# Homework Help: Is there a more rigorous way to prove this?

1. Mar 10, 2012

### Keshroom

1. The problem statement, all variables and given/known data
Show that d(v^2)/dt = 2 . (d^2r/dt^2) . (dr/dt)

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

2. Relevant equations

3. 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: Mar 10, 2012
2. Mar 10, 2012

### tiny-tim

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'

3. Mar 10, 2012

### Keshroom

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: Mar 10, 2012
4. Mar 10, 2012

### tiny-tim

5. Mar 10, 2012

### Keshroom

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

Last edited: Mar 10, 2012
6. Mar 10, 2012

### tiny-tim

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

7. Mar 10, 2012

--_--

haha thanks!