :wink:)
 
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')
 
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 :wink:
 
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.
 
  • #10
You still use er' = θeθ' and eθ' = -θer' , with the particlar formulas for (t2)' and (t2)''
 
  • #11
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:
  • #12
squenshl said:
R'''(t) = 0er +3(2)er + 3(2t)er' + t2er
= 6er + 6ter' + t2er

No, 6er' + 6ter'' + t2er''' :smile:
 

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