Is there a more rigorous way to prove this?

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Homework Help Overview

The discussion revolves around proving the equation d(v^2)/dt = 2 . (d^2r/dt^2) . (dr/dt), where v represents the velocity vector defined as the derivative of the position vector r with respect to time. Participants are exploring different methods of differentiation and the implications of vector notation in the context of this problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the chain rule and the product rule in differentiating the expression for v^2. There is also a focus on the correct interpretation of vector operations, particularly the distinction between scalar multiplication and the dot product.

Discussion Status

The conversation includes attempts to clarify the original poster's method and the validity of their reasoning. Some participants suggest rewriting the hint for clarity, while others emphasize the importance of correctly applying vector notation. There is ongoing exploration of the problem without a clear consensus on the most rigorous approach.

Contextual Notes

Participants note potential errors in the original question and express a need for clarification on notation, indicating that some foundational understanding may be lacking. The discussion is framed within the constraints of homework help, where complete solutions are not provided.

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Homework Statement


Show that d(v^2)/dt = 2 . (d^2r/dt^2) . (dr/dt)

HINT: v^2 = ||dr/dt||^2 = dr/dt . dr/dt

Homework Equations


The Attempt at a Solution



I did it another method:

d(v^2)/dt = d/dv(v^2) . dv/dt --------------chain rule
= 2v . dv/dt
since v=dr/dt and dv/dt = d^2r/dt^2
therefore = 2 . (d^2r/dt^2) . (dr/dt)

Is there a more rigorous method to solving this, using the 'HINT' it gave me. I can't quite figure it out.
 
Last edited:
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Hi Keshroom! :smile:

(try using the X2 button just above the Reply box :wink:)

Sorry, but this doesn't work at all. :redface:
Keshroom said:
Show that d(v^2)/dt = 2 . (d^2/dt^2) . (dr/dt)

HINT: v^2 = ||dr/dt||^2 = dr/dt . dr/dt

I did it another method:

d(v^2)/dt = d/dv(v^2) . dv/dt --------------chain rule
= 2v . dv/dt
since v=dr/dt and dv/dt = dr^2/dt^2

(your "." is ordinary multiplication of two scalars, unlike the question, where it's the dot-product of two vectors; but anyway:)

no, dv/dt is not d2r/dt2 :redface:

(except in one-dimensional motion)
Is there a more rigorous method to solving this, using the 'HINT' it gave me.

it's easier if you rewrite the hint as:

v2 = r' . r' :wink:
 
tiny-tim said:
Hi Keshroom! :smile:

(try using the X2 button just above the Reply box :wink:)

Sorry, but this doesn't work at all. :redface:(your "." is ordinary multiplication of two scalars, unlike the question, where it's the dot-product of two vectors; but anyway:)

no, dv/dt is not d2r/dt2 :redface:

(except in one-dimensional motion)it's easier if you rewrite the hint as:

v2 = r' . r' :wink:

Sorry I'm still learning with the notation and everything. Awww ok. You mind telling me how to prove this please?
thanks

btw there were errors in the question which i have edited now
 
Last edited:
tiny-tim said:
just differentiate r' . r' using the product rule :smile:

lol what...so simple. So it's just
r'r'' + r'r'' = 2r'r'' ?
 
Last edited:
no, it's r'.r'' + r'.r'' = 2r'.r'' :wink:
 
--_--

haha thanks!
 

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