Is there a more rigorous way to prove this?

• Keshroom
In summary: I was wondering how I could do that too :) In summary, the homework statement is trying to solve for v in terms of dr and dv, but there is a mistake in the equation.
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

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:
Hi Keshroom!

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

Sorry, but this doesn't work at all.
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

(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'

tiny-tim said:
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

Last edited:
tiny-tim said:
just differentiate r' . r' using the product rule

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''

--_--

haha thanks!

1. What is the purpose of using a more rigorous proof?

A more rigorous proof is used to provide a stronger and more convincing argument for the validity of a statement or theorem. It helps to eliminate any doubt or ambiguity in the proof and ensures that the result is true in all cases.

2. How does a more rigorous proof differ from a standard proof?

A more rigorous proof uses stricter logic and reasoning, and often involves more technical and complex mathematical techniques. It leaves no room for assumptions or shortcuts, and must be able to withstand rigorous scrutiny from other scientists and experts in the field.

3. Can a statement be considered proven without a rigorous proof?

While a standard proof may be sufficient for some purposes, a statement can only be considered truly proven if it has been rigorously proven. This is because a rigorous proof provides the highest level of certainty and eliminates any potential errors or flaws in the argument.

4. How can one determine if a proof is rigorous enough?

A proof is considered rigorous if it follows a logical and systematic approach, uses precise and unambiguous language, and is backed up by sound mathematical reasoning. It should also be able to withstand rigorous scrutiny and be reproducible by other scientists.

5. Are there any downsides to using a more rigorous proof?

The main downside of using a more rigorous proof is that it can be more time-consuming and complex to construct. It may also require a higher level of mathematical knowledge and expertise. However, the benefits of a rigorous proof far outweigh the potential drawbacks in terms of the strength and validity of the result.

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