Third derivative and polar coordinates

In summary, the second derivative of the position of a particle moving in the plane, in polar coordinates, is (r''-r(\vartheta')2)er + (r\vartheta''+2r'\vartheta')eo. To differentiate this, you need to know (er)' and (eθ)' and use the product rule on R'''(t). R'''(1) can be found using the same method, but with er' = θeθ' and eθ' = -θer'.
  • #1
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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([tex]\vartheta[/tex]')2)er + (r[tex]\vartheta[/tex]''+2r'[tex]\vartheta[/tex]')eo. o = [tex]\vartheta[/tex]

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, [tex]\vartheta[/tex](t) = t2
 
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  • #2
Hi squenshl! :smile:

(have a theta: θ :wink:)

Hint: what is (er)' ? what is (eθ)' ? :smile:
 
  • #3
er' = [tex]\vartheta[/tex]'e([tex]\vartheta[/tex])
e([tex]\vartheta[/tex])' = -[tex]\vartheta[/tex]'(er)
 
  • #4
(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:
 
  • #5
I keep getting different answers everytime.
I must be doing something wrong.
Can I just do the fact that er' = [tex]\vartheta'e_\vartheta[/tex] & e[tex]\vartheta[/tex] = -[tex]\vartheta'e_r[/tex]
and then just use the product rule on R''(t) to get R'''(t).
 
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  • #6
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:)
 
  • #7
I got R'''(t) = (r'' - r([tex]\vartheta'[/tex])2)'er + er'(r''-r([tex]\vartheta'[/tex])2) + (r[tex]\vartheta[/tex]'' + 2r'[tex]\vartheta'[/tex])'etheta + etheta'(r[tex]\vartheta[/tex]'' + 2r'[tex]\vartheta[/tex]') = (r'' - r([tex]\vartheta'[/tex])2)'er + [tex]\vartheta'[/tex]etheta(r''-r([tex]\vartheta'[/tex])2) + (r[tex]\vartheta[/tex]'' + 2r'[tex]\vartheta'[/tex])'etheta - [tex]\vartheta'[/tex]er(r[tex]\vartheta[/tex]'' + 2r'[tex]\vartheta[/tex]')
 
  • #8
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:
 
  • #9
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 & [tex]\vartheta(t)[/tex] = 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 [tex]\vartheta[/tex] = t2
r' = 2t [tex]\vartheta[/tex]' = 2t
r'' = 2 [tex]\vartheta[/tex]'' = 2
r''' = 0 [tex]\vartheta[/tex]''' = 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
 
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  • #12
squenshl said:
R'''(t) = 0er +3(2)er + 3(2t)er' + t2er
= 6er + 6ter' + t2er

No, 6er' + 6ter'' + t2er''' :smile:
 

1. What is the third derivative in polar coordinates?

The third derivative in polar coordinates is a mathematical concept that describes the rate of change of the third derivative of a function with respect to a polar coordinate, such as radius or angle. It is used to calculate the curvature and acceleration of a curve in polar coordinates.

2. How is the third derivative calculated in polar coordinates?

The third derivative in polar coordinates can be calculated using the chain rule and the polar coordinate transformation formula. It involves taking the derivative of the third derivative of the function with respect to the polar coordinate, multiplied by the derivative of the polar coordinate with respect to the original variable.

3. What is the physical significance of the third derivative in polar coordinates?

The third derivative in polar coordinates has physical significance in describing the rate of change of curvature and acceleration of a curve in polar coordinates. It can also be used to determine the inflection points and convexity/concavity of a polar curve.

4. Can the third derivative in polar coordinates be negative?

Yes, the third derivative in polar coordinates can be negative. This indicates that the curve is concave down or has a negative curvature at that particular point. The sign of the third derivative can change depending on the shape and direction of the curve at different points.

5. How is the third derivative used in real-world applications?

The third derivative in polar coordinates is used in various fields such as physics, engineering, and computer graphics. It can be used to analyze the motion and acceleration of objects in polar coordinate systems, design curved structures and surfaces, and create smooth and realistic animations in computer graphics.

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