Is there a more rigorous way to prove this?

[Keshroom]

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.

 

[tiny-tim]

Hi Keshroom!

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

Sorry, but this doesn't work at all.

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

(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' ​

 

[Keshroom]

Hi Keshroom!

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

Sorry, but this doesn't work at all. (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

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

v² = r' . r' ​

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

 

[tiny-tim]

just differentiate r' . r' using the product rule

 

[Keshroom]

just differentiate r' . r' using the product rule

lol what...so simple. So it's just
r'r'' + r'r'' = 2r'r'' ?

 

[tiny-tim]

no, it's r'.r'' + r'.r'' = 2r'.r''

 

[Keshroom]

--_--

haha thanks!